1. Chapter 3: Sensitivity Analysis and the Dual Problem & Shadow Prices
Department of Business Administration
SPRING
Operations Research
Chapter 3: Sensitivity Analysis and the Dual Problem & Shadow Prices

2. Chapter Topics Standard form Sensitivity Analysis Dual Problem
Example Problems

3. Linear Programming Problem: Standard Form
Standard form requires all variables in the constraint equations to appear on the left of the inequality (or equality) and all numeric values to be on the right-hand side.
Examples: (Equation)
x3  x1 + x2 must be converted to x3 - x1 - x2  0
x1/(x2 + x3)  2 becomes x1  2 (x2 + x3)
and then x1 - 2x2 - 2x3  0

4. Linear Programming Problem: Standard Form
Models are also transformed into Standard form.
Examples: (model)
Having defined the profit or cost function as well as constraints functions within the system eqn, standard form can be formulated as follows:
Z= 12x x2 Subject to: 3x1 + 2x2  500
4x1 + 5x2  800
x1 x2  0
Standard form:

5. Beaver Creek Pottery Example Sensitivity Analysis (1 of 4)
Sensitivity analysis determines the effect on the optimal solution of changes in parameter values of the objective function and constraint equations.
Changes may be reactions to anticipated uncertainties in the parameters or to new or changed information concerning the model.

6. Figure 3.1 Optimal Solution Point
Beaver Creek Pottery Example
Sensitivity Analysis (2 of 4)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 3.1 Optimal Solution Point

7. Figure 3.2 Changing the x1 Objective Function Coefficient
Beaver Creek Pottery Example
Change x1 Objective Function Coefficient (3 of 4)
Maximize Z = $100x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 3.2 Changing the x1 Objective Function Coefficient

8. Figure 3.3 Changing the x2 Objective Function Coefficient
Beaver Creek Pottery Example
Change x2 Objective Function Coefficient (4 of 4)
Maximize Z = $40x1 + $100x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure Changing the x2 Objective Function Coefficient

9. Objective Function Coefficient
Sensitivity Range (1 of 3)
The sensitivity range for an objective function coefficient is the range of values over which the current optimal solution point will remain optimal.
The sensitivity range for the xi coefficient is designated as ci.

10. Objective Function Coefficient Sensitivity Range (1 of 3)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
50x2 =Z -40x1
x2 =Z/50 -4/5x1
Slope is – 4/5
x2 = 120/3-4/3x1
Slope is – 4/3, now assume that
they are parallel each other so both can be equated in the following form;
(Z=C) -C/50=-4/3 C=66.67
50x2 =Z -40x1
x2 =Z/50 -4/5x1
Slope is – 4/5
x2 = 40/2-1/2x1
Slope is – 1/2
They are parallel each other so both can be equated in the following form;
(Z=C) -C/50=-1/2 C=25
This means that profit for bowl can vary between $ 25 and $ and the optimal point will not change.

11. Objective Function Coefficient Sensitivity Range (2 of 3)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
40x1 =Z -50x2
x1 =Z/40 -5/4x1
Slope is – 5/4
x1 = 120/4-3/4x2
Slope is – 3/4, now assume that
they are parallel each other so both can be equated in the following form;
(Z=C) -C/40=-3/4 C=30
40x1 =Z -50x2
x1 =Z/40 -5/4x2
Slope is – 5/4
x1 = 40-2x2
Slope is – 2
They are parallel each other so both can be equated in the following form;
(Z=C) -C/40=-2 C=80
This means that profit for mug can vary between $ 30 and $ 80 and the optimal point will not change.

12. Figure 3.4 Determining the Sensitivity Range for c1
Objective Function Coefficient
Sensitivity Range for c1 and c2 (3 of 3)
objective function Z = $40x1 + $50x sensitivity range for:
x1: 25  c1  x2: 30  c2  80
Figure 3.4 Determining the Sensitivity Range for c1

13. Figure 3.4 Determining the Sensitivity Range for c1
Objective Function Coefficient
Sensitivity Range for c1 and c2 (2 of 3)
objective function Z = $40x1 + $50x sensitivity range for:
x1: 25  c1  x2: 30  c2  80
If the slope of objective function increase to -4/3, the objective function
Becomes exactly parallel to the constraint line: 4x1 + 3x2 =120
C1/50=-4/3 C1=$ upper limit
C1/50=-1/2 C1=$ lower limit
Figure 3.4 Determining the Sensitivity Range for c1

14. Figure 3.5 Fertilizer Cost Minimization Example
Objective Function Coefficient
Fertilizer Cost Minimization Example ( 1of 3)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
sensitivity ranges:
4  c1  
0  c2  4.5
Figure Fertilizer Cost Minimization Example

15. Changes in Constraint Quantity Values Sensitivity Range (2 of 3)
The sensitivity range for a right-hand-side value is the range of values over which the quantity’s value can change without changing the solution variable mix, including the slack variables.
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
sensitivity ranges:
4  c1  
0  c2  4.5

16. Changes in Constraint Quantity Values Sensitivity Range (3 of 3)
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
x2 =Z/3-2x1
Slope is – 2
x2 = 8-4/3x1
Slope is – 4/3
They are parallel each other so both can be equated in the following form;
(Z=C) -C/3=-4/3 C=4
4  c1  
Minimize Z = $6x1 + $3x2
subject to:
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
x1 =Z/6-1/2x2
Slope is – 1/2
x1 = 6-3/4x2
Slope is – 3/4
They are parallel each other so both can be equated in the following form;
(Z=C) -C/6=-3/4 C=4.5
0  c2  4.5

17. Figure 3.6 Increasing the Labor Constraint Quantity
Changes in Constraint Quantity Values
Increasing the Labor Constraint (1 of 3)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure Increasing the Labor Constraint Quantity

18. Figure 3.7 Determining the Sensitivity Range for Labor Quantity
Changes in Constraint Quantity Values
Sensitivity Range for Labor Constraint (2 of 3)
Sensitivity range for:
30  q1  80 hr
Figure Determining the Sensitivity Range for Labor Quantity

19. Figure 3.8 Determining the Sensitivity Range for Clay Quantity
Changes in Constraint Quantity Values
Sensitivity Range for Clay Constraint (3 of 3)
Sensitivity range for:
60  q2  160 lb
Figure Determining the Sensitivity Range for Clay Quantity

20. Other Forms of Sensitivity Analysis
Topics (1 of 4)
Changing individual constraint parameters
Adding new constraints
Adding new variables

21. Figure 3.9 Changing the x1 Coefficient in the Labor Constraint
Other Forms of Sensitivity Analysis
Changing a Constraint Parameter (2 of 4)
Maximize Z = $40x1 + $50x2
subject to: x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure Changing the x1 Coefficient in the Labor Constraint

22. Other Forms of Sensitivity Analysis Adding a New Constraint (3 of 4)
Adding a new constraint to Beaver Creek Model: x x2  5 hours for packaging Original solution: 24 bowls, 8 mugs, $1,360 profit
To find out Optimal solution coordinates, we use the both constraints.
x1 = 40 -2x2
4(40 -2x2) + 3x2 =120
160-8x2 + 3x2 =120
-5x2 = -40
X2 = 8
x1 = 24
Z = $40 (24) + $50 (8)
Z = $ 1360 max daily profit possible
With the constraint for packing added, the new solution: 20 bowls, 10 mugs, $1,300 profit.
Exhibit 3.17

23. Other Forms of Sensitivity Analysis
Adding a New Variable (4 of 4)
Adding a new variable to the Beaver Creek model, x3, a third product, cups
Maximize Z = $40x1 + 50x2 + 30x3
subject to:
x1 + 2x x3  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).

24. Shadow Prices (Dual Variable Values)
Defined as the marginal value of one additional unit of resource.
The sensitivity range for a constraint quantity value is also the range over which the shadow price is valid.

25. The Dual Problem and Shadow prices
Every linear programming problem, called the primal problem, has a corresponding or symmetrical problem called the dual problem.
A profit-maximization primal problem has a cost-minimization dual problem, and vice versa. The solution of a dual problem yields the shadow prices.
They give the change in the value of the objective function per unit change in each constraint in the primal problem.
According to the duality theorem, the optimal value of the objective function is the same in the primal and in the corresponding dual problems.

26. Recall: Duality and Shadow prices
In a business application, a shadow price is the maximum price that management is willing to pay for an extra unit of a given limited resource. For example, if a production line is already operating at its maximum 40 hours limit, the shadow price would be the price of keeping the line operational for an additional hour. The value of the shadow price can provide decision makers powerful insight into problems.
For instance if you have a constraint that limits the amount of labor available to 40 hours per week, the shadow price will tell you how much you would be willing to pay for an additional hour of labor. If your shadow price is $10 for the labor constraint, for instance, you should pay no more than $10 an hour for additional labor. Labor costs of less than $10/hour will increase the objective value; labor costs of more than $10/hour will decrease the objective value. Labor costs of exactly $10 will cause the objective function value to remain the same.

27. Dual of the Profit Maximization Problem
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

28. Dual of the Profit Maximization Problem
In the dual problem we seek to minimize the shadow prices of inputs A, B, and C used by the firm. Defining VA, VB, as 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 dual objective function as minimize
C=7 VA+5 VB+2 VC
Thus the constraints of the dual problem can be written as follow:
1VA + 0.5VB  $30
1VA + 1VB + 0.5VC  $40 so VC=0 due to slack variable.
1VA + 0.5VB = $30
1VA + 1VB = $40 therefore 0.5VB=$10, VB=$20 and VA=$20
C=7 ($20)+5 ($20) +2 (0)=$240 this is the minimum cost.

29. Dual of the Cost Minimization Problem
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)
Maximize
Subject to
 = 14VP + 10VM + 6VV
1VP + 1VM + 1VV  $2
2VP + 1VM + 0.5VV  $3
VP, VM, VV  0

30. Dual of the Cost Minimization Problem
Since we know that from the solution of the primal problem that vitamin constraint is a slack variable, so that VV=0, subtracting the first from the second constraint, we can get the solution of the dual problem as follow:
2VP + 1VM =3
1VP + 1VM = 2 , therefore VP =$1 and VM =$1
The profit as follows:
 = 14($1)+ 10($1) + 6($0)
= $24.

31. Example Problem 1
Problem Statement (1 of 2)
Two airplane parts: no.1 and no. 2.
Three manufacturing stages: stamping, drilling, milling.
Decision variables: x1 (number of part no.1 to produce)
x2 (number of part no.2 to produce)
Model: Maximize Z = $650x x2
subject to:
4x x2  105 (stamping,hr)
6.2x x2  90 (drilling, hr)
9.1x x2  110 (finishing, hr)
x1, x2  0

32. Figure 3.10 Graphical Solution
Example Problem 1
Graphical Solution (2 of 2)
Maximize Z =
$650x1 + $910x2
subject to:
4x x2  105
6.2x x2  90
9.1x x2  110
x1, x2  0
s1 = 0, s2 = 0, s3 = hr
 c1  1,151.43
 q1  89.10
Figure Graphical Solution

33. Example Problem 2 Problem Statement
Southern Sporting Goods Company makes basketballs and footballs balls. Each product is produced from two resources—rubber and leather. The resource requirements for each product and the total resources available are as follows:
Each basketball produced results in a profit of $12, and each football earns $16 in profit.
Product
Resource Requirements per Unit
Rubber(lb)
Leather (ft.2)
Basketball 3 4
Football 2 5
Total resources available
500 lb.
800 ft.2

34. Example Problem 2 Problem Statement
a. Formulate a linear programming model to determine the number of basketballs and footballs to produce in order to maximize profit.
b. Transform this model into standard form.
c. Solve the model formulated in the Problem for Southern Sporting Goods Company graphically.
d. Identify the amount of unused resources (Le., slack) at each of the graphical extreme points.
e. What would be the effect on the optimal solution if the proflt for a basketball changed from $12 to $13? What would be the effect if the profit for a football changed from $16 to$15?
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 additional square feet of leather could be obtained?
For the linear programming model for Southern Sporting Goods Company, formulated in section a and solved graphically in section b:
g. Determine the sensitivity ranges for the objective function coeffıcients and constraint quantity values, using graphical analysis.
h. Determine the shadow prices for the resources and cxplain their meaning.

35. Example Problem 2 Problem Solution

36. Example Problem 2 Problem Solution

37. Example Problem 2 Problem Solution

38. Example Problem 2 Problem Solution
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=0
v2 = 16/5=3.20
(h)
Maximize Z = $12x1 + $16x2
subject to:
3x1 + 2x2  500 (rubber)
4x1 + 5x2  800 (leather)
x1, x2  0
A 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.

39. Example Problem 3 Problem Statement
An Aluminum Company produces three grades (high, medium, and low) of aluminum at two mills. Each mill has a different production capacity (in tons per day) for each grade, as follows:
The company has contracted with a manufacturing firm to supply at least 12 tons of high-grade aluminum, 8 tons of medium-grade aluminum, and 5 tons of low-grade aluminum. It costs United $6,000 per day to operate mill 1 and $7,000 per day to operate mill 2. The company wants to know the number of days to operate each mill in order to meet the contract at the minimum cost.
Mill
Aluminum Grade 1 2
High
Medium
Low 6 4 10

40. Example Problem 3 Problem Statement
a. Formulate a linear programming model for this problem.
b. Solve the linear programming model formulated in section a for United Aluminum Company graphically.
c. How much extra (i.e., surplus) high-, medium-, and low-grade aluminum does the company produce at the optimal solution?
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 day?
e. What would be the effect on the optimal solution if the company could supply only 10 tons of high-grade aluminum?
f. Identify and explain the shadow prices for each of aluminum grade contract requirements
g. Determine the sensitivity ranges for the objcctive function coefficients and for the constraint quantity values.

41. Example Problem 3 Problem Solution

42. Example Problem 3 Problem Solution

43. Example Problem 3 Problem Solution
Min Z= $12v1 + $8v2+ $5v3
subject to:
6v1 + 2v2 + 4v3  6000
2v1 + 2v2 + 10v3  7000
v1, v2 ,, v3 0
Since v3 is a slack variable, we set v3=0
6v1 + 2v2 =6000
v2 = v1
2v (3- 6 v1=16
v1=- 250, v1  0 so v1=0
v2 = 3000
(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.

44. Thanks