Chapter 7
Transportation, Assignment &
Transshipment Problems Part 1
ISE204/IE252
Prof. Dr. Arslan M. ÖRNEK
Remember from ISE 203:
A Transportation Example
Warehouse supply of
Television Sets:
Retail store demand
for television sets:
1 - Cincinnati
300
A - New York
150
2 - Atlanta
200
B - Dallas
250
3 - Pittsburgh
200
C - Detroit
200
Total
700
Total
600
Unit Shipping Costs:
From Warehouse
1
2
3
A
$16
14
13
To Store
B
$18
12
15
C
$11
13
17
2
Transportation Problem: Characteristics
■ A transportation problem aims to find the best way to fulfill the
demand of n demand points using the capacities of m supply
points.
■ A product is transported from a number of sources to a number
of destinations at the minimum possible cost.
■ Each source is able to supply a fixed number of units of the
product, and each destination has a fixed demand for the
product.
■ The linear programming model has constraints for supply at each
source and demand at each destination.
■ In a balanced transportation model supply equals demand.
3
7.1 Formulating Transportation Problems
How many tons of wheat to transport from each grain elevator to
each mill in order to minimize the total cost of transportation?
Grain Elevator
Supply
1. Kansas City
150
A. Chicago
200
2. Omaha
175
B. St. Louis
100
3. Des Moines
275
C. Cincinnati
300
Total
600 tons
Mill
Total
Demand
600 tons
Transport Cost from Grain Elevator to Mill ($/ton)
Grain Elevator
A. Chicago
B. St. Louis
C. Cincinnati
1. Kansas City
$6
$8
$ 10
2. Omaha
7
11
11
3. Des Moines
4
5
12
4
Transportation Model Example
Transportation Network Routes
5
Figure 6.1 Network of Transportation Routes for Wheat Shipments
Transportation Model Example
Linear Programming Model Formulation
xij = tons of wheat from each grain elevator, i, i = 1, 2, 3,
to each mill j = A,B,C
Minimize Z = $6x1A + 8x1B + 10x1C + 7x2A + 11x2B + 11x2C +
4x3A + 5x3B + 12x3C
subject to:
x1A + x1B + x1C = 150
x2A + x2B + x2C = 175
x3A + x3B + x3C = 275
x1A + x2A + x3A = 200
x1B + x2B + x3B = 100
x1C + x2C + x3C = 300
xij  0
6
Transportation Model Example
Optimal Solution
Figure 6.2 Transportation Network Solution
7
7.1 Formulating Transportation Problems
Example 1:
Powerco has three electric power plants that supply the electric
needs of four cities.
•The associated supply of each plant and demand of each city are
known.
•The cost of sending 1 million kwh of electricity from a plant to a
city depends on the distance the electricity must travel.
• Formulate an LP to minimize cost.
Solution:
Decision Variable:
x14 = Amount of electricity produced at plant 1 and sent to city 4
Since we want to minimize the total cost of shipping from plants to
cities;
Objective Function:
Minimize Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24
+14X31 +9X32 +16X33 +5X34
Solution (cont):
Since each supply point has a limited production capacity;
X11+X12+X13+X14 <= 35
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40
Supply Constraints
Since each demand point has a demand to satisfy;
X11+X21+X31 >= 45
X12+X22+X32 >= 20
Demand Constraints
X13+X23+X33 >= 30
X14+X24+X34 >= 30
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
Sign Constraints
Solution (cont):
LP formulation:
Network representation of Optimal Solution:
General Description of a Transportation Problem
1. A set of m supply points with a supply of at most si
units.
2. A set of n demand points with a demand of at least
dj units.
3. Each unit produced at supply point i and shipped to
demand point j incurs a variable cost of cij.
General Formulation of a Transportation Problem
xij = number of units shipped from supply point i to
demand point j
General Formulation of a Transportation Problem
• If a problem has these constraints and is a
maximization problem, then it is still a transportation
problem.
• If total supply equals to total demand, the problem is
said to be a balanced transportation problem:
General Formulation of a Balanced Transportation
Problem
It is desirable to formulate a problem as a balanced transportation
problem (due to the ease of solution procedures).
Balancing a Transportation Problem if total supply
exceeds total demand
If total supply exceeds total demand, we can balance
the problem by adding a dummy demand point.
Since shipments to the dummy demand point are not
real, they are assigned a cost of zero.
Balancing a TP if total supply exceeds total demand
Example 1: Suppose that in the Powerco problem, the demand for
city 1 were reduced to 40 million kwh.
To balance the Powerco problem, we would add a dummy
demand point (City 5) with a demand of Total Supply – Total
Demand = 125 - 120 = 5 million kwh.
From each plant, the cost of shipping 1 million kwh to the dummy
is 0.
Optimal Solution of the Balanced Powerco Problem
Transportation tableau
A transportation problem is specified by the supply, the demand,
and the shipping costs. So the relevant data can be summarized
in a transportation tableau.
Transportation tableau
Optimal transportation tableau for Powerco:
Balancing a transportation problem if total
supply is less than total demand
If total supply < total demand  The problem has no
feasible solution.
When total supply is less than total demand, it is
sometimes desirable to allow the possibility of leaving
some demand unsatisfied. In such a situation, a penalty
is often associated with unmet demand.
• Example 2 (Handling Shortages):
23
• Solution:
24
25
Sailco Corporation must determine how many sailboats should be
produced during each of the next four quarters (one quarter is
three months).
Demand : first quarter, 40 sailboats; second quarter, 60 sailboats;
third quarter, 75 sailboats; fourth quarter, 25 sailboats.
Sailco must meet demand on time. At the beginning of the first
quarter, Sailco has an inventory of 10 sailboats.
At the beginning of each quarter, Sailco must decide how many
sailboats should be produced during the current quarter.
For simplicity, we assume that sailboats manufactured during a
quarter can be used to meet the demand for that quarter.
During each quarter, Sailco can produce up to 40 sailboats at a
cost of $400 per sailboat. By having employees work overtime
during a quarter, Sailco can produce additional sailboats at a cost
of $450 per sailboat.
At the end of each quarter (after production has occurred and the
current quarter’s demand has been satisfied), a carrying or
holding cost of $20 per sailboat is incurred.
Formulate a balanced transportation problem to minimize the
sum of production and inventory costs during the next four
quarters.
Capacity of each OT supply point = 150 =
200 (Total demand) – 10 (initial
inventory) – 40 (regular time production
capacity)
7.2 Finding Basic Feasible Solution for TP
A balanced TP with m supply points and n demand
points is easier to solve than a regular LP, although it
has m + n equality constraints.
If a set of values for the xij’s satisfies all but one of the
constraints of a balanced transportation problem, then
the values for the xij’s will automatically satisfy the
other constraint.
This means that only m+n-1 constraints are linearly
independent.  m+n-1 basic variables
Methods to find the bfs for a balanced TP
There are three basic methods:
1. Northwest Corner Method
2. Minimum Cost Method
3. Vogel’s Method
1. Northwest Corner Method (NWC)
To find the bfs by the NWC method:
Begin in the upper left (northwest) corner of the
transportation tableau and set x11 as large as
possible (the limitations for setting x11 will be the
demand of demand point 1 and the supply of supply
point 1. Your x11 value can not be greater than the
minimum of this two values).
Example: Set x11=3 (meaning demand of demand
point 1 is satisfied by supply point 1).
5
6
2
3
5
2
3
3
2
6
2
X
5
2
3
35
After we check the east and south cells, we see
that we can go east (meaning supply point 1 still
has capacity to fulfill some demand).
3
2
X
6
2
X
3
2
3
3
2
X
3
3
2
X
X
2
3
36
After applying the same procedure, we see that we
can go south this time (meaning demand point 2
needs more supply by supply point 2).
3
2
3
X
2
1
2
X
X
3
2
3
X
3
X
2
1
X
2
X
X
X
2
37
Finally, we will have the following bfs:
x11=3, x12=2, x22=3, x23=2, x24=1, x34=2
3
2
3
X
X
X
2
X
1
X
2
X
X
38
Example:
39
Example:
Supply and
demand
equal, cross
only one,
not both!
40
Example:
Degenerate
solution
1
m+n-1 =
3+4-1=6
basic
variables
41
2. Minimum Cost Method
The Northwest Corner Method does not utilize
shipping costs. It can yield an initial bfs easily but the
total shipping cost may be very high.
The minimum cost method uses shipping costs in
order come up with a bfs that has a lower cost. To
begin the minimum cost method, first we find the
decision variable with the smallest shipping cost.
Then assign xij its largest possible value, which is the
minimum of si and dj.
Cross out the row or column, then continue with the
next minimum cost.
Example: Step 1: Select the cell with minimum cost.
2
3
5
6
5
2
1
3
5
10
3
12
8
8
4
4
6
6
15
Step 2: Cross-out column 2
6
5
3
2
5
5
3
1
2
2
8
12
X
6
4
8
3
4
15
6
44
Step 3: Find the new cell with minimum shipping
cost and cross-out row 2
2
3
5
6
5
2
1
3
5
X
2
8
3
10
8
X
4
4
6
15
6
45
Step 4: Find the new cell with minimum shipping
cost and cross-out row 1
2
3
5
6
X
5
2
1
3
5
X
2
8
3
5
8
X
4
4
6
15
6
46
Step 5: Find the new cell with minimum shipping
cost and cross-out column 1
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
4
6
10
5
X
X
4
6
47
Step 6: Find the new cell with minimum shipping
cost and cross-out column 3
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
6
6
4
X
X
X
6
48
Step 7: Finally assign 6 to last cell. The bfs is found
as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
4
X
X
6
X
6
X
X
49
3. Vogel’s Method
Begin with computing each row and column a penalty.
The penalty will be equal to the difference between the
two smallest shipping costs in the row or column.
Identify the row or column with the largest penalty.
Find the first basic variable which has the smallest
shipping cost in that row or column. Then assign the
highest possible value to that variable, and cross-out
the row or column as in the previous methods.
Compute new penalties and use the same procedure.
Example: Step 1: Compute the penalties.
6
7
15
Demand
Column Penalty
Supply
Row Penalty
10
7-6=1
15
78-15=63
8
80
78
15
5
5
15-6=9
80-7=73
78-8=70
51
Step 2: Identify the largest penalty and assign the
highest possible value to the least-cost variable.
6
7
Supply
Row Penalty
5
8-6=2
15
78-15=63
8
5
15
Demand
Column Penalty
80
78
15
X
5
15-6=9
_
78-8=70
52
Step 3: Identify the largest penalty and assign the
highest possible value to the variable.
6
7
5
Column Penalty
Row Penalty
0
_
15
_
8
5
15
Demand
Supply
80
78
15
X
X
15-6=9
_
_
53
Step 4: Identify the largest penalty and assign the
highest possible value to the variable.
6
0
7
5
Supply
Row Penalty
X
_
15
_
8
5
15
80
78
Demand
15
X
X
Column Penalty
_
_
_
54
Step 5: Finally the bfs is found as X11=0, X12=5, X13=5,
and X21=15
6
0
7
5
Supply
Row Penalty
X
_
X
_
8
5
15
80
78
15
Demand
X
X
X
Column Penalty
_
_
_
55