Page 92 Problem 5
Table of Contents…………………………………………………………………………………………………………….……………………2
Problem Statement3
Summary of Problem3
Table 1: Chandler Oil Company- Oil Information3
Table 2: Chandler Oil Company- Products Made from Oil3
Formulation of Problem3-6
1.) Define the decision variables4
Table 3: Summary of Problem4
2.) Provide explanatory information and assumptions4
3.) Formulate Objective Function (O.F) 5
4.) Formulate Constraints 5-6
5.) List All Equations 6-7
Table 4: Constraint Equation List6
Table 5: LP Form of Constraint Equation List6
Solution using WinQSB7-8
WinQSB Screen Shot 1: Original Input7
Table 6: WinQSB Variable and Name7
Table 7: WinQSB Constraint and Name7
WinQSB Screen Shot 2: Output8
Report to Manager9-10
Sensitivity Analysis10-15
WinQSB Screen Shot 3: Objective Function Graph for Coefficient of Oil 1 for Gas SA11
WinQSB Screen Shot 4: Table for Objective Function SA of Oil 1 for Gas11
WinQSB Screen Shot 5: RHS Sensitivity Analysis Graph for Oil 1 Availability13
WinQSB Screen Shot 6: RHS Sensitivity Analysis Table for Oil 1 Availability13
WinQSB Screen Shot 7: RHS Sensitivity Analysis Table for Demand Gas15
WinQSB Screen Shot 8: RHS Sensitivity Analysis Table for Demand Gas15
Simplex Tableau16-19
WinQSB Screen Shot 8: Simplex Tableau for Iteration #s1-816-17
Table 8: Original Simplex Tableau Table17
Table 9: Simplex Tableau for Iteration #1 18
Table 10: Simplex Tableau for Iteration #2 18
Acknowledgments19
PowerPoint Slide Printouts20
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 InformationOil / # 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 OilProduct (Blend) / Avg. Quality Level / Demand Created per $1 spent on Advertising / Selling Price per 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
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
Avg. Quality Level / Demand of Barrels Created per $1 spent on Advertising / Selling Price per BarrelProduct (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.
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.
Ø Daily Revenues from Blend Sales (Sales of Gas and Heating Oil)
= 25(x11 + x21) + 20 (x12 + x22)
Ø Daily Advertising Cost
= a1 + a2
Ø Daily Profit = Daily Revenues from Blend Sales - Daily Advertising Cost
Daily Profit = [25(x11 + x21) + 20 (x12 + x22)] – [a1 + a2]
Ø 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.
10x11+5x12x11+x21≥8 Simplify 2x11 – 3x21 0
Constraint 4: Heating oil must have an average quality level of at least 6.
10x12+5x22x12+x22≥6 Simplify 4x12 – x22 0
Constraint 5[1]: 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 / TypeMax 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 FormEquation / TypeMax 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
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
WinQSB Output:
• WINQSB Screen Shot 2: Output
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.
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 JobJ) . 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
Ø 1 Sensitivity Analysis for a Coefficient in the Objective Function that is a Basic Variable
Choice:
Variable / Name Givenx11 / 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)
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)