Chapter 6 Solutions - :Solutions 5 a 100 160 A 1 180 B 2 75...

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Chapter 6 – Distribution and Network Models:  Solutions: 5. a. Avery 1 Baker 2 Campbell 3 A 1 B 2 C 3 D 4 130 140 150 155 100 115 125 100 120 115 135 120 160 160 140 100 75 85 180 b. Let  x ij   = number of hours from consultant  i  assigned to client  j . Max 100 x 11 + 125 x 12 + 115 x 13 + 100 x 14 + 120 x 21 + 135 x 22 + 115 x 23 s.t. + 120 x 24 + 155 x 31 + 150 x 32 + 140 x 33 + 130 x 34 x 11 + x 12 + x 13 + x 14 160 x 21 + x 22 + x 23 + x 24 160 x 31 + x 32 + x 33 + x 34 140 x 11 + x 21 + x 31 = 180 x 12 + x 22 + x 32 = 75 x 13 + x 23 + x 33 = 100 x 14 + x 24 + x 34 = 85 x ij    0  for all   i j   Optimal Solution Hours Assigned Billing Avery - Client B  40 $  5,000 Avery - Client C 100   11,500 Baker - Client A  40    4,800 Baker - Client B  35    4,725 Baker - Client D  85   10,200 Campbell - Client A 140       21,700 Total Billing $57,925 c. New Optimal Solution Hours Assigned Billing Avery - Client A  40 $  4,000 Avery - Client C 100   11,500 Baker - Client B  75   10,125 Baker - Client D  85   10,200 Campbell - Client A 140       21,700 Total Billing $57,525 6 - 1
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13 th  Edition 9. a. 1 2 1 2 Jackson Client 1 Ellis Client 2 3 3 Smith Client 3 1 1 1 1 1 1 34 24 22 40 22 14 32 16 10 b. Min 10 x 11 + 16 x 12 + 32 x 13 + 14 x 21 + 22 x 22 + 40 x 23 + 22 x 31 + 24 x 32 + 34 x 33 s.t.    x 11 +    x 12 +    x 13 1    x 21 +    x 22 +    x 23   1       x 31 +    x 32 +    x 33 1    x 11 +    x 21 +    x 31 = 1    x 12 +    x 22 +    x 32 = 1 x 13 + x 23 +    x 33 = 1 x ij    0  for all  i j   Solution  x 12  = 1,  x 21  = 1,  x 33  = 1    Total completion time = 64 14 th  Edition  9. 9.     The linear programming formulation and optimal solution are shown. Let x 1A = Units of product A on machine 1 x 1B = Units of product B on machine 1   6 - 2
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Chapter 6 – Distribution and Network Models:  Solutions: x 3C = Units of product C on machine 3 Min x 1A 1.2 x 1B + +0.9 x 1C 1.3 x 2A + 1.4 x 2B 1.2 x 2C 1.1 x 3A x 3B + + + + s.t.
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