代写代考 17. a. Network Model

17. a. Network Model
Supply 450
7447 W18 C2
The linear programming formulation and solution are shown.

Copyright By PowCoder代写 加微信 powcoder

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 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com