Chapter 3: Sensitivity Analysis and the Dual Problem & Shadow Prices Department of Business Administration SPRING 2016-2017 Operations Research Chapter 3: Sensitivity Analysis and the Dual Problem & Shadow Prices

Chapter Topics Standard form Sensitivity Analysis Dual Problem Example Problems

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

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= 12x1 + 16 x2 Subject to: 3x1 + 2x2  500 4x1 + 5x2  800 x1 x2  0 Standard form:

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.

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

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

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 3.3 Changing the x2 Objective Function Coefficient

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.

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 $ 66.67 and the optimal point will not change.

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.

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 + $50x2 sensitivity range for: x1: 25  c1  66.67 x2: 30  c2  80 Figure 3.4 Determining the Sensitivity Range for c1

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 + $50x2 sensitivity range for: x1: 25  c1  66.67 x2: 30  c2  80 If the slope of objective function increase to -4/3, the objective function Becomes exactly parallel to the constraint line: 4x1 + 3x2 =120 C1/50=-4/3 C1=$ 66.67 upper limit C1/50=-1/2 C1=$ 25.00 lower limit Figure 3.4 Determining the Sensitivity Range for c1

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 3.5 Fertilizer Cost Minimization Example

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

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

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 3.6 Increasing the Labor Constraint Quantity

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 3.7 Determining the Sensitivity Range for Labor Quantity

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 3.8 Determining the Sensitivity Range for Clay Quantity

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

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: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 3.9 Changing the x1 Coefficient in the Labor Constraint

Other Forms of Sensitivity Analysis Adding a New Constraint (3 of 4) 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 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

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 + 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).

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.

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.

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.

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

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.

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

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.

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 = $650x1 + 910x2 subject to: 4x1 + 7.5x2  105 (stamping,hr) 6.2x1 + 4.9x2  90 (drilling, hr) 9.1x1 + 4.1x2  110 (finishing, hr) x1, x2  0

Figure 3.10 Graphical Solution Example Problem 1 Graphical Solution (2 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  1,151.43 137.76  q1  89.10 Figure 3.10 Graphical Solution

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

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.

Example Problem 2 Problem Solution

Example Problem 2 Problem Solution

Example Problem 2 Problem Solution

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.

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

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.

Example Problem 3 Problem Solution

Example Problem 3 Problem Solution

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 = 3000-3 v1 2v1 +6000-(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.

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