MSci603 Fall 08 Assignment 1-answer Key0

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Answer Key for Assignment #1 Q1 part a) Revised form is : MAX Z = -2x 1 + 2x 11 + x 2 – 4x 3 + 3x 4 subject to x 1 – x 11 + x 2 + 3x 3 + 2x 4 4 -x 1 + x 11 + x 3 - x 4 ≤ 1 2x 1 - 2x 11 + x 2 2 x 1 – x 11 + 2x 2 + x 3 + 2x 4 = 2 and x 1, , x A , x 2 , x 3 , x 4 0 Therefore the standard form will be: MAX Z = -2x 1 + 2x 11 + x 2 – 4x 3 + 3x 4 + 0s 1 + 0s 2 + 0s 3 subject to x 1 – x 11 + x 2 + 3x 3 + 2x 4 + s 1 = 4 -x 1 + x 11 + x 3 - x 4 + s 2 = 1 2x 1 - 2x 11 + x 2 + s 3 = 2 x 1 – x 11 + 2x 2 + x 3 + 2x 4 = 2 and x 1, , x 11 , x 2 , x 3 , x 4 , s 1 , s 2 , s 3 0 b. Basic C b x 1 X 11 x 2 x 3 x 4 s 1 s 2 s 3 a 1 RHS Division -M M -2M -M -2M -2M z 1 2 -2 -1 4 -3 0 0 0 0 0 s 1 0 1 -1 1 3 2 1 0 0 0 4 2 s 2 0 -1 1 0 1 -1 0 1 0 0 1 -- s 3 0 2 -2 1 0 0 0 0 1 0 2 -- a 1 0 1 -1 2 1 2 0 0 0 1 2 1

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c. Using Excel software following results will be achieved: Target Cell (Max) Cell Name Original Value Final Value \$I\$3 = 0 17 Adjustable Cells Cell Name Original Value Final Value \$B\$3 x1 0 0 \$C\$3 X11 0 4 \$D\$3 x2 0 0 \$E\$3 x3 0 0 \$F\$3 x4 0 3 Q2.a- I : number of units of insurance M : number of units of mortgage MAX 5I + 2M subject to 3I + 2M 2400 Underwriting constraint 1M 800 Administration constraint 2I 1200 Claims constraint I, M 0 b-
c) To solve algebraically, 1. Take the equation 2I = 1200. This will return I = 600. 2. Given I = 600, substitute the value into the equation 3I + 2M = 2400. b 3(600) + 2M = 2400 b 2M = 2400 – 1800 = 600 b M = 300 3. The value of optimal solution = 5(600) + 2(300) = 3600.

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