Homework #1 Solutions 3.1 a. Let X1 =the number of hours of process 1 used X2 =the number of hours of process 2 used The objective is to minimize the cost of producing the three products A, B and C. The constraints are on the demand of each product. Min 4X1 + X2 subject to 3X1 + X2 ≥ 10 (A’s demand) X1 + X2 ≥ 5 (B’s demand) X1 ≥ 3 (C’s demand) X1 , X2 ≥ 0 (nonnegativity) b. From the graph we can see that the isocost line just leaves the feasible region where the demand for C and B intersect. The point of intersection is (3,2). The cost at that point is 14. That is the minimum cost. Graphical Solution to 3.1
10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 X1 4 5 6
A B C Isocost=14
a. Let X1 =the number of type 1 trucks produced X2 =the number of type 2 trucks produced The objective is to maximize the profit of producing the two truck types. The constraints are capacities of the paint and assembly shops. Max 300X1 + 500X2 subject to 7X1 + 8X2 ≤ 5600 (Paint Shop) 4X1 + 5X2 ≥ 6000 (Assembly Shop)
X1 , X2 ≥ 0, integers (nonnegativity & integrality) 3.10 We want to minimize production and inventory costs while meeting demand. Only half of a week’s demand can be filled from that week’s production. I assume that the other half is available to meet the following weeks’ demand. Let pi=the number produced dur ing week i, i=1, 2, 3 Ai= the number available to meet demand during week i, i=1, 2, 3 Ii=the number in inventory at the end of week i, i=1, 2, 3
Min 13p1 + 14p2 + 15p3 + 2(I1 + I2 + I3 ) subject to A1 = 5 + .5p1 (Number available to meet demand for week 1) A1 ≥ 20 (We must meet demand for week 1) I1 = A1 -20 + .5 p1 (Balance constraint for week 1) A2 = I1 + .5p2 (Number available to meet demand for week 2) A2 ≥ 10 (We must meet demand for week 2) I2 = A2 -10 + .5 p2 (Balance constraint for week 2) A3 = I2 + .5p3 (Number available to meet demand for week 3) A2 ≥ 15 (We must meet demand for week 2) I3 = A3 -15 + .5 p3 (Balance constraint for week 2) All...
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