Linear Programming Model _Excel Solution Module 5 P5-18_P5-26_P5-30_Answer

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 5-26 A trauma center keeps ambulances at locations throughout the east side of a city in an attempt to mini- mize the response time in the event of an emergency. The times, in minutes, from the ambulance locations to the population centers are POPULATION CENTERS AMBULANCE 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 Find the optimal assignment of ambulances to popu- lation centers that will minimize the total emergency response time. 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 =

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Linear Programming Model _Excel Solution Module 5 P5-18_P5-26_P5-30 Complete A+ Answer