IE 416: Operations Research I

Table of Contents…………………………………………………………………………………………………………….……………………2 Problem Statement...... 3 Summary of Problem...... 3 Table 1: Chandler Oil Company- Oil Information...... 3 Table 2: Chandler Oil Company- Products Made from Oil...... 3 Formulation of Problem...... 3-6 1.) Define the decision variables...... 4 Table 3: Summary of Problem...... 4 2.) Provide explanatory information and assumptions...... 4 3.) Formulate Objective Function (O.F) 5 4.) Formulate Constraints 5-6 5.) List All Equations 6-7 Table 4: Constraint Equation List 6 Table 5: LP Form of Constraint Equation List 6 Solution using WinQSB...... 7-8 WinQSB Screen Shot 1: Original Input...... 7 Table 6: WinQSB Variable and Name 7

Table 7: WinQSB Constraint and Name 7

WinQSB Screen Shot 2: Output 8

Report to Manager...... 9-10 Sensitivity Analysis...... 10-15 WinQSB Screen Shot 3: Objective Function Graph for Coefficient of Oil 1 for Gas SA...... 11 WinQSB Screen Shot 4: Table for Objective Function SA of Oil 1 for Gas...... 11 WinQSB Screen Shot 5: RHS Sensitivity Analysis Graph for Oil 1 Availability...... 13 WinQSB Screen Shot 6: RHS Sensitivity Analysis Table for Oil 1 Availability...... 13 WinQSB Screen Shot 7: RHS Sensitivity Analysis Table for Demand Gas...... 15 WinQSB Screen Shot 8: RHS Sensitivity Analysis Table for Demand Gas...... 15 Simplex Tableau...... 16-19 WinQSB Screen Shot 8: Simplex Tableau for Iteration #s1-8...... 16-17 Table 8: Original Simplex Tableau Table...... 17 Table 9: Simplex Tableau for Iteration #1 ...... 18 Table 10: Simplex Tableau for Iteration #2 ...... 18 Acknowledgments 19

2 | P a g e PowerPoint Slide Printouts 20 Problem Statement

Chandler Oil Company has 5,000 barrels of oil 1 and 10,000 barrels of oil 2. The company sells two products: gasoline and heating oil. Both products are produced by combining oil 1 and oil 2. The quality level of each oil is as follows: oil 1- 10; oil 2- 5. Gasoline must have an average quality level of at least 8, and heating oil at least 6. Demand for each product must be created by advertising. Each dollar spent advertising gasoline creates 5 barrels of demand and each spent on heating oil creates 10 barrels of demand. Gasoline is sold for $25 per barrel, heating oil for $20. Formulate an LP to help Chandler maximize profit. Assume that no oil of either type can be purchased. Summary of Problem

•Table 1: Chandler Oil Company – Oil Information Chandler Oil Company - Oil Information Oil # of barrels Oil Quality Oil 1 5000 10 Oil 2 10000 5

•Table 2: Chandler Oil Company – Products Made from Oil Chandler Oil Company - Products Made from Oil Demand Created Selling Product (Blend) Avg. Quality Level per $1 spent on Price per Advertising Barrel Gas 8 5 $25 Heating Oil 6 10 $20

We will condense the tables above further in the report. Explanation will be given.

Dr. Parisay’s comments are in red. Formulation of Problem

3 | P a g e Chandler Oil Company must make two types of decisions: first how much money should be spent in advertising each of their products (blends): gas and heating oil, and second, how to blend each product from the available types of oil. For example, Chandler Oil Company must decide how many barrels of oil 1 should be used to produce gas.

1.) Define the decision variables ai = dollars spent daily on advertising blend i (i = 1,2) xij = barrels of oil i used daily to produce blend j (i = 1,2 ; j = 1,2)

Sign Restrictions: ai > 0 xij > 0

The definition of the decision variable implies: x11 + x12 = barrels of oil 1 used daily x11 + x21 = barrels of gas produced daily x21 + x22 = barrels of oil 2 used daily x12 + x22 = barrels of heating oil produced daily

Now that we have defined our decision variables, we can combine Table 1 and Table 2 to create a more efficient summary of our problem:

•Table 3: Summary of Problem Demand of Barrels Avg. Created per $1 Selling Price Quality spent on per Barrel Product Level Advertising (Blend) Decision Variables

8 5 $25 Gas x11 x21

6 10 $20 Heating Oil x12 x22

OIL 1 2 # of barrels 5000 10000 Oil Quality 10 5

2.) Provide explanatory information and assumptions

To simplify, let’s assume that gas and heating oil cannot be stored, so it must be sold on the day it is produced. This implies that for j = 1,2, the amount of blend produced daily should equal the daily demand for blend j.

4 | P a g e If the amount of blend produced daily exceeds daily demand, we would incur unnecessary production and purchase cost. Even though the problem did not provide us with the production and purchase cost, we know in a real world environment these costs are real.

If the amount of blend produced daily is less than daily demand, we fail to meet mandatory demand and incur unnecessary advertising cost.

Therefore, the goal of this problem is to maximize profit by using the right amounts of oil to produce the right number of blends.

3.) Formulate Objective Function (O.F)

The basic formula for profit is revenues minus cost. Profit = Revenue – Cost

Profit = Zmax

Therefore, we must define both revenue and cost in relation to this problem. After defining revenue and cost, we can determine profit (Zmax) by using the formula above.

 Daily Revenues from Blend Sales (Sales of Gas and Heating Oil) = 25(x11 + x21) + 20 (x12 + x22)

 Daily Advertising Cost = a1 + a2

 Daily Profit = Daily Revenues from Blend Sales - Daily Advertising Cost

Daily Profit = [25(x11 + x21) + 20 (x12 + x22)] – [a1 + a2]

 Simplify

Zmax = 25x11 + 25x21 + 20x12 + 20x22 –a1 – a2

4.) Formulate Constraints

Constraint 1: Maximum of 5,000 barrels of oil 1 are available for production

x11 + x12 < 5000

Constraint 2: Maximum of 10,000 barrels of oil 2 are available for production

x21 + x22 < 10,000

Constraint 3: Gasoline must have an average quality level of at least 8.

2x11 – 3x21 > 0

Constraint 4: Heating oil must have an average quality level of at least 6.

5 | P a g e 4x12 – x22 > 0

Constraint 51: Demand of gas is increased by 5 barrels for every dollar spent on advertising. x11 + x21 = 5a1

Constraint 61: Demand of heating oil is increased by 10 barrels for every dollar spent on advertising.

x12 + x22 = 10a2

5.) List all Equations

•Table 4: Constraint Equation List

Description Equation Type

Max Profit Zmax = 25x11 + 25x21 + 20x12 + 20x22 –a1 – a2 Objective Function

Oil 1 Avail. x11 + x12 < 5000 Constraint

Oil 2 Avail. x21 + x22 < 10,000 Constraint

Gas Quality 2x11 – 3x21 > 0 Constraint

H. Quality 4x12 – x22 > 0 Constraint

Demand Gas x11 + x21 = 5a1 Constraint

Demand H. x12 + x22 = 10a2 Constraint

•Table 5: LP Form of Constraint Equation List The following table needs modification. Do not define a1 and a2 in constraints as you have them as ad cost in OF. Replace with a3 and a4. The OF is not standard form. Refer to the format in your book.

Description Standard LP Form Equation Type

Max Profit Zmax = 25x11 + 25x21 + 20x12 + 20x22 –a1 – a2 Objective Function

Oil 1 Avail. x11 + x12 + S1 = 5000 Constraint

Oil 2 Avail. x21 + x22 + S2 =10,000 Constraint

Gas Quality 2x11 – 3x21 - e1 + a1 = 0 Constraint

H. Quality 4x12 – x22 - e2 + a2 = 0 Constraint

1 Remember, the problem stated that demand for each problem must be created by advertising. Also remember in 2.) we assumed that amount of blend produced daily should equal daily demand. So, here supply equals demand. In layman’s terms, the amount we produce will be equal to the demand we create through advertising.

6 | P a g e Demand Gas x11 + x21 - 5a1 = 0 Constraint

Demand H. x12 + x22 - 10a2 = 0 Constraint

We define slack variables in Oil 1 Avail. and Oil 2 Avail. Slack variable si (si=slack variable for the ith constraint), which is the amount of resource unused in the ith constraint. If constraint i of an LP is a < constraint we convert it to an equality constraint by adding a slack variable si.

We define excess variables and artificial variables in Gas Quality and H. Quality. If the ith constraint of an LP is a > constraint, then it can be converted to an equality by subtracting an excess variable ei, from the ith constraint. We add an artificial variable too so we have a basic variable for initial simplex tableau. Solution using WinQSB

Input into WINQSB:

• WinQSB Screen Shot 1: Original Input

Explanation of WinQSB Variables and Constraints:

•Table 6: WinQSB Variable and Name •Table 7: WinQSB Constraint and Name

7 | P a g e WinQSB Output:

• WINQSB Screen Shot 2: Output

8 | P a g e Report to Manager

If Chandler Oil Company wants to maximize its profit to $323,000 for the current production of gasoline and heating oil it should:

 Produce 5,000 barrels of gasoline by mixing 3,000 barrels of oil 1 with 2,000 barrels of oil 2

 Produce 10,000 barrels of heating oil by mixing 2,000 barrels of oil 1 with 8,000 barrels of oil 2

By these combinations we are able to meet the exact quality requirement for gasoline which is 8 and the exact quality requirement for heating oil which is 6. (You concluded this from the related constraints to be binding. If constraints were not binding you need to calculate the quality value.)

The production value of 5,000 barrels of gasoline will only remain if the profit per barrel of gas stays between $18.88-$83.17. (good job) A change in unit profit within this range will cause maximum profit to change while barrels of gasoline produced remains at 5,000 barrels. Anything outside this profit range will cause barrels of gas produced to change and maximum profit to change.

(good explanation) We must take into account both allowable ranges of profit for oil 1 for gas and oil 2 for gas since gas is a mixture of both. Although oil 1 for gas has a profit range of $16.83-$83.17 and oil 2 for gas has a profit range of $18.88-$112.25, we consider the correlation between both oil profit ranges. When taking this into consideration, we determine the $18.88-$83.17 range mentioned.

Using the same concept, the production value of 10,000 barrels of heating will only remain if the profit per barrel of heating oil stays between $5.46-$26.13. A change in unit profit within this range will cause maximum profit to change while barrels of heating oil produced remains at 10,000 barrels. Anything outside this profit range will cause barrels of heating oil produced to change and maximum profit to change.

We must take into account both allowable ranges of profit for oil 1 for heating oil and oil 2 for heating oil since heating oil is a mixture of both. Although oil 1 for heating oil has a profit range of $0-$28.17 and oil 2 for heating oil has a profit range of $5.46-$26.13, we consider the correlation between both oil profit ranges. When taking this into consideration, we determine the $5.46-$26.13 range mentioned.

9 | P a g e To reach the profit maximization mentioned, we must pay $1000 in advertisement for gas and $1000 in advertisement for heating oil to generate the demand for the 5,000 barrels of gasoline and 10,000 barrels of heating oil.

Our solution remains optimal if the range of oil 1 usage is from 2,500-15,000 barrels. For each additional barrel of oil 1 made available for use, we can increase our profit by $29.70. This in turn means that we are willing to buy extra barrels of oil 1, up to 15,000 barrels, for an extra cost of up to $29.70 for each barrel (which means we would break even) (This is managerial application of shadow price. We will cover it after midterm. Good Job) . However, as of now we only have room to store 5,000 barrels of oil 1 and we have used all this storage.

Our solution remains optimal if the range of oil 2 usage is from 3,333-20,000 barrels. For each additional barrel of oil 2 made available for use, we can increase our profit by $17.45. This in turn means that we are willing to buy extra barrels of oil 2, up to 20,000 barrels, for an extra cost of up to $17.45 for each barrel (which means we would break even). However, as of now we only have room to store 10,000 barrels of oil 2 and we have used all this storage. Sensitivity Analysis

 1 Sensitivity Analysis for a Coefficient in the Objective Function that is a Basic Variable

Choice:

Variable Name Given

x11 Oil 1 for Gas

We chose to do a sensitivity analysis on x11 (oil 1 for gas) in the objective function.

Why didn’t we choose advertising cost?

We knew we wanted to do an analysis on oil instead of advertising cost. We know that our supply of oil is fixed. We have assumed that supply equals demand, therefore, if all demand is met, then demand is also fixed. To maximize profit, all demand will be met. Demand is only generated through advertising. Because demand is only generated through advertising and our demand is fixed, then our advertising cost is also fixed. For example, each dollar spent advertising gasoline creates five barrels of demand for gas, meaning advertising cost per barrel of gasoline is $0.20. If you look at WinQSB Output 3 you will see this value in the shadow price column for Demand Gas. Thus, the problem would have to change completely for a sensitivity analysis on advertising cost to be valid.

Motivation for SA on x11 (oil 1 for gas)

10 | P a g e Currently, oil 1 for gas is tied for the highest unit profit of $25 c(j) with oil 2 for gas. However, it has the highest allowable max c(j) compared with oil 2 for gas. Like previously mentioned, we are looking at these variables being correlated. Although technically oil 2 for gas has a allowable max c(j) of $112.25, we have already established that the range for the oils for gas is $18.88-

$83.17. Therefore this variable x11 contains the $83.17 allowable max c(j) we are looking for. Of course, if we do change the unit cost of x11 from $16.83-$83.17 the solution value will remain at 3,000 barrels, but the value of the objective function will change. (good observation)

(This paragraph is a good effort, but not expected for this level of class) To describe the entering and leaving variables WinQSB says the following:

In a simplex iteration, a decision variable will enter the basis depending on a particular rule.

The typical decision rule is to choose the decision variable with the best Cj-Zj to enter

the basis. For the maximization problem, it is the variable with the most positive Cj-Zj. The decision rule to choose the leaving basic variable is the one with the minimum ratio of RHS/A (i,j) in the updated simplex tableau.”

With this definition in mind, we can see the entering and leaving variables in range 1 and range 3.

• WINQSB Screen Shot 3: Objective Function Graph for Coefficient of Oil 1 for Gas SA

11 | P a g e • WINQSB Screen Shot 4: Table for Objective Function SA of Oil 1 for Gas

 1 Sensitivity Analysis for a Coefficient in the Objective Function that is a Non- Basic Variable

o There are no non basic variables in the objective function.

 1 Sensitivity Analysis for RHS in Constraint that is Binding

12 | P a g e Choice:

Constraint Name Given x11 + x12 < 5000 Oil 1 Avail.

Motivation for SA on x11 + x12 < 5000 (oil 1 availability)

We chose to do a sensitivity analysis on oil 1 availability because it has the highest shadow price of $29.70. From the optimal solution we also chose the sensitivity analysis because it had the highest room for profit.

Oil 1 availability has an allowable max RHS of 15,000 barrels and a shadow price of $29.70. If we max out our RHS we get 15,000 x $29.70 = $445,500 increase in profit.

Oil 2 Availability has an allowable max RHS of 20,000 barrels and is the only other constraint with a relatively high shadow price of $17.45. If we max out our RHS on this constraint we get 20,000 x $17.45 = $349,000 increase in profit.

Because oil 1 availability has the higher potential of making more profit, we went with this choice. In reality, if we were to max out both RHS of oil 1 and oil 2 we would need to get more storage area for oil and generate more demand by advertisement. (good note)

Obviously, an infinite supply of either oil 1 or oil 2 is infeasible.

• WINQSB Screen Shot 5: RHS Sensitivity Analysis Graph for Oil 1 Availability

13 | P a g e • WINQSB Screen Shot 6: RHS Sensitivity Analysis Table for Oil 1 Availability

14 | P a g e  1 Sensitivity Analysis for a RHS in Constraint that is Binding

(This is a good effort; however, try to avoid performing SA on those RHS that does not have specific/clear meaning, such as here. RHS is zero which does not have a practical meaning. You got zero through some math operation for simplex method.)

Constraint Name Given

x11 + x21 - 5a1 = 0 Demand Gas Choice:

Motivation for SA on x11 + x21 - 5a1 = 0 (Demand Gas)

We chose to do a sensitivity analysis on Demand Gas because it has the highest shadow price of $.20 between the two binding constraints available. In terms of profit, it wouldn’t matter if we chose to increase allowable RHS for Demand Gas or Demand H because the same profit will be acquired:

Shadow Price of Demand Gas x Max RHS = Amount of Increased Profit Due to Demand Gas $.20 x 5,000 = $1000

Shadow Price of Demand H. x Max RHS = Amount of Increased Profit Due to Demand H. $.10 x 10,000 = $1000

However, according to the solution both demand for gas and demand for heating oil is at their max point. Gas is better because it pays a higher shadow price per barrel than heating oil. Theoretically this in turn is better for a company because having to produce, move, store, and sell 5,000 barrels would be easier than having to produce, move, store, and sell 10,000 barrels. So, you are putting in half the effort but making the same in profit.

Notice that the max allowable for “Demand Gas” is 5000. That is if you increase zero to 5000. Here zero does not have a specific meaning, just equates gas and ad $. Therefore, you cannot make the above conclusion. It is better not to perform SA.

15 | P a g e • WINQSB Screen Shot 7: RHS Sensitivity Analysis Graph for Demand Gas

• WINQSB Screen Shot 8: RHS Sensitivity Analysis Table for Demand Gas

(As mentioned above, avoid this conclusion. There will be a new set of solution.) From the Sensitivity Analysis, if our demand increases up to 5000 barrels, so will our profit. If our demand for gas is equal to 5000 we will have a profit maximization of $323,000 (as shown in WINQSB Screen Shot 2). However, once our demand goes over 5000 our profit will reduce

16 | P a g e because we cannot meet demand. When gas demand equals 8333 barrels our profit will reduce to $208,333 because more money has to be spent in advertising to create that demand. Any demand above 8333 barrels is infeasible.

Simplex Tableau

WinQSB Output:

• WINQSB Screen Shot 7: Simplex Tableau Iteration #s1-8

These shots are from WinQSB iterations of Simplex method and were not required. WinQSB’s presentation of simplex tableau is different from your textbook. You need to know your book’s style for quiz and exam.

17 | P a g e 18 | P a g e Simplex Tableau Hand Calculation explanation for “Oil 1 for Gas” column:

This table needs modification to follow your book’s style. For example, all coefficients in Row 0 should be negative. Though, the next step will not change. The term “iteration” and “step” should be used with care. There are some mistakes below on proper term to be used. Please update this after discussion of simplex method in class.

•Table 7:Original Simplex Tableau Table

19 | P a g e Note: WinQSB performs simplex method different than what is taught in the supplementary notes. Although it does use the Gauss-Jordan method of elementary row operation, its procedure is slightly different. You can search “Simplex Method” on the WinQSB software to obtain a 6 step procedure as to how WinQSB performs its iterations.

Iteration #1: If we look at the WinQSB output for iteration 1, we can see how the software used the Gauss-Jordan method of elementary row operation. The goal is to get every number in the “Oil 1 for Gas” column zero and the pivot term 1. Luckily, the pivot term (highlighted in blue) is already 1. Now we concentrate on making the other numbers zero. WinQSB achieved this in two iterations. In the first iteration, row 3 and row 4 were both multiplied by negative one in the original tableau. This is the result of those operations.

•Table 8: Simplex Tableau for Iteration #1

Iteration #2 this is a step: If you look at iteration #1, the only two numbers in the “Oil 1 for Gas” column left to make zero are in row 1 and row 3. In the second iteration, algebra was performed on row 1 and row 3 from the iteration #1 tableau. The following is the result of those operations. Now, we are completely done with this column.

20 | P a g e •Table 9: Simplex Tableau for Iteration #2

We repeat the method of choosing an entering variable, ratio testing, finding the pivot row, and finding the pivot term until every column has a number 1 value for its pivot terms and the remaining column numbers numbers are zero (not including the max row).

Acknowledgements

Bibliography:

Chang, Yih-Long, and Kiran Desai. WinQSB Version 2.0. New York: Wiley, 2003. Print

Winston, Wayne L. Operations Research Application and Algorithms. 4th Edition. New York;

Duxbury, 2003. Print

Software Used:

 Microsoft Word 2007 (Windows)

 Microsoft Excel 2007 (Windows)

 Win QSB, Version 2.0 (Windows)

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