Contents 1. the Simplex Method 1 1.1. Lecture 1: Introduction

Contents 1. The Simplex Method 1 1.1. Lecture 1: Introduction. 1 1.2. Lecture 2: Notation, Background, History. 3 1.3. Lecture 3: The simplex method 7 1.4. Lecture 4: An example. 10 1.5. Lecture 5: Complications 13 1.6. Lecture 6: 2-phase method and Big-M 15 1.7. Lecture 7: Dealing with a general LP 18 1.8. Lecture 8: Duality 22 1.9. Lecture 9: more duality 24 1.10. Lecture 10: Assorted facts on duality 26 1.11. Lecture 11: A problem for duality 29 1.12. Lecture 12: Post-optimal considerations and general Duality 31 1.13. Lecture 13: The Revised Simplex 34 1.14. Lecture 14: Filling the knapsack 38 1.15. Lecture 15: Knapsack, day 2 40 1.16. Lecture 16: Knapsack, day 3 43 1.17. Lecture 17: A branch and bound example in class 50 1.18. Lecture 18: Matrix games 51 1.19. Lecture 19: Mixed strategies 53 1.20. Lecture 20, 21: a game example 59 1.21. Lecture 22: Review for the test 62 1.22. Lecture 23: Midterm 64 1.23. Lecture 24, 25: Network Simplex, Transshipment (Chapter 19) 65 1.24. Lecture 26: Network Simplex, Transshipment: initial trees 69 1.25. Lecture 27: an example on network simplex 73 1.26. Lecture 28: Upper bounded Transshipment problems 75 1.27. Lecture 29: Upper bounded network II 76 1.28. Lecture 30: Network flows 78 1.29. Network simplex on maximum flows problems 78 1.30. Lecture 31: Network flows, part 2 80 1.31. Lecture 32: An example on maximum flows 82 1.32. Lecture 33: Applications of network simplex 83 1.33. Lecture 34: More applications of the network simplex 87 1.34. Lecture 35: Transportation problems 89 1.35. Lecture 36: An transport problem 94 1.36. Lecture 37: Transport example, second day 95 1.37. Lecture 38: Integer Programming 96 1.38. Lectures 39, 40: Gröbnerbases and Buchberger algorithm 98 1.39. Lecture 41: Gröbnerbases with Maple 100

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1.40. Lectures 42, 43, 44: Review of ﬁnal material, questions 104 421 COURSE NOTES

ULI WALTHER

Abstract. Textbook: V. Chvatal, “Linear Programming”. grade: 30% midterm, 30% homework, 40% ﬁnal. homework collected each Thursday in class or in my oﬃce by 3pm.

1. The Simplex Method What is linear programming? 1.1. Lecture 1: Introduction. 1.1.1. Some examples:

Example 1.1 (Knapsack problem). Looking for x1, . . . , xn such that

x1c1 + ... + xncn → max

x1a1 + ... + xnan ≤ A (the knapsack)

xi = 0, 1∀i

Example 1.2 (Transportation problem). Looking for x1,1, . . . , xm,n such that m n X X ci,jxi,j → min (fuel from i to j) i=1 j=1 n X xi,j = ri∀i (producers) j=1 m X xi,j = si∀j (consumers) i=1 xi,j ≥ 0∀i, j

Example 1.3 (matrix games). 2 players play against each other a version of stone-paper-scissors. The outcome is decided by how they choose their strategies, no random components. The question is how to choose the strategy. Minimize losses in worst case scenario? Maximize wins in best case? Maximize wins in worst case? Minimize losses in best case? When is a game fair? For example, 1 2 ULI WALTHER

a b c d A 0 2 -3 0 B -2 0 0 3 C 3 0 0 -4 D 0 -3 4 0 A,. . . ,d are the possible choices the players Bob and Alice have. Tabulated are the winnings of Alice. How should they play? Example 1.4 (Isoperimetric problem). Amongst all closed curves in the plane with circumference 1 meter, which curve encloses the largest area? (This we will not answer!) Example 1.5 (Garbage removal). Given a weighted graph, ﬁnd a cheapest (closed?) path that travels along all edges. We may talk about that. 1.1.2. Some geometric remarks. Let us solve 2x + 3y → max x + 3y ≤ 6 x ≤ 12 x − y ≤ 1 3x + 2y ≤ 6 x, y ≥ 0. (2,3) is the direction in which we maximize. The lines give a ﬁnite region with straight lines and corners as boundary. The max will be taken in a corner (unless the maximizing direction is perpendicular to a boundary → degeneracy). 421 COURSE NOTES 3

1.2. Lecture 2: Notation, Background, History.

Deﬁnition 1.6. A convex set C is a set for which all line segments between points of C are completely in C as well.

Show some examples of convex sets.

n n Deﬁnition 1.7. A point in R \ C (the complement of C in R ) belongs to n the boundary of C ⊆ R if one can get arbitrarily close to P while staying n in C.A closed set C ⊆ R is a set that contains its boundary. n A half space of R is the collection of points satisfying a single linear inequality. A polyhedron or polytope is the intersection of a ﬁnite number of half spaces. Polyhedra are convex, and closed.

Theorem 1.8. Let C be closed. A linear function f on C takes its maximum and minimum at a point on the boundary, or at inﬁnity. For polyhedra, maxima occur in corners or at inﬁnity.

Proof. If f is constant there is nothing to prove. n Let L be a line in R that goes through P ∈ C and pick it in such a way that f grows. Then either, we get at some point to the end of C. Then the maximum point on L is part of the boundary of C, which as C is closed is part of C. Or, we never get to the end of C, in which case f is unbounded on C. Hence, no matter what P is, there is always a better point than P on the boundary. (Or f is unbounded.) So the max is on the boundary, and we know that is part of C. 2

In principle: given any number (say m) of conditions (the inequalities) in (say) n < m variables, • choose n conditions, • read them as equalities, • solve for the xi, • check if the other inequalities are OK (admissible solution) • and if so, calculate the objective function. Once this has been done, compare all results and pick the best. In theory linear optimization (with linear constraints) is trivial. Problem: in praxis, one often has hundreds of variables to thousands of 2000 constraints. Let’s say n = 500, m = 2000. Then there are 500 = at least 10100 corners. If we could do 1010 per second (that is about 1,000,000 of what can be done) it would take so much time (1082 years) that a snail could go 1054 times from one end of the universe to the other. Thus, one needs a clever way of checking. Strategy: start in some corner, and move to a better one. This can in the 2000/500 example usually be done in a couple of hundreds of steps. 4 ULI WALTHER

Example 1.9 (A diet problem:).

3x1 + 24x2 + 13x3 + 9x4 + 20x5 + 19x6 → min price

x1 ≤ 4 (oatmeal)

x2 ≤ 3 (chicken)

x3 ≤ 2 (eggs)

x4 ≤ 8 (mil)k

x5 ≤ 2 (cherry pie)

x6 ≤ 2 (pork+beans)

110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260xx ≥ 2000 (calories)

4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 ≥ 55 (protein, g)

2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 ≥ 800 (calcium, mg)

xi ≥ 0∀i The ﬁrst six are dictated by taste, the other 3 by nutritionists. Without the taste constraints, one could have x6 = 10, all others zero (gives 1.90$). With them, 8 = x4, 2 = x5 works(gives 1.12$). But is it the cheapest? Explain: objective (or cost) function, linear equation (constraint), non- negativity constraints, optimal solution, optimal value, feasible/infeasible solution/problem, unbounded problem, decision variables. Explain how unbounded and degenerate problems and infeasible ones come about. Remark 1.10. George Dantzig (1947) made linear programming a science. Before, various people thought about it and recognized the importance (Fourier), but could not make it eﬃcient. Kantorovich (1939) had good ideas like Dantzig, but they were not published. 1975, 2 mathematicians became Nobel prize winners in economics. one a student of Dantzig. The simplex algorithm (next) works most of the time really well, but sometimes is awful. The probability of an awful case is zero. There does exist an algorithm (Khachian,1979) that is never awful, but it is almost always beat by the simplex algorithm. Means: with probability zero simplex loses. 421 COURSE NOTES 5

Homework 1 (for week 2). 2 (1) Find an example of an optimization problem on a region in R with a linear objective function f that is bounded on C but where there is no max in C. (This will by our theorem require a non-closed C). Prove that there is no max on C. (2) Find an example of an optimization problem with a closed C where the max of f is not on the boundary and not at inﬁnity. (By the theorem, this will require a non-linear f.) Prove that the max is where you claim it to be. (3) Solve the following linear program graphically and explain why the max is taken where you claim it is.

x1 ≤ 4

2x2 ≤ 12

3x1 + 2x2 ≤ 18

x1, x2 ≥ 0

3x1 + 5x2 → max (4) Solve the following linear program graphically and explain why the max is taken where you claim it is.

x2 ≤ 10

x1 + 2x2 ≤ 15

x1 + x2 ≤ 12

5x1 + 3x2 ≤ 45

x1, x2 ≥ 0

10x1 + 20x2 → max (5) The Southern Confederation of Kibbutzim (SCK) is a group of three kibbutzim (communal farming communities) in Israel. Overall plan- ning is done by the Coordinating Technical Office. The agricultural output for each kibbutz is limited by both the amount of available irrigable land and the quantity of water allocated for irrigation by the Water Commissioner, a national governmental official. These data are given in the following table. Kibbutz Usable land water allocation (acres) (acre feet) 1 400 600 2 600 800 3 300 375 The crops suited for this region include sugar beets, cotton, and sorghum, and these are the only ones considered for the upcoming season. These crops differ in their expected net return per acre, and their water needs. Additionally, the Ministry of Agriculture has set 6 ULI WALTHER

a maximum quota for the total acreage that can be used to each of these crops by the SCK as shown in the next table. Crop Max Quota Water consumption Net return (Acres) (Acre feet/Acre) ($/Acre) sugar beets 600 3 1000 cotton 500 2 750 sorghum 325 1 250 Using the variables x1, . . . , x9 to indicate how much of sugar beets, cotton and sorghum is grown by kibbutz 1,2 and 3 (see the table below), formulate a linear program whose optimal solution maximizes the total net return, but do not solve it. kibbutz 1 2 3 beets x1 x2 x3 cotton x4 x5 x6 sorghum x7 x8 x9 421 COURSE NOTES 7

1.3. Lecture 3: The simplex method.

2x1 + 3x2 + x3 ≤ 5(1.1)

4x1 + x2 + 2x3 ≤ 11

3x1 + 4x2 + 2x3 ≤ 8

xi ≥ 0

5x1 + 4x2 + 3x3 → max

Introduce slack variables x4, x5, x6 and z the objective function; x1, x2, x3 the decision variables. Then

z → max(1.2)

x4 = 5 − 2x1 − 3x2 − x3

x5 = 11 − 4x1 − x2 − 2x3

x6 = 8 − 3x1 − 4x2 − 2x3

z = 5x1 + 4x2 + 3x3

x1,...,6 ≥ 0

• Feasible solutions of (1.2) correspond to feasible solutions of (1.1) and conversely. • Optimal solutions correspond as well. Discuss this correspondence. Note, that it is very easy to get a feasible solution for (1.2), by putting all decision variables to zero. Initial solution is

x1 = 0, x2 = 0, x3 = 0, x4 = 5, x5 = 11, x6 = 8, z = 0.

Now try to improve. Jump from corner to corner. Try to increase x1 (in the spirit of the objective function). We need to satisfy

x1 ≤ 5/2, x1 ≤ 11/4, x1 ≤ 8/3.

The ﬁrst is the most stringent. Note, that then a slack variable (x4) becomes zero. The next solution is

x1 = 5/2, x2 = 0, x3 = 0, x4 = 0, x5 = 1, x6 = 1/2, z = 25/2.

3 This is a new point of intersection of three planes in R , the three directions 3 of R being x1, x2, x3 and the planes being x2 = 0, x3 = 0, 5 = 2x1 +3x2 +x3. In order to go one more step, we need a new system of equations like (1.2) where now x4 moved to the right and x1 to the left. In general, the ones on the right are those that are zero, on the left those that are not. Since x1 = 5/2 − 3x2/2 − x3/2 − x4/2, replace in (1.2) each x1 on the right by this 8 ULI WALTHER expression, discard the relation that expresses x4 and add the new relation:

x1 = 5/2 − 3x2/2 − x3/2 − x4/2

x5 = 1 + 5x2 + 2x4

x6 = 1/2 + x2/2 − x3/2 + 3x4/2

z = 25/2 − 7x2/2 + x3/2 − 5x4/2

Now it’s clear that we should not make x2 bigger, nor x4, but x3.(x2, x4 have a minus in z) How much can we increase it? The three constraints x1, x5, x6 ≥ 0 give three conditions: x3 ≤ 5, no constraint, and x3 ≤ 1. So we take x3 = 1 next. Then (from x3 = 1, x2 = x4 = 0)

x1 = 2, x2 = 0, x3 = 1, x4 = 0, x5 = 1, x6 = 0, z = 13.

This is the corner to x2 = 0, 2x1 + 3x2 + x3 = 5, 3x1 + 4x2 + 2x3 = 8. We also need a new system: (from x3 = 1 + x2 + 3x4 − 2x6, get rid of x3 on the right)

x3 = 1 + x2 + 3x4 − 2x6

x1 = 2 − x2 − 2x4 + x6

x5 = 1 + 5x2 + 2x4

z = 13 − 3x2 − x4 − x6 The last row indicates that 13 is the best possible value. Hence we are done. 6 Note: we have 3 = 20 corners, but the simplex only visits 3 of them. 1.3.1. Dictionaries. Given the problem n X cjxj → max j=1 n X ai,jxj ≤ bi ∀i ≤ m j=1

xj ≥ 0∀j ≤ n Pn ﬁrst introduce slack variables xn+i = bi − j=1 ai,jxj. This gives equations as constraints: n X (1.3) xn+i = bi − ai,jxj ∀i ≤ m j=1 Pn We also set z = j=1 cjxj. The left hand variables are called basic, the others non-basic. Why? Because the columns in the m × (m + n)-matrix to the system (1.3) form a basis for the column space. A feasible solution is then m+n nonnegative numbers that ﬁt (1.3). Each of these feasible solutions corresponds to a linear system that explicitly solves for m of the m + n variables, called a dictionary. 421 COURSE NOTES 9

In principle, all dictionaries say the same (how the variables are related to each other) but in very diﬀerent ways. In general, the suggestion is that the variables on the right can be chosen freely, and the ones on the left are implied. To be a dictionary, a system of equations must satisfy • each of its solutions is a solution to (1.3), • m of the variables x1, . . . , xn+m are expressed explicitly, and so is z. Note 1.11. A dictionary is obtained from a selection of basic variables by putting the columns to the selected variables up front and computing a row reduced form of the matrix of relations. Hence to each basis there is only one dictionary. There is a nice further property: setting all right hand side variables to zero may give a feasible solution. This is not always so (and only happens if all constant terms in all equalities are at least zero). Conversely, not all feasible solutions come from dictionaries (because dictionary solutions correspond to corners of the feasible region, never to the interior). Dictionaries that give feasible solutions we call feasible dictionaries and solutions that come out of dictionaries are called basic. The simplex only moves along basic feasible solutions coming out of feasible dictionaries. The process of construction a new dictionary is called pivoting on the column of the entering xi and the row of the exiting xi. 10 ULI WALTHER

1.4. Lecture 4: An example.

Example 1.12. A bicycle maker makes 3- and 5-speed bikes. The plant makes 100 frames per day. Tires, brakes, gears are made by a supplier. The maker has to put on finish, and assemble. There are 40 hours of finishing available and 50 hours of assembling per day. The profit is 12 bucks for a 3-speed, and 15 for a 5-speed. Number of hours per bicycle needed: 3-speed 5-speed finishing 1/3 1/2 assembly 1/4 2/3 How many of each type should be made to maximize profit? The objective function is z = 12x1 + 15x2. If we make x1 3-speeds and x2 5-speeds, we will need x1/3 + x2/5 to finish them and x1/4 + 2x2/3 to assemble them.

12x1 + 15x2 → max

x1/3 + x2/2 ≤ 40

x1/4 + 2x2/3 ≤ 50

x1 + x2 ≤ 100

x1, x2 ≥ 0

We introduce slack,

x3 = 40 − x1/3 − x2/2

x4 = 50 − x1/4 − 2x2/3

x5 = 100 − x1 − x2

z = 12x1 + 15x2

x1, x2 ≥ 0 z → max

The initial solution as usual is x1 = x2 = 0 with z = 0. x1 has a positive entry in the objective function, and it’s zero. So we make it larger, to min(3 · 40, 4 · 50, 1 · 100) = 100. Then x5 = 0 moves to the right, x1 to the left, and the new system is

x1 = 100 − x5 − x2

x3 = 20/3 − x2/6 + x5/3

x4 = 25 − 5x2/12 + x5/4

z = 1200 + 3x2 − 12x5 421 COURSE NOTES 11

Now we should get x2 into the basis and the constraints are x2 ≤ 100, x2 ≤ 40, x2 ≤ 60. Hence x3 will be kicked out of the basis. Kick out x3. We get

x1 = 60 + 6x3 − 3x5

x2 = 40 − 6x3 + 2x5

x4 = 25/3 + 5x3/2 − 7x5/12

z = 1320 − 18x3 − 6x5 (Note: I think these numbers are right, but please let me know if you dis- agree.) It is useful to point out that if one ﬁrst uses the other possibility to enter a variable, namely x2 (perhaps, because it looks more promising in terms of the objective function), then one actually goes through 3 iterations instead of 2. There is no way of knowing this. We shall discuss strategies a bit more in the upcoming days. 12 ULI WALTHER

Homework 2 (for week 3). (1) Farmer Jones has 100 acres of land to devote to wheat and corn and wishes to plan his planting to maximize the expected revenue. Jones has only $800 in capital to apply to planting the crops, and it costs $5 to plant an acre of wheat, and $10 for an acre of corn. Their other activities leave the Jones family only 150 days of labor for the crops. Two days are required for each acre of wheat and one day for an acre of corn. Past experience indicates a return of $80 for an acre of wheat and $60 for an acre of corn. How should farmer Jones use his land? (2) You are given the following information on steaks and potatoes: steak potatoes daily needs nutrient (grams) (grams) (grams) carbohydrates 5 15 ≥ 50 protein 20 5 ≥ 40 fat 15 2 ≤ 60 cost per serving $ 4 $ 2 Find the cheapest steak-and-potato diet fitting the nutritional re- quirements. (Note: if you introduce slack, be careful you put it on the correct side of the inequality - it must be added to the smaller side! Also, in this case, the origin “steak=0, potato=0” is not feasible. Because of this, you are given the feasible, non-yummy initial solution potato=30, steak=0. You therefore need to find first the correct table. What variables should be on the left? Those that are non-zero. This will be the potatoes, and 2 slack variables.) 421 COURSE NOTES 13

1.5. Lecture 5: Complications.

1.5.1. Lack of Uniqueness. It is possible that a problem has more than one optimal solution. Example 1.13.

x1 + x2 ≤ 1

x1 + x2 → max

x1, x2 ≥ 0. Of course, every point (a, 1 − a) with 0 ≤ a ≤ 1 is optimal. The simplex method gives

x3 = 1 − x1 − x2

z = x1 + x2 = 0

We add x1 to the basis and kick out x3 to get

x1 = 1 − x2 − x3

z = 1 − x3

So we detect the solution (1, 0, 0). But we might have chosen x2 to enter the basis, leading to (0, 1, 0). We conclude Lemma 1.14. There may be several (and then infinitely many) optimal solutions. If that is the case, it is possible that (depending on the choices for pivoting) different runs lead to different optimal solutions. 2 1.5.2. Existence. Does every LP (linear program) have an optimal solution? The simplex method is a way of obtaining better solutions from known solutions. In order to run one step, we need (given a valid dictionary) • a decision variable with positive impact on the objective function, • the possibility to increase that variable from zero to a positive value. If the first of these fails, then we must have an optimal solution in hand. The entering variable may not be constrained:

1.5.3. Unbounded problems. To ﬁnd a leaving variable, one chooses the one that is determined by the entering variable attaining maximal size while satisfying all constraints. If the entering variable can be chosen of arbitrary size, no exit variable is proposed. The test is that every entry in the pivot column is negative or zero. It means that there is no best solution - there are always better ones. Example 1.15.

x1 ≤ 4

z = 3x1 + 5x2

xi ≥ 0 14 ULI WALTHER

1.5.4. Degeneracy. If the second part above of a simplex step fails, we speak of degeneracy. It means that when setting the decision variables to zero, one or more of the basic ones is also zero. This is therefore an accident and would not happen in a random example, but happens reasonably often in praxis. Degeneration is annoying and there is really nothing one can do against it. It means, that a few steps have to be done without actually increasing the objective function. It is not clear, how many steps are required. Example 1.16.

x1 + x2 ≤ 0

x1 + x2 → max

x1, x2 ≥ 0. This gives 2 simplex steps without any improvement of the objective function. The basic variables that happen to be zero are called degenerate. Occasionally, degeneracy leads to cycling. This means, that after a number of steps caused by degeneracy the dictionary is the same again. Then of course we are stuck in a loop forever. This happens practically almost never. Here is an example. See page 31 in Chvatal. Theorem 1.17. Suppose that an LP is solved by the simplex and that the simplex never stops. Then there is cycling. Proof. There are only a ﬁnite number of ways of selecting basic variables (corners of the polyhedron). So there are a ﬁnite number of dictionaries. So if the simplex does not terminate, some dictionary must be used twice. This is cycling. 2 Fact 1.18. One can avoid cycling by a clever tie-breaking between leaving/entering variables. For example, one can decide always to take the entering/exiting variable with the smallest index. This guarantees that cycling is avoided. There are other methods. See page 34-37.

1.5.5. The perturbation technique. Disturb the m + i-th inequality by εm+i and assume that

0 < εm+n εm+n−1 · · · εm+1 1. Sums involving these quantities and integers are identiﬁable with words in the alphabet with the ε’s, and are ordered like one does in a dictionary:

2 + 12ε2 < 2 + ε1 the same way as AC comes after AB. This will eliminate degenerated corners by separating them, and so eliminate cycling. see page 34. 421 COURSE NOTES 15

1.6. Lecture 6: 2-phase method and Big-M. It is conceivable that in the ﬁrst dictionary (at the very beginning) the origin (all xi = 0) is infeasible. In that case, it is not clear there is any solution.

Example 1.19.

x1 − 3x2 ≤ −5

x1 + x2 ≤ 1

−2x1 + x2 ≤ 2

xi ≥ 0

z = x1 + 4x2 Even if there is one, it may be hard to ﬁnd one.

Example 1.20.

x1 − 3x2 ≤ −1

x1 + x2 ≤ 1

−2x1 + x2 ≤ 2

xi ≥ 0

z = x1 + 4x2

In that case, introduce a new variable x0:

x1 − 3x2 − x0 ≤ −1

x1 + x2 − x0 ≤ 1

xi ≥ 0

z = −x0 → max

Certainly, this system will have feasible solutions (for x0 large). Finding an optimal solution with z = −x0 = a answers whether the initial problem has a feasible solution: if a ≥ 0, setting x0 to zero gives a feasible solution. Otherwise, there is none. Here is the math. Start at

x3 = −1 − x1 + 3x2 + x0

x4 = 1 − x1 − x2 + x0

z = −x0 This is problematic because this is an infeasible dictionary. This is no accident, this will always be the case if we try to work on an infeasible dictionary to start with. But, a single well-chosen iteration is guaranteed to create a feasible dictionary for the modiﬁed problem: exit the least feasible variable (x3) and enter x0: (there is method to this - it is the only substitution that 16 ULI WALTHER will accomplish feasibility under all circumstances)

x0 = 1 + x1 − 3x2 + x3

x4 = 2 − 4x2 + x3

z = −1 − x1 + 3x2 − x3

Setting x1, x2, x3 = 0 gives a solution to the modiﬁed problem, but since we know that this does not solve the original problem, we have to keep working: enter x2, exit x0 (because the x0-condition is more stringent).

x2 = 1/3 − x0/3 + x1/3 + x3/3

x4 = 2/3 − 4x0/3 − 4x1/3 − x3/3

z = −x0 Now we do have a feasible dictionary, so this display suggests that z = 0 is the optimal value and it is achieved with x0 = x1 = x3 = 0, x2 = 1/3, x4 = 2/3. Forgetting x0, x3, x4 we get an initial solution for the original problem of x1 = 0, x2 = 1/3. Clearly it is feasible. Note 1.21. As soon as in the auxiliary problem a feasible solution is reached where x0 is non-basic, we can stop, because the objective function of the auxiliary problem is then zero, proving that forgetting x0 and all slack variables of the auxiliary problem gives a feasible solution for the original problem (in fact, we are at an optimal solution for the auxiliary problem). Hence whenever we can, we should have x0 exit the basis. This is known as 2-phase simplex: • Designed for infeasible origins in original problem. • Introduce x0 and form auxiliary system where origin is again infeasible • Find the least feasible constraint, this determines the exit variable. x0 will enter the basis. • After this initial simplex step the dictionary of the auxiliary problem will be feasible. Solve the auxiliary problem. • If in the optimal solution x0 > 0, the original problem is infeasible. Otherwise forget about x0 in the optimal solution, this will give a feasible solution to the original problem. Theorem 1.22 (Fundamental Theorem). Every LP in standard form sat- isﬁes • Either it has an optimal solution or it is unbounded or it is infeasible. • If it is feasible, it has a basic feasible solution. • If it has an optimal solution, it has a basic optimal solution. Proof. The two phase simplex proves the second item. If a basic feasible solution is at hand, the normal simplex either exhibits an optimal basic solution or discovers unboundedness. 421 COURSE NOTES 17

Homework 3 (for week 4). (1) Solve the following linear program by the 2-phase method:

2x1 + 3x2 + x3 → max

x1 + 4x2 + 2x3 ≥ 8

3x1 + 2x2 ≥ 6

x1, x2, x3 ≥ 0. (Recall: if a constraint has a ≥, introduce the slack on the right and solve for slack. The ﬁrst dictionary you then get is infeasible because the constants are negative. 2-phase suggests to introduce a new variable, x6, and subtract it on the left. Then solve an auxiliary problem maximizing −x6...) (2) Solve the following linear program by the 2-phase method:

3x1 + 2x2 → min

2x1 + x2 ≥ 10

−3x1 + 2x2 ≤ 6

x1, x2 ≥ 0. (For this one, note that minimizing z is the same as maximizing −z. Introduce slack on the correct side, and use an auxiliary variable to get 2-phase started.) 18 ULI WALTHER

1.7. Lecture 7: Dealing with a general LP. We have so far solved problems where

• problems are of max-type, • we maximize, • constraints are of less-or-equal type.

The 2-phase method allows us to deal with problems where the constants are negative. In general, we could have several bad things:

• min-problems, • greater-or-equals, • equalities, and both these could come with negative right hand sides. Here is the strategy how to deal with them.

(1) Make the problem a max-problem, put all constants on the right, all variables on the left. (2) Make each right hand side non-negative (by multiplying with -1). (3) In each equality, A = B, introduce a auxiliary variable A + xA=B = B. One for each equation. Each of these gets coeﬃcient −M in z where we imagine M to be very large. (4) In each greater-or-equal constraint A ≥ B, introduce 2 new variables: A − xA≥B + xA≥B = B. The variable xA≥B gets a −M in z as well, the variable xA≥B gets a zero in z. (5) Introduce slack in all ≤ constraints. (6) In the basis for the ﬁrst dictionary are • all slack variables • all xA=B variables • all xA≥B. They make up a feasible dictionary. (7) Solve the problem with the simplex (and the new z.)

Example 1.23.

x1 + x2 → min

x1 + 2x2 = 3

x1 − 2x2 ≤ −4

x1 + 5x2 ≤ 1

x1, x2 ≥ 0 421 COURSE NOTES 19

First make it a maximize, and make all constants positive.

−x1 − x2 → max

x1 + 2x2 = 3

−x1 + 2x2 ≥ 4

x1 + 5x2 ≤ 1

x1, x2 ≥ 0 Now get a new variable to deal with the equality, and introduce slack:

−x1 − x2 − Mx3 → max

x1 + 2x2 + x3 = 3

−x1 + 2x2 ≥ 4

x1 + 5x2 + x4 = 1

x1, x2, x3, x4 ≥ 0 For the ≥ constraint, get the 2 new variables:

−x1 − x2 − Mx3 − Mx6 → max

x1 + 2x2 + x3 = 3

−x1 + 2x2 − x5 + x6 = 4

x1 + 5x2 + x4 = 1

xi ≥ 0

Now the basic variables are the slack x4, the extra variable x3 in the equation constraints, and the second variables in all ≥ constraints x6.

x3 = 3 − x1 − 2x2

x6 = 4 + x1 − 2x2 + x5

x4 = 1 − x1 − 5x2

xi ≥ 0

z = −x1 − x2 − M(3 − x1 − 2x2)

−M(4 + x1 − 2x2 + x5)

= −7M − x1 + x2(4M − 1) − Mx5 which is a feasible dictionary. Now do iterations as usual. First, x2 in, x4. It turns out, that the new z is

z = −7M + 4(−M − 1)x1/5 + (−4M + 1)x4/5 − Mx5. This shows that after this last change of basis we have an optimal solution. It has the values

x1 = x4 = x5 = 0, x2 = 1/5, x3 = 13/5, x6 = 18/5. The problem is, that z has an enormously bad value, since M is huge. It means that the best we can do is terrible, and the reason is that x3 and x6 show in the optimal solution. Because of that, the original problem cannot 20 ULI WALTHER have a feasible solution: if it had one, x3 and x6 could both be zero and we could much improve the z here! So a z with a negative M-coeﬃcient indicates an infeasible original problem. A z without any M in it constitutes an optimal solution of the original problem (after dumping the newly introduced variables x3, . . . , x6 of course). z cannot show up with a positive M ever because it would mean that x3 or x6 are negative. This is called the Big-M-method. 1.7.1. Strategies of Selection. Thus far we have always taken as entering the variable that seems most promising from the point of view of the cost function. But this may be foolish, due to the fact that the same problem reformulated can make a diﬀerent suggestion. Consider Example 1.24.

3x1 + x2 + 3x3 ≤ 30

2x1 + 2x2 + 3x3 ≤ 40

xi ≥ 0

z = 4x1 + 3x2 + 6x3

The basic variables are (x4, x5), (x3, x5), (x2, x3) if we use steepest increase in z. 0 Now assume that someone uses a different scale. Say, x1 = 2x1. Then the system becomes Example 1.25. 0 6x1 + x2 + 3x3 ≤ 30 0 4x1 + 2x2 + 3x3 ≤ 40 0 x1, x2, x3 ≥ 0 0 z = 8x1 + 3x2 + 6x3 0 Suddenly, the first change will go from (x4, x5) to (x1, x5) despite the fact that there is no geometric difference between the 2 examples. This is bad. We should like a method that is independent of the scale we use. But it’s much worse actually. There are examples of Klee and Minty that are not very large (n equations) where the simplex when run as we have seen it goes through 2n − 1 corners. This is terrible. Maybe the steepest increase rule is not so good, because it only looks at one part of the problem. Instead one could ask for the actual increase in the objective function. Then Klee-Minty goes in one step. 2 problems: first, the selection process is actual work (for each possible enter variable one needs to check what the result in the objective function would be, instead of just looking at a number). Second, there are examples for this strategy that are just as bad as steepest increase with Klee-Minty. 421 COURSE NOTES 21

One can use other variations. But it seems: for all strategies in the simplex there are terrible examples. But they are extremely rare. Most of the time, the simplex is very fast. There are no complete explanations for this, but some partial ones too long to give them here (see references page 46). Perhaps the most important thing is to use a strategy that rules out cycling (like for example always take the entering and exiting variable to be those with smallest subscript). 22 ULI WALTHER

1.8. Lecture 8: Duality. Consider

x1 − x2 − x3 + 3x4 ≤ 1

5x1 + x2 + 3x3 + 8x4 ≤ 55

−x1 + 2x2 + 3x3 − 5x4 ≤ 3

xi ≥ 0

4x1 + x2 + 5x3 + 3x4 → max Each feasible solution gives an “approximation” (bad, maybe) for the optimal z. Called “lower bounds”. How about upper bounds, i.e., numbers that we know z cannot exceed? For example, add row 2 and row 3:

4x1 + 3x2 + 6x3 + 3x4 ≤ 58. Comparing this with the objective function, we know that z ≤ 58, without having the slightest idea whether the system is even feasible. Perhaps other linear combinations would give even better estimates? Sup- pose we take 3 numbers y1, y2, y3 and use it to make a linear combination: y1(row 1) + y2(row 2) + y3(row 3). This is after some reordering

(1.4) (y1 + 5y2 − y3)x1

+(−y1 + y2 + 2y3)x2

+(−y1 + 3y2 + 3y3)x3

+(3y1 + 8y2 − 5y3)x4 ≤ y1 + 55y2 + 3y3. We are trying to find those y-values for which the LHS is coefficient by coefficient at least the objective function. Note that we must use nonnegative values for y because otherwise the constraints flip the relation sign.Thus we require

y1 + 5y2 − y3 ≥ 4

−y1 + y2 + 2y3 ≥ 1

−y1 + 3y2 + 3y3 ≥ 5

3y1 + 8y2 − 5y3 ≥ 3

yi ≥ 0 Now if we had such y, what would we do? We’d look at the right hand side of (1.4) and evaluate. This gives a number that z cannot exceed. Clearly we’d like this to be small to get a good upper bound. We add this to the inequalities above to get a new LP:

y1 + 55y2 + 3y3 → min . This is the dual problem. The one we started with is called primal. 421 COURSE NOTES 23

Note that the dual is somehow the transpose of the primal. Explicitly, if the primal is A · ~x ≤ ~b ~x ≥ ~0 z = ~c · ~x → max then the dual is AT · ~y ≥ ~c ~y ≥ ~0 w = ~b · ~y → min If the primal has m constraints and n variables, the dual has n constraints and m variables. Each variable in one system corresponds precisely to one constraint in the other. Consider now a feasible solution ~y of the dual problem. Then of course AT ~y ≥ ~c. Multiplying by any feasible solution ~xT , ~xT AT ~y ≥ ~xT ~c. On the other hand, A~x ≤ ~b or ~xT AT ≤ ~bT because ~x is feasible. Therefore ~bT ~y ≥ xT AT ~y ≥ ~xT ~c. This is true for all choices of feasible solutions ~x, ~y and in particular the optimal ones. Hence (1.5) z ≤ w. Consider now the feasible solution ~x = (0, 14, 0, 5) with z = 29. Also, consider the feasible solution ~y = (11, 0, 6). It gives w = 29. How interesting. This says (1) the optimal (maximal) z is at least 29, (2) the optimal (minimal) w is at most 29 (3) by (1.5), z is at most 29 and w is at least 29. The conclusion is that 29 is the optimal value for both the primal and the dual. 24 ULI WALTHER

1.9. Lecture 9: more duality. Here is another weird thing. It turns out that if we introduce slack and solve the primal system, the ﬁnal dictionary is

x2 = 14 − 2x1 − 4x3 − 5x5 − 3x7

x4 = 5 − x1 − x3 − 2x5 − x7

x6 = 1 + 5x1 + 9x3 + 21x5 + 11x7

z = 29 − x1 − 2x3 − 11x5 − 6x7

It is spectacular that the coeﬃcients of x5, x6, x7 in the optimal solution are exactly (minus) the optimal solution for the dual system. This is not an accident and in fact the crucial idea of the proof for the next theorem: Theorem 1.26. Let the primal be A~x ≤ ~b, ~x ≥ ~0,~cT ~x → max, let it have the origin feasible. If the primal system has an optimal solution with optimal value z then the dual has an optimal solution with the same optimal value.

Proof. Suppose the last dictionary gives m+n ∗ X z = z + ckxk k=1 where z∗ is a number, the optimal solution. By deﬁnition of z∗, z∗ = Pn Pn j=1 cjxj. Keep in mind that xn+i = bi − j=1 ai,jxj. So n n+m ∗ X X z = z + cjxj + cixi j=1 i=n+1 n n+m n ∗ X X X = z + cjxj + ci(bi−n − ai−n,jxj) j=1 i=n+1 j=1 m n m ∗ X X X = (z + bicn+i) + (cj − ai,jcn+i)xj i=1 j=1 i=1 The relevant thing in this display is to realize that it is simply a rewriting for z. In particular, it expresses z for any choice of x (even non-feasible ones!). In particular, put all x equal to zero. Then z = 0 (inspect the original T ∗ Pm formulation c x → max!) and hence z = − i=1 bicn+i. ∗ ∗~ ∗ It remains to show that yi = −cn+i is feasible because then ~y b = z . ∗ Since ci ≤ 0 by optimality of the last dictionary, yi is nonnegative. If we T put all x except xj to zero, then z = c x is simply cjxj. On the right ∗ Pm hand side, we saw above that z = − i=1 bicn+i so the ﬁrst bracket is zero. Pm Left on the right is therefore (cj − i=1 ai,jcn+i)xj. Equating the two and Pm Pm ∗ cancelling xj, cj = cj − i=1 ai,jcn+i and so cj ≤ i=1 ai,jyi (recall the last 421 COURSE NOTES 25

∗ dictionary is optimal and so the cj are negative). So y is feasible for the dual. 2 Example 1.27. Back to the bicycle maker of Example 1.12:

12x1 + 15x2 → max

20x1 + 30x2 ≤ 2400

15x1 + 40x2 ≤ 3000

x1 + x2 ≤ 100

x1, x2 ≥ 0 (this is in minutes)! The dual problem is

2400y1 + 3000y2 + 100y3 → min

20y1 + 15y2 + y3 ≥ 12

30y1 + 40y2 + y3 ≥ 15

y1, y2, y3 ≥ 0 The optimal value of the problem was 1320. By considering that the optimal value is a profit, it becomes clear that yi must be a profit as well, measuring the usefulness of a resource of type i (assembling, finishing, or a raw frame). In the optimal solution, x1 = 60, x2 = 40. All finishing time and all frames are used, but there is extra (500 minutes) of assembling time. The optimal y-values are 3/10, 0 and 6. Thus we have 30 cents pro finishing minute, none pro assembling minute, and $ 6 per frame. This is of course not what we actually make, but what we might make provided we had more work. In other words, if we had another frame and the same number of people in assembly and painting, we’d make another 6 bucks (and the numbers of how many 3- and 5-speed bikes we make would change of course). Or, if we had another minute of painting, we could get another 30 cents. Had we another minute of assembly, we would still only make 1320 bucks, because we already have some assembly time extra. Of course if we had a unlimited extra number of frames, we could not make an unlimited extra amount of money, because the painters and the finishers could not do the work. But at least for a while, the increase would be 6/frame. We may later consider how much we may increase profits by increasing a resource with a nonzero profit promise before we run into problems in other resources. This also indicates to what prices the management should be interested to buy extra resources of a certain type. For example, if they can get bike frames from someone else at the current price plus 3 bucks, this would be good because they’d still make 3. ∗ yi is sometimes called marginal value of the resource. (Refers to margin of profit.) It represents the difference of price in a resource on the common market vs to what the manager thinks it is worth to him. 26 ULI WALTHER

1.10. Lecture 10: Assorted facts on duality.

1.10.1. Eﬀectivity. If we get to an example where the matrix is say 100 by 10, then empirically the number of iterations is proportional to (number of rows) times log(number of columns) = 100 · 2.5 = 250. The dual problem would have around 10 · log(100) = 50 iterations. Since solving either one exhibits the solution to the given one, one ought to do the one with fewer rows. Lemma 1.28. The dual of a dual is the primal.

Proof. This is pretty clear. 2 Fact 1.29. table page 60. Note also another curious thing. The primal constraints that are made equalities by the optimal solution correspond to dual nonzero variables. The primal constraints that are inequalities correspond to dual zero variables. This is actually pretty clear. The optimal y-values represent a linear combination of the inequalities that best estimates the primal objective function. So, the “harshest” inequalities will show up in this estimate, the more lax ones won’t.

Proposition 1.30 (Complementary slackness). Suppose x1, . . . , xn and y1, . . . , yn are feasible solutions of the primal and the dual problem respectively. They are (simultaneously) optimal if and only if m X ai,jyi = cj or xj = 0 or both i=1 and n X ai,jxj = bi or yi = 0 or both j=1 for all i, j.

∗ Proof. In the duality theorem we proved that yi is negative the coefficient of xn+i in z of the optimal dictionary. Now, if xn+i is in the basis for the final optimal dictionary of the primal, ∗ then it will not show up in the expression for z , so its coefficient cn+i is ∗ zero. So yi = 0. If on the other hand xn+i does not show up in the optimal basis, then of course it must be zero itself in the optimal solution. ∗ ∗ We conclude that for each index i, either xn+i or yi is zero. That is the ∗ same as saying that either yi = 0 or the i-th primal constraint is an equality (or both). That proves the first part. To see the second part, just exchange the position of the dual and the primal problem (which is possible because the dual of the dual is the primal). 2 421 COURSE NOTES 27

Recall, that each yi was matched up with a slack in the primal problem, and naturally every xj is matched with a slack in the dual. This proposition says that for all these pairs of variables, at least one of them must be zero. (Or both.) What is the point? Suppose we have just a bunch of proposals for ∗ ∗ x1, . . . , xn, but no y-values. We want to know whether the x’s are optimal. (With the y-s, this would be easy because we just plug them into the primal and the dual objective functions, and see if the values are the same, see the duality theorem.) Pick hence all the x’s that are not zero, and write down the corresponding constraint from the dual problem. With slack, this gives a bunch of equations. Solve them and test for complementary slack. If it fails, the x’s were not optimal. There is a drawback to this, it only works if the x’s form a non-degenerate basic feasible solution (otherwise the system for the y’s does perhaps not solve uniquely.) . It seems likely to me, that non-degenerate and feasible are suﬃcient. 28 ULI WALTHER

Homework 4 (for week 5). (1) Problem 5.4 in the book.(Explain how you got the solution, do not just state what it is. Note that there are hints on page 64.) More importantly, note that the solution given to Exercise 1.6 at the end of the book is wrong – it has the wrong number of variables. (2) Construct and graph a problem with the property that it has no feasible solutions. Graph the dual as well and show that it is unbounded. (Use 2 constraints in 2 variables for the primal.) (3) Construct and graph a problem in 2 variables with 2 constraints that has no feasible solutions and its dual has no feasible solutions either. (4) Make up a feasible problem in 2 variables with 2 constraints that has more than 1 optimal solution. Investigate, what eﬀect this mul- tiplicity has for the dual problem. Try to prove your claim. (To get an idea, take the primal and see what happens to the dual if you change the objective function ever so slightly.) 421 COURSE NOTES 29

1.11. Lecture 11: A problem for duality.

Example 1.31. A forester has 100 acres of hardwood timber. Felling and letting the area regenerate would cost $ 10 per acre immediately, and bring a return of $ 50 per acre. An alternative idea is felling, followed by replanting pine. That would cost $ 50 immediately per acre, and brings $ 120 per acre later. However, only $ 4000 starting capital is available. (1) Set up and solve the primal problem of maximizing proﬁt. (2) Determine the solution to the dual. (3) Suppose the forester could take a loan of $ 100 (where she would have to pay back $ 180 later). Should she do that? (4) If the forester could invest $ 100 in bonds that would return $ 145, should she do that? (5) Suppose she could also fell combined with conifer planting. Let’s say this would cost a dollars per acre to do it and returns A dollars later on. Under what circumstances (i.e., conditions on a, A) should she do some of that?

First we set up the problem:

40x1 + 70x2 → max

x1 + x2 ≤ 100(acres)

10x1 + 50x2 ≤ 4000(money)

xi ≥ 0

Use (0,0) as initial solution:

z = 40x1 + 70x2

x3 = 100 − x1 − x2

x4 = 4000 − 10x1 − 50x2

xi ≥ 0 enter x1, exit x3:

z = 40(100 − x2 − x3) + 70x2

= 4000 + 30x2 − 40x3

x1 = 100 − x2 − x3

x4 = 4000 − 10(100 − x2 − x3) − 50x2

= 3000 − 40x2 + 10x3

xi ≥ 0 30 ULI WALTHER enter x2, exit x4:

z = 4000 + 30(75 + x3/4 − x4/40) − 40x3

= 6250 − 65x3/2 − 3x4/4

x1 = 100 − (75 + x3/4 − x4/40) − x3

= 25 − 3x3/4 + x4/40

x2 = 75 + x3/4 − x4/40

xi ≥ 0

Recall, yi is the coefficient of (−slacki)in z. In this optimal table, both slacks ∗ ∗ are zero. For example, y1 = −(−65/2), y2 = −(−3/4). ∗ ∗ ∗ Also, x1 = 25, x2 = 75, z = 6250. This finishes parts 1 and 2. If she takes a loan of $ 100, she could expect an additional profit of 3/4 times $ 100, because 3/4 is the marginal price on available capital. So this would not be so great, because 1.8 > 1.75. On the other hand, if she invests a little money (say $t, then she will have (from the forest) have an expected 3t/4 dollars less in revenue. Hence, for small amounts of investments that return 45% she should not go for it. (Large amounts will have more damaging effects on the forest revenues!) Now if she could plant conifers, let’s say she uses t acres for that. Then her loss compared to the optimal table above would be (32.5 + 0.75a)t dollars. On the other hand, she would later make tA dollars. Hence, she needs to see whether A > 32.5 + 0.75a for the conifer business to make sense. Note: 40x1 + 70x2 is the profit, not the revenue. So, in the loan question, it’s really $1.75, not $0.75 that we compare to. 421 COURSE NOTES 31

1.12. Lecture 12: Post-optimal considerations and general Dual- ity.

1.12.1. Post-optimal Considerations. Suppose we have solved a problem, and then the conditions or the objective function change. What to do? Let

(1.6) A~x ≤ ~b ~x ≥ ~0 ~cT · ~x → max be the given system with optimal solution ~x∗.

T (1) Change in ~c: c~0 · ~x → max: The optimal solution to the initial system (1.6) is going to be still feasible for the changed system. If the change in ~c was not big, the optimal solution to the new system should be “close” to ~x∗. Idea: start with ~x∗ for new system. (2) Change in ~b: A~x ≤ ~b0 In this case, ~x∗ may not be feasible anymore. Trick: look at the dual system. The last dictionary to (1.6) reveals the dual optimal solution, and this dual optimal solution ~y∗ is still feasible for the changed dual system. Idea: solve this one with initial solution ~y∗ and use the final dictionary to find the new optimal solution for the changed primal system. (3) Change in A: A0~x ≤ ~b Two steps: first consider an intermediate problem

A0~x ≤ A~0~x∗ ~x ≥ ~0 ~cT · ~x → max

This has ~x∗ as feasible solution. Solve it starting at ~x∗. Then from this intermediate to the changed system is a step of the sort “change ~b”. Thus, take the optimal solution of the dual intermediate problem and use it as a start solution to the changed dual. Compute an optimal dual and read oﬀ the optimal changed primal. (4) Changes in more than one of A,~b,~c: First change ~c, then ~b, then A according to the steps above. 32 ULI WALTHER

1.12.2. General Duality. The LP n X cjxj → max j=1 n X ai,jxj ≤ bi (i ∈ I) j=1 n X ai,jxj = bi (i ∈ E) j=1

xj ≥ 0 (j ∈ R) has dual m X biyi → min i=1 m X ai,jyi ≥ cj (j ∈ R) i=1 m X ai,jyi = cj (j ∈ F ) j=i

yi ≥ 0 (i ∈ I) here, F are the free x’s, R the ones restricted to nonnegative values, I all indices to inequalities and E all indices to equations in the primal. The meaning of the dual is as before: any collection of numbers that Pm ﬁt all dual constraints produce a i=1 yibi-value that bounds the objective Pn function of the primal. Conversely, a feasible ~x produces a j=1 xici that bounds the dual objective function from below. We have a correspondence for the same index labeling the following things in (D) and (P) respectively: In (D) In (P) restricted variables inequality constraints free variables equality constraints inequality constraints restricted variables equality constraints free variables Again, duals of duals are primals. The default way of constructing the dual of a max-problem is as follows: 0 (1) Write every inequality of the form xi ≤ 0 as xi ≥ 0 replacing xi by 0 −xi if necessary. (2) Write all other inequalities (diﬀerent from those of the previous item) as ≤. Including those that say x2 ≥ 4 etc. (3) apply the formalism given at the beginning of this section. If one has to dualize a min-problem, 421 COURSE NOTES 33

0 (1) Write every inequality of the form xi ≤ 0 as xi ≥ 0 replacing xi by 0 −xi if necessary. (2) Write all other inequalities (diﬀerent from those of the previous item) as ≥. Including those that say x2 ≤ 4 etc. (3) apply the formalism given at the beginning of this section backwards. For example, let’s dualize

5x1 + 3x2 + x3 = −8

4x1 + 2x2 + 8x3 ≤ 23

6x1 + 7x2 + 3x3 ≥ 1

x1 ≤ 4, x3 ≤ 0

3x1 + 2x2 + 5x3 → min Step one:

5x1 + 3x2 − x3 = −8

4x1 + 2x2 − 8x3 ≤ 23

6x1 + 7x2 − 3x3 ≥ 1

x1 ≤ 4, x3 ≥ 0

3x1 + 2x2 − 5x3 → min Step 2:

−4x1 − 2x2 + 8x3 ≥ −23

6x1 + 7x2 − 3x3 ≥ 1

−x1 ≥ −4

5x1 + 3x2 − x3 = −8

x3 ≥ 0

3x1 + 2x2 − 5x3 → min so R = {1, 2, 3}, F = {4}, I = {3}. This implies, E = {1, 2}. And dualize:

−4y1 + 6y2 − y3 + 5y4 = 3

−2y1 + 7y2 + 0y3 + 3y4 = 2

8y1 − 3y2 + 0y3 − 1y4 ≤ −5

y1, y2, y3 ≥ 0

−23y1 + 1y2 − 4y3 − 8y4 → max The duality theorem still holds: if both (P) and (D) are feasible, their optimal values re the same. Also, one can modify the usual simplex to a dual simplex method such that the ﬁnal dictionary gives both the solution to (P) and to (D). 34 ULI WALTHER

1.13. Lecture 13: The Revised Simplex. The key idea is that updating may be made more eﬃcient. It is not true that this new method is always more eﬃcient. But heuristically, it is for practical problems.

Example 1.32. Suppose we are solving

x2 ≤ 10

x1 + 2x2 ≤ 15

x1 + x2 ≤ 12

5x1 + 3x2 ≤ 45

x1, x2 ≥ 0

10x1 + 20x2 → max Then the ﬁrst dictionary has 4 equations, saying (essentially) that

x2 + x3 = 10

x1 + 2x2 + x4 = 15

x1 + x2 + x5 = 12

5x1 + 3x2 + x6 = 45

“Essentially” refers to the fact that x3, x4, x5, x6 are on the left, all other entries get moved to the right. After one iteration, we have again 4 equalities, this time x1, x3, x4, x6 are on the left. But, they are the same equations. In fact, all dictionaries encode the same system just diﬀering in which variables are on the left.

Consequence: A dictionary solving for xi1 , . . . , xim may be computed from the original system by • writing down the matrix of the original system, • putting the columns i1, . . . , im up front, • computing the rref. If we write B for the elements in the basis (left side) and N for the decision (non-basic) ones and A for the matrix describing the initial system and AB,AN and ~xB, ~xN for the two components, we have

A · ~x = ~b is equivalent to ~ AB~xB + AN ~xN = b and so −1~ −1 ~xB = AB b − AB AN ~xN .

The matrix AB is invertible because of a little argument given on page 100 of the book. If one also splits ~c into the basis and the decision part, the 421 COURSE NOTES 35 dictionary to ~cT ~x → max A~x = ~b ~x ≥ ~0 z = ~cT · ~x T T = ~cB · ~xB + ~cN · ~xN takes the form −1~ −1 ~xB = B b −B AN ~xN | {z∗ } ~xB T −1~ T T −1 z = ~cBB b +(~cN − ~cBB AN )~xN . | {z∗ } zB −1~ Note that B b is the vector of numbers that tells the values of ~xB at this moment. To perform one step in the revised simplex, we need to know what exactly the entering/exiting variable is.

1.13.1. The entering one: It can be any one that has a positive entry in z, T −1 so z must be computed. That requires the solution of ~v = cBB , followed by computation of ~cN − ~vAN . 1.13.2. The exiting one: It is found by ﬁnding the variable that ﬁrst hits zero when the entering variable is increased. Of course, only B-variables ∗ are considered. Let’s say the current value of ~xB is ~xB, and the entering ∗ −1 ∗ variable is xj. Note that ~xB = ~xB − B AN ~xN starts at ~xB (when xj is still ∗ ~ ~ zero) and becomes ~xB − xjd where d is the column that corresponds to xj −1 in B AN . Note also, that because of that d~ can be represented as B−1~a where ~a is really the xj-column of A. So we need to compute d~ from ~a and then see ∗ ~ when a component of ~xB − xjd becomes zero. Let’s say that happens to the variable corresponding to xi. These were 2 necessary computations that revised simplex needs which are unnecessary in standard simplex.

1.13.3. Updating: To update now the dictionary we need to do nothing however: we just have to write down the new choice of B (which is the old B ∗ ∗ ~ without column i but with column j), and the new ~xB obtained as ~xB − xjd with d~ and xj as computed above. ∗ So, the revised simplex at any stage only recalls the current ~xB, and what variables are in B. All other things are computed only when needed. This ought to be better than standard simplex, because it is a bit like computing only one row and one column of a dictionary. 36 ULI WALTHER

Algorithm 1.33 (One revised step). Input: The basis B = AB, and the ∗ current value ~xB. ∗ Output: The next B and ~xB T T (1) Solve ~v B = ~cB. T (2) Choose an entering index: ﬁnd j ∈ N such that ~v Aj < (~cN )j. T T (This makes the j-th component of ~cN − ~v AN strictly positive!) If no such j exists, the current solution is optimal. Set ~a = Aj. (3) Solve Bd~ = ~a. ∗ ∗ ~ ~ (4) Find the largest value xj for t such that ~xB − td ≥ 0 component- wise. If no such t exists, the problem is unbounded. (To obtain large ∗ ~ values of the objective function, use ~xB − td for large t.) If such t can be found, the variable xi whose row becomes zero is the exiting variable. ∗ ∗ ∗ ~ ∗ (5) Replace ~xB by ~xB − xj d, get rid of xi in the basis, and put in xj . Each step involves the solution of two m × m-systems (in Steps 1,3). Each iteration in the usual simplex amounts to computing the row reduced echelon form of an m × (m + n) matrix. This can be thought of as solving n linear m × m-systems at once. In all likelihood then the revised simplex should beat the usual one. The revised simplex eﬃciency depends on the “sparsity” of the matrix A. If the percentage of nonzero entries in A is α (0 ≤ α ≤ 1), then revised 2 simplex is useful if n > (1 + 1−α )m. (See Judin/Goldstein.) 421 COURSE NOTES 37

Homework 5 (for week 6). (1) Problem 5.6 from the text. (2) Theorem 5.5 in the text states what we have learned informally in class: the y-components of the dual optimal solution to an LP in standard form talk about the marginal values of the resources. This means, that if the amount of resource number i (that is, bi) is increased a tiny bit ti, then the optimum of the objective function ∗ ∗ z increases by yiti (this is of course only helpful, if yi 6= 0 which happens only provided that the corresponding slack variable xm+i is zero, i.e., the resource is maxed out in the optimal solution – recall the complementary slack theorem!). This actually is only guaranteed to work if the optimal solution is not degenerate. Construct an example with a degenerate optimal solution where Theorem 5.5 fails. (Hint: take 2 constraints in 2 variables only. Pick them in a non-random way and then choose a good objective function.) Explain how “degenerate” is required to make the example work. (3) (5 Extra points) 5.8 in the text. 38 ULI WALTHER

1.14. Lecture 14: Filling the knapsack. In many cases, fractional solutions are no good. For example, one cannot really produce 3.7 bikes or sell 13/7 of a Mercedes. The class of problems in which constraints and answers are integers as opposed to reals are called integer LP’s. In principle, one could solve the real LP and then take a lattice point that is close to the optimal corner. The problem is, there may not be one close that is feasible. And the closest may not be optimal. Example 1.34. Here is an example where the real and integral optimum are maximally apart.

x/5.5 + y/1.5 ≤ 1 x, y ≥ 0 x/5.5 + y/1.4 → max The real optimum is at (0, 1.5) while the integral optimum is at (5, 0). It is extremely hard to ﬁnd for huge m, n the actual integral optimum. We shall consider a few integer LP’s, the knapsack is the ﬁrst. In standard notation, the knapsack problem is A · ~x ≤ ~b

xi ≥ 0 xi ∈ Z ~cT · ~x → max Typically, all entries in A and ~b are non-negative (no “anti-gravity” and no “customs”). Note, by the way, that A is only one row and ~b is just a number. We hence write just b. It is obvious that we can then assume that all ci are positive (because if any xi has a negative ci we’d just never think of taking it – this would be different if A or b were allowed to have negative entries). One may in several ways think of “value” in this context. For example, • order the articles by increasing volume, • or by decreasing objective coefficient • or by decreasing ci/ai (some sort of a value-density, called efficiency), • or other versions of this. We shall sort them by efficiency. So

c1/a1 ≥ c2/a2 ≥ · · · ≥ cn/an. ∗ ∗ We note immediately that any solution ~x for which A · ~x + ak ≤ b cannot be optimal, because we could increase xk by one and still be feasible. So for all k we have in an optimal solution ∗ A · ~x + ak > b. 421 COURSE NOTES 39

Every feasible solution that has this property (optimal or not) we call sensible. Since optimal solutions must be sensible, let’s ﬁgure out how to deal with sensible solutions. For example,

33x1 + 49x2 + 51x3 + 22x4 ≤ 120

4x1 + 5x2 + 5x3 + 2x4 → max There are 13 sensible solutions, they are illustrated on page 202. Recall the notion of a graph?

Deﬁnition 1.35. A graph consists of a collection of vertices {v1, . . . , vN } together with a collection of edges {(vi1 , vj1 ),..., (viM , vijM )}. Edges are unordered pairs, may occur repeatedly (multiple edges), and may join a vertex with itself (loops). Vertices may be isolated.

A path in a graph is a sequence of edges {(vl1 , vl2 ), (vl2 , vl3 ),..., (vlk−1, vlk )} such that each edge ends with the vertex the next one starts with. A circuit is a path whose starting and ending vertex are the same. The picture on page 202 is a special graph, called a tree. A tree is a connected graph without any circuits (meaning that for each choice of 2 vertices there is exactly one way of getting from one to the other). Actually, the picture is that of a rooted tree. That means, one of the “ends” of the tree has been chosen as root, and all other ends are “leaves”. Every vertex in a tree is also called a node, some occur at branchings, some just in the middle of a branch. The meaning of our tree is that if you start at the root of the tree and follow the path to any leaf, at each node you are forced to make a decision about the value of a variable. Namely, at node i (counting the root as node 1) you specify what xi should be in the sensible solution. Note, that at the last node (the leaf) there is no choice really, you just take as many x4’s as the knapsack can hold). The picture is drawn in such a way that higher branches coming out of the same node correspond to higher values of the xj chosen at that node. Now, the topmost branch corresponds to the choice of values $ j−1 ! % X xj = b − aixi /aj i=1 making successively x1, x2,. . . as big as possible. In fact, all sensible solutions can be thought of as words in a strange alphabet: if for example ~x = (1, 0, 0, 3), write the word x1x4x4x4. The tree is made in such a way that the sensible solutions are ordered alphabetically from top to bottom. Let’s think of the tree as some kind of Oxford Martian Dictionary, OMD. 40 ULI WALTHER

1.15. Lecture 15: Knapsack, day 2.

Deﬁnition 1.36. The incumbent is the currently best, known solution.

In a sense, one could just list all sensible solutions and see what z they give (there are only ﬁnitely many sensible solutions – see the homework). But, as for dictionaries for usual simplex methods, the number of sensible solutions is likely to be prohibitive. So we ought to look through the list with a bit of cleverness. The strategy is to start with the top leaf mentioned above, and scanning through the branchings looking for better solutions. Here is the algorithm for scanning through all branches: Algorithm 1.37 (moving to the next sensible solution in the OMD). Input: A leaf in the tree, (x1, . . . , xn). Output: The next leaf in the OMD.

(1) Set k = max{i : xi > 0} the index of the last letter in the word to ~x. (2) Replace xk by xk − 1 (that frees up some space in the knapsack). (3) For j > k set

$ j−1 ! % X xj = b − aixi /aj i=1

which maxes out the variables after xk in order of the alphabet.

Now some of the branches we move along are going to be truly hopeless. Let’s compare the z-value of the currently best and the new solution from this algorithm. ∗ Let’s say the incumbent is ~x , the current branch is (x1, . . . , xk, ∗,..., ∗), and we are inspecting the new proposal ~x = (x1 = x1,..., xk−1 = xk−1, xk = ∗ xk − 1, xk+1,..., xn). Assume that ~x produces z = M. The way the variables are ordered implies that

ck+1/ak+1 ≥ · · · ≥ cn/an.

So the value of the tail (xk+1,..., xn) is no more than

n n n X X ci ck+1 X cixi = aixi ≤ aixi. ai ak+1 i=k+1 i=k+1 i=k+1 421 COURSE NOTES 41

Since ~x is feasible, n X aixi ≤ b, i=1 and we have also n k X X aixi ≤ b − aixi i=k+1 i=1 and so n k k ! X X ck+1 X c x ≤ c x + b − a x . i i i i a i i i=1 i=1 k+1 i=1 If now k k ! X ck+1 X (1.7) c x + b − a x ≤ M i i a i i i=1 k+1 i=1 the entire branch (x1,..., xk, ∗ ..., ∗) (no matter what numbers we give to ∗ xk+1,..., xn) will be no better than ~x . The point is, that this test can be performed without giving a value to xk+1,..., xn. Also, the more branches have been inspected, the better M is and the easier branches will be ruled out. Running along the example, we have an initial solution ~x∗ = (3, 0, 0, 0). Here, M = 12. The index of the last letter in the word to ~x∗ is 1, so k = 1. Change x1 from 3 to 2. Test the inequality: 5 LHS = 8 + (120 − 66) ≈ 13.5 49 As M = 12, this branch might be better than the current best known one. The top leaf in this branch is (2, 1, 0, 0) with value 13. So we now have ~x∗ = (2, 1, 0, 0). This time, k = 2 is the index of the last letter. So we set now x1 = 2, x2 = 1 − 1 = 0. The inequality says now that 5 LHS = 8 + (120 − 66) ≈ 13.3 51 would have to be bigger than 13. It appear that this could be true, but we are mislead: any integral solution will result in integer M (because ~c is integral) and there is no integer that is less than 13.3 and bigger than 13 at the same time. So this branch is useless. We don’t have to think any further about solutions that start (2, 0, ∗, ∗). So we reduce further, again k = 1. So the next proposal is x1 = 1. To see if that branch is useful, check the inequality: 5 LHS = 4 + (120 − 33) ≈ 12.9 49 42 ULI WALTHER does not beat 13, so we don’t have to worry. Similarly x1 = 0 (one further reduction with k = 1) gives 5 LHS = 0 + (120 − 0) ≈ 12.2 49 which again does not beat 13. We conclude that (2, 1, 0, 0) is the best solution. 421 COURSE NOTES 43

1.16. Lecture 16: Knapsack, day 3. Recall the test that allowed us to throw out entire branches: k k ! X ck+1 X (1.8) c x + b − a x ≤ M i i a i i i=1 k+1 i=1 Ignoring branches that are bad according to the inequality test can be thought of as pruning them.

Lemma 1.38. If a branch (x1,..., xk, ∗,..., ∗) is pruned (because of inequality (1.7) ), then all branches from the same immediate junction un- derneath will also be pruned. (This is saying that all branches of the form (x1,..., xk − d, ∗,..., ∗) will be pruned for all choices of d ≥ 0.)

Proof. If (x1,..., xk, ∗,..., ∗) is pruned, we have k k ! X ck+1 X c x + b − a x ≤ M i i a i i i=1 k+1 i=1

In order to prune (x1,..., xk − d, ∗,..., ∗), we need/want k−1 k−1 ! X ck+1 X c x + c (x − 1) + b − a x − a (x − 1) ≤ M. i i k k a i i k k i=1 k+1 i=1 The LHS of the given inequality is bigger than the LHS of the desired in- ck+1 equality by ck −ak . This is nonnegative by the assumption on decreasing ak+1 eﬃciency (and because ak is a positive number). The RHS in both inequalities is the same. Hence the given inequality implies the wanted one for xk − 1. Iterating this thought, we can prune away xk − d for all d ≥ 0 2 Algorithm 1.39 (Branch and bound, xi ordered by decreasing eﬃciency). (1) Integrality: Make the problem integral by multiplying the constraint (and ~c) with a number that clears all denominators. (2) Initialize: Set M = 0, k = 0, ~x∗ = ~0. (3) Find the top leaf of the current branch: For j = k + 1, . . . , n set $ j−1 ! % X xj = b − aixi /aj . i=1 Then set k = n. Pn Pn ∗ (4) Test the branch: If i=1 cixi > M, set M = i=1 cixi, and ~x = ~x. (5) (a) Backtrack to next branch: If k = 1, stop the algorithm. Other- wise replace k by k − 1. (b) If xk = 0, return to Step 5a, otherwise replace xk by xk − 1. (6) Test value of branch: If k k ! X ck+1 X c x + b − a x ≤ M i i a i i i=1 k+1 i=1 FAILS then return to Step 3, otherwise return to Step 5. 44 ULI WALTHER

If some of the coeﬃcients are not integers (for example, if one skips Step 1), one must use as right hand side in the test inequality · · · ≤ M. But, it is more eﬃcient to clear denominators. 421 COURSE NOTES 45

Homework 6 (for week 7). (1) Consider the following LP:

2x1 − x2 + x3 − x4 − x5 + 2x6 + 0x7 + x8 = 4

x1 − 2x2 − x3 + x4 − 2x5 − 2x6 + 0x7 − x8 = −7

x1 + 0x2 − x3 + 0x4 + 2x5 + x6 − x7 + x8 = 0

xi ≥ 0

x1 + 2x2 − x3 − x4 + 2x5 + x6 − 3x7 + x8 → max The initial solution ~x = (0, 0, 0, 0, 1, 0, 7, 5) is known. Use Matlab and the revised simplex method to ﬁnd the optimal solution. Hints: • Enter the matrix of the system A.. • For each iteration follow the steps of Algorithm 1.33 in these notes. • It is useful to enter for each iteration the basis matrix B and then use inv(B) to solve the systems in step 1 and 3 of the algorithm. (2) Suppose that in the IP

Ax = a1x1 + ... + anxn ≤ b xi ∈ Z c1x1 + ... + cnxn → max all entries for A are positive. Prove that there are only a ﬁnite number of feasible solutions. (Note that “feasible” includes “integral” in this case.) (3) Consider the following problem

2x1 + 2x2 + x3/2 ≤ 2

−4x1 − 2x2 − 3x3/2 ≤ 3

x1 + 2x2 + x3/2 ≤ 1

6x1 + x2 + 2x3 → max

Assume that slack is introduced with variable names x4, x5, x6. After the simplex method has been used to solve this problem, the ﬁnal dictionary is

x1 = 2 + 2x1 + 2x2 + 2x3 − x4 + 0x5 + x6

x3 = 2 + 2x1 + 2x2 + 2x3 + 2x4 + 0x5 − 4x6

x5 = 2 + 2x1 + 2x2 + 2x3 − x4 + 0x5 − 2x6

z = 2 − 2x1 − 2x2 − 2x3 − 2x4 − 0x5 − 2x6 Each 2 stands for a number that was lost after the final simplex tableau had been printed. What numbers go into these boxes? (Note: I don’t want you to go and solve the problem from scratch. The answers can be figured out from the final tableau. Explain, how. Hint: the final equations are linear combinations of the input.) 46 ULI WALTHER

1.16.1. Branch and bound on binary problems. Suppose each item can be taken only once or not at all. Example 1.40.

6x1 + 3x2 + 5x3 + 2x4 ≤ 10

x3 + x4 ≤ 1

−x1 + x3 ≤ 0

−x2 + x4 ≤ 0

xj ∈ {0, 1}

9x1 + 5x2 + 6x3 + 4x4 → max

One splits the problem into 2 pieces, according to x1 = 0 and x1 = 1. This gives a rooted tree just as the integer (not binary) case of before, but now only 2 branches grow out of a node (at most). The variable that causes the branching (here, x1) is the branching variable. This time we use a different way of estimating the usefulness of a branch. This is what is used in general (if there are more than 1 constraint) and in a way generalizes the previous estimation process. Before, we determined the value of a branch by taking the first k values as given by the node, and filling up the “rest” of the backpack with xk+1, be this an integral amount or not. So, we are really looking at a rational version of a subproblem. If one ignores the integrality condition (and uses 0 ≤ xj ≤ 1) and uses the simplex to solve the rational problem, one gets ~x = (5/6, 1, 0, 1) and z = 33/2. Obviously, whatever binary solution we come up with, its value cannot exceed 33/2. In fact, it can’t exceed 16 because the objective vector is integral. For BIP’s, and general IP’s, one does just that. In our example, consider the two branches x1 = 0, 1. The first has an associated subproblem

3x2 + 5x3 + 2x4 ≤ 10

x3 + x4 ≤ 1

x3 ≤ 0

−x2 + x4 ≤ 0

xj ∈ {0, 1}

5x2 + 6x3 + 4x4 → max

One can now consider this same problem except that 1 ≥ xj ≥ 0 and solve it with the simplex. Whatever comes out of that will be an upper bound for the best result this branch can have with binary variables. We ﬁnd that x1 = 0 leads to an optimal solution ~x = (0, 1, 0, 1) with z = 9, and x1 = 1 gives ~x = (1, 4/5, 0, 4/5) with z = 81/5. The former is (lucky shot) a binary solution. So we actually proved that the initial problem is feasible. The only way to get solutions is trial and error (in clever ways). We store (0, 1, 0, 1) as an incumbent. We can now always 421 COURSE NOTES 47 disregard branches where the rational optimum is not at least 9 (since the binary optimum cannot be more than the rational). Another eﬀect is that no branch of x1 = 0 needs to be explored any further, because we know the best possible value for each of them. The other branch has z = 81/5 which may well contain better binary solutions. So consider x1 = 1. We do a further branching, x2 = 0 and x2 = 1. This induces the two LP’s

5x3 + 2x4 ≤ 4

x3 + x4 ≤ 1

x3 ≤ 1

x4 ≤ 0

xj ∈ {0, 1}

9 + 6x3 + 4x4 → max and

5x3 + 2x4 ≤ 1

x3 + x4 ≤ 1

x3 ≤ 1

x4 ≤ 1

xj ∈ {0, 1}

14 + 6x3 + 4x4 → max

Both branches are called a descendant of x1 = 1. It produces a rational optimum of (1, 0, 4/5, 0) with z = 69/5; the other branch (1, 1, ∗, ∗) gives ~x = (1, 1, 0, 1/2) with z = 16. Both of these optima exceed the known solution value z = 9, so they might both be good branches. Unfortunately, we did not get any integer solution, so we cannot prune either branch. We must go down one more level. Between the two currently open problems, the second is more promising, so we explore now (1, 1, 0, ∗) and (1, 1, 1, ∗) (but keep in mind that (1, 0, ∗, ∗) still needs to be done). We now have the two systems

2x4 ≤ 1

x4 ≤ 1

xj ∈ {0, 1}

14 + 4x4 → max 48 ULI WALTHER

(with x3 = 0) and

2x4 ≤ −4

x4 ≤ −2

x4 ≤ 1

xj ∈ {0, 1}

20 + 4x4 → max

(with x3 = 1). It is clear that the second problem has no feasible solution (rational or otherwise). So we may dismiss that part of the current branch. The first problem has rational optimum of ~x = (1, 1, 0, 1/2) with value 16. Now we have 2 branches to investigate, (1, 0, ∗, ∗) and (1, 1, 0, ∗). The one we chose (based on heuristic experience) is the one where more values occur. This is the one that was created later. For that one we could have (1, 1, 0, 0) and (1, 1, 0, 1). The former is feasible with z = 14, the latter is infeasible. Thus, our incumbent gets updated, it is not ~x = (1, 1, 0, 0) with z = 14. We now return to the branch (1, 1, ∗, ∗). It had conceivably value 13, which was fine before we changed the incumbent. Now the entire branch cannot beat the incumbent, so we prune it. It follows that (1, 1, 0, 0) is the optimal binary solution. Note: we have 16 = 24 possible solutions to look at. Of these, we actually looked only at two explicitly, (1, 1, 0, 0) and (1, 1, 0, 1). We also saw (0, 1, 0, 1), but this was not because we had to investigate a branch of depth 4, but because it came out as a rational optimal solution. We also had to solve 5 LP’s. This is kind of typical. One rarely goes all the way in the branching process (except for a couple of times at the end) but computes instead a reasonable number of rational LP’s. This is considered ok, because the simplex has polynomial computation time (linear in rows, logarithmic in variables), while the number of potential feasible solutions of an BLP is 2n, which is monstrous. Algorithm 1.41 (Branch and bound a binary IP). (1) Initialize: Set z∗ = −∞. Apply the three tests given after the algorithm to the problem. If any of them apply, stop right there. Otherwise, label the problem as “current subproblem”. (2) Branching: Of the current subproblems, pick one that was created most recently. Break ties by choosing those with larger bound z∗. Branch off 2 new subproblems, setting the branching variable to 0 and 1. Dicard the subproblem that was just branched. (3) Bounding: For both new subproblems, compute its rational optimum. Round down the value to an integer. (4) Fathoming: For both subproblems run the three tests below. Discard any subproblem to which any test applies. (5) Optimality: The algorithm stops if no current subproblems are on the list. The optimal solution is the current incumbent. If there is no incumbent, the problem is infeasible. 421 COURSE NOTES 49

Algorithm 1.42 (Fathom and update). Input: An incumbent ~x∗, a current value z∗ (from the incumbent), a subproblem. Output: An updated incumbent, current bound, and subproblem list. (1) If the LP of the subproblem is infeasible, discard the subproblem from the subproblem list and return to the main algorithm. (2) If the LP of the subproblem has an optimal solution not exceeding the current bound, discard the subproblem from the current subproblem list end return to the main algorithm. (3) If the optimal solution of the LP to the current subproblem is integral, then • if its value beats the current bound, replace the incumbent by this optimal solution and the current bound by the value of the new incumbent. • if its value does not beat the current bound, ignore it. 50 ULI WALTHER

1.17. Lecture 17: A branch and bound example in class.

5x1 + 6x2 + 7x3 + 6x4 + 6x5 ≤ 28

7x1 + 8x2 + 11x3 + 10x4 + 9x5 → max xi ∈ N

First, order them by decreasing eﬃciency ci/ai:

6y1 + 7y2 + 6y3 + 5y4 + 6y5 ≤ 28

10y1 + 11y2 + 9y3 + 7y4 + 8y5 → max yi ∈ N Now run branch and bound. Initialize by z = 0, ~y = (0, 0, 0, 0, 0). (1) Top leaf (4, 0, 0, 0, 0), z = 40, update z, ~y. New branch: (3, ∗, ∗, ∗, ∗), 11 possible value 30 + 7 (28 − 18) = 44. Keep branch. (2) Top leaf (3, 1, 0, 0, 0), z = 41, update z, ~y. New branch: (3, 0, ∗, ∗, ∗), 9 possible value: 30 + 6 (28 − 18) = 45. Keep branch. (3) Top leaf: (3, 0, 1, 0, 0), z = 39, don’t update. New branch: (3, 0, 0, ∗, ∗), 7 possible value: 30 + 5 (28 − 18) = 44. Keep branch. (4) Top leaf: (3, 0, 0, 2, 0), z = 44, update z, ~y. New branch: (3, 0, 0, 1, ∗), 8 value: 37 + 6 (28 − 24) = 42. Prune branch. Hence by lemma, also prune (3, 0, 0, 0, ∗). Hence new branch to try is (2, ∗, ∗, ∗, ∗) with 11 possible value 20 + 7 (28 − 12) = 45. Keep branch. (5) Top leaf: (2, 2, 0, 0, 0), z = 42, don’t update. New branch: (2, 1, ∗, ∗, ∗), 9 value: 21 + 6 (28 − 19) = 44. Drop branch. By lemma, also prune 11 (2, 0, ∗, ∗, ∗). Next branch: (1, ∗, ∗, ∗, ∗), value: 10 + 7 (28 − 6) = 44. Also prune. By lemma, also prune (0, ∗, ∗, ∗, ∗). Conclusion: (3, 0, 0, 2, 0) is optimal solution. 421 COURSE NOTES 51

1.18. Lecture 18: Matrix games. Example 1.43. Mimi1 and Uli play the heads and tails as follows. Each has a coin, hidden under the hand, on the table. On “go” each player shows the coin. If two coins or two heads show, Uli gets a penny, if a head and a tail show, Mimi gets a penny. Is there anything that should be done to maximize Uli’s profit? The answer is “sort of”. If he plays the same way all the time (say, “heads”), then Mimi will catch on and always play “tails”. Also, if Uli always plays “tails”, Mimi will start playing “heads” and again she wins. So it seems there is no way for Uli to draw even. The point is that he should play one or the other in a random way. Then (assuming that Mimi is not able to read his mind2), in the long run nobody should come out ahead. A game of this sort is called 2-player 0-sum game, because what one wins the other loses (no money goes to the house). One can play it with several more players. One can also relax the condition of a 0-sum game, which really means that a fictitious player “house” sits on the table and loses if the players combined win, and wins if they combined lose. We assume that each player can choose from a finite number of strategies and we tabulate the results of both choosing something in a matrix A. One player is the row player, the other the column player. The entry ai,j gives the winnings of the row player if the row player chooses strategy i and the column player strategy j. Note: it is quite possible that A is not square. There are several questions one can ask in such a situation. (1) Is this a fair game? (2) What does “fair” mean? (3) How should one play this game? How not? Let us consider Example 1.44. The table

P1 P2 P3 R1 0, 9 0.4 0.2 R2 0.3 0.6 0.8 R3 0.5 0.7 0.2 contains the likelihood that ground-air missiles of type Ri take out attacking planes of type Pj. The defenders (R) want to optimize the hits, the attackers want to minimize it. In our example, the defenders want to minimize the risk (this is not necessarily the case, they also might have other things in mind). From the point of the defenders, they pick a weapon and then a certain type of plane comes along. If R chooses a pure strategy Ri he will only use one rocket. Depending on what Pj comes along, the result will be

1www.lems.brown.edu/˜mimi 2perhaps a faulty assumption 52 ULI WALTHER determined by ai,j. It’s good for R if this is big. So he will choose the row such that choose i such that min(ai,j) → max . j This number maxi minj(ai,j) we call α, it is the worst case scenario shooting performance for the best rocket. Similarly, β = minj maxi(ai,j) is the worst thing that can happen to the planes if the best plane has been chosen. For example, we have α = 0.3, β = 0.7. Let’s say α is the smallest number in row l and β the largest in column k. Suppose each of R and P plays the “safe strategy” of minimal losses. That is, R uses row l and P uses column k. Then the really happening thing is al,k = 0.6 which must (by deﬁnition) lie between α and β. Deﬁnition 1.45. Suppose α = β. Then we write ν for this number, ν is the value of the game. Proposition 1.46. The matrix A has a saddle, (that is, there exists an element al,k that is simultaneously the smallest in its row and the largest in its column) if and only if α = β. 2 In this case, once R has chosen his best rocket, and P his best plane, neither player will have a better option to play given that the other does not change his strategy. This means that if the matrix has a saddle, then both players will want to choose this row (and column) as their best (most defensive) strategy and so α is the value of the game. 421 COURSE NOTES 53

1.19. Lecture 19: Mixed strategies. If A has no saddle, it is not so clear what would happen. If R chooses to play the row that maximizes the row-minimum, and P the column that minimizes the column-maximum, both might be positively surprised by the actual result. The players will resort to mixed strategies, that is random strategies according to a certain distribution. Let ~x be a vector of length equal to the number of rockets, and ~y one of length equal to the number of planes. Assume that both have nonnegative components and the sum of the components is 1 in each vector. They are called stochastic strategies.

Deﬁnition 1.47. We call solution of a 2-person 0-sum game a pair of stochastic strategies (~x∗, ~y∗) for R and P that has the property

~xT · A · ~y∗ ≤ ~x∗T · A · ~y∗ ≤ ~x∗T · A · ~y for all stochastic strategies ~x,~y. This property means that if R plays according to strategy ~x∗ then for player P the reasonable thing to do is to use strategy ~y∗, and conversely.

Let us consider the two optimal solutions for each player:

x1 + x2 + x3 = 1 min(0.9x1 + 0.3x2 + 0.5x3, 0.4x1 + 0.6x2 + 0.7x3, 0.2x1 + 0.8x2 + 0.2x3) → max

xi ≥ 0

Write this as a standard form LP:

x1 + x2 + x3 = 1

z − 0.9x1 − 0.3x2 − 0.5x3 ≤ 0

z − 0.4x1 − 0.6x2 − 0.7x3 ≤ 0

z − 0.2x1 − 0.8x2 − 0.2x3 ≤ 0

xi, z ≥ 0 z → max

This is equivalent because the optimal z will be as large as the smallest of the three expressions in the min-function of the previous system. It turns out that this system has optimal solution ~x∗ = (19/52, 29/52, 4/52) with with optimal value z∗ = 139/260. The other player has a system of the form

y1 + y2 + y3 = 1 max(0.9y1 + 0.4y2 + 0.2y3, 0.3y1 + 0.6y2 + 0.8y3, 0.5y1 + 0.7y2 + 0.2y3) → min

yi ≥ 0 54 ULI WALTHER

This can be rewritten as

y1 + y2 + y3 = 1

w − 0.9y1 − 0.4y2 − 0.2y3 ≥ 0

w − 0.3y1 − 0.6y2 − 0.8y3 ≥ 0

w − 0.5y1 − 0.7y2 − 0.2y3 ≥ 0

yi, w ≥ 0 w → min

This has optimal solution ~y∗ = (9/26, 12/26, 5/26) with optimal value 139/260. What a coincidence. The minimal loss one player can enforce is the maximal win the other can enforce. John von Neumann proved the following generalization of the previous proposition:

Theorem 1.48 (Minimax theorem). Every 2-person 0-sum game has a solution.

Proof. Let’s say the matrix A is m × n. One can describe the problem of ﬁnding an optimal strategy for R as the solution to the LP

m X xi = 1 i=1 xi ≥ 0 m X min ai,jxi → max j i=1

This can be rewritten to

m X z − ai,jxi ≤ 0 for j = 1, . . . , n i=1 m X xi = 1 i=1 xi ≥ 0 z → max

The solutions are the same because saying in the ﬁrst system that one maximizes the minimum of a bunch of quantities is the same as ﬁnding the largest number that does not exceed any of them, which is what the second system says. 421 COURSE NOTES 55

Similarly, a strategy is optimal for the player P is it satisﬁes

n X yj = 1 j=1 yj ≥ 0 n X max ai,jyj → min i j=1 which can be written as

n X w − ai,jyj ≥ 0 for i = 1, . . . , m j=1 n X yj = 1 j=1 yj ≥ 0 w → min

The trick is to see that these are dual systems. To see this, note that the set I of inequalities in the primal (row player) is {1, . . . , n}, the set E of equations is {n + 1}, the set R of restricted variables is {1, . . . , m} and the set of free variables is {m + 1}. Applying the dualization process we get an inequality for each restricted variable, an equality for each free one and a restriction for each inequality. This implies that their optimal solutions are the same and hence for the optimal vectors ~x∗, ~y∗ we have

n m X ∗ X ∗ max ai,jyj = min ai,jxi . i j j=1 i=1

Now this equality will prove the theorem once we show that

n X ∗ T ∗ max ai,jyj = max ~x A~y . i ~x j=1 and m X ∗ ∗T min ai,jxi = min ~x A~y j ~y i=1 (These equalities mean that if one player plays optimal then the other player can play to optimal results with a pure strategy. It does not say that he should do that, because if that pure strategy is not optimal, the one player 56 ULI WALTHER

could improve results by changing.) Let us look at the second one.

n m ! ∗T X X ∗ ~x A~y = yj ai,jxi i=1 i=1 n m ! X X ∗ ≥ yj min ai,jxi j j=1 i=1 m X ∗ = min ai,jxi j i=1 The reverse inequality follows because the min over all ~y is no larger than any number we get plugging in a standard unit vector for ~y, in particular it is not smaller than their minimum. Pn ∗ T ∗ The equality maxi j=1 ai,jyj = max~x ~x A~y follows in the same way by switching the roles of ~x and ~y. 2

Suppose ~x∗ is R’s best strategy. It is possible that only a few of the m strategies of R are actually used in the optimal one. We call these useful. Similarly we call a strategy ~x dominated by strategy ~x0 if component-wise 0 xi ≤ xi. Being dominated indicates that R would never want to chose strategy ~x because ~x0 promises in all cases better results. A strategy ~y of 0 0 player P is dominated by ~y if component-wise yj ≥ yj. It turns out (no proof) that if one player uses his optimal strategy, the expected outcome of the game is always the same no matter how the other player mixes together his useful strategies. If however the other player uses not useful strategies, the other player will start losing. This is saying that as long as you play optimal, you can actually tell your opponent and it won’t hurt you. (It will however quite likely prevent you from making unexpected wins, unless your opponent is a moron.) A game is called fair if the minimax value in the theorem is zero. An obvious case of fair games is an anti-symmetric matrix A, ai,j = −aj,i.

1.19.1. Finding the optimal solution. If the matrix has a saddle, it is easy – just ﬁnd it. If not, then what? In general, one has to go and solve the LP. In case one or both players see Hirche have only 2 strategies, one can draw some pictures.♣ In eﬀect, one solves then the associated LP graphically. 3 5 4 6 5 6 3 8 Example 1.49. Suppose the game matrix is 8 7 9 7 4 2 8 3 One can then cancel column 4 because it is dominated. After that, one can cancel rows 1,2,4 because they are dominated. Of the left over (8, 7, 9) one can cancel the dominated columns 1 and 3 and so the value of the game 421 COURSE NOTES 57 is 7 and the optimal strategies are (0, 0, 1, 0) and (0, 1, 0, 0). These are pure because A has a saddle in a3,2. Even if there is no saddle, cancelling rows and columns is useful to decrease complexity of the ensuing LP. 58 ULI WALTHER

Homework 7 (for week 8).

(1) 13.4 from the book. (Note: the “z(k − ai)” in the formula is not the product of z and k−ai but the z that corresponds to the integer k−ai. Hint: imagine the knapsack is ﬁlled in an optimal way, and you take one item out of it while simultaneously shrinking the knapsack by the volume of the item you took. Is the result an optimally packed knapsack? Now inspect all the possible ways you could have taken out one item.) (2) Solve the knapsack

34x1 + 23x2 + 41x3 + 31x4 + 27x5 + 33x6 ≤ 168 xi ∈ N 74x1 + 23x2 + 89x3 + 67x4 + 59x5 + 71x6 → max (3) 13.5 from the book with the following modiﬁcation: I’d like you to write down the ﬂow chart of the algorithm that you invent. Roughly, this algorithm will for each k reduce the computation of z(k) to that of a few z(l)’s with l < k, and work its way up from k = 1 to k = b. 421 COURSE NOTES 59

1.20. Lecture 20, 21: a game example. Example 1.50. Consider a 2 by 3 chess board. Player 1 puts down a 1 by 2 domino, player 2 a 1 by 1 domino. Player 1 gets a buck if the large covers the small domino, player 2 wins otherwise. We use the numeration ∗ 7 ∗ 6 of choices for player 1 as in Exercise 15.2. Let ~x ∈ R and ~y ∈ R be the optimal solutions of both players. The symmetries of the problem suggest that x1 = x2 = x3 = x4, x5 = x7. The indices for the strategies of the 1 2 3 second player mean the following: . We infer y = y = y = y 4 5 6 1 3 4 6 and y2 = y5. 1 1 -1 -1 -1 -1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 -1 The matrix of winnings for player 1 is -1 -1 -1 -1 1 1 1 -1 -1 1 -1 -1 -1 1 -1 -1 1 -1 -1 -1 1 -1 -1 1 Thus the LP for player 1 is   −2x1 − x6  −2x5 + x6     −2x1 − x6  min   → max  −2x1 − x6     −2x5 + x6  −2x1 − x6 4x1 + 2x5 + x6 = 1

0 ≤ xi ≤ 1 Which simpliﬁes to

4x1 + 2x5 + x6 = 1

min(−2x1 − x6, −2x5 + x6) → max

Substituting x6 = 1 − 4x1 − 2x5 we want min(−2x1 − 1 + 4x1 + 2x5, −2x5 + 1 + 4x1 − 2x5) = min(−1 + 2x1 + 2x5, −4x1 − 4x5 + 1) to be large. Write A = x1 + x5 and get min(−1 + 2A, −4A + 1) → max. Note that 0 ≤ x1 ≤ 1/4 and 0 ≤ x5 ≤ 1/2. So we are trying to maximize the minimum of −1 + 2A, −4A + 1 over the part of the rectangle where 4x1 + 2x5 ≤ 1. That is really a triangle then. On the triangle, A varies from 0 to 1/2. By making a picture one can see that to the left of A = 1/3 the minimum is the second term, to the right of A = 1/3 the minimum is the ﬁrst term, and at A = 1/3 the 2 lines meet and form the largest possible value for the minimum. Note that this does not specify what x1 and x5 are, just their sum 1/3. We conclude that the optimal strategy for player 1 calls for a distribution (x1, x1, x1, x1, 1/3 − x1, 1 − 4x1 − 2(1/3 − x1) = 1/3 − 2x1, 1/3 − 60 ULI WALTHER x1). Of course, x1 may only vary between 0 and 1/6. One solution is (1/6, 1/6, 1/6, 1/6, 1/6, 0, 1/6). Another is (0, 0, 0, 0, 1/3, 1/3, 1/3). For player 2, the LP is

4y1 + 2y2 = 1

0 ≤ yi ≤ 1

max(−2y1, −2y1, −2y1, −2y1, −2y2, −4y1, −2y2) → min

This amounts to max(−4y1, 2(−1+4y1)) → min where y1 is allowed between 0 and 1/4. The max happens at equality when y1 = 1/6. Then y2 = 1/6 as well. Thus, player 2 should play all options equally. The value of the game comes out as −1/3. Note: it is not clear that 1/3 must be the answer just based on “if the domino is put, there are 4 of 6 left open”. If the board is 1 by 3 then this kind of calculation suggests that the row player wins in 2 of 3 cases. However, the column player would be stupid to ever play the middle square (it guarantees that he loses). So he will pick one of the 2 outer ones and hence make it a fair game instead of one that he expects to lose. My theory on this is as follows. If the first players “domino” fits perfectly into the playing area in the sense that one could tile the whole field with dominos of type one, then the game as stated here (with the row player win- ning if the domino covers the little domino) has expected positive outcome for the row player in (number of total squares – size of domino of player 1) out of (number of total squares) cases. And this determines the value of the game for the row player as (2 times size of his domino minus the size of the field) divided by the size of the field. The reason is that the row player can guarantee at least this by using each of the tiles in the tiling with equal frequency, while the column player can guarantee at most that amount by playing his single field randomly over all squares. If on the other hand the row player has no tiling it is quite possible that the value of the game goes down. (The column player by using random squares can always ensure to win at least (size of field minus 2 times size of row domino) divided by the size of the field!) But the other direction seems to be vulnerable. It might be interesting for example to calculate through the case of a row domino that is 2 by 2 on a field that is 3 by 3 and see if the value is (9 − 2 ∗ 4)/9 as area counting would predict. The question becomes even more interesting if the column domino is also not good for tiling. In fact, compare the homework question on this: row domino is 2 by 2, column domino is 1 by 2, field is 3 by 3. The row player wins if the two dominos cover each other. The area count predicts that once the row domino is put down, the probability of the column domino to be inside the row domino is 4 (4 ways to put it inside) in 12 (12 ways to put the column domino on the field overall). 421 COURSE NOTES 61

This would suggest that the row player wins 1/3 and loses 2/3 of all cases putting the value of the game at -1/3. I don’t know if that is the value. Modiﬁed version: row player wins if the two dominos have a ﬁeld in common. Then I really have no idea. On another line, it is interesting to consider the case where the win and loss revenues are exchanged (that amounts to multiplying the matrix of winnings by -1). In our example above, the result is the same as before. I suspect this is because equal distribution of both players is optimal. In the case where the board is 1 by three, before the value was zero. Now it is 1 because the column player can enforce the single domino to be covered by the double one. So reversing the win does not always reverse the outcome. 62 ULI WALTHER

1.21. Lecture 22: Review for the test. • dictionaries, updating • optimality detection • unboundedness detection • 2-phase, big-M, perturbation • setting up general LP’s • finding dual problems • duality theorem • economic significance of dual • complementary slackness • post-optimal modifications • revised simplex • branch and bound for xi ∈ N, {0, 1} • pruning criteria 421 COURSE NOTES 63

Homework 8 (for week 9). (1) 15.1 from the text. (2) 15.5 from the text. (3) 15.3 from the text. 64 ULI WALTHER

1.22. Lecture 23: Midterm. 421 COURSE NOTES 65

1.23. Lecture 24, 25: Network Simplex, Transshipment (Chapter 19). Problem: given a highway network, sources, sinks, and nodes. Find the cheapest way of delivery. Definition 1.51. Directed graph, arcs, head and tail of an arc, weighted graph. A schedule for shipping is it takes away the right amount at each source, it brings the right amount to each sink, and the flow sum in each node is zero. Of course, no arc has negative shipment. Typical example: figure 19.2. One makes an incidence matrix of size (number of nodes=n) times (number of arcs=m), such that each arc xi,j has 2 indices, and the matrix has a 1 in ak,(i,j) if there is an arc from node i to node j and k = j, and a minus 1 if k = i. Of course, most of the time k is neither i nor j and then the entry in ak,(i,j) is zero. Then one can write the transshipment problem as A~x = ~b

x(i,j) ≥ 0 ~cT ~x → min ~b contains for each vertex the amount it demands (=negative, if source). Theoretically, this can now be done with the simplex. But, this is wasteful because it is not using the special structure of the matrix and the problem. Pn Basic assumption: We shall assume in this chapter that i=1 bi = 0, that is, total demand equals total supply. This means that the entries of the lowest row of A and of ~b can be determined from the other rows (because in each column the sum of all elements is zero). Forgetting the last row we get the truncated incidence matrix A˜. 1.23.1. Trees. Recall trees in general from the knapsack section. Our highway system has subgraphs which may or may not be trees. If a subgraph has no circuits (so that is is acyclic), and if it connects every vertex to any other (disregarding the matter of orientation for a moment) it is a spanning tree. Such a tree is maximal (see homework) in the sense that there is no subgraph of the highway system N that is a) a tree and b) strictly contains the spanning tree. Suppose we are given a feasible schedule S. Then deleting all roads not used in S gives a subgraph of N. If that’s a tree, we call is a feasible tree solution. The point is, that in order to know the feasible solution, one only needs to know the tree. The particular values carried on each road follow. Here is why. Any tree T has (at least) one vertex in which fewer than two arcs of the tree meet (this is also proved in the homework). Pick such a vertex v of N. Since we know there is a feasible schedule that uses only the roads in T , and there is only one road going out of v, this road xi,j must carry exactly as much stuff as the vertex produces. So we know, how much this road carries, say t tons. Delete the vertex from N, from T , and make 66 ULI WALTHER it disappear from the LP by substituting t for xi,j and then pretend v and all roads out of it never existed. This gives a system in one fewer variables, for which we still have a feasible tree schedule. We may hence repeat what we just did and determine and eliminate a second vertex. Keep going and you’ll find all values for all roads. In the book this is proved in a different way. Now let’s see whether our schedule might be optimal. The shipping of stuff makes a difference in the price we’ll sell at. For example, if we have 2 suppliers v1, v3 and 1 user v2 with roads x(1,2) and x(3,2) with prices c(1,2) = 3 and c(3,2) = 5 zillion dollars per pound of stuff, the people who live in v2 ought to pay 3 zillion more per pound than those in v1 but 5 zillion more that the folks in v3. This is saying that the fair asking price p2 at v2 satisfies p2 = 3 + p1 = 5 + p2, so that p1 and p2 are not equal. We also can’t quite figure out what p1 etc. should be, but as soon as one is fixed, the other are determined too. So let’s say the price at the last vertex vn is zero, so the other prices will become relative prices. In praxis, it is no problem whatsoever to find those prices, just go along the roads in T (and since T has no circuits, there will be no confusion about the prices). Now consider a competitor that buys the stuff at our fair price in node i, carries it along xi,j to vj and sells it there. If there is a way of doing this for less than we think the fair price is, we are demonstrably stupid – we ought to have used that arc in our tree in the first place. Again, it’s easy to check through all arcs (only those not in T of course!!!) to see whether one of them suggests we are stupid. If no such improving arc can be found, we are done and have an optimal tree. If an improving arc is found, it is the entering arc. We ought to use it as much as we can. Thus we need to consider the ripple effect of carrying t tons along the new arc on the rest of the network. It’s a bit like the butterfly and the earthquake: sometimes the constraint for t can be quite a ways away from the arc we are introducing. The new arc will cause a circuit to spring up in the schedule, and along this circuit and nowhere else t will have an impact. Enlarging t forever will have the effect of driving at least one of the loads along the circuit to become negative. This is illegal, so t can’t grow beyond when that arc gets a zero load. This means that with t tons on the new arc, the new schedule is a tree again. It should be pointed out that when we say “tree” then one or more of the arcs may carry a zero load. We must distinguish between such arcs, and those not in T . For an example, consider the one in the text. We have

Example 1.52. An example (from the book) for the network simplex. 421 COURSE NOTES 67

Suppose we have 7 nodes, numbered v1, . . . , v7. The supply and demand v v v v v v v at each node are 1 2 3 4 5 6 7 . This is the demand vec- 0 0 6 10 8 −9 −15 tor. Let’s say we have roads as indicated by the following table. -1 -1 -1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 -1 -1 -1 -1 0 0 1 0 0 1 0 10001000001000 01000101000000 0 0 1 0 0 0 1 -1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 -1 -1 -1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 -1 -1 T The cost vector is ~c = (c13, c14, c15, c21, c23, c24, c25, c54, c61, c62, c63, c67, c72, c75). We want to solve A~x = ~b, ~x ≥ ~0, ~cT ~x → min. Initially we pick the tree x6,1 = 9, x15 = 8, x14 = 1, x24 = 9, x23 = 6, x72 = 15. Next we ﬁnd out fair price diﬀerentials. We put y7 = 0 and solve the system

y2 − y7 = 23

y3 − y2 = 60

y4 − y2 = 28

y4 − y1 = 18

y5 − y1 = 29

y1 − y6 = 44

That solves easily to y2 = 23, y3 = 83, y4 = 51, y1 = 33, y5 = 62, y6 = −11, y7 = 0. Now we test if use of another route could improve the prices somewhere. One can see that x75 could be useful. So we put t tons on that road and see what the consequences are: x6,1 = 9, x15 = 8 − t, x14 = 1 + t, x24 = 9 − t, x23 = 6, x72 = 15 − t. So t should be 8. So the new tree is x6,1 = 9, x75 = 8, x14 = 9, x24 = 1, x23 = 6, x72 = 7. Now we do the next iteration. First ﬁnd the fair price diﬀerential: y2 = 23, y3 = 83, y4 = 51, y1 = 33, y5 = 59, y6 = −11, y7 = 0. Note that one price became cheaper. As the next useful road we take x21. This leads to t = 1 and the new tree x6,1 = 9, x75 = 8, x14 = 10, x21 = 1, x23 = 6, x72 = 7. The fair prices are now y2 = 23, y3 = 83, y4 = 49, y1 = 31, y5 = 59, y6 = −13, y7 = 0. No arc promises improvement. The solution is optimal. Algorithm 1.53 (A step in then network simplex). Input: A feasible tree solution T . Output: An improved feasible tree if it exists; an optimality proof for T otherwise. 68 ULI WALTHER

(1) Determine fair price diﬀerentials along the tree T by starting at some (any will do) vertex v and using the costs along T to determine fair prices on the way. (2) Check all roads xi,j not in T whether they can be used by a competitor buying at node i and selling at node j to beat our fair price. (3) If no such road can be found, T is optimal. (4) If any such road is found, choose any one of the promising roads and add it to the tree T . This results in a unique circuit. Giving xi, j load t determine the resulting loads along the circuit. (No other road of T will be aﬀected!) (5) Pick the largest t that does not result in negative loads, and cancel the arc of T that carries load zero for the maximal t. This gives a new improved feasible tree solution. 421 COURSE NOTES 69

1.24. Lecture 26: Network Simplex, Transshipment: initial trees. 1.24.1. Degeneracy and cycling. Degeneracy happens as before. Can’t help it. But cycling can be avoided as we learned before by clever choice of leaving arcs. We see to that later. 1.24.2. A starting tree schedule. Not all transshipment problems are feasible. (This is a problem if the roads are partially one-way. In a 2-way road system all transshipment problems on connected graphs are solvable.) Certainly, if we make up illusory roads, things become easier. Hence let’s invent roads that go from some node w to all sinks, and from all sources to w. Then obviously one can make very easily a tree that is feasible by shipping through this dreamroad system to each sink what they want and from each source what they have extra. We hence install a version of the big-M method. We pretend the arcs are there, but we give them M as cost and imagine M huge. If we solve this auxiliary problem, then three possible outcomes are in the mist: • The optimal solution uses only real existing arcs. Then this is the optimal solution for the original problem. • The optimal solution uses some imaginary roads, but has all their carrying loads equal to zero. Then forget about the imaginary roads, the left over driving instructions are an optimal schedule. (Note: the leftovers may not be a tree, they could make a forest of little bushes so to speak.) • The optimal solution really uses some dream roads. This means that the price will contain M or multiples. Thus, no real existing feasible schedule exists, the problem is unsolvable. One can modify this as follows. First solve an auxiliary problem where each dream road has price 1 and all others price 0. Cases: • The optimal solution uses only real existing arcs. Then this is a feasible solution for the original problem. • The optimal solution uses some imaginary roads, but has all their carrying loads equal to zero. Then forget about the imaginary roads, the left over driving instructions are a feasible schedule. (Note: the leftovers may not be a tree, they could make a forest of little bushes so to speak.) • The optimal solution really uses some dream roads. This means that no real existing feasible schedule exists, the problem is unsolvable. What ought one to do if the second case happens and we are dealt a feasible forest? The chicken way is to put carrying load zero to some unused roads until we have a tree. The clever way goes like this. Let T be the optimal tree to the auxiliary problem and let x(a,b) be a road that is imaginary and shows up in T with load zero. Recall that in order to prove optimality of T we computed these fair prices at each node called yi. Decompose the set of all nodes into two 70 ULI WALTHER sets: those with yi ≤ ya (called Va) and those (called V>a with yi > ya. The former set contains va (obvious) , the latter vb (because the best way of getting from va to vb is along x(a,b) which costs 1).

Proposition 1.54. Consider the node set Va with roads and costs inherited from the original problem. Do the same with V>a. Then

(1) The auxiliary optimal tree T induces an auxiliary optimal tree on Va and on V>b. By forgetting the dream roads one obtains feasible trees (or forests) for the actual subproblems. (2) Solving the two subproblems produces an optimal solution to the large problem on V . The consequence is that without further computations one can break each subproblem into smaller pieces until all dream roads of T have been forgotten and on each subproblem T induces a tree rather than a forest. Moreover, solving these (much) easier small problems we solve the big one as well.

Proof. The idea of the proof is that obviously no real road leads from any point in Va to any point in V>a. (True because any two points linked by a real road must have the same auxiliary price, which they don’t if one comes from Va and one from V>a). Consider now the sum of all sinks and all sources in V>a. If we can show that this is zero, we have shown that as a country the region V>a has a balanced economy and no access to import. Hence they don’t export either and we may consider Va and V>a as separated by an iron curtain. Why then is Va. Suppose a road going the other way is used in T . Say it goes from vi ∈ V>a to vj ∈ Va. That means of course by our deﬁnition of fair prices that yi = yj because in the auxiliary problem real roads are free. But yi = yj is impossible as one is at least a + 1 and the other at most a. We conclude that T contains no road from V>a to Va. (note the diﬀerence in conclusion: there are no real roads from Va to V>a, but there may be such roads while they are not used the other way). This proves the proposition. 2

1.24.3. Avoidance of cycling. An iteration step is called degenerate if the t that talks about the load on the entering road in the tree is zero. One avoids cycling if the entering arc is always chosen to lead away from the node v1. (Theorem 19.1). One can arrange this always by cleverly choosing the exiting road. Network simplex is very eﬃcient.

1.24.4. Open question. Q: If the entering arc maximizes yj − ii − ci,j, can cycling happen? A tree is called “strongly feasible” if xij = 0 implies that the arc is directed away from the root. 421 COURSE NOTES 71

The one has Theorem (1.19 in th book): If al trees are strongly feasible then cycling is impossible. Proof idea: One shows that the tree to the auxiliary problem is ok. Then given a strongly feasible tree, either is the entering arc strongly feasible, or it is wrong way with load zero. If the arc has the wrong direction, deletie it and separate tree T into R ∪ S. If an arc from R to S exists, use it to replace the bad arc. If no such arc exists, the problem decomposes into two subproblems. 72 ULI WALTHER

Homework 9 (for after spring break). (1) Solve the following transshipment problem. The matrix A is the one used in problem 19.1. The vector ~b is the vector ~b used in problem 19.1.i with the modiﬁcations that b1 = −1 instead of −2 and b8 = 2 instead of 3. The cost vector is the one from problem 19.1.i. (2) 19.10 (3) Prove that if T is a tree (or a forest) then there is at least one vertex that has only one edge coming out of it. (Hint: show that if each vertex has more than one edge coming out of it then there must be a circuit.) (4) Prove: if G is a graph on n vertices and T a spanning tree for G, then the number of edges in T is n − 1. (A spanning tree is a tree in which each vertex has at least one edge coming out of it.) Prove more generally that if F is a forest of k trees in G such that each vertex of G is used in at least one edge of F then the number of edges of F is n − k. (Hint: use the previous problem.) 421 COURSE NOTES 73

1.25. Lecture 27: an example on network simplex. Example 1.55. The example on page 354 without boundary conditions. 74 ULI WALTHER

Homework 10 (for week 10?). • 19.2 • Let G be a connected graph drawn on a sheet of paper. Assume that G has no crossing edges. Imagine perhaps that G represents a political world map. That is, the edges represent the borders between countries or the shores of a country to the sea. Let F be the number of faces (all countries+ocean), e the number of edges, and v the number of vertices. There is a simple equation relating those 3. Find it and prove it. Hint: investigate first the case when G is a tree. That will tell you the equation. Now let G be any graph and see what happens to the quantities v, e, f if you erase one edge. Be careful: you must not erase an edge that disconnects the graph. (If you do, the equation does not hold any more.) Explain why you can erase an edge without disconnecting the graph as long as the graph is not a tree. Note: you may have heard of “regular polyhedra”. These are 5 in number, have been known to the ancient Greeks and go by the name of tetrahedron, cube=hexahedron, octahedron, dodecahedron and icosahedron. (These are ordered by the number of faces, the names mean 4,6,8,12,20.) One can interpret them as maps, and for them the equation holds as well. This was known for probably 3000 years. That it works for all connected graphs drawn in the plane without crossing edges (so-called “planar graphs”) was probably also known for a long time but was first officially proved by the great Leonhard Euler in the 18th century. 421 COURSE NOTES 75

1.26. Lecture 28: Upper bounded Transshipment problems. Suppose we have a transshipment problem where each road has a maximal capacity. Then the usual network simplex may not work because we may not get a new tree really. What one should do is to consider as “in the tree” the roads on which the load is neither zero nor the maximum allowed for that road. Example 1.56. Let’s say we have 6 nodes with roads as in Figure 21.1 of the text. (Both pictures are needed for production/cost/schedule). We now have 3 kinds of arcs: those in the tree, those with maximal load (“saturated arcs”) and zero load arcs. Figure 21.2 shows a feasible solution. Dashed means saturated, solid means in tree. As in the usual network simplex we compute relative fair prices, given in 21.3. One can see that there are improving arcs, for example x13. Using this arc produces t = 8 and a new solution given in 21.5. Next one can use arc x54 to enter. This time t = 6 because the arc x43 may not carry more than 9. The result is depicted in 21.7. Now the new entering arc is going to be x21 which results in t = 2 and picture 21.9. Note that the tree did not change in this iteration, although the solution did. Weird! At this point no arc satisfies the “cheapening condition” yj > yi +cij. On the other hand, the schedule is not optimal. The reason is that the condition yj > yi + cij only tests whether it is useful to increase the load along arc ij. It is however quite conceivable that one ought to decrease the load along a certain arc in order to get improvement. This property would be measured by the condition yi + cij > yj while xij is saturated. (Recall that before we needed to check whether yj > yi + cij for those arcs that had load zero!) The arc 23 fits that condition. It means, that using arc 23 is wasteful. (In an unconstrained network, this would not have happened because then we could have chosen x43 and x12 differently. Recall, that we did this problem as an example in class last time!) So let us decrease x23, by t = 2 due to x42. We get the picture of 21.11 where x42 left the tree. Next x64 should be decreased which leads to saturation of x62 at t = 1. The picture is 21.13 and x62 left the tree. Now one can see that decreasing x56 would be a good idea. This can be done by 6 so that x56 actually goes completely from its upper to its lower bound. The tree does not change and the new solution is in 21.15. At this point, yj ≤ yi + cij for all xij = 0 (no improvement possible by increasing load of an arc), and yj ≥ yi + cij whenever xij = uij (which says that no improvement is to be expected from decreasing any load). The current solution must be optimal. 76 ULI WALTHER

1.27. Lecture 29: Upper bounded network II. Algorithm 1.57 (Upper bounded network simplex). Input: A feasible tree solution T (that is, the tree and the saturated arcs). Output: An improved feasible tree if it exists; an optimality proof for T otherwise. (1) Determine fair price differentials along the tree T by starting at some (any will do) vertex v and using the costs along T to determine fair prices on the way. (2) Check all roads xi,j not in T whether they can be used by a competitor buying or not buying at node i and selling at node j to beat our fair price. That means: if xij = 0, check yj > yi + cij; if xij = uij, check yj < yi + cij. (3) If no such road can be found, T is optimal. (4) If any such road is found, choose any one of the promising roads and add it to the tree T . This results in a unique circuit. Giving xi,j load t determine the resulting loads along the circuit. (No other road of T will be affected!) (5) Pick the largest t that does not result in negative or oversaturated loads, and cancel the arc of T that carries load zero (or is saturated) for the maximal t. This gives a new improved feasible tree solution. The two conditions mean that each road that can be used as a shortcut is already at maximal operation capacity, while each road that is wasteful is already at zero usage. As always, we call a pivot (the choice of an entering road) with t = 0 degenerate. These kinds of iterations don’t change the cost function. Occasionally, an iteration does not give a constraint. (If some of the roads have unbounded capacity.) This requires that some cij are negative. It means that somewhere in the graph is a circuit which has negative cost. Carrying stuff around it in large amounts gives arbitrarily small costs. We are not going going to worry about this much. We also kind of ignore degeneracy. One can avoid it by clever choice of the iteration steps (see page 363).

1.27.1. Initial solutions. To get an initial solution, one does as in the un- restricted network: pick a node as root, introduce arcs into the root from all nodes of negative demand and out of the root to all nodes with positive or zero demand. Costs on real roads are zero, those on made-up roads are 1. An initial solution is easy to ﬁnd, for example each node could carry everything it has into the root. If the optimal auxiliary solution has only reals roads, it is a feasible solution to the original problem. If it really uses some dream roads, the original problem is infeasible. If the optimal auxiliary tree has some dream roads with zero load, one can decompose the original problem and the optimal auxiliary tree induces a feasible tree for all subproblems. 421 COURSE NOTES 77

In fact, infeasibility is detected by that optimal auxiliary tree and the conclusion is: the original problem is infeasible if and only if there is a region that needs more than all roads into it can carry. One can show the existence of such a region from the optimal auxiliary schedule: the region S in question is found as follows. Let xij be any imaginary road of T (the optimal tree of the auxiliary problem) with positive carrying load. The region with over- demand is the set of all nodes where the fair price associated to T is at least yj (the j being the index of the node that the dream road xij goes into). Basically, the argument is that if this collection of vertices could import on real roads as much as they need to eat, the road xij would not have shown up in an optimal auxiliary schedule. 1.27.2. Integrality. Theorem 1.58. Every feasible transshipment problem with integral constraints and integral demands has an integral optimal solution. Proof. This is clear from the fact that all algorithms we have seen in network questions move along integer-valued tree solutions. Why are tree solutions integral? Recall that in a tree there are leaves (vertices of degree 1). Along the arc in the tree to a leaf there is no choice whatsoever what the value of the flow is: it must be the demand at the leaf. That was an integer. Thus one may “feed that vertex”, reduce the number of vertices, and restate the question. This leads to the unique (and by construction integral) schedule associated to our tree. 2 A consequence is the so-called wedding theorem of DénesKönig. Theorem 1.59. If there are n girls and n boys and each boy knows exactly k girls and every girl knows exactly k boys, then a grand wedding can be arranged where each person knows her/his future spouse. A basic assumption not stated in the text is that k be non-zero. Proof. Make a network with a node for each person and an arc from a boy to a girl if they know each other. Consider girls as sinks of demand 1 and boys as source of supply 1. Since each node has degree exactly k, one can make a feasible schedule carrying 1/k along each arc. By the integrality theorem, there is an optimal integral solution. Such a solution consists of matching a boy to a girl he knows. (Note: the actual costs are irrelevant!) 78 ULI WALTHER

1.28. Lecture 30: Network flows. A network flow is a upper bounded transshipment problem with one source and one sink. A flow is a schedule that respects this, i.e., it has input equal to output for all other nodes. A cut in a network is collection of nodes, including the source, but not including the sink. The capacity of a cut is the sum of the capacities of all roads leading out of the cut. Suppose one has a certain flow in the network. Then it is clear that the volume of the flow (the amount that leaves the source) is always a lower bound for the capacity of any cut, because at the end of the day that many units have arrived in the sink, which is outside the cut. Theorem 1.60. The volume of the largest flow equals the capacity of the smallest cut. So in fact there is a cut whose capacity is exactly the volume of the largest flow. Considering the graph in which the given network is augmented with a road from the sink to the source with infinite capacity the network algorithm shows that if the capacities are all integral, then there is an integral optimal flow. (Of course, it is possible that there are non-integral optimal flows. It just says there is at least one integral optimal one.)