The document describes a transportation model for optimally allocating shipments from multiple origins to destinations. It includes requirements like origin capacities and destination demands, assumptions of homogeneous items and fixed shipping costs. A scenario and table illustrate origins, destinations, costs and supplies/demands. Linear programming formulations and the North-West Corner Method are presented to solve transportation problems and minimize costs.

1. 2-1
Transportation Model

2. TRANSPORTATION MODELTRANSPORTATION MODEL
 Requirements -
 List of origins and each one’s capacity.
 List of destinations and each one’s demand.
 Unit cost of shipping.
 Assumptions -
 Items to be shipped are homogeneous.
 Shipping cost per unit is the same.
 Only one route between origin and destination.
 Demand and supply is equal.

3. TRANSPORTATION MODEL SCENARIOTRANSPORTATION MODEL SCENARIO
D
(demand)
B
(demand)
C
(demand)
A
(demand)
1
(supply)
2
(supply)
3
(supply)

4. A TRANSPORTATION TABLEA TRANSPORTATION TABLE
Warehouse
4 7 7 1
100
12 3 8 8
200
8 10 16 5
150
450
450
80 90 120 160
A B C D
1
2
3
Factory
Factory 1
can supply 100
units per
period
Total
supply
capacity
per
period
Total demand
per period
Demand
Warehouse B can use 90
units per period

5. NETWORK PRESENTATIONNETWORK PRESENTATION
• Transportation problem can be represented as a network. Circles
represent origins and destinations, and the arcs between them
represent the decision variables, i.e. the amounts shipped.
Supply
Origin
Demand
Destination
D1
D2
D3
4
10
6
8
16
6
14 18
10
100
300
300
200
300
200
S1
S2
S3

6. LINEAR PROGRAMMING FORMULATIONLINEAR PROGRAMMING FORMULATION
• Let xij be unknown number of units shipped from origin i to
destination j (i,j = 1,2,3)
x11 + x21 + x31 > 200
x12 + x22 + x32 > 300
x13 + x23 + x33 > 200
Demand constraintsSupply constraints
x11 + x12 + x13 < 100
x21 + x22 + x33 < 300
x31 + x32 + x33 < 300
min z = 4x11 + 10x12 + 6x13
+ 8x21 + 16x22 + 6x23
+ 14x31 + 18x32 + 10x33
st.
xij > 0
Non-negative constraints

7. TRANSPORTATION PROBLEMTRANSPORTATION PROBLEM
Des Moines
(100 units)
capacity
Cleveland
(200 units)
required
Boston
(200 units)
required
Evansville
(300 units)
capacity
Ft. Lauderdale
(300 units)
capacity
Albuquerque
(300 units)
required

8. From
(Sources)
To
(Destinations)
Albuquerque Boston Cleveland
Des Moines
Evansville
Fort Lauderdale
$5
$8
$9
$4
$4
$7
$3
$3
$5
TRANSPORTATION PROBLEMTRANSPORTATION PROBLEM

9. Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque
(A)
Boston
(B)
Cleveland
(C)
Factory
Capacity
5 4 3
3
57
48
9
UNIT SHIPPING COST: 1 UNIT, FACTORY TOUNIT SHIPPING COST: 1 UNIT, FACTORY TO
WAREHOUSEWAREHOUSE

10. Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque
(A)
Boston
(B)
Cleveland
(C)
Factory
Capacity
300 200 200 700
300
300
100
TOTAL DEMAND & TOTAL SUPPLYTOTAL DEMAND & TOTAL SUPPLY

11. Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque
(A)
Boston
(B)
Cleveland
(C)
Factory
Capacity
300 200 200 700
300
300
100
5 4 3
3
57
48
9
TRANSPORTATION TABLETRANSPORTATION TABLE

12. • Start in the upper left-hand cell and allocate units to shipping
routes as follows:
– Exhaust the supply (factory capacity) of each row before moving
down to the next row.
– Exhaust the demand (warehouse) requirements of each column
before moving to the next column to the right.
– Check that all supply and demand requirements are met.
NORTH - WEST CORNER METHODNORTH - WEST CORNER METHOD

13. Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque
(A)
Boston
(B)
Cleveland
(C)
Factory
Capacity
300 200 200 700
300
300
100
5 4 3
3
57
48
9
100
200 100
100 200
NORTH - WEST CORNER METHOD (CONTD.)NORTH - WEST CORNER METHOD (CONTD.)
Total = $4,200

14. Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque
(A)
Boston
(B)
Cleveland
(C)
Factory
Capacity
2000
700
800
500
5 4 3
3
57
48
9
500
600 200
400 300
400 900 700
Total Cost: 10,200
LEAST COST METHODLEAST COST METHOD

15. Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque
(A)
Boston
(B)
Cleveland
(C)
Factory
Capacity
2000
700
800
500
5 4 3
3
57
48
9
100
800
400 900 700
Total Cost: 9,100
400
700
SAME EXAMPLE WITH NWMSAME EXAMPLE WITH NWM

16. EXCEL TEMPLATEEXCEL TEMPLATE