assign1.sol

assign1.sol - CO 350 Assignment 1 Winter 2010, Solutions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/28/2011 for the course CO 250 taught by Professor Guenin during the Fall '10 term at Waterloo.

Page1 / 6

assign1.sol - CO 350 Assignment 1 Winter 2010, Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online