2. An Assignment problem is a particular case of a
transportation problem where the sources are assignees and
the destination are tasks.
Each assignee is to be assigned exactly one task
Such kind of problem of assignment arises because the
resources that are such as men, machine, etc, have varying
degree of efficiency for performing different activities(these
have to be judges on the basis of time, cost or profit which
also different for different task)
The problem is : How should the assignments be made so as to
optimize the given objective?
Some of the assignment problem which we are going to face
off : Workers to Machine, Job Seeker to Job Providers,
Salesman to Different sales area,Clerks to various check
counters, classes to rooms, vehicle to routes, Investment to
returns, contract to bidders, Flight Crews Scheduling etc.
Compiled By : Prof Amit Kumar
3. a) Enumeration Method : In this method , a list
of all possible assignment among the given
resources (men, machine etc.) and
activities(jobs, sales area, worker etc) is
prepared. If there are nth worker/jobs , then
there are n! possible assignment
b) Simplex Method : In this technique,
assignment can be formulated as 0 and 1
Integer Programming Problem, by simplex
table.( A complex one solve manually)
Compiled By : Prof Amit Kumar
4. Transportation Problem : An assignment
problem can be form into transportation
problem. But here only n assignment possible,
which make the assignment problem
degenerate.(so it’s not a good technique at all
for assignment cases)
Hungarian Method* : It is developed by
Hungarian Mathematician D.Konig ; its
provide us with an efficient method of finding
the optimal solution , without having direct
comparison. It works on the principle of
reducing the given cost matrix(opportunity
loss) to a matrix of opportunity cost.
Compiled By : Prof Amit Kumar
5. The project work consists of four major jobs for
which an equal number of contractors have
submitted tenders. The tenders amount quoted
(in lakh of rupees) is given in the matrix.
Find the assignment which minimizes the total cost
of the project when each contractor has to be
assigned at least one job by hungarian method.
Compiled By : Prof Amit Kumar
Jobs
A B C d
Contrac
tors
C1 10 24 30 15
C2 16 22 28 12
C3 12 20 32 10
C4 9 26 34 16
6. Step 1 : No. of
Assignee (row)= No.
of Task (Column)
Step 2 : Check
whether the case
minimize or
maximize**
Step 3 : Row
Reduction
Step 4 : Column
Reduction-After we
get Opportunity
Cost table
Step 5 : Now
eliminate(cut) Zero
by horizontal or
vertical lines
Compiled By : Prof Amit Kumar
7. Step 1 : No. of row = No. of column ;
if not we have to introduce dummy
make it equal(all values of dummy is
Zero)
Step 2 : Case of Minimization can be
directly start using step 3 ; But in case
of maximization cases one more step
involved that is step2A in which
highest value from the whole table
is get subtracted so the opportunity
loss table made
Compiled By : Prof Amit Kumar
8. Step 3 : Row Reduction (choosing minimum
from each row and then subtracted from their
respective row)
Step 4 : Column Reduction (after Row
reduction from the reduced table , choosing
minimum from each column and then
subtracted from their respective column)
Step 5 : Now zero elimination by using
vertical or horizontal lines (maximum lines
upto = max. Row = max. Column)(Try to use
less lines to cut)
Step 6** : Now a moment when all lines(max
lines) were used , we called the moment of
Optimal solution/assignment, assign square
boxes to each zero so that assignment process
can start.
Compiled By : Prof Amit Kumar
9. Problem 2 : A construction company has requested
bids for subcontracts on five different projects.
Five companies have responded and their Bids
are represented below :
Determine the minimum cost assignment of
subcontracts to bidders , assuming that each
bidder can receive only one contract (Answer :
Rs.1,55,000)
Compiled By : Prof Amit Kumar
Bid Amount (‘000s Rs)
I II III IV V
B
I
D
D
E
R
S
1 41 72 39 52 25
2 22 29 49 65 81
3 27 39 60 51 40
4 45 50 48 52 37
5 29 40 45 26 30
10. Problem 3 : In the modification of a plant layout of a factory four new
machines M1, M2, M3 and M4 are to be installed in a machine shop. There
are five vacant places A,B,C,D and E available. Because of limited space,
machine M2 cannot be placed at C and M3 cannot be placed at A. The Cost
of locating the machine at a place (in 000’s INR) is as follow :
Find the optimal Assignment schedule and which location will remain vacant.
(Rs.32000)
Compiled By : Prof Amit Kumar
Location
A B C D E
M
A
C
H
INE
M1 9 11 15 10 11
M2 12 9 --- 10 9
M3 --- 11 14 11 7
M4 14 8 12 7 8
11. Steps Involved :
Reduce the whole table by the highest value ,
then we get opportunity loss table.
After getting opportunity loss table, if there is
dummy required then its must be provided,
thereafter we will apply Hungarian method.
On assigning the task from the optimal table,
pick up the value from the original table
Compiled By : Prof Amit Kumar
