Chapter 8 b625aswc08.pptx

# Chapter 8 b625aswc08.pptx - Chapter 8 Linear Programming...

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Chapter 8 Linear Programming: Sensitivity Analysis and Interpretation of Solution 1

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Sensitivity Analysis Recall in the formulation of the linear programming problem that we have an objective (to min cost or max profit, for example) and we have limits or constraints on what we can do. We have decisions to make on what we can control (choosing values for the decision variables) and we have some things we have no control over (the parameters of the problem). Sensitivity analysis is essentially about understanding how the values of the decision variables change given a change in a parameter. 2
Sensitivity Analysis In sensitivity analysis we compare every change to the original or base solution to a problem. For example, before I worked with a problem of a company making bowls and mugs. We found a solution. Sensitivity analysis is esentially a scenario analysis game. We can play many scenarios, but we play each 1 at a time and we also compare a scenario with the base solution. Some call this a postoptimality analysis Let’s use the problem we worked on as the basis for our work here. 3

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Here is a basic linear programming problem we developed before. We want to Maximize profit Z= 40b + 50m, but we have constraints 1b + 2m <= 40 (labor), 4b + 3m <= 120 (clay), b, m >= 0. b and m are the decision variables and represent the amount of each item the company produces. Some of the parameters are the profit contributions of each item the firm makes. In the profit function these contributions are coefficients on the decision variables (here we have values 40 and 50). 4
There are also parameters in the contraints. Recall that there was only 40 hours of labor available and only 120 pounds of clay available. The parameters are called the “right-hand-side” values of the constraints. So, in sensitivity analysis we can check on the following scenarios: 1) Changing an objective function coefficient, 2) Changing a limiting value on a constraint (a “right- hand-side” value, or Remember: do 1 at a time and compare to base soultion. 5

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Both constraints m b (40, 0) (0, 20) (0, 40) (30, 0) You may recall from our previous work using the problem of bowls and mugs that the constraints look like this and the shaded area is the feasible region. The next slide shows us the corner points and the profits at each corner. (24, 8) 6
Corner points Our three corner points are (30, 0), (0, 20) and (24, 8). Profit is 40b + 50m. Profit at each point (30, 0) 1200 (found by 40[30] + 50[0] = 1200 + 0) (0, 20) 1000 (24, 8) 1360. Corner (0, 0) should also be checked - profit is 0 there. 1360 is highest profit. Make 24 bowls and 8 mugs. 7

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Both constraints m b (40, 0) (0, 20) (0, 40) (30, 0) Remember profit was Z = 40b + 50m and the best point was (24, 8) with profit = 1360. So the profit line can be written 1360 = 40b + 50m or m =(1360/50) – (40/50)b. This line has m intercept (0, 1360/50) or (0, 27.2) and b intercept (1360/40, 0) or (34, 0). I put in this profit line here. This is the optimal solution.
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