Operational Research assignment problem

Assignment problems
Operational Research-
Level 4
Prepared by T.M.J.A.Cooray
Department of Marthematics

Introduction
 This is a special type of transportation
problem in which each source should have
the capacity to fulfill the demand of any of
the destinations.
 In other words any operator would be able
perform any job regardless of his
skills,although the cost( or the time taken)
will be more if the job does not match with
operator’s skill.

MA 402--assignment problem 2

General format of assignment problem
Let m be the number of jobs as well as the operators,
and tij be the processing time of the job i if it is assigned
to the operator j. Here the objective is to assign the jobs
to the operators such that the total processing time is
minimized.
Operators
1 2 … j … m
1 t11 t12 t1j t1m
2
Job .
i ti1 tij tim
.
m tm1 t t
MA 402--assignment problemmj
m2 tmm 3

 Examples of assignment problem

Row entity Column entity Cell entity
jobs operators Processing time
Programmer program Processing time

operators machine Processing time

Drivers Routes Travel time
Teachers Subjects Students pass
percentage

Assignment problem as a zero-one
( Binary) programming problem .
 Min Z= c11x11++cijXij+.+cmmXmm = m m
Min Z = ∑∑ Cij X ij
 Subject to x11+…………...+x1m =1 i =1 j =1

x21+…………...+x2m =1 m

∑X ij = 1 for i = 1,....m
…….. j =1

xm1+…………...+xmm =1
m
x11+…………...+xm1 =1 ∑X ij = 1 for j = 1,....m
i =1
x12+…………...+xm2 =1
………………..
x1m+…………...+xmm =1
xij.=0 or 1 for i=1,2….m and
j=1,2…..m. MA 402--assignment problem 5

Types of assignment problems

 As in transportation problems assignment
problems also can be balanced ( with equal
number of rows and columns) or unbalanced.
 When it is unbalanced the necessary number
of row/s or column/s are added to balance it.
That is to make a square matrix.
 The values of the cell entries of the dummy
rows or columns will be made equal to zero.


Example : Assign the 5 operators to the 5
jobs such that the total processing time is
minimized.

Operator 1 2 3 4 5
job

1 10 12 15 12 8
2 7 16 14 14 11
3 13 14 7 9 9
4 12 10 11 13 10
5 8 13 15 11 15


Hungarian method
 Consists of two phases.
 First phase: row reductions and column
reductions are carried out.
 Second phase :the solution is optimized in
iterative basis.


Phase 1: Row and column
reductions
 Step 0: Consider the given cost matrix
 Step 1: Subtract the minimum value of
each row from the entries of that row, to
obtain the next matrix.
 Step 2: Subtract the minimum value of
each column from the entries of that
column , to obtain the next matrix.
 Treat the resulting matrix as the input
for phase 2.


Phase 2: Optimization
 Step3: Draw a minimum number of lines to cover
all the zeros of the matrix.
 Procedure for drawing the minimum number of
lines:
 3.1 Row scanning

1 Starting from the first row ,if there’s only one zero
in a row mark a square round the zero entry and
draw a vertical line passing through that zero.
Otherwise skip the row.
2.After scanning the last row, check whether all the
zeros are covered with lines. If yes go to step 4.
Otherwise do column scanning. Ctd

 3.2 Column scanning.
1. Starting from the first column: if there’s
only one zero in a column mark a
square round the zero entry and draw a
horizontal line passing through that
zero. otherwise skip the column.
2.After scanning the last column, check
whether all the zeros are covered with
lines. If yes go to step 4. Otherwise do
row scanning. ctd 


 Step 4: check whether the number of squares
marked is equal to the number of rows/columns of
the matrix.
 If yes go to step 7. Otherwise go to step 5.
 Step 5: Identify the minimum value of the undeleted
cell values ,say ‘x’. Obtain the next matrix by the
following steps.
5.1 Copy the entries covered by the lines ,but not on
the intersection points.
5.2 add x to the intersection points
5.3 subtract x from the undeleted cell values.
Step 6: go to step 3.
Step 7: optimal solution is obtained as marked by the
squares MA 402--assignment problem 12

Maximization problem
 If the problemis a maximization
problem ,convert the problem into a
minimization problem by multiplying by -1.
 Then apply the usual procedure of an
assignment problem.


Example : Assign 4 sales persons to four
different sales regions such that the total
sales is maximized.
Sales
region
1 2 3 4

Sales person

1 10 22 12 14

2 16 18 22 10

3 24 20 12 18

4 16 14 24 20

Modified data , after multiplying the cell
entries by -1.

Sales
region
1 2 3 4

Sales person

1 -10 -22 -12 -14

2 -16 -18 -22 -10

3 -24 -20 -12 -18

4 -16 -14 -24 -20

After step 1

Sales
region
1 2 3 4

Sales person

1 12 0 10 8

2 6 4 0 12

3 0 4 12 6

4 8 10 0 4

After step 2

Sales
region
1 2 3 4

Sales person

1 12 0 10 4

2 6 4 0 8

3 0 4 12 2

4 8 10 0 0

Phase 2

Sales
region
1 2 3 4

Sales person

1 12 0 10 4

2 6 4 0 8

3 0 4 12 2

4 8 10 0 0

 Note that the number of squares is equal to the
number of rows of the matrix. solution is feasible
and optimal.
 Result: Salesman Sales region Sales
1 2 22
2 3 22
3 1 24
4 4 20


Branch and Bound algorithm for the assignment
problem
 Terminology:
 K-level number in the branching tree
 For root node k=0
 σ-assignment made in the current node of a branching tree
 Pσk –assignment at level k of the branching tree
 A-set of assigned cells up to the node Pσk from the
root node
 Vσ - lower bound of partial assignment A up to P σk
 
∑Cij + ∑ ∑min C

i∈X  j∈


 Such that Vσ = i , j∈A Y 


 Cij is the cell entity of the cost matrix
 X rows which are not deleted up to node P σk
from the root node in the branching tree.
 Y columns which are not deleted up to node
P σk
from the root node in the branching tree.



Branching guidelines
 1.At level k,the row marked as k of the
assignment problem,will be assigned with
the best column of the assignment problem.
 2.if there is a lower bound ,then the terminal
node at the lower most level is to be
considered for further branching
 3.stopping rule:if the minimum lower bound
happens to be at any one of the terminal
nodes at the (n-1)th level ,the optimality is
reached.

Example : Assign the 5 operators to the 5
jobs such that the total processing time is
minimized.
Operator 1 2 3 4 5
job

1 10 12 15 12 8

2 7 16 14 14 11

3 13 14 7 9 9

4 12 10 11 13 10

5 8 13 15 11 15


Pφ0

P111 51 P121 44 P131 49 P14 1 44 P151 40
1
lower bound for P 11

σ = {(11)}, A = {(11)}, X = {2,3,4,5}, Y = {2,3,4,5}

V11 =c11 + ∑( ∑ C
min
i = , 3, 4 , 5
2 j = , 3, 4 , 5
2
ij )

=10 +(11 +7 + + ) =49
10 11

Pφ0

P111 49 P12 1 44 P131 49 P141 44 P151 40

P212 43 P222 50 P232 49 P242 47
2
lower bound for P21
σ = {( 21)}, A = {( 21), (15)}, X = {3,4,5}, Y = {2,3,4}
V21 =15 +21 +
c c ∑∑ Cij )
( min
i = 4,5
3, j =, 3, 4
2

=++ + + ) =
8 7 (7 10 11 43

Pφ0

P111 49 P12 1 44 P131 49 P141 44 P151 40

P212 43 P222 50 P232 49 P242 47

P323 51 P333 43
P343 47

P424 43 P444 48

 The optimum allocation will be
 Job operator time
1 5 8
2 1 7
3 3 7
4 2 10
5 4 11
 43


Example :ROW SCANNING.
Operator 1 2 3 4 5
job

1 10 12 15 12 8

2 7 16 14 14 11

3 13 14 7 9 9

4 12 10 11 13 10

5 8 13 15 11 15


Example : Assign the 5 operators to the 5 jobs
such that the total processing time is
minimized.
Operator
Operator 1 2 3 4 5
job

1 10 12 15 12 8

2 7 16 14 14 11

3 13 14 7 9 9

4 12 10 11 13 10

5 8 13 15 11 15


minimized.
Operator
Operator 1 2 3 4 5
job

1 2 4 7 4 0

2 0 9 7 7 4

3 6 7 0 2 2

4 2 0 1 3 0

5 0 5 7 4 8


minimized.
Operator
Operator 1 2 3 4 5
job

1 2 4 7 2 0

2 0 9 7 5 4

3 6 7 0 0 2

4 2 0 1 1 0

5 0 5 7 2 8


minimized.
Operator
Operator 1 2 3 4 5
job

1 2 4 6 1 0

2 0 9 6 4 4

3 7 8 0 0 3

4 2 0 0 0 0

5 0 5 6 1 8


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