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# assign1.sol - CO 350 Assignment 1 Winter 2010 Solutions Due...

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CO 350 Assignment 1 – Winter 2010, Solutions Due: Friday January 15 at 10:25 a.m. Question # Max. marks Part marks 1 7 2 4 1 1 1 1 3 3 4 8 2 2 2 2 5 4 2 2 6 4 1 3 7(Bonus) (2) Total 30 + (2) 1. A widget plant makes 800 type-A widgets, 600 type-B widgets, and 400 type-C widgets every day. Each of these products can be sold either with or without a wireless card. The total number of widgets (of any of the 3 types) that can be ﬁtted with wireless cards during a normal working day is 800. In addition, up to 220 widgets (of any of the 3 types) can be ﬁtted with wireless cards on overtime, at a higher cost. The net proﬁts are as follows: Without Regular time install of Overtime install of wireless-card wireless-card wireless-card Type-A widget \$10 \$20 \$15 Type-B widget \$20 \$30 \$25 Type-C widget \$20 \$50 \$35 The objective is to ﬁnd a schedule that maximizes the total net proﬁt. Give a formulaton as a linear programming problem. Remarks: Distribution of the points: 1 for setup, 1 for objective function, 1 for non-negativity constraints, and 4 for the other constraints. Solution: We use the following 9 nonnegative variables: type A type B type C number of widgets with no wireless card x 1 x 2 x 3 number of widgets with cards installed on regular time y 1 y 2 y 3 number of widgets with cards installed on overtime z 1 z 2 z 3 We write the objective function as 10 x 1 + 20 x 2 + 20 x 3 + 20 y 1 + 30 y 2 + 50 y 3 + 15 z 1 + 25 z 2 + 35 z 3 We have three constraints for the total number of widgets produced of each type: Type A: x 1 + y 1 + z 1 = 800 Type B: x 2 + y 2 + z 2 = 600 Type C: x 3 + y 3 + z 3 = 400 . 1

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There are two more constraints for the total number of widgets that can be ﬁtted with wireless cards in normal time, and overtime: y 1 + y 2 + y 3 800 z 1 + z 2 + z 3 220 . The LP formulation is: max 10 x 1 + 20 x 2 + 20 x 3 + 20 y 1 + 30 y 2 + 50 y 3 + 15 z 1 + 25 z 2 + 35 z 3 s.t. x 1 + y 1 + z 1 = 800 x 2 + y 2 + z 2 = 600 x 3 + y 3 + z 3 = 400 y 1 + y 2 + y 3 800 z 1 + z 2 + z 3 220 x 1 , x 2 , x 3 , y 1 , y 2 , y 3 , z 1 , z 2 , z 3 0 Note: There are of course other correct solutions. In particular, it is possible to make a formulation using only the variables y 1 ,y 2 ,y 3 ,z 1 ,z 2 ,z 3 , since it is possible to express the x j variables in terms of them. In this way one can get the following LP formulation: max 10 y 1 + 10 y 2 + 30 y 3 + 5 z 1 + 5 z 2 + 15 z 3 (+28 , 000) subject to y 1 + z 1 800 y 2 + z 2 600 y 3 + z 3
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assign1.sol - CO 350 Assignment 1 Winter 2010 Solutions Due...

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