chapter 4

# chapter 4 - Linear Programming Applications Chapter 4...

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Linear Programming Applications Chapter 4 Linear Programming Applications 2. a. Let x 1 = units of product 1 produced x 2 = units of product 2 produced Max 30 x 1 + 15 x 2 s.t. x 1 + 0.35 x 2 100 Dept. A 0.30 x 1 + 0.20 x 2 36 Dept. B 0.20 x 1 + 0.50 x 2 50 Dept. C x 1 , x 2 0 Solution: x 1 = 77.89, x 2 = 63.16 Profit = 3284.21 b. The dual price for Dept. A is \$15.79, for Dept. B it is \$47.37, and for Dept. C it is \$0.00. Therefore we would attempt to schedule overtime in Departments A and B. Assuming the current labor available is a sunk cost, we should be willing to pay up to \$15.79 per hour in Department A and up to \$47.37 in Department B. c. Let x A = hours of overtime in Dept. A x B = hours of overtime in Dept. B x C = hours of overtime in Dept. C Max 30 x 1 + 15 x 2 - 18 x A - 22.5 x B - 12 x C s.t. x 1 + 0.35 x 2 - x A 100 0.30 x 1 + 0.20 x 2 - x B 36 0.20 x 1 + 0.50 x 2 - x C 50 x A 10 x B 6 x C 8 x 1 , x 2 , x A , x B , x C 0 x 1 = 87.21 x 2 = 65.12 Profit = \$3341.34 Overtime

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Chapter 4 Dept. A 10 hrs. Dept. B 3.186 hrs Dept. C 0 hours Increase in Profit from overtime = \$3341.34 - 3284.21 = \$57.13 6. Let x 1 = units of product 1 x 2 = units of product 2 b 1 = labor-hours Dept. A b 2 = labor-hours Dept. B Max 25 x 1 + 20 x 2 + 0 b 1 + 0 b 2 s.t. 6 x 1 + 8 x 2 - 1 b 1 = 0 12 x 1 + 10 x 2 - 1 b 2 = 0 1 b 1 + 1 b 2 900 x 1 , x 2 , b 1 , b 2 0 Solution: x 1 = 50, x 2 = 0, b 1 = 300, b 2 = 600 Profit: \$1,250 8. Let x 1 = the number of officers scheduled to begin at 8:00 a.m. x 2 = the number of officers scheduled to begin at noon x 3 = the number of officers scheduled to begin at 4:00 p.m. x 4 = the number of officers scheduled to begin at 8:00 p.m. x 5 = the number of officers scheduled to begin at midnight x 6 = the number of officers scheduled to begin at 4:00 a.m. The objective function to minimize the number of officers required is as follows: Min x 1 + x 2 + x 3 + x 4 + x 5 + x 6 The constraints require the total number of officers of duty each of the six four-hour periods to be at least equal to the minimum officer requirements. The constraints for the six four-hour periods are as follows: Time of Day 8:00 a.m. - noon x 1 + x 6 5 noon to 4:00 p.m. x 1 + x 2 6 4:00 p.m. - 8:00 p.m. x 2 + x 3 10 8:00 p.m. - midnight x 3 + x 4 7 midnight - 4:00 a.m. x 4 + x 5 4 4:00 a.m. - 8:00 a.m. x 5 + x 6 6 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 0
Linear Programming Applications Schedule 19 officers as follows: x 1 = 3 begin at 8:00 a.m. x 2 = 3 begin at noon x 3 = 7 begin at 4:00 p.m. x 4 = 0 begin at 8:00 p.m. x 5 = 4 begin at midnight x 6 = 2 begin at 4:00 a.m. 11. Let x ij = units of component i purchased from supplier j Min 12 x 11 + 13 x 1 2 + 14 x 13 + 10 x 21 + 11 x 22 + 10 x 23 s.t. x

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chapter 4 - Linear Programming Applications Chapter 4...

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