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August 26, 2009

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ASSIGNMENT PROBLEM

It is a matching problem Can be used for allocation

Warehouses to MarketsFactories to MarketsFactories to Warehouses

Operator to machinesMachines to JobsSalesmen to market territoriesEtc.,

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Example Time matrix is given below. Each market should be supplied by one

warehouse only.

MKT1 MKT2 MKT3 MKT4

Warehouse 1 14 5 8 10

Warehouse 2 2 12 6 5Warehouse 3 7 8 3 9

Warehouse 4 2 4 6 7

We must determine which warehouse to supplywhich market tominimize the total time.

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We may try all possible combinations

There are 4 ! Number of solutions. Find the total time for each solution

and choose the best ???MKT1 MKT2 MKT3 MKT4



Warehouse 3 7 8 3 9

Warehouse 4 2 4 6 7

We may apply common sense but may not be consistent and may not guaranteeoptimality

We need smarter ways of solving this problem optimizing algorithm

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Algorithm

Step 1:

(a) Find the minimum element in each rowof the time matrix. Form a new matrix bysubtracting this time element from the

respective row. (b) Find the minimum time in each column

of the new matrix, and subtract this fromeach column.

This is the reduced time matrix.

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Example: Step 1(a)

MKT1 MKT2 MKT3 MKT4



Warehouse 3 7 8 3 9

Warehouse 4 2 4 6 7

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 5


Warehouse 3 4 5 0 6

Warehouse 4 0 2 4 5

Row Reduction

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Example: Step 1(b)

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 2


Warehouse 3 4 5 0 3

Warehouse 4 0 2 4 2

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 5


Warehouse 3 4 5 0 6

Warehouse 4 0 2 4 5

Column Reduction

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Example: Final Solution

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 2


Warehouse 3 4 5 0 3

Warehouse 4 0 2 4 2

Optimal assignment,1,1,1,1

41332412==== xxxx

How did weknow which 0sto choose?!

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FINAL ALLOCATION AND THE

EFFECTMKT1 MKT2 MKT3 MKT4

Warehouse 1 5

Warehouse 2 5

Warehouse 3 3

Warehouse 4 2

What is the effect ?

Find the value of the solution in terms of total time

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WHAT IS THE VALUE OF THE

SOLUTION X 12 = 5

X 24 = 5 X 33 = 3

X41 = 2

TOTAL TIME = 15 Can you find a solution better than this ?

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PRACTICE PROBLEM

A DEPARTMENT HAS 4 JOBS AND 4

MEN. ROWS REPRESENT JOBS 1 To 4

COLUMNS REPRESENT MEN A TO D THE TIME MATRIX IS GIVEN.

CHOOSE THE BEST JOB-MANCOBINATIONS.

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8 26 17 11

13 28 4 26

38 19 18 15

19 26 24 10

Job 1

Job 2

Job 3

Job4

M/C1 M/C2 M/C3 M/C 4

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THE PROBLEM

AIM : TO MINIMIZE THE TOTAL TIME

TAKEN TO COMPLETE ALL JOBS. THERE ARE 4! NUMBER OF POSSIBLE

ALLOCATIONS. REQUIREMENT :

A SYSTEMATIC METHOD.

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HUNGARIAN METHODPHASE 1STEPS

Row operation

Find the smallest number in each row and Subtractit from the remaining elements in that row.

Column operationFind the minimum element in each column andsubtract it from the other elements in therespective column

Check if allocation is possibleSee if we have the number of independent zeros

equal to number of rows or columns.

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8 26 17 11

13 27 4 26

38 19 18 15

19 26 24 10

0 18 9 3

9 23 0 22

23 4 3 0

9 16 14 0

0 14 9 3

9 19 0 22

23 0 3 09 12 14 0

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COMMENTS

HERETHE INDEPENDENT ASSINGNMENTS ARE:

A-1 (8)B-3 (19)C-2 (4)D-4 (10)

CHECK THE ORIGINAL MATRIX AND CHECK THEEFFECTIVENESS (41)

IF THE INDEPENDENT ASSINGNMENTS ARE < n, WHERE n ISEQUAL TO THE NUMBER OF ALLOCATIONS REQUIRED

GO TO PHASE 2

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WILL IT BE THIS SIMPLE ALWAYS ?

Consider an example

Same situation, but the numbers areslightly different

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Example We must determine which warehouse to supply

which market minimize the total time. Timematrix is given below.

Each market should be supplied by onewarehouse only.

MKT1 MKT2 MKT3 MKT4



Warehouse 3 7 8 3 9


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Hungarian Algorithm Phase1

Step 1:

(a) Find the minimum element in each rowof the time matrix. Form a new matrix by

subtracting this time from each row. (b) Find the minimum time in each column

of the new matrix, and subtract this from

each column.

This is the reduced time matrix.

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Example: Step 1(a)

MKT1 MKT2 MKT3 MKT4



Warehouse 3 7 8 3 9


MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 2


Warehouse 3 4 5 0 6

Warehouse 4 0 2 4 8

Row Reduction

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Example: Step 1(b)

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 0


Warehouse 3 4 5 0 4

Warehouse 4 0 2 4 6

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 2


Warehouse 3 4 5 0 6

Warehouse 4 0 2 4 8

Column Reduction

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THE PROBLEM ?

Only 3 allocations are possible

We need 4 allocations

Go to Phase 2

Hungarian Algorithm Phase 2

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Hungarian Algorithm Phase 2

Step 2: Draw the minimum number oflines that are needed to cover all the zerosin the reduced cost matrix.

If m (number of assignments required)lines are available, then an optimal

solution is available among the coveredzeros in the matrix.

Otherwise, continue to Step 3.How do we

find theminimumnumber of

lines?

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Example: Step 2

MKT1 MKT2 MKT3 MKT4

Warehouse 1 9 0 3 0


Warehouse 3 4 5 0 4

Warehouse 4 0 2 4 6

We have 3 lines, but

we need 4 lines.Now continue with Step 3

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Hungarian Algorithm

Step 3: Find the smallest nonzero element

(say, k) in the reduced time matrix that isuncovered by the lines.

Subtractk

from each uncovered element,and add k to each element that is covered

by two lines.

Return to Step 2.

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Example: Step 3

MKT1 MKT2 MKT3 MKT4

Warehouse 9 0 3 0

Warehouse 0 10 4 1

Warehouse 4 5 0 4

Warehouse 0 2 4 6

MKT1 MKT2 MKT3 MKT4

Warehouse 10 0 3 0

Warehouse 0 9 3 0

Warehouse 5 5 0 4

Warehouse 0 1 3 5

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Example: Step 2 (again)

MKT1 MKT2 MKT3 MKT4

Warehouse 1 10 0 3 0Warehouse 2 0 9 3 0

Warehouse 3 5 5 0 4

Warehouse 4 0 1 3 5

There are 4 lines.We can get the optimal assignment.

We stop

Zero Assignment

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Example: Final Solution

MKT1 MKT2 MKT3 MKT4


Warehouse 2 0 9 3 0

Warehouse 3 5 5 0 4

Warehouse 4 0 1 3 5

Optimal assignment1,1,1,1

24413312==== xxxx

How did we

know which 0sto choose?!

HOW TO DRAW THE LINES

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HOW TO DRAW THE LINES

SYSTEMATICALLY ? CONSIDER ONE MORE EXAMPLE

CONSIDER THE FOLLOWING

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CONSIDER THE FOLLOWING

EXAMPLE There are 5 operators and 5 jobs

Time required by each operator to

complete each job is given

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11 17 8 16 20

9 7 12 6 15

13 16 15 12 16

21 24 17 28 26

14 10 12 11 15

Job1 Job2 Job3 Job 4 JOb5

M5

M3

M4

M2

M1

3 9 0 8 12

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3 9 0 8 12

3 1 6 0 91 4 3 0 4

4 7 0 11 9

4 0 2 1 5

2 9 0 8 8

2 1 6 0 5

0 4 3 0 0

3 7 0 11 5

3 0 2 1 1

PHASE 2 0F ASSIGNMENTS

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PHASE 2 0F ASSIGNMENTS

STEPS MARK ALL ROWS HAVING NO

ASSINGNMENT MARK COLUMNS WITH ZEROS IN MARKED

ROWS.

MARK ROWS THAT HAVE ASSIGNMENTS INMARKED COLUMN.

REPEAT STEPS 1 TO 3 TILL THE CHAIN OF

MARKING ENDS. DRAW LINES THROUGH UNMARKED ROWS

AND MARKED COLUMNS.

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2 9 0 8 8

2 1 6 0 5

0 4 3 0 0

3 7 0 11 5

3 0 2 1 1

CHECK

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CHECK

IF THE PROCEDURE HAS BEEN CARRIED OUT CORRECTLY:

THERE SHOULD BE AS MANY LINES AS THEREARE ASSIGNMENTS IN THE MAXIMAL ASSIGNMENT

EVERY ZERO WILL HAVE AT LEAST ONE LINETHROUGH IT.

THIS METHOD YIELDS THE MINIMUM NUMBER OF LINES THATWILL PASS THROUGH ALL ZEROS.

HAVING DRAWN THE SET OF LINES IN STEPS 1 TO 4, DO THE

FOLLOWING.

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0 7 0 6 6

2 1 8 0 5

0 4 5 0 0

1 5 0 9 3

3 0 4 1 1

IMPORTANT STEP TO FIND

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ADDITIONAL ALLOCATION EXAMINE ELEMENTS THAT DO NOT

HAVE A LINE THROUGH THEM.

SELECT THE SMALLEST OF THESEAND SUBTRACT IT FROM ALL THEELEMNTS THAT DO NOT HAVE A LINE

THROUGH THEM. ADD THIS SMALLEST ELEMENT TO

EVERY ELEMENT THAT LIES AT THEINTERSECTION OF TWO LINES.

LEAVE THE REMAINING ELEMENTS OF

THE MATRIX UNCHANGED.

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0 7 0 6 6

2 1 8 0 5

0 4 5 0 0

1 5 0 9 3

3 0 4 1 1

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FINAL SOLUTION

X11 = 1 ( 11)

X24 = 1 (6) X35 = 1 (16)

X43 = 1 (17 ) X52 = 1 (10)

Total Value = 60

PRACTICE PROBLEMS

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PRACTICE PROBLEMS

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MANJOBS

1 2 3 4 5

A 10 5 13 15 16

B 3 9 18 13 6

C 10 7 2 2 2

D 7 11 9 7 12

E 7 9 10 4 12

ANSWER : A-2, B-1, C-5, D-3, E-4 ( CHECK)

> A, B, C, D, E ARE THE TOWNS REACHING WHICH CARS ARE TOBE HIRED.> a b c d e ARE CAR DEPOTS FROM WHERE CARS CAN BE HIRED

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> a, b, c, d, e ARE CAR DEPOTS FROM WHERE CARS CAN BE HIRED.EACH DEPOT CAN ONLY GIVE CAR TO REACH ONE TOWN.

MATRIX- DISTANCE BETWEEN DEPOT AND TOWNLINK THE DEPOT AND THE TOWN TO MINIMISE TOTAL DISTENCE

DEPOTS

a b c d e

A 160 130 175 190 200

B 135 120 130 160 175

C 140 110 155 170 185

D 50 50 90 80 110

E 55 35 70 80 105

ANS : A-e, B-c, C-b, D-a, E-d( CHECK)

TOWNS

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