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This example demonstrates the transformation of a linear programming model into standard form, sensitivity analysis, computer solution, and shadow prices.
The Xecko Tool Company is considering bidding on a job for two airplane wing parts . Each wing part must be processed through three manufacturing stagesstamping, drilling, and finishingfor which the company has limited available hours. The linear programming model to determine how many of part 1 ( x 1 ) and part 2 ( x 2 ) the company should produce in order to maximize its profit is as follows :
maximize Z = $650 x 1 + 910 x 2
4 x 1 + 7.5 x 2 105 (stamping, hr.)
6.2 x 1 + 4.9 x 2 90 (drilling, hr.)
9.1 x 1 + 4.1 x 2 110 (finishing, hr.)
x 1 , x 2
Solve the model graphically.
Indicate how much slack resource is available at the optimal solution point.
Determine the sensitivity ranges for the profit for wing part 1 and the stamping hours available.
Solve this model by using Excel.
The slack at point B , where x 1 = 5.97 and x 2 = 10.82, is computed as follows:
4(5.97) + 7.5(10.82) + s 1 = 105 (stamping, hr.)
s 1 = 0 hr.
6.2(5.97) + 4.9(10.82) + s 2 = 90 (drilling, hr.)
s 2 = 0 hr.
9.1(5.97) + 4.1(10.82) + s 3 = 110 (finishing, hr.)
s 3 = 11.35 hr.
The sensitivity range for the profit for part 1 is determined by observing the graph of the model and computing how much the slope of the objective function must increase to make the optimal point move from B to C . This is the upper limit of the range and is determined by computing the value of c 1 that will make the slope of the objective function equal with the slope of the constraint line for drilling, 6.2 x 1 + 4.9 x 2 = 90:
The lower limit is determined by computing the value of c 1 that will equate the slope of the objective function with the slope of the constraint line for stamping, 4 x 1 + 7.5 x 2 = 105:
485.33 c 1 1,151.43
The upper limit of the range for stamping hours is determined by first computing the value for q 1 that would move the solution point from B to where the drilling constraint intersects with the x 2 axis, where x 1 = 0 and x 2 = 18.37:
The lower limit of the sensitivity range occurs where the optimal point B moves to C , where x 1 = 8.87 and x 2 = 7.15:
Summarizing, 89.10 q 1 137.76.