172A08PS2Ans

# 172A08PS2Ans - Economics 172A Introduction to Operations...

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Economics 172A: Introduction to Operations Research Winter 2008 Problem set 2 Due Thursday, March 13 at start of class (no late papers) Instructions Unless otherwise noted, you are required to supply complete answers and explain how you got them. Simply stating a numerical answer is insufficient. For graphs, clearly label the graph and identify what’s on it. 1. Consider the problem choose x (a scalar) to solve maximize 3x subject to x 5 x b x 0 where b (also a scalar) > 0. (a) Graphically or by educated guess, whichever you prefer, compute x*(b), the optimal value of x, as a function of the parameter b. Your answer must tell what the optimal value of x is for any value of b > 0; that is, it must be a clearly specified function of b. (Hint: Given that b > 0, could there ever be a solution with x* = 0? Which constraint would you expect to determine the solution when b < 5? When b > 5?) (a) Clearly x*(b) > 0 for all b > 0. For b > (<) 5, the second (first) constraint is redundant. One of these constraints must always be binding, so x*(b) = b if b 5, and x*(b) = 5 if b > (or ) 5. (b) Write the dual, using y i to represent the dual control variable that is the shadow price of the ith constraint in the primal. (b) Choose y 1 , y 2 to solve minimize 5y 1 + by 2 subject to y 1 + y 2 3 y 1 0 y 2 0. (c) Graphically or by educated guess, whichever you prefer, compute y 1 *(b) and y 2 *(b), the optimal values of y 1 and y 2 , as functions of the parameter b. Your answer must tell what the optimal values of y 1 and y 2 are for any value of b. (When b = 5, identify all the possible optimal values of y 1 *(b) and y 2 *(b).) (c) If b < (or ) 5, y 1 *(b) = 0 and y 2 *(b) = 3. If b > (or ) 5, y 1 *(b) = 3 and y 2 *(b) = 0. If b = 5, any y 1 and y 2 with y 1 + y 2 = 3 and y 1 0 and y 2 0 is optimal, including but not limited to the values specified above.

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(d) Use the Duality Theorem to show that your solutions to the primal in (a) and the dual in (c) are both optimal for all values of b. (d) The primal objective function value is 3b if b 5 and 15 if b > 5. The dual objective function value is 3b if b 5 and 15 if b > 5. Since the solutions are feasible for the primal and the dual, and the objective function values are equal in each case, the Duality Theorem shows that they are both optimal in each case. (e) Verify directly that your solutions to the primal in (a) and the dual in (c) satisfy Complementary Slackness, saying clearly what Complementary Slackness requires. (e) Complementary Slackness requires that if a primal (dual) control variable is strictly positive, then the associated dual (primal) constraint must be binding; and that if a primal (dual) constraint is slack (non-binding), then the associated dual (primal) control variable must be zero. Checking: (i) y 1 *(b) + y 2 *(b) = 3 for all b, so CS in the dual constraint is satisfied; (ii) if b < 5, x*(b) < 5 but y 1 *(b) = 0, and x*(b) = b, so CS in the primal constraints are satisfied; and (iii) if b > 5, x*(b) = 5, and
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## This note was uploaded on 04/08/2008 for the course ECON 172A taught by Professor Crawford during the Winter '08 term at UCSD.

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172A08PS2Ans - Economics 172A Introduction to Operations...

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