University:
California State University, Northridge
Course:
MGT 360 | Management and Organizational Behavior
Academic year:

2024
Views:

316

Pages:

25
Author:

enirehtaknlm

Sensitivity Analysis: Computer Solution Software packages such as Lingo and Microsoft Excel provide the following LP information:  Information about the objective function: • its optimal value • coefficient ranges (ranges of optimality)  Information about the decision variables: • their optimal values • their reduced costs  Information about the constraints: • the amount of slack or surplus • the dual prices • right-hand side ranges (ranges of feasibility) Reduced Cost   The reduced cost associated with a variable is equal to the dual value for the non-negativity constraint associated with the variable. In general, if a variable has a non-zero value in the optimal solution, then it will have a reduced cost equal to 0. Relevant Cost and Sunk Cost     A resource cost is a relevant cost if the amount paid for it is dependent upon the amount of the resource used by the decision variables. Relevant costs are reflected in the objective function coefficients. A resource cost is a sunk cost if it must be paid regardless of the amount of the resource actually used by the decision variables. Sunk resource costs are not reflected in the objective function coefficients.   Cautionary Note on the Interpretation of Dual Values Resource cost is sunk The dual value is the maximum amount you should be willing to pay for one additional unit of the resource. Resource cost is relevant The dual value is the maximum premium over the normal cost that you should be willing to pay for one unit of the resource. Example 1 (Again)  LP Formulation Max 5x1 + 7x2 s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1, x2 > 0 Example 1  Range of Optimality for c1 and c2 Adjustable Cells Cell $B$8 $C$8 Final Reduced Name Value Cost X1 5.0 0.0 X2 3.0 0.0 Constraints Final Cell Name Value $B$13 #1 5 $B$14 #2 19 $B$15 #3 8 Dual Value 0 2 1 Objective Coefficient 5 7 Allowable Increase 2 0.5 Constraint Allowable R.H. Side Increase 6 1E+30 19 5 8 0.33333333 Allowable Decrease 0.33333333 2 Allowable Decrease 1 1 1.66666667 Example 1  Dual Values Adjustable Cells Cell $B$8 $C$8 Final Reduced Name Value Cost X1 5.0 0.0 X2 3.0 0.0 Constraints Final Cell Name Value $B$13 #1 5 $B$14 #2 19 $B$15 #3 8 Dual Value 0 2 1 Objective Coefficient 5 7 Allowable Increase 2 0.5 Constraint Allowable R.H. Side Increase 6 1E+30 19 5 8 0.33333333 Allowable Decrease 0.33333333 2 Allowable Decrease 1 1 1.66666667 Example 1  Range of Feasibility Adjustable Cells Cell $B$8 $C$8 Final Reduced Name Value Cost X1 5.0 0.0 X2 3.0 0.0 Constraints Final Cell Name Value $B$13 #1 5 $B$14 #2 19 $B$15 #3 8 Dual Value 0 2 1 Objective Coefficient 5 7 Allowable Increase 2 0.5 Constraint Allowable R.H. Side Increase 6 1E+30 19 5 8 0.33333333 Allowable Decrease 0.33333333 2 Allowable Decrease 1 1 1.66666667 Example 2: Olympic Bike Co. Olympic Bike is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from special aluminum and steel alloys. The anticipated unit profits are $10 for the Deluxe and $15 for the Professional. The number of pounds of each alloy needed per frame is summarized on the next slide. Example 2: Olympic Bike Co. A supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy weekly. Deluxe Professional Aluminum Alloy 2 4 Steel Alloy 3 2 How many Deluxe and Professional frames should Olympic produce each week? Example 2: Olympic Bike Co.  Model Formulation • Verbal Statement of the Objective Function Maximize total weekly profit. • Verbal Statement of the Constraints Total weekly usage of aluminum alloy < 100 pounds. Total weekly usage of steel alloy < 80 pounds. • Definition of the Decision Variables x1 = number of Deluxe frames produced weekly. x2 = number of Professional frames produced weekly. Example 2: Olympic Bike Co.  Model Formulation (continued) Max 10x1 + 15x2 (Total Weekly Profit) s.t. (Aluminum Available) (Steel Available) 2x1 + 4x2 < 100 3x1 + 2x2 < 80 x1, x2 > 0 Example 2: Olympic Bike Co.  6 7 8 9 10 11 12 13 14 Partial Spreadsheet Showing Solution A Bikes Made B C Decision Variables Deluxe Professional 15 17.500 Maximized Total Profit Constraints Aluminum Steel Amount Used 100 80 D 412.500 <= <= Amount Avail. 100 80 Example 2: Olympic Bike Co.  Optimal Solution According to the output: x1 (Deluxe frames) = 15 x2 (Professional frames) = 17.5 Objective function value = $412.50 Example 2: Olympic Bike Co.  Range of Optimality Question: Suppose the profit on deluxe frames is increased to $20. Is the above solution still optimal? What is the value of the objective function when this unit profit is increased to $20? Example 2: Olympic Bike Co.  Sensitivity Report Adjustable Cells Cell $B$8 $C$8 Name Deluxe Profess. Constraints Cell Name $B$13 Aluminum $B$14 Steel Objective Allowable Final Reduced Value Cost Coefficient Increase 15 0 10 12.5 17.500 0.000 15 5 Allowable Decrease 2.5 8.333333333 Final Shadow Constraint Allowable Value Price R.H. Side Increase 100 3.125 100 60 80 1.25 80 70 Allowable Decrease 46.66666667 30 38 Example 2: Olympic Bike Co.  Range of Optimality Answer: The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 7.5 and 22.5. Because 20 is within this range, the optimal solution will not change. The optimal profit will change: 20x1 + 15x2 = 20(15) + 15(17.5) = $562.50. 39 Example 2: Olympic Bike Co.  Range of Optimality Question: If the unit profit on deluxe frames were $6 instead of $10, would the optimal solution change? 40 Example 2: Olympic Bike Co.  Range of Optimality Adjustable Cells Cell $B$8 $C$8 Name Deluxe Profess. Constraints Cell Name $B$13 Aluminum $B$14 Steel Final Reduced Objective Allowable Value Cost Coefficient Increase 15 0 10 12.5 17.500 0.000 15 5 Allowable Decrease 2.5 8.333333333 Final Value 100 80 Allowable Decrease 46.66666667 30 Dual Value 3.125 1.25 Constraint Allowable R.H. Side Increase 100 60 80 70 41 Example 2: Olympic Bike Co.  Range of Optimality Answer: The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 7.5 and 22.5. Because 6 is outside this range, the optimal solution would change. 42 Example 2: Olympic Bike Co.  Range of Feasibility and Sunk Costs Question: Given that aluminum is a sunk cost, what is the maximum amount the company should pay for 50 extra pounds of aluminum? 43 Example 2: Olympic Bike Co.  Range of Feasibility and Sunk Costs Adjustable Cells Cell $B$8 $C$8 Name Deluxe Profess. Constraints Cell Name $B$13 Aluminum $B$14 Steel Final Reduced Objective Allowable Value Cost Coefficient Increase 15 0 10 12.5 17.500 0.000 15 5 Allowable Decrease 2.5 8.333333333 Final Value 100 80 Allowable Decrease 46.66666667 30 Dual Value 3.125 1.25 Constraint Allowable R.H. Side Increase 100 60 80 70 44 Example 2: Olympic Bike Co.  Range of Feasibility and Sunk Costs Answer: Because the cost for aluminum is a sunk cost, the shadow price provides the value of extra aluminum. The shadow price for aluminum is the same as its dual value (for a maximization problem). The shadow price for aluminum is $3.125 per pound and the maximum allowable increase is 60 pounds. Because 50 is in this range, the $3.125 is valid. Thus, the value of 50 additional pounds is = 50($3.125) = $156.25. 45 Example 2: Olympic Bike Co.  Range of Feasibility and Relevant Costs Question: If aluminum were a relevant cost, what is the maximum amount the company should pay for 50 extra pounds of aluminum? 46 Example 2: Olympic Bike Co.  Range of Feasibility and Relevant Costs Answer: If aluminum were a relevant cost, the shadow price would be the amount above the normal price of aluminum the company would be willing to pay. Thus if initially aluminum cost $4 per pound, then additional units in the range of feasibility would be worth $4 + $3.125 = $7.125 per pound. 47