department of business administration fall 200 7 -0 8 management science by asst. prof. sami fethi...

Department of Business Administration

FALL 2007-08Management Science

byby

Asst. Prof. Sami FethiAsst. Prof. Sami Fethi

© 2007 Pearson Education

2© 2007/08, Sami Fethi, EMU, All Right Reserved. Operations Research

Ch 3: Computer Solution and Sensitivity analysis

Chapter TopicsChapter Topics

Standard formStandard form

Sensitivity AnalysisSensitivity Analysis

Dual ProblemDual Problem

Example ProblemsExample Problems



Linear Programming Problem: Standard FormLinear Programming Problem: Standard Form

Standard form requires all variables in the constraint Standard form requires all variables in the constraint equations to appear on the left of the inequality (or equations to appear on the left of the inequality (or equality) and all numeric values to be on the right-hand equality) and all numeric values to be on the right-hand side.side.

Examples:Examples: (Equation) (Equation)

xx3 3 x x11 + x + x22 must be converted to x must be converted to x33 - x - x1 1 - x- x22 0 0

xx11/(x/(x22 + x + x33) ) 2 becomes x 2 becomes x1 1 2 (x 2 (x22 + x + x33) )

and then xand then x11 - 2x - 2x22 - 2x - 2x33 0 0



Linear Programming Problem: Standard FormLinear Programming Problem: Standard Form

Models are also transformed into Models are also transformed into Standard form.Standard form.

Examples:Examples: (model) (model)

Having defined the profit or cost function as well as constraints functions within Having defined the profit or cost function as well as constraints functions within

the systemthe system eqn, standard form can be formulated as followseqn, standard form can be formulated as follows::

Z= 12Z= 12xx11 + 16 + 16 xx2 2 Subject to: 3xSubject to: 3x1 1 + 2+ 2xx2 2 500500

4x4x1 1 + 5+ 5xx2 2

800800

xx11

xx22 0 0

Standard formStandard form::



Beaver Creek Pottery ExampleSensitivity Analysis (1 of 4)

Sensitivity analysis determines the effect on the optimal Sensitivity analysis determines the effect on the optimal solution of changes in parameter values of the objective solution of changes in parameter values of the objective function and constraint equations.function and constraint equations.

Changes may be reactions to anticipated uncertainties in Changes may be reactions to anticipated uncertainties in the parameters or to new or changed information the parameters or to new or changed information concerning the model.concerning the model.



Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Figure 3.1 Optimal Solution Point

Beaver Creek Pottery ExampleSensitivity Analysis (2 of 4)




subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Figure 3.2 Changing the x1 Objective Function Coefficient

Beaver Creek Pottery ExampleChange x1 Objective Function Coefficient (3 of 4)




subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Figure 3.3 Changing the x2 Objective Function Coefficient

Beaver Creek Pottery ExampleChange x2 Objective Function Coefficient (4 of 4)



The sensitivity range for an objective function coefficient is The sensitivity range for an objective function coefficient is the range of values over which the current optimal solution the range of values over which the current optimal solution point will remain optimal.point will remain optimal.

The sensitivity range for the xThe sensitivity range for the xii coefficient is designated coefficient is designated

as cas ci.i.

Objective Function CoefficientSensitivity Range (1 of 3)



objective function Z = $40x1 + $50x2 sensitivity range for:

x1: 25 c1 66.67 x2: 30 c2 80

Figure 3.4 Determining the Sensitivity Range for c1

Objective Function CoefficientSensitivity Range for c1 and c2 (2 of 3)



Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 164x1 + 3x2 24

x1, x2 0

sensitivity ranges: 4 c1 0 c2 4.5

Objective Function CoefficientFertilizer Cost Minimization Example (3 of 3)

Figure 3.5 Fertilizer Cost Minimization Example



Changes in Constraint Quantity ValuesSensitivity Range (1 of 4)

The sensitivity range for a right-hand-side value is the The sensitivity range for a right-hand-side value is the range of values over which the quantity’s value can range of values over which the quantity’s value can change without changing the solution variable mix, change without changing the solution variable mix, including the slack variables.including the slack variables.


Ch 3: Computer Solution and Sensitivity analysis Changes in Constraint Quantity ValuesIncreasing the Labor Constraint (2 of 4)


subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Figure 3.6 Increasing the Labor Constraint Quantity


Ch 3: Computer Solution and Sensitivity analysis Changes in Constraint Quantity ValuesSensitivity Range for Labor Constraint (3 of 4)

Sensitivity range for:30 q1 80 hr

Figure 3.7 Determining the Sensitivity Range for Labor Quantity


Ch 3: Computer Solution and Sensitivity analysis Changes in Constraint Quantity ValuesSensitivity Range for Clay Constraint (4 of 4)

Sensitivity range for: 60 q2 160 lb

Figure 3.8 Determining the Sensitivity Range for Clay Quantity



Changing individual constraint parametersChanging individual constraint parameters

Adding new constraintsAdding new constraints

Adding new variablesAdding new variables

Other Forms of Sensitivity AnalysisTopics (1 of 4)


Ch 3: Computer Solution and Sensitivity analysis Other Forms of Sensitivity AnalysisChanging a Constraint Parameter (2 of 4)


subject to: 1x1 + 2x2 40 4x2 + 3x2 120

x1, x2 0

Figure 3.9 Changing the x1 Coefficient in the Labor Constraint



Adding a new constraint to Beaver Creek Model: 0.20x1+ 0.10x2 5 hours for packaging

Original solution: 24 bowls, 8 mugs, $1,360 profit

Exhibit 3.17

Other Forms of Sensitivity AnalysisAdding a New Constraint (3 of 4)

To find out Optimal solution coordinates, we use the both constraints.x1 = 40 -2x2

4(40 -2x2) + 3x2 =120160-8x2 + 3x2 =120-5x2 = -40X2 = 8x1 = 24

Z = $40 (24) + $50 (8)Z = $ 1360 max daily profit possible



Adding a new variable to the Beaver Creek model, x3, a third product, cups

Maximize Z = $40x1 + 50x2 + 30x3

subject to:

x1 + 2x2 + 1.2x3 40 hr of labor

4x1 + 3x2 + 2x3 120 lb of clay

x1, x2, x3 0

Solving model shows that change has no effect on the original solution (i.e., the model is not sensitive to this change).

Other Forms of Sensitivity AnalysisAdding a New Variable (4 of 4)



Defined as the marginal value of one additional unit of Defined as the marginal value of one additional unit of resource.resource.

The sensitivity range for a constraint quantity value is The sensitivity range for a constraint quantity value is also the range over which the shadow price is valid.also the range over which the shadow price is valid.

Shadow Prices (Dual Variable Values)Shadow Prices (Dual Variable Values)



The Dual Problem and The Dual Problem and Shadow pricesShadow prices

Every linear programming problem, called the Every linear programming problem, called the primal problem, has a corresponding or primal problem, has a corresponding or symmetrical problem called the dual problem. symmetrical problem called the dual problem.

A profit-maximization primal problem has a A profit-maximization primal problem has a cost-minimization dual problem, and vice cost-minimization dual problem, and vice versa. The solution of a dual problem yields the versa. The solution of a dual problem yields the shadow prices. shadow prices.

They give the change in the value of the They give the change in the value of the objective function per unit change in each objective function per unit change in each constraint in the primal problem. constraint in the primal problem.

According to the duality theorem, the optimal According to the duality theorem, the optimal value of the objective function is the same in value of the objective function is the same in the primal and in the corresponding dual the primal and in the corresponding dual problems.problems.



Dual of the Profit Maximization Dual of the Profit Maximization ProblemProblem

Maximize

Subject to

= $30QX + $40QY

1QX + 1QY 7

0.5QX + 1QY 5

0.5QY 2

QX, QY 0

(objective function)

(input A constraint)

(input B constraint)

(input C constraint)

(nonnegativity constraint)

Minimize

Subject to

C = 7VA + 5VB + 2VC

1VA + 0.5VB $30

1VA + 1VB + 0.5VC $40

VA, VB, VC 0



Dual of the Profit Maximization Dual of the Profit Maximization ProblemProblem

In the dual problem we seek to minimize the shadow prices of In the dual problem we seek to minimize the shadow prices of inputs A, B, and C used by the firm. Defining Vinputs A, B, and C used by the firm. Defining VAA, V, VBB, as the , as the shadow prices of inputs and C as the total imputed values of the shadow prices of inputs and C as the total imputed values of the fixed quantities of inputs available to the firm, we can write the fixed quantities of inputs available to the firm, we can write the dual objective function as minimizedual objective function as minimize

C=7 VC=7 VAA+5 V+5 VBB+2 V+2 VCC

Thus the constraints of the dual problem can be written Thus the constraints of the dual problem can be written as follow:as follow:

1V1VAA + 0.5V + 0.5VBB $30 $30

1V1VAA + 1V + 1VBB + 0.5V + 0.5VCC $40 so V $40 so VCC=0 due to slack variable. =0 due to slack variable.

1V1VAA + 0.5V + 0.5VBB = $30 = $30

1V1VAA + 1V + 1VBB = $40 therefore 0.5V = $40 therefore 0.5VBB=$10, V=$10, VBB=$20 and V=$20 and VAA=$20=$20

C=7 ($20)+5 ($20) +2 (0)=$240 this is the minimum C=7 ($20)+5 ($20) +2 (0)=$240 this is the minimum cost.cost.



Dual of the Cost Minimization Dual of the Cost Minimization ProblemProblem

Maximize

Subject to

= 14VP + 10VM + 6VV

1VP + 1VM + 1VV $2

2VP + 1VM + 0.5VV $3

VP, VM, VV 0

Minimize

Subject to

C = $2QX + $3QY

1QX + 2QY 14

1QX + 1QY 10

1QX + 0.5QY 6

QX, QY 0

(objective function)

(protein constraint)

(minerals constraint)

(vitamins constraint)

(nonnegativity constraint)



Dual of the Cost Minimization Dual of the Cost Minimization ProblemProblem

Since we know that from the solution of the Since we know that from the solution of the primal problem that vitamin constraint is a primal problem that vitamin constraint is a slack variable, so that Vslack variable, so that VVV=0, subtracting the =0, subtracting the first from the second constraint, we can get first from the second constraint, we can get the solution of the dual problem as follow:the solution of the dual problem as follow:2V2VPP + 1V + 1VMM =3 =3

1V1VPP + 1V + 1VMM = 2 , therefore V = 2 , therefore VPP =$1 and V =$1 and VMM =$1 =$1

The profit as follows:The profit as follows:

= 14($1)+ 10($1)= 14($1)+ 10($1) + 6($0)+ 6($0)

= $24. = $24.



Two airplane parts: no.1 and no. 2.Two airplane parts: no.1 and no. 2. Three manufacturing stages: stamping, drilling, milling.Three manufacturing stages: stamping, drilling, milling. Decision variables: xDecision variables: x11 (number of part no.1 to produce) (number of part no.1 to produce) xx22 (number of part no.2 to produce) (number of part no.2 to produce)

Model: Maximize Z = $650xModel: Maximize Z = $650x11 + 910x + 910x22

subject to:subject to: 4x4x11 + 7.5x + 7.5x22 105 (stamping,hr) 105 (stamping,hr) 6.2x6.2x11 + 4.9x + 4.9x22 90 (drilling, hr) 90 (drilling, hr) 9.1x9.1x11 + 4.1x + 4.1x22 110 (finishing, hr) 110 (finishing, hr) xx11, x, x22 0 0

Example Problem 1Problem Statement (1 of 2)



Maximize Z = $650x1 + $910x2

subject to: 4x1 + 7.5x2 105 6.2x1 + 4.9x2 90 9.1x1 + 4.1x2 110 x1, x2 0

s1 = 0, s2 = 0, s3 = 11.35 hr

485.33 c1 1,151.43137.76 q1 89.10

Figure 3.10 Graphical Solution

Example Problem 1Graphical Solution (2 of 2)



Example Problem 2Problem Statement

Southern Sporting Goods Southern Sporting Goods Company makes Company makes basketballs and footballs. basketballs and footballs. Each product is produced Each product is produced from two resources—from two resources—rubber and leather. The rubber and leather. The resource requirements for resource requirements for each product and the total each product and the total resources available are as resources available are as follows:follows:

Each basketball produced Each basketball produced results in a profit of $12, results in a profit of $12, and each football earns and each football earns $16 in profit. $16 in profit.

ProductResource Requirements per

Unit

Rubber(lb) Leather (ft.2)

Basketball 3 4

Football 2 5

Total resources available

500 lb. 800 ft.2




a. Formulate a linear programming model to determine the number of a. Formulate a linear programming model to determine the number of basketballs and footballs to produce in order to maximize profit. basketballs and footballs to produce in order to maximize profit.

b. Transform this model into standard form. b. Transform this model into standard form. c. Solve the model formulated in the Problem for Southern Sporting c. Solve the model formulated in the Problem for Southern Sporting

Goods Company graphically.Goods Company graphically. d. Identify the amount of unused resources (Le., slack) at each of the d. Identify the amount of unused resources (Le., slack) at each of the

graphical extreme points. graphical extreme points. e. What would be the effect on the optimal solution if the proflt for a e. What would be the effect on the optimal solution if the proflt for a

basketbali changed from $12 to $13? What would be the effect if the basketbali changed from $12 to $13? What would be the effect if the profit for a footbali changed from $16 to$15? profit for a footbali changed from $16 to$15?

f. What would be the effect on the optirnal solution if 500 additional f. What would be the effect on the optirnal solution if 500 additional pounds of rubber could be obtained? What would be the effect if 500 pounds of rubber could be obtained? What would be the effect if 500 additional square feet of leather could be obtained?additional square feet of leather could be obtained?

For the linear programming model for Southern Sporting Goods For the linear programming model for Southern Sporting Goods Company, formulated in section a and solved graphically in section b: Company, formulated in section a and solved graphically in section b:

g. Determine the sensitivity ranges for the objective function g. Determine the sensitivity ranges for the objective function coeffıcients and constraint quantity values, using graphical analysis. coeffıcients and constraint quantity values, using graphical analysis.

h. Determine the shadow prices for the resources and cxplain their h. Determine the shadow prices for the resources and cxplain their meaning.meaning.



Example Problem 2Problem Solution




(h)Maximize Z = $12x1 + $16x2

subject to: 3x1 + 2x2 500 (rubber)

4x1 + 5x2 800 (leather)

x1, x2 0A profit-max primal problem has a cost-min dual problem and vice-versa.The soln. Of a dual problem yields the shadow prices. They give the change in the value of the obj. Function per unit change in each constraint in the primal problem.

Min C= $500v1 + $800v2

subject to: 3v1 + 4v2 12 2v1 + 5v2 16 v1, v2 0

3v1 + 4v2 12

v2 = 3-(3/4) v1

2v1 + 5(3-(3/4) v1=16

v1=-4/7, v1 0 so v1=0v2 = 16/5=3.20




An Aluminum Company produces An Aluminum Company produces three grades (high, medium, and three grades (high, medium, and low) of aluminum at two milis. low) of aluminum at two milis. Each miii has a different Each miii has a different production capacity (in tons per production capacity (in tons per day) for each grade, as follows:day) for each grade, as follows:

The company has contracted with The company has contracted with a manufacturing flrm to supply at a manufacturing flrm to supply at least 12 tons of high-grade least 12 tons of high-grade aluminum, 8 tons of meclium-aluminum, 8 tons of meclium-grade aluminum, and 5 tons of grade aluminum, and 5 tons of low-grade aluminum. It costs low-grade aluminum. It costs United $6,000 per day to operate United $6,000 per day to operate miii 1 and $7,000 per day to miii 1 and $7,000 per day to operate miii 2. The company wants operate miii 2. The company wants to know the number of days to to know the number of days to operate each miii in order to meet operate each miii in order to meet the contract at the minimum cost. the contract at the minimum cost.

Mill

Aluminum Grade 1 2

High

Medium

Low

6

2

4

2

2

10




a. Formulate a linear programming model for this problem. a. Formulate a linear programming model for this problem. b. Solve the linear programming model formulated in section b. Solve the linear programming model formulated in section

a for United Aluminum Company graphically. a for United Aluminum Company graphically. c. How much extra (i.e., surplus) high-, medium-, and low-c. How much extra (i.e., surplus) high-, medium-, and low-

grade aluminum does the company produce at the optimal grade aluminum does the company produce at the optimal solution? solution?

d. What would be the effect on the optimal solution if the d. What would be the effect on the optimal solution if the cost of operating mill 1 increased frorn $6,000 to $7,500 per cost of operating mill 1 increased frorn $6,000 to $7,500 per day? day?

e. What would be the effect on the optimal solution if the e. What would be the effect on the optimal solution if the company could supply only 10 tons of high-grade aluminum? company could supply only 10 tons of high-grade aluminum?

f. Identify and explain the shadow prices for each of f. Identify and explain the shadow prices for each of aluminum grade contract requirementsaluminum grade contract requirements

g. Determine the sensitivity ranges for the objcctive function g. Determine the sensitivity ranges for the objcctive function coefficients and for the constraint quantity values.coefficients and for the constraint quantity values.




(f)Min C= $6000x1 + $7000x2

subject to: 6x1 + 2x2 12 (high)

2x1 + 2x2 8 (medium)

4x1 + 10x2 5 (low) x1, x2 0 A cost-min primal problem has a profit-max dual problem and vice-versa.The soln. Of a dual problem yields the shadow prices. They give the change in the value of the obj. Function per unit change in each constraint in the primal problem.

Min Z= $12v1 + $8v2+ $5v3

subject to: 6v1 + 2v2 + 4v3 6000 2v1 + 2v2 + 10v3 7000 v1, v2 ,, v3 0Since v3 is a slack variable, we set v3=06v1 + 2v2 =6000

v2 = 3000-3 v1

2v1 +6000-(3- 6 v1=16

v1=- 250, v1 0 so v1=0v2 = 3000



End of Chapter

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