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Unformatted text preview: IE 335 Operations Research  Optimization Solutions to Homework 2 Fall 2011 Problem 7 7.(a) We define the decision variables C and D such that C = number of chairs D = number of desks Now, we can define the optimization model as follows: maximize 40 D + 25 C subject to C ≥ 2 D 4 D + 3 C ≤ 20 C,D ≥ 0 and integer 7.(b) The optimal solution is C = 4 and D = 2 thus resulting in a profit of 40 × 2 + 25 × 4 = 180 7.(c) Yes. The optimal value contour intersects the feasible set at exactly one point, i.e. (4,2) and therefore we have a unique optimal solution. See the graph below. 7.(d) There are several ways to render this problem infeasible. One way would be to add the following constraint: D ≥ 3 Hence the model is infeasible since there are no points which satisfy all the constraints. 7.(e) For now, we assume fractional solutions are acceptable, and so we ignore any integrality constraints. Therefore, we can formulate and solve the model as a linear program (LP) instead of a mixedintegerprogram (MIP). The GAMS LST file has been included which verifies the optimal values for the objective function, objval = 180, C = 4 and D = 2. 1 GAMS Rev 230 WEXVIS 23.0.2 x86/MS Windows 09/05/11 18:26:11 Page 1 G e n e r a l A l g e b r a i c M o d e l i n g S y s t e m C o m p i l a t i o n 1 variables 2 objval "Objective function value" 3 C "# manufactured chairs" 4 D "# manufactured desks"; 5 6 free variables 7 objval; 8 9 positive variables 10 C, D; 11 12 equations 13 obj "Total profit" 14 marketing "Marketing restrictions" 15 cap_wood "Available units of wood"; 16 17 obj .. 18 objval =e= 40*D + 25*C; 19 20 marketing .....
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This note was uploaded on 12/14/2011 for the course IE 335 taught by Professor Jeanphilippe,r during the Fall '08 term at Purdue University.
 Fall '08
 Jeanphilippe,R
 Operations Research, Optimization

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