17. a. Network Model
Supply 450
7447 W18 C2
The linear programming formulation and solution are shown.
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LINEAR PROGRAMMING PROBLEM
MIN 4X14 + 7X15 + 8X24 + 5X25 + 5X34 + 6X35 + 6X46 + 4X47 + 8X48 + 4X49 + 3X56 + 6X57 + 7X58 + 7X59
1) X14+X15<=450
2) X24+X25<=600
3) X34+X35<380
4) X46+X47+X48+X49-X14-X24-X34=0 5) X56+X57+X58+X59-X15-X25-X35=0 6) X46+X56=300
7) X47+X57=300 8) X48+X58=300 9) X49+X59=400
OPTIMAL SOLUTION
Optimal Objective Value 11850.00000
Variable X14 X15 X24 X25 X34 X35 X46 X47 X48 X49 X56 X57 X58 X59
Constraint 1
Value Reduced Cost 450.00000 0.00000 0.00000 2.00000 0.00000 4.00000 600.00000 0.00000 250.00000 0.00000 0.00000 0.00000 0.00000 2.00000 300.00000 0.00000 0.00000 0.00000 400.00000 0.00000 300.00000 0.00000 0.00000 3.00000 300.00000 0.00000 0.00000 4.00000
Slack/Surplus Dual Value 0.00000 -1.00000 0.00000 -1.00000 130.00000 0.00000 0.00000 9.00000 0.00000 9.00000 0.00000 13.00000 0.00000 9.00000 0.00000 5.00000 0.00000 6.00000
There is an excess capacity of 130 units at plant 3.
Distribution and Network Models
Boston 150
NewYork 100
100 TupperLake Portsmouth
P h il a d el p h ia
¡ê 300 ¡ê 100 = 0 = 0 = 150 = 100 = 150
Min 7x13 + 5x14 + 3x23 + 4x24 + 8x35 + 5x36 + 7x37 + 5x45 + 6x46 + 10x47 s.t.
x13 + x14 -x13
Optimal Solution:
x23 + x24 - x23
+x35 +x36 +x37 - x24
+ x45 + x46 + x47 + x45
x37 xij 3 0 for all i and j
Value x14 250
Objective Function: 4300
x36 0 x37 150
x45 150 x46 100
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