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172A-PS2 - Econ172A ProblemSet#2:SensitivityAnalysis...

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Econ 172A Problem Set #2: Sensitivity Analysis Winter 2010 1. Consider the linear programming problem , 0 max , 2 3 s.t. 4 16 8 x y f x y x y x y x a. Solve this problem graphically. x * = 8, y * = 2, f ( x *, y *) = 22, S 1 * = 0, S 2 * = 0 Sensitivity analysis: Consider changing some of the coefficients (one at a time) in the original problem. , 0 max , 2 3 s.t. 4 16 8 x y f x y x a y x b y c x d b. For each of the coefficients, a , b , c and d , find the sensitivity limits. If sensitivity analysis cannot be used on any coefficient, explain why. a : The slope of a level curve of the objective function is –2/(3 + a). The slope of the constraints are –1/4 and undefined (vertical). The same corner will remain optimal provided: x 2 4 8 12 16 y 4
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1 2 1 5 | | | | 3 4 Notation: The absolute value of the slope of a level curve greater than the absolute value of the slope of the first constraint. 2 0 3 0 3 3 5 LC C LC a m m a a m a a   The solution will be x * = 8, y * = 2, f ( x *, y *) = 16 + (3 + a )2, S 1 * = 0, S 2 * = 0. b : Sensitivity analysis cannot be used because one of the coefficients on the LHS of a constraint is changing. To find out how our solution would change we would need to reformulate and then solve a new linear program. c : The solution will remain at the “same” corner if we increase the RHS of Constraint 1 (regardless of by how much). It will also remain at the same corner if we decrease the RHS of Constraint 1 until its x -intercept occurs at (8, 0). c –8 The intersection of the two constraints will be x = 8, y = (8 + c )/4. The solution will be x * = 8, y * = (8 + c )/4, f ( x *, y *) = 16 + 3[(8 + c )/4], S 1 * = 0, S 2 * = 0. d : The solution will remain at the “same” corner if we increase the RHS of Constraint 2 to 16 where it will barely become redundant. It will also remain at the “same” corner if we decrease the RHS of Constraint 2 to 0. –8 d 8 The intersection of the two constraints will be x = 8 + d , y = (8 – d )/4.
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