# Linear programming model excel solution module 5 p5 18 p5 26 p5 30

Problem 5-18 (a) 5-18 A food distribution company ships fresh spinach from its four packing plants to large East-coast cities. The shipping costs per crate, the supply and demand are shown in the table at the bottom of this page. (a) Formulate a model that will permit the company to meet its demand at the lowest possible cost. (b) The company wishes to spread out the source for each of its markets to the maximum extent possible. To accomplish this, it will accept a 5% increase in its total transportation cost from part (a). What is the new transportation plan, and what is the new cost? Shipments: To Flow balance equations From Atlanta Boston Charlestown Dover Flow out Location Flow in Flow out Net flow Sign RHS Eaglestown Eaglestown = -8000 Farrier Farrier = -10000 Guyton Guyton = -5000 Hayesville Hayesville = -9000 Flow in Atlanta = 8000 Boston = 9000 Unit costs: To Charlestown = 10000 From Atlanta Boston Charlestown Dover Dover = 5000 Eaglestown \$6.0 \$7.0 \$7.5 \$7.5 Farrier \$5.5 \$5.5 \$4.0 \$7.0 Guyton \$6.0 \$5.0 \$6.5 \$7.0 Hayesville \$7.0 \$7.5 \$8.5 \$6.5
Total cost =
Problem 5-18 (b) 5-18 A food distribution company ships fresh spinach from its four packing plants to large East-coast cities. The shipping costs per crate, the supply and demand are shown in the table at the bottom of this page. (a) Formulate a model that will permit the company to meet its demand at the lowest possible cost. (b) The company wishes to spread out the source for each of its markets to the maximum extent possible. To accomplish this, it will accept a 5% increase in its total transportation cost from part (a). What is the new transportation plan, and what is the new cost?
Shipments: To Flow balance equations From Atlanta Boston Charlestown Dover Flow out Location Flow in Flow out Net flow Sign RHS Eaglestown Eaglestown = -8000 Farrier Farrier = -10000 Guyton Guyton = -5000 Hayesville Hayesville = -9000 Flow in Atlanta = 8000 Boston = 9000 Unit costs: To Charlestown = 10000 From Atlanta Boston Charlestown Dover Dover = 5000 Eaglestown \$6.0 \$7.0 \$7.5 \$7.5 Farrier \$5.5 \$5.5 \$4.0 \$7.0 5% Allowable Cost Increase < = 184275 Guyton \$6.0 \$5.0 \$6.5 \$7.0 Hayesville \$7.0 \$7.5 \$8.5 \$6.5 Max ship = Problem P 5-26 LOCATIONS EAST NORTH- EAST SOUTH- EAST CENTRAL Site 1 12 8 9 13 Site 2 10 9 11 10 Site 3 11 12 14 11 Site 4 13 11 12 9 Ambulance Locations Population Centers Flow balance equations East North-East South-East Central Flow out Location Flow in Flow out Net flow Sign RHS Site 1 0.0 0.0 1.0 0.0 1.0 Site 1 0.0 1 -1 = -1 Site 2 0.0 1.0 0.0 0.0 1.0 Site 2 0.0 1 -1 = -1 Site 3 1.0 0.0 0.0 0.0 1.0 Site 3 0.0 1 -1 = -1 Site 4 0.0 0.0 0.0 1.0 1.0 Site 4 0 1 -1 = -1 Flow in 1.0 1.0 1.0 1.0 Site 1 1 1 = 1 Site 2 1 1 = 1 Ambulance Locations Population Centers Site 3 1 1 = 1 East North-East South-East Central Site 4 1 1 = 1 Site 1 12.0 8.0 9.0 13.0 Site 2 10.0 9.0 11.0 10.0 Site 3 11.0 12.0 14.0 11.0 Site 4 13.0 11.0 12.0 9.0 Total Emergency Response Time 38 Problem 5-30 The city of Six Mile, South Carolina, is considering making several of its streets one way. What is the maximum number of cars per hour that can travel from east (node 1) to west (node 8)? The network is shown in Figure 5.20. ……………………………………………………………………………………………. Flows: To Flow balance equations From Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 Node 7 Node 8 Flow out Node Flow in Flow out Net flow Sign RHS Node 1 Node 1 = 0.0 Node 2 Node 2 = 0.0 Node 3 Node 3 = 0.0 Node 4 Node 4 = 0.0 Node 5 Node 5 = 0.0 Node 6 Node 6 = 0.0 Node 7 Node 7 = 0.0 Node 8 Node 8 = 0.0 Flow in Capacities: To From Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 Node 7 Node 8 Node 1 0 2 5 1 0 0 0 0 Node 2 0 0 0 0 2 0 0 0 Node 3 0 0 0 0 2 2 0 0 Node 4 1 0 0 0 0 3 0 4 Node 5 0 2 1 0 0 0 2 0 Node 6 0 0 2 0 0 0 0 4 Node 7 0 0 0 0 2 0 0 2 Node 8 1000 0 0 0 0 0 2 0 Max flow =