Ch. 6 assignment

Ch. 6 assignment
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Unformatted text preview: John Latus Professor Saffarian MGMT 3413 A. ( Material) 6x1 + 4x2 =48 6x + 4(0) = 48 6(0) + 4x = 48 6x = 48 4x = 48 X1 = 8 x2 = 12 (Labor) 4x + 8x = 80 4x + 8(0) = 80 4(0) + 8x = 80 4x = 80 8x = 80 X 1= 20 x2 = 10 2(6x +4x = 48) 1(4x + 8x = 80) 12x + 8x = 96 - 4x + 8x = 80 8x = 16 X1 = 2 4(2) + 8x2 = 80 8 + 8x2 = 80 8x2 = 72 X2 = 9 Z= 4(2) + 3(9) Z= 35 1. X1=2 X2=9 Z=35 2. No slack 3. No surplus 4. No redundant Constraints B. Maximize Z=2x1+10x2 Durability 10x+4x=40 Strength x+6x=24 Time x+2x=14 10x1+4(0) = 40 X1=4 10(0) +4x2=40 X2=10 X1+6(0) = 24 X1=24 1(0) +6x2=24 X2=4 X1+2(0) =14 X1=14 1(0) +2x2=14 X2=7 D-S [1.5(10x1+4x2=40) x1+6x2=24] = 14x1=36 X1=2.57 X2=3.57 D-T [10x1+4x2=40- 2(x1+2x2=14)]= 8x1=12 X1=1.5 X2=6.25 S-T [x1+6x2=24 (x1+2x2=14)= 4x2=10 X2= 2.5 X1=9 Z= 2(2.57)+10(3.57)= 40.84 Z= 2(1.5)+10(6.25)= 65.5 (optimum) Z= 2(9)+10(2.5)= 43 1. X1=1.5 X2=6.25 Z=65.5 2. No slack 3. The Strength Constraint has a surplus of 15 4. No redundant constraints C. Maximize Z=6a+3b Material ...
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