Assignment

Assignment problemsAssignment problemsOperational Research- Operational Research-

Level 4Level 4

Prepared by T.M.J.A.CoorayPrepared by T.M.J.A.Cooray

Department of MarthematicsDepartment of Marthematics

MA 402--assignment problemMA 402--assignment problem 22

Introduction Introduction

This is a special type of transportation This is a special type of transportation problem in which each source should have problem in which each source should have the capacity to fulfill the demand of any of the capacity to fulfill the demand of any of

the destinations.the destinations. In other words any operator would be able In other words any operator would be able

perform any job regardless of his perform any job regardless of his skills,although the cost( or the time taken) skills,although the cost( or the time taken) will be more if the job does not match with will be more if the job does not match with

operator’s skill.operator’s skill.


Let Let m be the number of jobsm be the number of jobs as well as as well as the operatorsthe operators, , and and ttijij be the processing time of the job i if it is assigned be the processing time of the job i if it is assigned

to the operator j.to the operator j. Here the objective is to assign the jobs Here the objective is to assign the jobs to the operators such that the total processing time is to the operators such that the total processing time is

minimizedminimized.. OperatorsOperators

JobJob

11 22 …… jj …… mm

11 tt1111 tt1212 tt1j1j tt1m1m

22

..

ii tti1i1 ttijij ttimim

..

mm ttm1m1 ttm2m2 ttmjmj ttmmmm

General format of assignment problem


Examples of assignment problemExamples of assignment problem

Row entityRow entity Column entityColumn entity Cell entityCell entity

jobsjobs operatorsoperators Processing timeProcessing time

Programmer Programmer programprogram Processing timeProcessing time

operatorsoperators machinemachine Processing timeProcessing time

Drivers Drivers RoutesRoutes Travel timeTravel time

Teachers Teachers Subjects Subjects Students pass Students pass percentage percentage


Assignment problem as a zero-one Assignment problem as a zero-one ( Binary) programming problem .( Binary) programming problem .

Min Z= cMin Z= c1111xx1111++c++cijijXXijij+.+c+.+cmmmmXXmm mm ==

Subject to xSubject to x1111+…………...+x+…………...+x1m1m =1 =1 xx2121+…………...+x+…………...+x2m2m =1 =1 …… …….... xxm1m1+…………...+x+…………...+xmmmm =1 =1 xx1111+…………...+x+…………...+xm1m1 =1 =1 xx1212+…………...+x+…………...+xm2m2 =1 =1 ……………… ……………….... xx1m1m+…………...+x+…………...+xmmmm =1 =1 xxijij.=0 or 1 .=0 or 1 for i=1,2….m and j=1,2…..m.for i=1,2….m and j=1,2…..m.

mjforX

miforX

XCZMin

m

iij

m

jij

m

i

m

jijij

,....11

,....11

1

1

1 1


Types of assignment problemsTypes of assignment problems

As in transportation problems assignment As in transportation problems assignment problems also can be balanced ( with equal problems also can be balanced ( with equal number of rows and columns) or unbalanced.number of rows and columns) or unbalanced.

When it is unbalanced the necessary number When it is unbalanced the necessary number of row/s or column/s are added to balance it. of row/s or column/s are added to balance it. That is to make a square matrix.That is to make a square matrix.

The values of the cell entries of the dummy The values of the cell entries of the dummy rows or columns will be made equal to zero.rows or columns will be made equal to zero.


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

OperatorOperator

jobjob11 22 33 44 55

11 1010 1212 1515 1212 88

22 77 1616 1414 1414 1111

33 1313 1414 77 99 99

44 1212 1010 1111 1313 1010

55 88 1313 1515 1111 1515


Hungarian methodHungarian method

Consists of two phases.Consists of two phases. First phase:First phase: row reductions and column row reductions and column

reductions are carried out.reductions are carried out. Second phaseSecond phase :the solution is optimized in :the solution is optimized in

iterative basis.iterative basis.


Phase 1: Row and column Phase 1: Row and column reductionsreductions

Step 0:Step 0: Consider the given cost matrix Consider the given cost matrix Step 1:Step 1: Subtract the minimum value of Subtract the minimum value of

each row from the entries of that row, to each row from the entries of that row, to obtain the next matrix.obtain the next matrix.

Step 2Step 2: Subtract the minimum value of : Subtract the minimum value of each column from the entries of that each column from the entries of that column , to obtain the next matrix.column , to obtain the next matrix.

Treat the resulting matrix as the input Treat the resulting matrix as the input for phase 2.for phase 2.


Phase 2: OptimizationPhase 2: Optimization Step3:Step3: Draw a minimum number of lines to cover Draw a minimum number of lines to cover

all the zeros of the matrix.all the zeros of the matrix. Procedure for drawing the minimum number of Procedure for drawing the minimum number of

lines:lines: 3.1 Row scanning3.1 Row scanning

1 Starting from the first row ,if there’s 1 Starting from the first row ,if there’s only one zeroonly one zero in a row mark a square round the zero entry and in a row mark a square round the zero entry and draw a vertical line passing through that zero. draw a vertical line passing through that zero. Otherwise skip the row.Otherwise skip the row.

2.After scanning the last row, check whether all the 2.After scanning the last row, check whether all the zeros are covered with lines. If yes go to step 4. zeros are covered with lines. If yes go to step 4. Otherwise do column scanning. CtdOtherwise do column scanning. Ctd


3.2 Column scanning3.2 Column scanning..

1. Starting from the first column: if there’s 1. Starting from the first column: if there’s only one zeroonly one zero in a column mark a in a column mark a square round the zero entry and draw a square round the zero entry and draw a horizontal line passing through that horizontal line passing through that zero. otherwise skip the column.zero. otherwise skip the column.

2.After scanning the last column, check 2.After scanning the last column, check whether all the zeros are covered with whether all the zeros are covered with lines. If yes go to step 4. Otherwise do lines. If yes go to step 4. Otherwise do row scanning. ctd row scanning. ctd


Step 4:Step 4: check whether the number of squares marked check whether the number of squares marked is equal to the number of rows/columns of the matrix. is equal to the number of rows/columns of the matrix.

If yes go to step 7. Otherwise go to step 5.If yes go to step 7. Otherwise go to step 5. Step 5:Step 5: Identify the minimum value of the undeleted Identify the minimum value of the undeleted

cell values ,say ‘x’. Obtain the next matrix by the cell values ,say ‘x’. Obtain the next matrix by the following steps.following steps.

5.1 5.1 Copy the entries covered by the lines ,but not on Copy the entries covered by the lines ,but not on the intersection points.the intersection points.

5.2 5.2 add x to the intersection points add x to the intersection points

5.35.3 subtract x from the undeleted cell values. subtract x from the undeleted cell values.

Step 6:Step 6: go to step 3. go to step 3.

Step 7:Step 7: optimal solution is obtained as marked by the optimal solution is obtained as marked by the squares squares


Maximization problem Maximization problem

If the problem is a maximization If the problem is a maximization problem ,problem ,convert the problem into a convert the problem into a minimization problem by multiplying by -1.minimization problem by multiplying by -1.

Then apply the usual procedure of an Then apply the usual procedure of an assignment problem.assignment problem.


Example : Example : Assign 4 sales persons to four Assign 4 sales persons to four different sales regions such that the total different sales regions such that the total

sales is maximized.sales is maximized.

Sales Sales regionregion

Sales personSales person

11 22 33 44

11 1010 2222 1212 1414

22 1616 1818 2222 1010

33 2424 2020 1212 1818

44 1616 1414 2424 2020


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



11 22 33 44

11 -10-10 -22-22 -12-12 -14-14

22 -16-16 -18-18 -22-22 -10-10

33 -24-24 -20-20 -12-12 -18-18

44 -16-16 -14-14 -24-24 -20-20


After step 1After step 1



11 22 33 44

11 1212 00 1010 88

22 66 44 00 1212

33 00 44 1212 66

44 88 1010 00 44


After step 2After step 2



11 22 33 44

11 1212 00 1010 44

22 66 44 00 88

33 00 44 1212 22

44 88 1010 00 00


Phase 2Phase 2



11 22 33 44

11 1212 00 1010 44

22 66 44 00 88

33 00 44 1212 22

44 88 1010 00 00


Note that the number of squares is equal to the Note that the number of squares is equal to the number of rows of the matrix. solution is feasible number of rows of the matrix. solution is feasible and optimal.and optimal.

Result:Result: Salesman Salesman Sales region Sales region Sales Sales

11 22 2222

22 33 2222

33 11 2424

44 44 2020


Branch and Bound algorithm for the assignment Branch and Bound algorithm for the assignment problemproblem

Terminology:Terminology: K-level number in the branching treeK-level number in the branching tree For root node k=0For root node k=0 --assignment made in the current node of a branching treeassignment made in the current node of a branching tree

PPk k –assignment at level k of the branching tree–assignment at level k of the branching tree

A-set of assigned cells up to the node PA-set of assigned cells up to the node Pk k from the from the

root node root node VV

- - lower bound of partial assignment A up to Plower bound of partial assignment A up to Pk k

Such that VSuch that V = =

Aji Xi Yjij CC

,

min


CCijij is the cell entity of the cost matrix is the cell entity of the cost matrix

X rows which are not deleted up to node PX rows which are not deleted up to node Pk k

from the root node in the branching tree.from the root node in the branching tree. Y columns which are not deleted up to node Y columns which are not deleted up to node

PPk k

from the root node in the branching tree.from the root node in the branching tree.


Branching guidelines Branching guidelines

1.At level k,the row marked as k of the 1.At level k,the row marked as k of the assignment problem,will be assigned with assignment problem,will be assigned with the best column of the assignment problem.the best column of the assignment problem.

2.if there is a lower bound ,then the terminal 2.if there is a lower bound ,then the terminal node at the lower most level is to be node at the lower most level is to be considered for further branchingconsidered for further branching

3.stopping rule:if the minimum lower bound 3.stopping rule:if the minimum lower bound happens to be at any one of the terminal happens to be at any one of the terminal nodes at the (n-1)nodes at the (n-1)thth level ,the optimality is level ,the optimality is reached.reached.


OperatorOperator

jobjob11 22 33 44 55

11 1010 1212 1515 1212 88

22 77 1616 1414 1414 1111

33 1313 1414 77 99 99

44 1212 1010 1111 1313 1010

55 88 1313 1515 1111 1515

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


P0

P121 P13

1 P141 P15

1 P111

}5,4,3,2{},5,4,3,2{)},11{()},11{(

111

YXA

Pforboundlower

49)1110711(10

)min(5,4,3,2 5,4,3,2

1111

i j

ijCcV

51 44 49 44 40


P0

P121 P13

1 P141 P15

1 P111

}4,3,2{},5,4,3{)},15(),21{()},21{(

221

YXA

Pforboundlower

43)11107(78

)min(5,4,3 4,3,2

211521

i j

ijCccV

49 44 49 44 40

P212 P23

2 P242 43

50P222 49 47


P0

P121 P13

1 P141 P15

1 P111

49 44 49 44 40

P212 P23

2 P242 43

50P222 49 47

P323 P33

3 P343 51 43 47

P424 P44

4 43 48


The optimum allocation will be The optimum allocation will be Job operator time Job operator time 1 5 81 5 8 2 1 72 1 7 3 3 73 3 7 4 2 104 2 10 5 4 115 4 11 4343


OperatorOperator

jobjob11 22 33 44 55

11 1010 1212 1515 1212 88

22 77 1616 1414 1414 1111

33 1313 1414 77 99 99

44 1212 1010 1111 1313 1010

55 88 1313 1515 1111 1515

Example :ROW SCANNING.Example :ROW SCANNING.


OperatorOperator

jobjob11 22 33 44 55

11 1010 1212 1515 1212 88

22 77 1616 1414 1414 1111

33 1313 1414 77 99 99

44 1212 1010 1111 1313 1010

55 88 1313 1515 1111 1515

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


OperatorOperator

jobjob11 22 33 44 55

11 22 44 77 44 00

22 00 99 77 77 44

33 66 77 00 22 22

44 22 00 11 33 00

55 00 55 77 44 88



OperatorOperator

jobjob11 22 33 44 55

11 22 44 77 22 00

22 00 99 77 55 44

33 66 77 00 00 22

44 22 00 11 11 00

55 00 55 77 22 88



OperatorOperator

jobjob11 22 33 44 55

11 22 44 66 11 00

22 00 99 66 44 44

33 77 88 00 00 33

44 22 00 00 00 00

55 00 55 66 11 88


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