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1© 2003 Thomson© 2003 Thomson/South-Western/South-Western Slide

Slides Prepared bySlides Prepared by

JOHN S. LOUCKSJOHN S. LOUCKSSt. Edward’s UniversitySt. Edward’s University


Chapter 3 Chapter 3 Linear Programming: Sensitivity Analysis Linear Programming: Sensitivity Analysis

and Interpretation of Solutionand Interpretation of Solution Introduction to Sensitivity AnalysisIntroduction to Sensitivity Analysis Graphical Sensitivity AnalysisGraphical Sensitivity Analysis Sensitivity Analysis: Computer SolutionSensitivity Analysis: Computer Solution Simultaneous ChangesSimultaneous Changes


Standard Computer OutputStandard Computer Output

Software packages such as Software packages such as The Management ScientistThe Management Scientist andand

Microsoft ExcelMicrosoft Excel provide the following LP information: provide the following LP information: Information about the objective function:Information about the objective function:

• its optimal valueits optimal value•coefficient ranges (ranges of optimality)coefficient ranges (ranges of optimality)

Information about the decision variables:Information about the decision variables:•their optimal valuestheir optimal values•their reduced coststheir reduced costs

Information about the constraints:Information about the constraints:•the amount of slack or surplusthe amount of slack or surplus•the dual pricesthe dual prices•right-hand side ranges (ranges of feasibility)right-hand side ranges (ranges of feasibility)


Standard Computer OutputStandard Computer Output In Chapter 2 we discussed:In Chapter 2 we discussed:

•objective function valueobjective function value•values of the decision variablesvalues of the decision variables•reduced costsreduced costs•slack/surplusslack/surplus

In this chapter we will discuss:In this chapter we will discuss:•changes in the coefficients of the objective changes in the coefficients of the objective

functionfunction•changes in the right-hand side value of a changes in the right-hand side value of a

constraintconstraint


Sensitivity AnalysisSensitivity Analysis Sensitivity analysisSensitivity analysis (or post-optimality analysis) (or post-optimality analysis)

is used to determine how the optimal solution is is used to determine how the optimal solution is affected by changes, within specified ranges, in:affected by changes, within specified ranges, in:•the objective function coefficientsthe objective function coefficients•the right-hand side (RHS) valuesthe right-hand side (RHS) values

Sensitivity analysis is important to the manager Sensitivity analysis is important to the manager who must operate in a dynamic environment who must operate in a dynamic environment with imprecise estimates of the coefficients. with imprecise estimates of the coefficients.

Sensitivity analysis allows him to ask certain Sensitivity analysis allows him to ask certain what-ifwhat-if questionsquestions about the problem. about the problem.


Example 1Example 1 LP FormulationLP Formulation

Max 5Max 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6 22xx11 + 3 + 3xx22 << 19 19 xx11 + + xx22 << 8 8

xx11, , xx22 >> 0 0


Example 1Example 1 Graphical SolutionGraphical Solution

88

77

66

55

44

33

22

11

1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

xx11 + + xx22 << 8 8Max 5x1 + 7x2

xx11 << 6 6Optimal: Optimal: xx11 = 5, = 5, xx22 = 3, = 3, zz = 46 = 46


Objective Function CoefficientsObjective Function Coefficients Let us consider how changes in the objective Let us consider how changes in the objective

function coefficients might affect the optimal function coefficients might affect the optimal solution.solution.

The The range of optimalityrange of optimality for each coefficient for each coefficient provides the range of values over which the provides the range of values over which the current solution will remain optimal.current solution will remain optimal.

Managers should focus on those objective Managers should focus on those objective coefficients that have a narrow range of coefficients that have a narrow range of optimality and coefficients near the endpoints of optimality and coefficients near the endpoints of the range.the range.


Example 1Example 1 Changing Slope of Objective FunctionChanging Slope of Objective Function

88

77

66

55

44

33

22

11

1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 xx11

FeasibleFeasibleRegionRegion

11 22

3344

55

xx22


Range of OptimalityRange of Optimality Graphically, the limits of a range of optimality Graphically, the limits of a range of optimality

are found by changing the slope of the objective are found by changing the slope of the objective function line within the limits of the slopes of the function line within the limits of the slopes of the binding constraint lines.binding constraint lines.

The slope of an objective function line, Max The slope of an objective function line, Max cc11xx11 + + cc22xx22, is -, is -cc11//cc22, and the slope of a constraint, , and the slope of a constraint, aa11xx11 + + aa22xx22 = = bb, is -, is -aa11//aa22..


Example 1Example 1

Range of Optimality for Range of Optimality for cc11 The slope of the objective function line is -The slope of the objective function line is -cc11//cc22. The . The slope of the first binding constraint, slope of the first binding constraint, xx11 + + xx22 = 8, is -1 = 8, is -1 and the slope of the second binding constraint, and the slope of the second binding constraint, xx11 + + 33xx22 = 19, is -2/3. = 19, is -2/3.Find the range of values for Find the range of values for cc11 (with (with cc22 staying 7) staying 7) such that the objective function line slope lies such that the objective function line slope lies between that of the two binding constraints:between that of the two binding constraints:

-1 -1 << - -cc11/7 /7 << -2/3 -2/3 Multiplying through by -7 (and reversing the Multiplying through by -7 (and reversing the

inequalities):inequalities): 14/3 14/3 << cc11 << 7 7


Example 1Example 1

Range of Optimality for Range of Optimality for cc22

Find the range of values for Find the range of values for cc22 ( with ( with cc11 staying 5) such that the objective function line staying 5) such that the objective function line slope lies between that of the two binding slope lies between that of the two binding constraints:constraints:

-1 -1 << -5/ -5/cc22 << -2/3 -2/3

Multiplying by -1: 1 Multiplying by -1: 1 >> 5/ 5/cc22 >> 2/3 2/3Inverting, Inverting, 1 1 << cc22/5 /5 << 3/2 3/2

Multiplying by 5: Multiplying by 5: 5 5 << cc22 << 15/2 15/2


Example 1Example 1

Range of Optimality for Range of Optimality for cc11 and and cc22

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$8 X1 5.0 0.0 5 2 0.333333333$C$8 X2 3.0 0.0 7 0.5 2

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$13 #1 5 0 6 1E+30 1$B$14 #2 19 2 19 5 1$B$15 #3 8 1 8 0.333333333 1.666666667


Right-Hand SidesRight-Hand Sides Let us consider how a change in the right-hand Let us consider how a change in the right-hand

side for a constraint might affect the feasible side for a constraint might affect the feasible region and perhaps cause a change in the region and perhaps cause a change in the optimal solution.optimal solution.

The The improvementimprovement in the value of the optimal in the value of the optimal solution per unit solution per unit increaseincrease in the right-hand side in the right-hand side is called the is called the dual pricedual price..

The The range of feasibilityrange of feasibility is the range over which is the range over which the dual price is applicable.the dual price is applicable.

As the RHS increases, other constraints will As the RHS increases, other constraints will become binding and limit the change in the become binding and limit the change in the value of the objective function.value of the objective function.


Dual PriceDual Price Graphically, a dual price is determined by adding Graphically, a dual price is determined by adding

+1 to the right hand side value in question and +1 to the right hand side value in question and then resolving for the optimal solution in terms then resolving for the optimal solution in terms of the same two binding constraints. of the same two binding constraints.

The dual price is equal to the difference in the The dual price is equal to the difference in the values of the objective functions between the values of the objective functions between the new and original problems.new and original problems.

The dual price for a nonbinding constraint is 0.The dual price for a nonbinding constraint is 0. A A negative dual pricenegative dual price indicates that the objective indicates that the objective

function will function will notnot improve if the RHS is increased. improve if the RHS is increased.


Relevant Cost and Sunk CostRelevant Cost and Sunk Cost A resource cost is a A resource cost is a relevant costrelevant cost if the amount if the amount

paid for it is dependent upon the amount of the paid for it is dependent upon the amount of the resource used by the decision variables. resource used by the decision variables.

Relevant costs Relevant costs areare reflected in the objective reflected in the objective function coefficients. function coefficients.

A resource cost is a A resource cost is a sunk costsunk cost if it must be paid if it must be paid regardless of the amount of the resource regardless of the amount of the resource actually used by the decision variables. actually used by the decision variables.

Sunk resource costs areSunk resource costs are not not reflected in the reflected in the objective function coefficients.objective function coefficients.


A Cautionary NoteA Cautionary Noteon the Interpretation of Dual Priceson the Interpretation of Dual Prices

Resource cost is sunkResource cost is sunkThe dual price is the maximum amount you The dual price is the maximum amount you should be willing to pay for one additional unit of should be willing to pay for one additional unit of the resource.the resource.

Resource cost is relevantResource cost is relevantThe dual price is the maximum premium over The dual price is the maximum premium over the normal cost that you should be willing to pay the normal cost that you should be willing to pay for one unit of the resource.for one unit of the resource.


Example 1Example 1 Dual PricesDual Prices Constraint 1Constraint 1: Since : Since xx11 << 6 is not a binding 6 is not a binding

constraint, constraint, its dual price is 0.its dual price is 0. Constraint 2Constraint 2: Change the RHS value of the : Change the RHS value of the

second second constraint to 20 and resolve for constraint to 20 and resolve for the optimal point the optimal point determined by the last two determined by the last two constraints: constraints: 22xx11 + 3 + 3xx22 = 20 and = 20 and xx11 + + xx22 = 8. = 8.

The solution is The solution is xx11 = 4, = 4, xx22 = 4, = 4, zz = 48. = 48. Hence, the Hence, the dual price = dual price = zznewnew - - zzoldold = 48 - 46 = 2. = 48 - 46 = 2.


Example 1Example 1 Dual PricesDual Prices

Constraint 3Constraint 3: Change the RHS value of the third : Change the RHS value of the third constraint to 9 and resolve for the optimal point constraint to 9 and resolve for the optimal point determined by the last two constraints: 2determined by the last two constraints: 2xx11 + + 33xx22 = = 19 and 19 and xx11 + + xx22 = 9. = 9.

The solution is: The solution is: xx11 = 8, = 8, xx22 = 1, = 1, zz = 47. = 47. The dual price is The dual price is zznewnew - - zzoldold = 47 - 46 = 1. = 47 - 46 = 1.


Example 1Example 1 Dual PricesDual Prices

Adjustable CellsAdjustable CellsFinalFinal ReducedReduced ObjectiveObjective AllowableAllowable AllowableAllowable

CellCell NameName ValueValue CostCost CoefficientCoefficient IncreaseIncrease DecreaseDecrease$B$8$B$8 X1X1 5.05.0 0.00.0 55 22 0.333333330.33333333$C$8$C$8 X2X2 3.03.0 0.00.0 77 0.50.5 22

ConstraintsConstraintsFinalFinal ShadowShadow ConstraintConstraint AllowableAllowable AllowableAllowable

CellCell NameName ValueValue PricePrice R.H. SideR.H. Side IncreaseIncrease DecreaseDecrease$B$13$B$13 #1#1 55 00 66 1E+301E+30 11$B$14$B$14 #2#2 1919 22 1919 55 11$B$15$B$15 #3#3 88 11 88 0.333333330.33333333 1.666666671.66666667


Range of FeasibilityRange of Feasibility The The range of feasibilityrange of feasibility for a change in the right for a change in the right

hand side value is the range of values for this hand side value is the range of values for this coefficient in which the original dual price coefficient in which the original dual price remains constant.remains constant.

Graphically, the range of feasibility is determined Graphically, the range of feasibility is determined by finding the values of a right hand side by finding the values of a right hand side coefficient such that the same two lines that coefficient such that the same two lines that determined the original optimal solution determined the original optimal solution continue to determine the optimal solution for continue to determine the optimal solution for the problem.the problem.


Example 1Example 1 Range of FeasibilityRange of Feasibility

Adjustable CellsAdjustable CellsFinalFinal ReducedReduced ObjectiveObjective AllowableAllowable AllowableAllowable

CellCell NameName ValueValue CostCost CoefficientCoefficient IncreaseIncrease DecreaseDecrease$B$8$B$8 X1X1 5.05.0 0.00.0 55 22 0.333333330.33333333$C$8$C$8 X2X2 3.03.0 0.00.0 77 0.50.5 22

ConstraintsConstraintsFinalFinal ShadowShadow ConstraintConstraint AllowableAllowable AllowableAllowable

CellCell NameName ValueValue PricePrice R.H. SideR.H. Side IncreaseIncrease DecreaseDecrease$B$13$B$13 #1#1 55 00 66 1E+301E+30 11$B$14$B$14 #2#2 1919 22 1919 55 11$B$15$B$15 #3#3 88 11 88 0.333333330.33333333 1.666666671.66666667


Olympic Bike is introducing two new lightweight Olympic Bike is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be bicycle frames, the Deluxe and the Professional, to be made from special aluminum and steel alloys. The made from special aluminum and steel alloys. The anticipated unit profits are $10 for the Deluxe and anticipated unit profits are $10 for the Deluxe and $15 for the Professional. The number of pounds of $15 for the Professional. The number of pounds of each alloy needed per frame is summarized below. A each alloy needed per frame is summarized below. A supplier delivers 100 pounds of the aluminum alloy supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy weekly. and 80 pounds of the steel alloy weekly.

Aluminum AlloyAluminum Alloy Steel Steel AlloyAlloy

Deluxe Deluxe 2 3 2 3 Professional 4 Professional 4 2 2

How many Deluxe and Professional frames should How many Deluxe and Professional frames should Olympic produce each week?Olympic produce each week?

Example 2: Olympic Bike Co.Example 2: Olympic Bike Co.


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Model FormulationModel Formulation

•Verbal Statement of the Objective FunctionVerbal Statement of the Objective Function Maximize total weekly profit.Maximize total weekly profit.•Verbal Statement of the ConstraintsVerbal Statement of the Constraints

Total weekly usage of aluminum alloy Total weekly usage of aluminum alloy << 100 100 pounds.pounds. Total weekly usage of steel alloy Total weekly usage of steel alloy << 80 pounds. 80 pounds.•Definition of the Decision VariablesDefinition of the Decision Variables xx11 = number of Deluxe frames produced weekly. = number of Deluxe frames produced weekly.

xx22 = number of Professional frames produced = number of Professional frames produced weekly.weekly.


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Model Formulation (continued)Model Formulation (continued)

Max 10Max 10xx11 + 15 + 15xx22 (Total (Total Weekly Profit) Weekly Profit)

s.t. 2s.t. 2xx11 + 4 + 4xx2 2 << 100 (Aluminum 100 (Aluminum Available)Available)

33xx11 + 2 + 2xx2 2 << 80 80 (Steel (Steel Available)Available)

xx11, , xx2 2 >> 0 0


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Partial Spreadsheet Showing Problem DataPartial Spreadsheet Showing Problem Data

A B C D1 Amount2 Material Deluxe Profess. Available3 Aluminum 2 4 1004 Steel 3 2 80

Material Requirements


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Partial Spreadsheet Showing SolutionPartial Spreadsheet Showing Solution

A B C D67 Deluxe Professional8 Bikes Made 15 17.5009

10 412.5001112 Constraints Amount Used Amount Avail.13 Aluminum 100 <= 10014 Steel 80 <= 80

Decision Variables

Maximized Total Profit


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Optimal SolutionOptimal Solution

According to the output:According to the output: xx11 (Deluxe frames) (Deluxe frames) = 15= 15 xx22 (Professional frames) (Professional frames) = 17.5= 17.5 Objective function value Objective function value = =

$412.50$412.50


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Range of OptimalityRange of Optimality

QuestionQuestionSuppose the profit on deluxe frames is Suppose the profit on deluxe frames is

increased to $20. Is the above solution still increased to $20. Is the above solution still optimal? What is the value of the objective optimal? What is the value of the objective function when this unit profit is increased to function when this unit profit is increased to $20?$20?


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Sensitivity ReportSensitivity Report


Cell Name Value Cost Coefficient Increase Decrease$B$8 Deluxe 15 0 10 12.5 2.5$C$8 Profess. 17.500 0.000 15 5 8.333333333


Cell Name Value Price R.H. Side Increase Decrease$B$13 Aluminum 100 3.125 100 60 46.66666667$B$14 Steel 80 1.25 80 70 30



AnswerAnswerThe output states that the solution remains The output states that the solution remains

optimal as long as the objective function optimal as long as the objective function coefficient of coefficient of xx11 is between 7.5 and 22.5. Since is between 7.5 and 22.5. Since 20 is within this range, the optimal solution will 20 is within this range, the optimal solution will not change. The optimal profit will change: 20not change. The optimal profit will change: 20xx11 + 15+ 15xx22 = 20(15) + 15(17.5) = $562.50. = 20(15) + 15(17.5) = $562.50.



QuestionQuestionIf the unit profit on deluxe frames were $6 If the unit profit on deluxe frames were $6

instead of $10, would the optimal solution instead of $10, would the optimal solution change?change?


Range of OptimalityRange of OptimalityAnswerAnswer

The output states that the solution remains The output states that the solution remains optimal as long as the objective function optimal as long as the objective function coefficient of coefficient of xx11 is between 7.5 and 22.5. Since 6 is between 7.5 and 22.5. Since 6 is outside this range, the optimal solution would is outside this range, the optimal solution would change.change.

Example 2: Olympic Bike Co.Example 2: Olympic Bike Co.


Range of Optimality and 100% RuleRange of Optimality and 100% Rule The The 100% rule100% rule states that simultaneous changes states that simultaneous changes

in objective function coefficients will not change in objective function coefficients will not change the optimal solution as long as the sum of the the optimal solution as long as the sum of the percentages of the change divided by the percentages of the change divided by the corresponding maximum allowable change in the corresponding maximum allowable change in the range of optimality for each coefficient does not range of optimality for each coefficient does not exceed 100%.exceed 100%.


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Range of Optimality and 100% RuleRange of Optimality and 100% Rule

QuestionQuestionIf simultaneously the profit on Deluxe frames was If simultaneously the profit on Deluxe frames was raised to $16 and the profit on Professional frames raised to $16 and the profit on Professional frames was raised to $17, would the current solution be was raised to $17, would the current solution be optimal?optimal?AnswerAnswerIf If cc11 = 16, the amount = 16, the amount cc11 changed is 16 - 10 = 6 . The changed is 16 - 10 = 6 . The maximum allowable increase is 22.5 - 10 = 12.5, so maximum allowable increase is 22.5 - 10 = 12.5, so this is a 6/12.5 = 48% change. If this is a 6/12.5 = 48% change. If cc22 = 17, the = 17, the amount that amount that cc22 changed is 17 - 15 = 2. The maximum changed is 17 - 15 = 2. The maximum allowable increase is 20 - 15 = 5 so this is a 2/5 = allowable increase is 20 - 15 = 5 so this is a 2/5 = 40% change. The sum of the change percentages is 40% change. The sum of the change percentages is 88%. Since this does not exceed 100%, the optimal 88%. Since this does not exceed 100%, the optimal solution would not change.solution would not change.


Range of Feasibility and 100% RuleRange of Feasibility and 100% Rule The The 100% rule100% rule states that simultaneous changes states that simultaneous changes

in right-hand sides will not change the dual in right-hand sides will not change the dual prices as long as the sum of the percentages of prices as long as the sum of the percentages of the changes divided by the corresponding the changes divided by the corresponding maximum allowable change in the range of maximum allowable change in the range of feasibility for each right-hand side does not feasibility for each right-hand side does not exceed 100%.exceed 100%.


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Range of Feasibility and Sunk CostsRange of Feasibility and Sunk Costs

QuestionQuestionGiven that aluminum is a sunk cost, what Given that aluminum is a sunk cost, what

is the maximum amount the company should is the maximum amount the company should pay for 50 extra pounds of aluminum? pay for 50 extra pounds of aluminum?



AnswerAnswerSince the cost for aluminum is a sunk cost, Since the cost for aluminum is a sunk cost,

the shadow price provides the value of extra the shadow price provides the value of extra aluminum. The shadow price for aluminum is aluminum. The shadow price for aluminum is the same as its dual price (for a maximization the same as its dual price (for a maximization problem). The shadow price for aluminum is problem). The shadow price for aluminum is $3.125 per pound and the maximum allowable $3.125 per pound and the maximum allowable increase is 60 pounds. Since 50 is in this range, increase is 60 pounds. Since 50 is in this range, then the $3.125 is valid. Thus, the value of 50 then the $3.125 is valid. Thus, the value of 50 additional pounds is = 50($3.125) = $156.25.additional pounds is = 50($3.125) = $156.25.


Example 2: Olympic Bike Co.Example 2: Olympic Bike Co. Range of Feasibility and Relevant CostsRange of Feasibility and Relevant Costs

QuestionQuestionIf aluminum were a relevant cost, what is the If aluminum were a relevant cost, what is the maximum amount the company should pay for 50 maximum amount the company should pay for 50 extra pounds of aluminum? extra pounds of aluminum? AnswerAnswerIf aluminum were a relevant cost, the shadow If aluminum were a relevant cost, the shadow price would be the amount above the normal price price would be the amount above the normal price of aluminum the company would be willing to pay. of aluminum the company would be willing to pay. Thus if initially aluminum cost $4 per pound, then Thus if initially aluminum cost $4 per pound, then additional units in the range of feasibility would be additional units in the range of feasibility would be worth $4 + $3.125 = $7.125 per pound.worth $4 + $3.125 = $7.125 per pound.


Example 3Example 3 Consider the following linear program:Consider the following linear program:

Min 6Min 6xx11 + 9 + 9xx22 ($ cost) ($ cost)

s.t. s.t. xx11 + 2 + 2xx2 2 << 8 8 1010xx11 + 7.5 + 7.5xx2 2 >> 30 30 xx2 2 >> 2 2

xx11, , xx22 >> 0 0


Example 3Example 3 The Management ScientistThe Management Scientist Output Output

OBJECTIVE FUNCTION VALUE = 27.000OBJECTIVE FUNCTION VALUE = 27.000 VariableVariable ValueValue Reduced CostReduced Cost xx11 1.500 1.500 0.000 0.000 xx22 2.000 2.000

0.0000.000 Constraint Constraint Slack/SurplusSlack/Surplus Dual PriceDual Price

1 2.500 0.0001 2.500 0.000 2 0.000 -0.600 2 0.000 -0.600

3 0.000 -4.500 3 0.000 -4.500


Example 3Example 3 The Management ScientistThe Management Scientist Output (continued) Output (continued)

OBJECTIVE COEFFICIENT RANGESOBJECTIVE COEFFICIENT RANGES VariableVariable Lower Limit Lower Limit Current ValueCurrent Value Upper LimitUpper Limit xx11 0.000 6.000 0.000 6.000

12.00012.000 xx22 4.500 9.000 4.500 9.000 No LimitNo Limit

RIGHTHAND SIDE RANGESRIGHTHAND SIDE RANGES Constraint Constraint Lower LimitLower Limit Current ValueCurrent Value Upper LimitUpper Limit 1 5.500 8.000 1 5.500 8.000 No LimitNo Limit 2 15.000 30.000 2 15.000 30.000

55.000 55.000 3 0.000 2.000 3 0.000 2.000

4.0004.000


Example 3Example 3 Optimal SolutionOptimal Solution

According to the output: According to the output: xx11 = 1.5 = 1.5

xx22 = 2.0 = 2.0Objective function value = 27.00Objective function value = 27.00


Example 3Example 3 Range of OptimalityRange of Optimality

QuestionQuestionSuppose the unit cost of Suppose the unit cost of xx11 is decreased to is decreased to

$4. Is the current solution still optimal? What is $4. Is the current solution still optimal? What is the value of the objective function when this unit the value of the objective function when this unit cost is decreased to $4?cost is decreased to $4?



OBJECTIVE COEFFICIENT RANGESOBJECTIVE COEFFICIENT RANGES VariableVariable Lower Limit Lower Limit Current ValueCurrent Value Upper LimitUpper Limit xx11 0.000 6.000 0.000 6.000

12.00012.000 xx22 4.500 9.000 4.500 9.000 No LimitNo Limit

RIGHTHAND SIDE RANGESRIGHTHAND SIDE RANGES Constraint Constraint Lower LimitLower Limit Current ValueCurrent Value Upper LimitUpper Limit 1 5.500 8.000 1 5.500 8.000 No LimitNo Limit 2 15.000 30.000 2 15.000 30.000

55.000 55.000 3 0.000 2.000 3 0.000 2.000

4.0004.000




optimal as long as the objective function optimal as long as the objective function coefficient of coefficient of xx11 is between 0 and 12. Since 4 is is between 0 and 12. Since 4 is within this range, the optimal solution will not within this range, the optimal solution will not change. However, the optimal total cost will be change. However, the optimal total cost will be affected: 6affected: 6xx11 + 9 + 9xx22 = 4(1.5) + 9(2.0) = $24.00. = 4(1.5) + 9(2.0) = $24.00.



QuestionQuestionHow much can the unit cost of How much can the unit cost of xx22 be be

decreased without concern for the optimal decreased without concern for the optimal solution changing?solution changing?



OBJECTIVE COEFFICIENT RANGESOBJECTIVE COEFFICIENT RANGES VariableVariable Lower Limit Lower Limit Current ValueCurrent Value Upper LimitUpper Limit xx11 0.000 6.000 0.000 6.000 12.000 12.000 xx22 4.500 9.000 No Limit 4.500 9.000 No Limit

RIGHTHAND SIDE RANGESRIGHTHAND SIDE RANGES Constraint Constraint Lower LimitLower Limit Current ValueCurrent Value Upper LimitUpper Limit 1 5.500 8.000 No Limit1 5.500 8.000 No Limit 2 15.000 30.000 2 15.000 30.000 55.000 55.000 3 0.000 2.000 3 0.000 2.000

4.0004.000




optimal as long as the objective function optimal as long as the objective function coefficient of coefficient of xx22 does not fall below 4.5. does not fall below 4.5.


Example 3Example 3 Range of Optimality and 100% RuleRange of Optimality and 100% Rule

QuestionQuestionIf simultaneously the cost of If simultaneously the cost of xx11 was raised was raised

to $7.5 and the cost of to $7.5 and the cost of xx22 was reduced to $6, was reduced to $6, would the current solution remain optimal?would the current solution remain optimal?


Example 3Example 3 Range of Optimality and 100% RuleRange of Optimality and 100% Rule

AnswerAnswerIf If cc11 = 7.5, the amount = 7.5, the amount cc11 changed is 7.5 - changed is 7.5 -

6 = 1.5. The maximum allowable increase is 12 6 = 1.5. The maximum allowable increase is 12 - 6 = 6, so this is a 1.5/6 = 25% change. If - 6 = 6, so this is a 1.5/6 = 25% change. If cc22 = = 6, the amount that 6, the amount that cc22 changed is 9 - 6 = 3. The changed is 9 - 6 = 3. The maximum allowable decrease is 9 - 4.5 = 4.5, so maximum allowable decrease is 9 - 4.5 = 4.5, so this is a 3/4.5 = 66.7% change. The sum of the this is a 3/4.5 = 66.7% change. The sum of the change percentages is 25% + 66.7% = 91.7%. change percentages is 25% + 66.7% = 91.7%. Since this does not exceed 100% the optimal Since this does not exceed 100% the optimal solution would not change.solution would not change.



QuestionQuestionIf the right-hand side of constraint 3 is If the right-hand side of constraint 3 is

increased by 1, what will be the effect on the increased by 1, what will be the effect on the optimal solution?optimal solution?



OBJECTIVE COEFFICIENT RANGESOBJECTIVE COEFFICIENT RANGES VariableVariable Lower Limit Lower Limit Current ValueCurrent Value Upper LimitUpper Limit xx11 0.000 6.000 0.000 6.000 12.000 12.000 xx22 4.500 9.000 No Limit 4.500 9.000 No Limit

RIGHTHAND SIDE RANGESRIGHTHAND SIDE RANGES Constraint Constraint Lower LimitLower Limit Current ValueCurrent Value Upper LimitUpper Limit 1 5.500 8.000 No Limit1 5.500 8.000 No Limit 2 15.000 30.000 2 15.000 30.000 55.000 55.000 3 0.000 2.000 3 0.000 2.000

4.0004.000



AnswerAnswerA dual price represents the improvement A dual price represents the improvement

in the objective function value per unit increase in the objective function value per unit increase in the right-hand side. A negative dual price in the right-hand side. A negative dual price indicates a deterioration (negative improvement) indicates a deterioration (negative improvement) in the objective, which in this problem means an in the objective, which in this problem means an increase in total cost because we're minimizing. increase in total cost because we're minimizing. Since the right-hand side remains within the Since the right-hand side remains within the range of feasibility, there is no change in the range of feasibility, there is no change in the optimal solution. However, the objective optimal solution. However, the objective function value increases by $4.50.function value increases by $4.50.


HomeworkHomework

Homework ProblemsHomework Problems

3-73-73-83-83-103-103-133-133-143-14


End of Chapter 3End of Chapter 3

ch03

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