Review of Linear Programming Single Period Models

ISDS 440 Dr. Z. GoldsteinNotes on Special Linear Programming Models 1 1. Introduction The objective of the set of notes on special linear programming models is to introduce several linear models not presented in your previous visit of this topic. The cases studied demonstrate the ability of linear programming to solve variety of real problems, even when at first glance the models seem to be non-linear. Before we proceed, let us warm up with a few types of the linear programming models you have already become familiar with. Review of linear programming single period models Please set up linear models and solve the following problems using Excel: (1) Classic hand-makes three models of elegant Christmas candles. Santa models requires.10 day of molding,.35 day of decorating, and .08 days of packaging and produce $16 of profit per unit sold. Corresponding values for Christmas trees are .10, .15, .03 and $9, while those of Gingerbread houses are .25, .40, .05, and $27. Classic wants to maximize profit on what it makes over the next 21 working days with its 1 molder, 3 decorators, and one packager, assuming that everything made can be sold. Formulate an operations management linear programming model to choose an optimal production plan, and solve your model.(2) Rozac is a drug produced of 4 chemicals. For tomorrow there is a need to for 1000lb of the drug. There are three active ingredients in Rozac: A, B, and C. By weight, at least 8% of the drug must consists of A, at least 4% of B, and at least 2% of C. The cost per pound of each chemical and the amount of each active ingredient in 1 lb of each chemical are given below. It is necessary that at least 300lb of chemical 3 be used, but not more than 320 lbs of chemical 1 and 2 combined. Formulate a linear programming model that would determine the most economical blend that satisfies tomorrow’s demand. Use Excel to solve.Chemical Cost/lb A B C 1 $8 .03 .02 .01 2 10 .06 .04 .01 3 11 .10 .03 .04 4 14 .12 .09 .04(3) Floataway Tour has $400,000 that may be used to purchase new rental boats for hire during the summer. The boats can be purchased from two different manufacturers. Pertinent data concerning the boats are summarized below:Price ($) Daily Boat Manufacturer Per boat Seating Profit($) Speedhawk Sleekboat 6000 3 70 Sillverbird Sleekboat 7000 5 50 Catman Racer 5000 2 80 Classy Racer 9000 6 1102 Floataway would like to purchase at least 50 boats. To maintain good business relationships with both manufacturers, Float-away will purchase the same number of boats from Sleekboat as from Racer. At the same time, the number of Speedhawks and the number of Catmans should not differ from one another by more than 20. Floataway wishes to have a total seating capacity of at least 200. Formulate this problem as a linear programming model. Use all the data provided.(4) Reductions in the defense budget are causing problems for Williams, Osborne, and Evans (WOE), a leading supplier of CI systems. WOE is faced with the need to downsize its labor force, while, at the same time, reduce waste and improve its competitive position. Its problem is to develop a mix of labor skills and functions which will not only be adequate to perform ongoing work but also meet certain headcount or cost reduction goals. Several different types of employees need to be considered with respect to headcount reduction: Managers, Engineers, technicians, and other (non-technical) staff. The following table summarizes the types of people currently at WOE, along with other relevant data. Type Direct Time Overhead Time Current Pay (weekly) Pay (weekly) Headcount Manager $320 per person 1280 per person 150 Engineer 800 -“- ---- 6000 Technician 600 -“- ---- 3000 Other 10 -“- 245 per person 900WOE has decided to meet several guidelines when conducting the reduction plan: a. The workforce in each category should be reduced by at most 50%. b. The number of managers should be reduced by at least 40%. c. Technical personnel (Engineers and Technicians combined) should be at least 5 times larger than non technical personnel (Managers and Others combined). d. For every engineer and technician combined laid off, at least two managers and other profession employees combined must be laid off. For example, if 2 engineers and 1 technician are fired, then at least 6 managers and “others” employees combined must be fired. e. Total payments for overhead time should not be more than 10% of the total payments for direct time. Set up a linear programming model that minimizes the total weekly pay and recommends how many employees of each kind to keep in the company.Excel solutions to the “Reminder” problems3 2. Linear Models with special characteristics In this chapter we introduce you to the new linear models. In each section we shall explain what’s unique about the problem, then demonstrate the formulation, and finally ask you to actively participate by solving some similar problems by yourself.Time Phased Models The models we dealt with by now supported single stage decisions (each decision was independent of previous decisions). However, many problems involve a series of decisions made over time where the outcome resulting from one decision affects the next decision. In these cases consecutive decisions are dependent. The models that deal with a sequence of decisions over time are called “Time Phased models”.In time-phased problems, aside from the regular constraints that involve activity levels for each period, we need to write constraints called “balance constraints” that connect activity levels in consecutive periods.Example 1 – Multi-period Investment problem: Investment plans involve the amount invested in each time period in order to maximize the cash at hand at the end of the planning horizon. Funds available for investment may include cash available at different points in time and money released from previous investments when matured. Interest paid at the end of period “t” is available for investment at the beginning of period “t+1”, generating relationship expressed by a ‘balance constraint’. Observe the problem next.At the beginning of month 1 FINCO has $400K in cash. At the beginning of month 1, 2, 3, 4 FINCO receives revenues from outside source, after which it pays bills (see table below). Any money leftover may be invested for one month at 0.1% interest rate per month; for two months at 0.5% per month; for three months at 1% per month; or for four months at 2% per month. Use LP to determine the investment strategy that maximizes cash on hand at the beginning of month 5. State the optimal solution. Revenues Bills Month 1 $400K $600K Month 2 $800K $500K Month 3 $300K $500K Month 4 $300K $250KSolution: The balance constraints for this problems have the form: Cash available at the beginning of month‘t’ = Investments in month‘t’ + bills paid in month’t’. This is true because it isn’t optimal to leave dollars available at the beginning of any given month unused. Such a constraint is called a balanced constraint (because funds flowing into month t are equal to funds used in this month). Each month has a balance constraint as demonstrated next.4 Define: Xij = the amount invested at the beginning of month i that matures at the end of month j. The objective is to maximize the cash at hand at the end of month 4.Max 1.08X14+1.03X24+1.01X34+1.001X44 Subject to X11+X12+X13+X14+600 = 800 400+400 X22+X23+X24+500= 800+1.001X11 X33+X34+500 = 1.01X12+1.001X22+300 X44+250 = 1.03X13+1.01X23+1.001X33+300Principal + interest returned on the money invested for one month at the beginning of month 1.Excel Solutions for examples 1, 2, 5Example 2: Multi-period production problem. Planning production for several periods ahead may involve changing demand and/or changing cost patterns. Under such conditions keeping production levels unchanged may result in unnecessary costs. To minimize costs dynamic solutions may offer more cost savings than rigid solutions. By this we mean allowing production levels to change from one period to another may be optimal. Since we allow production to exceed demand at any given period (if production costs at period‘t+1’ are expected to rise we might want to increase production at period‘t’ and keep the excess amount produced stocked for future demand) the inventory carried from one period to the next needs to be considered as part of the amount made available to meet future demand. This relationship creates the balance constraint for this kind of problems. Observe the following examples: Sailco Corp. wants to determine how many boats should be produced during each of the next four quarters. The demand during each quarter as well as the maximum production per quarter using regular time, and production cost per boat are shown in the table below:Quarter 1 2 3 4Demand 40 boats 60 boats 75 boats 25 boatsRegular Time Production capacity 40 boats 50 boats 40 boats 60 boatsReg. Time Production cost per boat $400 $450 $390 $375 Sailco must meet demand on time. Sailco currently has 10 boats available for sales in quarter 1. We assume that sailboats manufactured during a quarter can be sold during that quarter, or at any quarter thereafter. Sailco can use overtime at a cost of $450 per sailboat. At the end of each quarter (after production has occurred and the current quarter demand has been satisfied, a holding cost of 10% of the regular time unit production cost is incurred. Use linear programming to minimize the sum of production and inventory costs by finding an optimal production schedule.5 SolutionLet Xt and Yt be the number of sailboats produced in regular time and in overtime respectively. Let It represent the inventory carried over from period t to period t+1.The objective is to minimize Total Cost over the four quarters = Production cost in regular time + Production Cost in overtime + Inventory holding cost.Production cost in regular time = 400X1+450X2+390X3+375X4Production cost in overtime = 450Y1+450Y2+450Y3+450Y4Holding cost = 40I1 + 45I2+39I3+37.5I4 (notice that to have inventory left over at the end of the fourth quarter can’t be optimal, because this inventory has no value; this would be different if the planning horizon was rolling forward, in which case the units left over in quarter 4 could be sold. So in our case I4 = 0, thus will not be further included in the model). The constraints are presented next:Maximum production in each quarter: X1 40; X2 50; X3 40; X4 60Balance constraints: As you may recall, balance constraints connect consecutive time periods. In our case, inventory carried from period “t” to period “t+1” generates the link between consecutive periods. Basically the balance constraints state thatBoats available for sale at “t” = Boats sold at “t” + inventory left over at the end of “t”.Quarter 1: 10 + X1 + Y1 = 40 + I1 (Explanation: 10 boats plus boats manufactured in regular time and overtime combined are available for sales. The supply level is balanced by the demand for 40 boats plus I1 units carried over to the next period, if production exceeds the demand (note that the supply cannot fail to meet demand because shortage is not allowed; this is achieved by the non-negativity of I1). The rest of the balance constraints follow the same line of reasoning.Quarter 2: I1 + X2 + Y2 = 60 + I2 I1 is available for sale in quarter 2Quarter 3: I2 + X3 + Y3 = 75 + I3Quarter 4: I + X + Y = 25 + I I4 is considered equal to zero and will 3 4 4 4 not be included in the Solver inputNon-negativity constraints: Xt 0; Yt 0; It 0Excel Solution – See link aboveSailco allows shortages. If Sailco could evaluate shortage costs, it is possible it would decide to allow shortages in some quarters (provided the accumulated demand of the four quarters is eventually met by the end of the fourth quarter). This might happen if despite 6 the shortage penalties total cost could be reduced. The variable ‘I’ that was used to represent inventory is now expanded to represents both inventory and shortage. To do this ‘I’ must be allowed to take on negative values (i.e. if I2 = -3 then at the end of quarter 2 a shortage of 3 boats is suggested). The problem with this requirement is that simplex works with non-negative variables only. In the next section we’ll deal with variables whose sign is unrestricted in a formal manner, and return to the Sailco problem allowing for shortages.Variables with unrestricted sign and sum of absolute values. When variables can assume both positive and negative values (such variables are called unrestricted sign – URS) the solution procedure (called Simplex) is unable to perform the optimization, because it works only with positive values. In this case we use the following transformation:Define X = X+ – X-, where X+ and X- are both nonnegative. Note that X can be either positive (when X- < X+), equals to zero (when X- = X+), or negative (when X- > X+). Each time X appears in the model replace it with X+ – X-. What happens is that with two nonnegative variables, X+ and X-, we express the value of an URS variable, and simplex can be used. The X- and X+ transformation is also utilized in transforming a model that includes absolute values of URS variables into a simple linear programming model. Since the absolute value of a variable is not a linear function linear programming can’t be used as a solution approach. Using X- and X+ we can express the absolute value as a linear function and than use the linear programming modeling as usual. Specifically, instead of using |Xi| + - + - in the model we write Xi + Xi . As before for the URS variables we use Xi = Xi – Xi See the generic example below.Example 3: Original model Transformed model + - + - Min | X1|+| X2| Min X1 + X1 + X2 + X2 + - + - Subject to 2X1+ X2 100 Subject to 2(X1 – X1 ) + X2 – X2 100 + - + - X1+3X2  50 X1 – X1 + 3(X2 – X2 )50 + - + - X1 and X2 are URS. X1 , X1 , X2 , X2  0Example 4: Sailco with planned shortage. With respect to the Sailco problem let us add the following information. Sailco is willing to tolerate some shortages at the end of each quarter, if it is found economically viable. Customers who order boats are willing to wait until the boats are produced, and this may happen in future quarters. However, at the end of the fourth quarters all the demand must be satisfied. The shortage cost for each boat at the end of a quarter is $55. How many boats should be produced each quarter in regular time and in over time; how many boats should be carried over to the next quarter, and when; and what shortage should Sailco plan for each quarter?7 Solution Let ‘I’ represent the difference between the demand per quarter and the number of boats made available. ‘I’ is an urs variable since it can represent both an actual inventory left (i.e. I >0) and a shortage realized (I<0) at the end of a quarter. If we define I+ as an actual - + - inventory, and I as the amount of shortage at the end of a quarter, then I =I - I . The Sailco problem can now be re-formulated. Minimize Total Cost over the four quarters = Production cost in regular time + Production Cost in overtime + Inventory holding cost + Shortage cost =(400X1+450X2+390X3+375X4) + (450Y1+450Y2+450Y3+450Y4)+ + + + - - - (40I1 + 45I2 +39I3 ) + (55I1 +55I2 +55I3 ). + - Note that I4 =0 because it is non-optimal to leave inventory at the very end, and also I4 =0 because it is required that all the demand is met by the end of the fourth quarter. These two observations may change under different scenarios, as demonstrated in the next example. Subject to:Maximum production in each quarter: X1 40; X2 50; X3 40; X4 60Balance constraints: + - Quarter 1: 10 + X1 + Y1 = 40 + I1 - I1+ - + - Quarter 2: I1 - I1 + X2 + Y2 = 60 + I2 - I2+ - + - Quarter 3: I2 - I2 + X3 + Y3 = 75 + I3 - I3+ - Quarter 4: I3 - I3 + X4 + Y4 = 25 +, - Non-negativity: Xi, Yi, Ii Ii .Comment: If short demand cannot be met later in time (lost sales) the balance constraints are effected as follows: Quarter 1: No change + + - Quarter 2: I1 + X2 + Y2 = 60 + I2 - I2 + + - Quarter 3: I2 + X3 + Y3 = 75 + I3 - I3 + Quarter 4: I3 + X2 + Y2 = 25Example 5: Least absolute deviation (LAD) linear regression An interesting application where absolute values are included in a linear programming model is the Minimum Deviation Linear Regression model. This is an alternative approach to constructing a linear regression equation. The classical linear regression approach minimizes the sum of squared deviations between the actual points and the regression predicted points. An alternative approach is to minimize the sum of absolute deviations, which will be less sensitive to outliers. We use the idea developed in the previous example to solve the following problem.Build the LAD regression for the following data: Y 2 3 4 5 8 X 1 2 4 6 7 8 Since we are looking for the line that minimizes the sum of absolute deviations (i.e.  ei where ei is the deviation between the actual value of Yi and the corresponding point on the regression line (b0+b1Xi), we need to determine the coefficients b0, b1, that minimize this sum Thus, the variables are ei, b0, b1, and the constraints are the equations of the form Yi = b0 + b1Xi + ei, which define the distance ei between each Yi and the corresponding value on the straight line. However, since ei as well as b0 and b1 could be + - + - negative or nonnegative, the constraints are transformed into Yi = b0 – b0 + (b1 – b1 ) Xi + + - ei – ei . Adding the transformation discussed above concerning absolute values we have the following linear programming model.+ - + - + - + - + - Minimize  ei = Min [e1 + e1 + e2 + e2 + e3 + e3 + e4 + e4 + e5 + e5 ] + - + -) + - Subject to b0 - b0 + 1(b1 -b1 + e1 -e1 = 2 (The equation at point (X=1, Y=2)) + - + - + - b0 - b0 + 2(b1 -b1 ) + e2 -e2 = 3 (The equation at point (X=2, Y=3)) + - + - + - b0 - b0 + 4(b1 -b1 ) +- e3 -e3 = 4 + - + - + - b0 - b0 + 6(b1 -b1 ) + e4 -e4 = 5 + - + - + - b0 - b0 + 7(b1 -b1 ) + e5 -e5 = 8+ - + - + - b0 , b0 , b1 , b1 , ei and ei are nonnegative.The optimal solution obtained by Solver is: + - + - b0 = 1.33, b0 = 0; b1 = .667, b1 = 0; + - + - + - + - + - e1 = 0, e1 = 0; e2 = .333, e2 = 0; e3 = e3 = 0, e4 =, e4 = 0 .333; e5 = 2, e5 = 0.The regression equation is: y = 1.333 + 0.667x9 8 7 6 5 Y 4 YRegression 3 2 1 0 0 2 4 6 89 Comment A linear regression model based on the minimization of the sum of squared errors (the standard linear regression model) is effected more by large errors thereby tries to eliminate them more “religiously” than the absolute value based model discussed here. In our example the last point is placed somewhat out-of-line, generating a larger error than the other points. Now, the standard regression model is Y = 1.0154 + .8462X, and the errors for this models are 0.1385, 0.2923, -0.4, -1.092, 1.0615. Indeed, the standard regression does reduce the last point error From 2 to 1.0615, but the other errors are now generally larger. Excel solution10 Example 6: The ordinal regression The classical regression methods apply when all the data are in numeric form. Sometimes, however, data are found in a more subjective or qualitative form. Consider, for example, a marketing study in which you are trying to understand how consumers determine their preferences for various products in a class, e.g., breakfast foods. The data from a consumer survey might consist simply of preference statements of the form "I prefer Wheaties to Cheerios." There is available additional independent quantitative data on each breakfast food describing its characteristics, e.g., sweetness, nutritional value, and price. The desire is to determine how ranking by consumers depends upon the product characteristics. We’ll use the following car preference example to explain the idea behind the ordinal regression. Five different cars, denoted by B, C, F, L, and M, have been evaluated by a consumer panel. These evaluations are in the form of eight different pair-wise comparisons by different members of the panel. Clearly, these preferences are qualitative. The results are as follows:Comparison Less More Number Preferred Car Preferred Car 1 C L 2 M B 3 F L 4 F C 5 B F 6 L M 7 M C 8 M FFor example, the first comparison indicates that some consumers preferred an L to a C, whereas the last comparison indicates that some consumers preferred an F to an M. Although one could, it is not required that these pairwise rankings be consistent or transitive. Indeed, if these rankings are collected from a number of people, we would expect some inconsistency among them. For example, you can note an inconsistency among rankings (3), (6), (2), and (5). Ranking (3) says F is inferior to L; however, rankings (6), (2), and (5) suggest that L is inferior to F. We suspect that consumer preferences are affected by the four characteristics: price, weight, length, and mpg. These characteristics for the five cars are listed below:Characteristics Price Weight Length Car (1,000S) (1,000S) (feet) Mpg B 9 4 15 17 C 6 3 13 21 F 7 2.8 13.5 20 L 10 4.2 15.5 16 M 8 3.9 14 18Develop a scoring formula of the form: 11 Score = Wprice PRICE + WweightWeight + WlengthLength + WMPGMPG, such that more preferred cars score higher than less preferred cars? Such a formula would help in identifying the relative importance of various characteristics. The following is an approach for developing such a scoring formula.Define: Sj = the over all score of car j Sij = the score car j scores on characteristic ‘i’ Wi = the regression coefficient of characteristic i. Since each coefficient could be either negative or non-negative we define Wi = WPi – WNi, where both WP and WN are nonnegative. Ep = the prediction error in comparison ‘p’. Since the error could be either negative or non-negative we define Ep = EPp – ENp where both ‘EP’ and ‘EN’ are non-negative.Base on the comparisons previously made, at comparison i=1 car L is preferred to car C. A proper formula should give to car L a higher score than to car C. If car L does not score higher than car C, a penalty is imposed at the amount of the error made, that is E1 = S1C – S1L. Our objective is to minimize the total penalty, so we seek a set of weights Wi that do just that.The linear programming model follows.Min EN1 + EN2 +…+ EN8 Subject to For convenience, we define the score equation per car first: SB = 9WPrice + 4WWeight + 15WLength + 17WMPGSC = 6 WPrice + 3 WWeight + 13 WLength + 21WMPGSF = 7 WPrice + 2.8 WWeight + 13.5 WLength + 20WMPGSL = 10 WPrice + 4.2 WWeight + 15.5 WLength + 16WMPGSM = 8 WPrice + 3.9 WWeight + 14 WLength + 18WMPG Now we set up the score relationships based on the pairwise comparisons provided: SL – SC + EN1 – EP1 = 0 SF – SB + EN5 – EP5 = 0 SB – SM + EN2 – EP2 = 0 SM – SL + EN6 – EP6 = 0 SL – SF + EN3 – EP3 = 0 SC – SM + EN7 – EP7 = 0 SC – SF + EN4 – EP4 = 0 SF – SM + EN8 – EP8 = 0For example consider comparison 1: If SL<= SC a penalty of EN1 = Sc – SL is created while EP1 = 0. If SC >= SL (as should be expected) EP1 = SC – SL while no penalty is imposed (EN1 = 0).Finally we provide a normalizing constraint that takes care of different scoring scales for the different characteristics, that if not adjusted might affect the weights’ determination. (+) If for a given comparison S i is the score of characteristic ‘i’ for the more preferred car, (-) (+) (-) and S i is the counter part of the less preferred car, we calculate the value [S i – S i] over all the pairwise comparisons for characteristic ‘i’ and add all of them, then multiply by Wi. We repeat this operation for each characteristic and finally add all the products. Then we set the sum equal to 1, which constitutes the normalizing constraint. 12 Specifically, in our example the constraint becomes (Sprice,L - Sprice,C + Sprice,B – Sprice,M+ …)Wprice + (Sweight,L - Sweight,C + Sweight,B – Sweight,M+ …)Wweight (Slenght,L – Slenght,C + Slenght,B – Slenght,M+ …)Wlenght (SMPG,L - SMPG,C + SMPG,B – SMPG,M+ …)WMPG = 1Substituting the figures we have:(10-6+9-8+10-7+6-7+7-9+8-10+6-8+7-8)Wprice + (4.2-3+4-3.9+4.2-2.8+3-2.8+2.8-4+3.9-4.2+3-3.9+2.8-3.9)Wweight+ (15.5-13+15-14+15.5-13.5+13-13.5+13.5-15+14-15.5+13-14+13.5-14)Wlength+ (16-21+17-18+16-20+21-20+20-17+18-16+21-18+20-18)WMPG = 1. In summary this constraint is: 0Wprice – 0.6Wweigth + 0.5Wlength + 1WMPG = 1 Note the following range of definition for the variables: Wi is URS; Si is URS; ENi and EPi are non-negative.When running this LP model we get the following solution:Wprice = 3, Wweight = 2.5, Wlength = -0.5, WMPG = 2.75.The resulting regression equation is: SCORE = 3Price + 2.5Weight – .5Length + 2.75MPG Excel SolutionComments 1. Note that car length is not considered an important factor in determining a car score for the sake of comparison, when the other three characteristics are included. 2. We can calculate the Goodness of Fit (somewhat analogues to the coefficient of determination) for this model as follows: 1 Goodness of Fit  Z 1 0 ( N  K )Z0 is the objective value. N is the number of comparisons made. K is the number of regression coefficients estimated. In our example we have1 This is the objective value. The Goodness of Fit   .88 minimum sum of errors is 0.5. See 1 .5 8  4 the Solver solution.3. Multiple optimal solutions exist to this problem. This means that more than one set of weights provides the same objective value, and creates the same goodness of fit. We show above only one of the possible optimal solutions.13 Maximize the minimum value of several functions There are several applications where the objective function is of the form MaxMin or MinMax. Let us discuss these types of objective functions. Assume there are two functions F1 and F2, where F1 = 1X1+2X2, and F2 = 4X1+1X2. Let us select the pair X1=1, X2=1, and determine the minimum between F1 and F2 for this selection. F1 = 1(1)+2(1) = 3; F2 = 4(1)+1(1) = 5. The minimum of F1 and F2 is 3 (we write min(F1, F2) = 3). Let us change the selection to X1=2, X2=1: F1 = 1(2)+2(1) = 4; F2 = 4(2)+1(1) = 9. Now min(F1, F2) = 4. For each selection of X1 and X2 the minimum of F1 and F2 may change. We are looking for the values of X1 and X2 that maximize the minimum of F1 and F2, (or sometimes minimize the maximum of F1 and F2). Let us look at some examples that demonstrate the application of this concept, and also show how to make the non-linear objective function of the type “Min Max” “Max Min” a linear functionExample 7 The Highway Patrol of a southern state wishes to divide the effort of its on-duty 25 officers among highway segments in each territory to maximize speeding reduction. The potential speed reduction per officer differs from segment to segment. The following data was collected for the 8 highway segment (j = 1, …,8)Highway segment (j) 1 2 3 4 5 6 7 8Upper bound (Uj) 4 8 5 7 6 5 6 4 Reduction potential (Rj) 11 3 4 14 2 19 10 13Uj is the maximum number of officers that can be assigned to segment j. Rj is the speed reduction (in mile/hour) per one officer assigned to segment j. Variables: Xj = number of officers per week assigned to patrol segment j.Let’s assume for now we decided to maximize the total speed reduction. This might look reasonable because then (we believe) all the segments will participate in reducing the speed as much as possible. The linear model follows: Maximize 11X1 + 3X2 + … + 13X8 Subject to X1 + X2 + …+X8 25 X1 4, X2 8, … X8 4, and: X1, X2, …,X8 are nonnegative.14 The optimal solution to this model is: Xj = 4, 0, 0, 7, 0, 5, 5, and 4. One can see the imbalance of this solution. Some segments get no officer assignment at all while others get at least 4 officers. If resources are to be allocated more evenly, we turn to the Min Max or Max Min objective function for help; doing so we focus on the least favorable element in the model, and try to improve it as much as possible.In the context of this problem the least favorable element is the smallest actual speed reduction. Of course the speed reduction depends on how many police officers are assigned to each segment (denoted by X1, X2,…). So, the smallest speed reduction is equal to the minimum of (4X1, 8X2,…, 4X8). We would like to improve the smallest speed reduction as much as possible, that is, to maximize the minimum of (4X1, 8X2,…, 4X8). Therefore our objective is to Maximize the minimum of (4X1, 8X2… 4X8) Subject to X1 + X2 + …+X8 25 X1 11, X2 3… X8 13, where Xj are nonnegative.The objective function is non-linear, and this must be taken care of first. This is how you do it: Define the minimum of all the functions participating in the current max min type objective function as ‘f’. Since f is the minimum of 4X1, 8X2, …, then f  4X1, and f 8X2, and f 5X3,…and so on for all the 8 functions. We add these 8 constraints to the previous model, and change the objective function to: Maximize f. Maximize f Maximize the minimum called f Subject to: X1 + X2 + … +X8 25 X1 11, X2 3, …, X8 13 f  4X1, f  8X2, …, f  4X8 f and all Xj are nonnegative.This model is now linear and can be solved as usual.15 Review problems Solutions1. FINCO must determine the investment strategy for the next three years. At present (Time 0) $1,000,000 is available for investment. There are five possible investments FINCO is looking at. The cash flow associated with investing $1 in each investment is given below.Year 0 1 2 3 Investment A -1 .5 1 0 B 0 -1 .5 1 C -1 1.2 0 0 D -1 0 0 1.9 E 0 0 -1 1.5For example: Investing $1 in investment A at the beginning of year 0 (now), yields $.5 at the beginning of year 1 and $1 at the beginning of year 2. Also, note that investments B and E are not available now.To ensure that its portfolio is diversified, FINCO requires that at most $75,000 be placed in any single investment. In addition, FINCO can earn 8% annually by keeping un-invested cash in money market funds. Return from an investment may be immediately re-invested. The only cash considered for investment at any time is the cash on hand. Formulate a linear programming model that maximizes the cash on hand at the end of year 3.2. Global Minimum manufactures bikini-swinging suits. Business is highly seasonal, with expected demand being 850, 2800, 100, and 500 dozen suits over the four quarters of next year. The company currently carry 2000 dozen bikini suits and can produce 1200 dozen suits per quarter, but inventories must be built up to meet larger demands at a holding cost of $15 per dozen per quarter. Global wants to meet demand while minimizing this inventory cost. (a) Formulate a linear programming model to choose an optimal production plan for next year. Assume no unsold bikini suits left over from this year can be sold next year; therefore no inventory is required at the end of next year. (b) Re-solve this problem assuming inventory at the end of the last quarter of any given year is carried over to the first quarter of the next year, if the current demand pattern remains the same for the long run.3. Control Box. The following figure shows the ceiling locations of three sensors in a new factory relative to a coordinate system (in feet) with origin at the lower left. A control box will be located along the long (lower in the figure) wall with fiber-optic cables running rectilinearly to each sensor (see the example bellow). (a) Designers want to place the Box to minimize the total cable required. Formulate and solve a linear programming model to choose an optimal placement for the control box. (b) Re-solve this problem this time minimizing the length of the longest cable. Y (5,8)(25,10) (5,8)X16 4. For the data provided below… (a) Find the Least Absolute Deviation line (b) Find the line for which the maximum deviation is minimized.(x,y) = (1, 10); (3, 15); (4, 11); (8, 14); (12, 17)5. Pinnacle. The Pinnacle Steel Company provides structural steel for construction projects. Typical projects are shopping centers, warehouses, and manufacturing plants. The two major operations performed by Pinnacle are: (A) cut I-beams length and (B) punch bolt holes near the end of the cut beams. The time spent making cuts is related largely to the number of cuts being made, the cross- section of the cuts measured in square cm, and the number of holes. Pinnacle wants to improve its estimating formulae used in quoting the time to do a proposed Job. Below are tabulated some statistics for several recent completed jobs for the total time in minutes to do the cutting and punching. Pinnacle would like to derive a linear estimating formula based on these three explanatory factors. Derive a LAD regression estimating formula for Pinnacle.Total Number of Total Cut Total Holes Job Time Cuts Cross-Section 1 452 50 11,200 170 2 675 84 21,300 240 3 787 110 19,200 225 4 180 25 7500 52 5 1578 210 35,220 540 6 835 95 19,590 3086. Operators. At the beginning of the fall semester, the director of the computer facility of a certain university is confronted with the problem of assigning different working hours to his operators. Because all the operators are currently enrolled in the university, his main concern making certain that operators’ work schedule do not interfere with their study.There are six operators (four-man and two women). They all have different wage rates because of differences in their experience with computers and in their programming ability. The following table shows the wage rates, along with the maximum number of hours each can work each day.Maximum hours of availability Operator Mon. Tue. Wed. Thurs. Fri. 7. Kevin $8.50 6 0 6 0 6 8. Don $8.30 0 6 0 6 0 9. Harry $6.90 4 8 4 0 4 10. Sean $6.20 5 5 5 0 5 11. Helen $7.10 3 0 3 8 0 12. Kathy $7.90 0 0 0 6 2 Because of the tight budget, the director has to minimize cost. His decision is that the highest operator pay per week should be as small as possible. To clarify, if Kevin and Don are scheduled to work X1 and X2 hours a week respectively, their weekly pay is 8.5X1 and 8.3X2. The maximum of these two should be minimized. Of course, this approach should be applied to all the workers. Yet each operator should work enough hours so his/her knowledge of operation is not impaired. This level is set arbitrarily at eight hours per week for the male operators and seven hours a week for the female operators. 17 The computer facility is to be opened for operation from 8 AM to 10 PM Monday through Friday with exactly one operator on duty during these hours. On Saturdays and Sundays, the computer is to be operated by other staff. Formulate an LP model so the director can determine the number of hours he should assign to each operator on each day.7. For the data provided in problem 4, find the location (X,Y) that minimizes the travel distance between it and the farthest point, if travel must be made along horizontal and vertical roads [that is Min Max{|X-Xi| + |Y-Yi|}].8. Re-solve example 2 in this set of notes, with the following change; (i) Overtime cannot be used. (ii) Shortage is allowed in each quarter at a shortage cost of $600 per unit. Note that when shortage is allowed the inventory at the end of a quarter can take negative values, making it an URS variable. (iii) Demand per quarter is 40, 60, 75, 45 boats. (iv) Production capacity per quarter is 70, 50, 40, 90 boats.9. Shoemaker of America forecasts the following demand for each of the next four months: 1- 5000; 2 - 6000; 3 - 5000; 4 - 9000 pairs. It takes a shoemaker 15 minutes to produce the pair of shoes. Each shoemaker works 150 hours per months plus up to 40 overtime hours per month. A shoemaker is paid a regular salary of $2000 per month plus $50 per hour for overtime. At the beginning of each month, Shoemakers can either hire or fire workers. It costs the company $1500 to hire a worker and $1900 to fire a worker. The monthly holding cost per pair of shoes is 3% of the cost of producing a pair of shoes with regular-time labor. The raw materials in a pair of shoes cost $10. Formulate a linear programming model that minimizes the cost of meeting the demands of the next four months on time. At the beginning of months 1, Shoemaker has 13 workers.10. Donovan Enterprises produces electric mixers. During the next four quarters, the following demand for mixers must be met on time: Quarter 1-4000; quarter 2-2000; quarter 3-3000; quarter 4- 10,000. Each of Donovan’s workers works three-quarters of the year and gets one-quarter off. Thus, the worker may work in quarters 1, 2, and 4 and get quarter 3 off. Each worker is paid $30,000 per year (if working) and can produce up to 500 mixers during a quarter. At the end of each quarter, Donovan incurs the holding cost of $30 per mixer in inventory. Formulate a LP model that minimizes Donovan’s costs (labor and inventory) of meeting the next year’s demand (on time). At the beginning of quarter 1 1600 mixers are available. 11. Bumles Inc. uses part of its capacity in its Mexican plant to make custom hand – painted teapots sold in that local market. One teapot takes .5 hours of a painter’s time. Bumles has 30 painters available. The plant is used to make the teapots on Thursday, Friday, and Saturday each week. During the remainder of the week the productive capacity is devoted to another production line. Not all the 30 painters will necessarily be engaged, but each painter who is engaged is available to work any part of 8-hour day, 2 days a week. A painter can be assigned to any 2-day schedule, and is paid for a total of 16 hours of regular time work, no matter what part of that time he or she actually spend producing teapots. If there is not enough production to keep all the workers assigned to a particular day busy for the entire day, the slack time is spent on cleaning the plant and similar activities. If labor costs are not taken into account, the revenue, net of other costs, from selling a teapot is $15. Predicted demand is 100, 300, 600 on Thursday, Friday, and Saturday respectively, and must be satisfied on the day on which it occurs or it is lost. A penalty on shortage (due to lost demand) is charged as follows: $1 per unit on Thursday, $3 on Friday, and $5 on Sunday. Production on a given day can be used to satisfy demand that day or later in the week (i.e. teapots produced on Thursday can be inventoried and sold on Friday and Saturday). It costs $.50 to carry a teapot per day. Left over after Saturday sales must be shipped somewhere else, so there is no initial inventory on Thursday morning 18 and no final inventory on Saturday evening. Find the optimal production schedule. 12. Pat Statsdud, a student ranking near the bottom of the class, decided that a certain amount of studying could actually improve his chances to pass the course. Pass or fail grade is determined by the score in the final exam. Pat wanted to predict his score in the final exam. He believed there was a linear relationship between the final score and two explanatory variables: Avg. assignment score, midterm score. He collected data which included the final score, the assignment score and the midterm score for a sample of students. Based on the sample create a regression equation using the minimum sum of absolute errors approach.Final Assignment Midterm Grade 23 75 40 F 49 75 93 C 34 65 63 D 91 100 87 A 43 100 73 C 29 90 43 D 31 100 33 D 30 50 37 F 75 70 78 B13. The general manager of the Cleveland Indians baseball team is in the process of determining which minor-league players to draft. He is aware that his team needs home-run hitters and would like to find a way to predict the number of home-runs a player will hit. He gathers a random sample of players and records the number of homeruns he hit in the last two years; his age; and the number of years in the minor league. With this data he turns to a group of experts in the area of baseball (journalists, retired players), and ask them to compare pairs of players of their choice. The quantitative data and the comparison information are shown below. (a) Develop a scoring equation that can help the manager sign-in the best player. (b) Which of the following player is expected to be the best for the team if signed-in?14. Rozac – revisited (Challenge problem): Consider the Rozac problem presented before (see problem 2 of the “Reminder” problems). Suppose Rozac production is now planned for two successive time periods. At the first period there is a demand for 1000 lbs of the product and in the second period there is a demand of 1200 lbs. Purchase cost for the four chemicals per pound in the first period are as before (8, 10, 11, 14 dollars), but in the second period are expected to change to 9, 16, 13, 22 dollars. Chemicals are to be purchased now to meet demand of at least the first period. Extra amounts can be either stored for production in the next period or used to produce Rozac now. If stored as separate chemicals additional inventory cost are incurred of $1, $1.5, $2.75, and $3.5 per pound for each chemical stored respectively. If extra Rozac is produced at the first period, the extra amount is stored at a cost of $1.5 per pound for use at the second period. The rest of the conditions presented in the Rozac problem apply to this problem for each period separately. Specifically, in each period at least 300 lb of chemical 3 must be purchased, and in each period the total amount of chemical 1 and 2 combined purchased cannot exceed 320 lb. By weight at least 8%, 4%, and 2% of active ingredients A, B, and C respectively need to be included in the mix. How much chemical of each kind should be purchased and when, and how much should be produced in each period?15. Pat Statdud – revisited (Challenge problem): Pat got excited by the potential he discovered in making impressive predictions using regression models. He wanted to surprise his statistics instructor with an informed guess of the grading equation he was using to decide students’ grades. He knew his instructor was using a certain formula to calculate total scores based on the scores in the “Final”, “Midterm”, and “Assignments”. But the 19 weights each component in the scoring formula had remained unknown. The letter grade for the course would be determined by the total score which runs between 0 and 100, and by the following chart: A: 90 – 100; B: 75 – 89; C: 60 – 74; D: 50 – 59; F: Less than 50. Pat decided to consult Kurt, one of the better students in the course; Kurt advice was to collect information from friends who have already taken this course, which included their scores in the final, the midterm, the assignments, and the course letter grade they received (they could not provide their total score, because the instructor kept it for himself). See the data in problem 12. (a) With this Pat needed to formulate a linear programming model that would suggest a good scoring formula for the prediction of the total score, given the score in the final, in the midterm and in the course assignments. (b) If Pat’s scores in the midterm and the course assignments were 51% and 70% respectively, what is the minimum score in the final she must have in order to pass the course?20

Review of Linear Programming Single Period Models

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