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2_Tutorial_Exercises_Solution

# 2_Tutorial_Exercises_Solution - TTA Problems Transportation...

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TTA Problems Transportation Chapter 6 Q6 Solution Let ij X be the amount shipped from i to j where i=A, B, C and j= 1, 2, 3 Min Z = \$(6 1 A X + 9 2 A X +7 3 A X +12 1 B X +3 2 B X +5 3 B X +4 1 C X +8 2 C X +11 3 C X ) Subject to : 1 A X + 2 A X + 3 A X 130 (Supply Constraint) 1 B X + 2 B X + 3 B X 70 1 C X + 2 C X + 3 C X 100 1 A X + 1 B X + 1 C X = 80 (Demand Constraint) 2 A X + 2 B X + 2 C X = 110 3 A X + 3 B X + 3 C X = 60 ij X 0 (Non-negativity) By using the computer to solve the above problem, we get 1 A X = 0 2 A X = 20 3 A X = 60 1 B X = 0 2 B X = 70 3 B X = 0 1 C X = 80 2 C X = 20 3 C X = 0 Z = \$1,290 Q7 Solution Let ij X be the amounts of steel shipped from i to j where i=A (Bethlehem), B (Birmingham), C (Gary) and j=1 (Detroit), 2 (St. Louis), 3 (Chicago), 4 (Norfolk) Min Z = 14 1 A X +9 2 A X +16 3 A X +18 4 A X +11 1 B X +8 2 B X + 1000 3 B X +16 4 B X +16 1 C X +12 2 C X +10 3 C X +22 4 C X Subject to : 1 A X + 2 A X + 3 A X + 4 A X 150 (Production Constraint) 1 B X + 2 B X + 3 B X + 4 B X 210 1 C X + 2 C X + 3 C X + 4 C X 320 1 A X + 1 B X + 1 C X = 130 (Demand Constraint) 2 A X + 2 B X + 2 C X = 70 3 A X + 3 B X + 3 C X = 180 4 A X + 4 B X + 4 C X = 240 X B3 = 0 (prohibition of shipment from Birmingham to Chicago) ij X 0 (Non-Negativity)

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