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# H1 - IE 335 Operations Research Optimization Solutions to...

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Unformatted text preview: IE 335 Operations Research - Optimization Solutions to Homework 1 Fall 2010 Problem 1 Refer to Section 2.1 in the book 1-3.a Decision variables represent the decision to be taken in the optimization model. Since we need to decide on the quantities produced, this implies that x 1 and x 2 are the decision variables for this problem. 1-3.b Input parameters are quantities in the model not affected by decisions. Therefore, the input parameters in this problem are t 1 ,t 2 ,T,c 1 ,c 2 and b . 1-3.c It would be advantageous to minimize production cost. Since c i is the cost of producing a lot in production line i , the total cost is c 1 x 1 + c 2 x 2 and therefore the objective function is: minimize c 1 x 1 + c 2 x 2 (1) 1-3.d We identify the constraints on the decision variables from the problem statement: Statement Constraint b lots in total x 1 + x 2 = b in a total of at most T hours x 1 t 1 + x 2 t 2 ≤ T integer number x 1 & x 2 x 1 ,x 2 integer Since production quantities cannot be negative, we add nonnegativity constraints on x 1 and x 2 . The final optimization model is defined below: minimize c 1 x 1 + c 2 x 2 subject to x 1 + x 2 = b x 1 t 1 + x 2 t 2 ≤ T x 1 ,x 2 ≥ 0 and integer (2) Problem 2 1-8.a (3 possibilities for parameter 1) × (3 possibilities for parameter 2) ×···× (3 possibilities for parameter n ) = a total of 3 n sets of parameters 1-8.b 10 3 10 secs or 16.40 hours 15 3 15 secs or 166.07 days 20 3 20 secs or 110.56 years 30 3 25 secs or 6.5 million years 1-8.c It is apparent that the “try all combinations” method is too time consuming and inefficient when n > 10. 1 Problem 3 3.(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 (3) 3.(b) The optimal solution is C = 4 and D = 2 thus resulting in a profit of 40 × 2 + 25 × 4 = 180 3.(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. 3.(d) There are several ways to render this problem infeasible. One way would be to add the following constraint: D ≥ 3 (4) Hence the model is infeasible since there are no points which satisfy all the constraints. 3.(e) For now, we assume fractional solutions are acceptable, and so we ignore any integrality constraints....
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H1 - IE 335 Operations Research Optimization Solutions to...

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