Transportation
and
Assignment Models
Transportation Models
- Transportation problem involve determining how to
optimally (minimum shipping cost) transport goods
from the sources to destinations.
- Usually we are given the capacity of goods at
each source and the requirements at each
destination
- Typically the objective is to minimize total
transportation and production costs
• Consider the case given below. There are
sources (Albuquerque, Boston and Cleveland)
and there are destinations (Des Moines
Evansville and Fort Lauderdale) with costs,
and constraints for capacity and demand.
Shipping Cost ($) per truckload
To
From
Des Moines
Evansville
Fort Lauderdale
Demand
Albuquerque
5
8
9
300
Boston
4
4
7
200
Minimize Z =5x11 + 4x12 + 3x13 + 8x21 + 4x22 + 3x23 + 9x31 + 7x32 + 5x33
Constraints:
For Supply nodes;
x11 + x12 + x13  100
x21 + x22 + x23  300
x31 + x32 + x33  300
Supply cannot be bigger than capacity.
For Demand nodes;
x11 + x21 + x31 ≥ 300
x12 + x22 + x32 ≥ 200
x13 + x23 + x33 ≥ 200
Demand cannot be less than the required.
all xij and xji ≥0
Cleveland
3
3
5
200
Capacity
100
300
300
700
Basic feasible solution can be obtained by three
methods, they are
• North - west corner method,
• Least - cost cell method,
• Vogel's Approximation Method, generally
known as VAM.
After getting the basic feasible solution, do
optimality test to check whether the solution is
optimal or not.
There are two methods of doing optimality test,
they are
• Stepping Stone Method,
• Modified Distribution Method, generally
known as MODI method.
Properties of a Basic feasible
Solution
• It must satisfy availability constraints and
requirement constraints.
• It should satisfy non negativity constraint.
• Total number of allocations must be equal to
(m + n – 1), where 'm' is the number of rows
and 'n' is the number of columns.
Initial Solution with NORTH-WEST
corner method
Albuquerque
Des Moines
Evansville
Fort
Lauderdale
Demand
300
Boston
Cleveland
5
4
3
8
4
3
9
7
5
200
200
Capacity
100
300
300
700
Step1
Start from the left hand side top corner or cell and make
allocations depending on the availability and requirement
constraint.
Step 2
By satisfying availability and requirement constraints fill the
other cells.
Z=1005 + 2008+1004+1007+2005=500+1600+400+700+1000=4200 $
Initial Solution with LEAST COST
method
• Identify the lowest cost cell in the given matrix. In this particular example it
is 3. Make allocations to this cell. After filling search for lowest cost cell.
Proceed this way until all allocations are made.
Z=3009+2004+1003+1003=2700+800+300+300=4100 $
Initial Solution with Vogel’s
Approximation Method (VAM)
• Step 1: For each row and column find the difference between
the two lowest unit shipping costs.
Step 2
Assign as many units as possible to the lowest-cost square in the
row and column selected.
Step 3
Eliminate the column or row that has been satisfied.
Z= 1005+2004+1003+2009+1005=500+800+300+1800+500=3900 $
STEPPING STONE METHOD of
optimality test
Once, we get the basic feasible solution for a transportation problem,
the next duty is to test whether the solution we got is an optimal
solution or not?
• Steps to test unused squares;
• Select an unused square,
• Allocate +①unit to unused square and locate -① and +①
alternatively to corners of the selected closed path,
• Calculate the improvement index ,
• If improvement index is negative allocate as much as you can to that
unused square,
• Repeat the allocation till the improvement index is ≥0 for all unused
squares.
Application of the Method:
Since there is a decrease in the cost, we will allocate as much as we can to
Fort Lauderdale – Albuquerque. The amount is the minimum of the
numbers that we are assigning -① in the cycle.
Z=1005 + 1008+2004+1009+2005=500+800+800+900+1000=4000 $
The solution obtained may or may not be optimal. To check we return to first step to check the unused squares,
D to B
D to C
E to C
F to B
IDB
IDC
IEC
IFB
+DB-DA+AE-EB
+DC-DA+FA-FC
+EC-EA+FA-FC
+FB-EB+EA-FA
+4-5+8-4
+3-5+9-5
+3-8+9-5
+7-4+8-9
+3
+2
-1
+2
An improvement can be done by allocating to EC, since the minimum number in the
square is 100; we allocate 100 to EC square.
Z=1005 + 2009+2004+1003+1005=500+1800+800+300+500=3900 $
We will check the Improvement index in each cell again.
D to B
D to C
E to A
F to B
IDB
IDC
IEC
IFB
+DB-DA+FA-FC+EC-EB
+DC-DA+FA-FC
+EA-FA+FC-EC
+FB-EB+EC-FC
+4-5+9-5+3-4
+3-5+9-5
+8-9+5-3
+7-4+3-5
+2
+2
+1
+1
In the table, no more negative improvement index, so the solution is optimal.
Example
A company has three factories X, Y, and Z and three warehouses A, B, and C. It
is required to schedule factory production and shipments from factories to
warehouses in such a manner so as to minimize total cost of shipment and
production. Unit variable manufacturing cost (UVMC) and factory capacities and
warehouse requirements are given below:
From / To
A
B
C
Capacity
10
4
11
12
5
8
9
7
6
70
X
50
Y
Z
Demand
40
50
60
30
150
a) Load the table with North-west method.
b) Load the table with least cost method.
c) Load the table with Vogel’s approximation method, (VAM).
d) Solve the question by stepping stone algorithm.
Example
The demand and capacity are given.
From / To
Sarajevo
Mostar
Zenica
Tuzla
Demand
210
Travnik
Bihac
Capacity
5
7
15
4
2
8
6
3
10
160
100
a) Load the table with North-west method.
b) Load the table with least cost method.
c) Load the table with Vogel’s approximation method, (VAM).
d) Solve the question by stepping stone algorithm.
120
200
150
470
Assignment Model
• Allocating one person to one job and minimizing the
time or cost of problem is solved by ASSIGNMENT
MODEL.
• The assignment problem refers to the class of LP
problems that involve determining the most efficient
assignment of resources to tasks
• The objective is most often to minimize total costs or
total time to perform the tasks at hand
Consider the following assignment case;
Three repair person Adams, Brown and Cooper are tried to be assigned to repair a radio(1), a tooster oven (2) and a broken coffee
table(3) by the owner of the repair shop. The owner of the shop try to minimize the cost of assignment.
PERSON
Adams
Brown
Cooper
1
11
8
9
PROJECT($)
2
14
10
12
3
6
11
7
1
Adams
Adams
Brown
PROJECT ASSIGNMENT
2
3
Labor Cost
Brown Cooper
11+10+7
Cooper Brown
11+12+11
Adams Cooper
8+14+7
Brown
Cooper
Cooper
Cooper
Adams
Brown
Adams
Brown
Adams
8+12+6
9+14+11
9+10+6
Total Cost
28
34
29
26
34
25
There are n!=4!=432 1=24 number of solutions. If there is 5 assignment
there will be 5!=120 different combinations.Therefore , instead of writing all the
combinations, an algorithm (Hungarian Method) will be used to find the
solution.
Hungarian Method
• The Hungarian method is an efficient method of finding the
optimal solution to an assignment problem without having to
make direct comparisons of every option
• It operates on the principle of matrix reduction
• By subtracting and adding appropriate numbers in the cost table
or matrix, we can reduce the problem to a matrix of opportunity
costs
• Opportunity costs show the relative penalty associated with
assigning any person to a project as opposed to making the best
assignment
• We want to make assignment so that the opportunity cost for
each assignment is zero
• There are five steps to find the optimum
result.
Step1. Subtract the smallest element in each row from the other elements of the row.
PERSON
Adams
Brown
Cooper
1
11
8
9
PROJECT
2
14
10
12
3
6
11
7
...subtract 6 
...subtract 8 
...subtract 7 
1
5
0
2
PROJECT
2
8
2
5
3
0
3
0
Step2. Subtract the smallest element in each column from the other elements of the column.
PERSON
Adams
Brown
Cooper
1
5
0
2
subtract 0
PROJECT
2
8
2
5
subtract 2
3
0
3
0
subtract 0

1
5
0
2
PROJECT
2
6
0
3
3
0
3
0
Step3. Draw minimum number of lines to cover the zeros. If the lines drawn are equal to the number of rows or columns
(because of square matrix), we can make assignment.
1
5
0
2
PROJECT
2
6
0
3
3
0
3
0
Number of lines is 2 which is less than number of rows or column. So solution is not
optimum.
Step4. If the lines drawn are less than the number of rows or columns,
then we cannot make assignment. Hence the following procedure is to be
followed:
- The cells covered by the lines are known as Covered cells.
- The cells, which are not covered by lines, are known as uncovered
cells.
- The cells at the intersection of horizontal line and vertical lines are
known as Crossed cells.
(a) Identify the smallest element in the uncovered cells.
(i) Subtract this element from the elements of all other uncovered
cells.
(ii) Add this element to the elements of the crossed cells.
(iii) Do not alter the elements of covered cells.
(b) Once again cover all the zeros by minimum number of horizontal
and vertical lines.
(c) Once the lines drawn are equal to the number of rows or columns,
assignment can be made.
Step5. Assigning the zeros: Once you find a single zero in the first row; assign that cell by enclosing the element of the cell by a
square and cancel the zeros corresponding column. Continue this procedure until all assignments are made. Sometimes we may
not find single zero and find more than one zero in a row or column. It indicates that the problem has an alternate solution. We can
write alternate solutions.
PROJECT
PROJECT
1
2
3
1
3
0
0
4
0
1
0
5
0
3
0
0

PROJECT
3
1
2
3
4
0
1
0
5
0
3
0
0
4
0
1
0
5
0

(ii)
(i)
PROJECT
2

(iii)
So corresponding costs are,
PERSON
Adams
Brown
Cooper
1
$11
8
9
PROJECT
2
$14
10
12
3
$6
11
7
So the total cost is 6+10+9=$25.
1
2
3
3
0
0
4
0
1
0
5
0
(iv)
Example
There are 3 jobs A, B, and C and three machines X, Y, and Z. All
the jobs can be processed on all machines. The time required for
processing job on a machine is given below in the form of matrix.
Make allocation to minimize the total processing time.
Machines (time in hours)
Jobs
A
X
11
Y
16
Z
21
B
20
13
17
C
13
15
12