• Methods to solve LPP :

1. Graphical Method - for only two variables.

2. Simplex Method - Universal method.

3. Assignment Method - Special method.

4. Transportation Method - Special method.

Note : Methods (2), (3) and (4) are iterative

methods.

Assignment Method

(Special Method)

• Assignment Problems

and

• Methods to solve such

problems

C11 C12 . . C1n

C21 C22 . . C2n

. . . . .

. . . . .

Cm1 . . . Cmn

Jobs (Activities)

1 2 . . n

1

2

.

.

m

Persons (R

esources)

CasesCases

• m = n

• m ≠ n

• Cij

• Pij

• Cij = Cost associated with assigning ith

resource to jth activity

Assignment ProblemAssignment Problem

A. Balanced Minimization → m = n with Cij

B. Unbalanced Minimization →m ≠ n with Cij

C. Balanced Maximization → m = n with Pij

D. Unbalanced Maximization →m ≠ n with Pij

Categories of Assignment ProblemsCategories of Assignment Problems

C11 C12 . . C1n

C21 C22 . . C2n

. . . . .

. . . . .

Cn1 . . . Cnn

Jobs (Activities)

1 2 . . n

1

2

.

.

n

Persons (R

esources)

• Xij = assignment of ith resource to jthactivity• Assignments are made on one to one basis

A.A. Balanced Minimization ProblemBalanced Minimization Problem

Formulation of Assignment Problem as LPP

1 1

.n n

i j

Min Z Cij Xij= =

= ∑∑

1

/ 1 ( )n

j

s t Xij for all i I=

= −∑

1

1 ( )n

i

Xij for all j II=

= −∑0 1 & .All Xij or for all i j=

Methods to solve Assignment ProblemsMethods to solve Assignment Problems

4. Hungarian Method

1. Enumeration Method

2. Integer Programming Method

3. Transportation Method

C11 C12

C21 C22

1 2

1

2

1. 1. Enumeration MethodEnumeration Method

• For n x n Problem---- n!

2! (=2)• No. of Possible solutions =

• P1J1 , P2J2 OR P1J2 , P2J1

• For 4 x 4 Problem---- 4! = 24

• For 5 x 5 Problem----5! = 120

• All Xij = 0 or 1 → 0-1 Integer programming.

2. 2. Integer Programming MethodInteger Programming Method

• Difficult to solve manually.

• For n x n Problem → Variables = n x n. Constraints = n+n = 2n

• For 5 x 5 Problem → Variables = 25 Constraints = 10

• Formulate the problem in Transportation Problem format.

3.3. Transportation MethodTransportation Method

C11 C12 . . C1n

C21 C22 . . C2n

. . . . .

. . . . .

Cn1 . . . Cnn

1 2 . . n

12..n

1

1

.

1

1 1 . . 1

4.4. Hungarian MethodHungarian Method

1. Row Deduction

2. Column Deduction

3. Assign zeros-If all assignments are over,

Then STOP

Else Go To Step 4

4. Adopt Tick Marking Procedure

5. Modify Cij and Go To Step 3

(Mr. D. Konig - A Hungarian Mathematician)

Steps :Steps :

1 2 3

1

2

3

Prob. 1Prob. 1 Balanced Minimization Problem Balanced Minimization Problem

Perform Row DeductionPerform Row Deduction

19 28 31

11 17 16

12 15 13

0 9 12

0 6 5

0 3 1

1 2 3

1

2

3

Row DeductionRow Deduction

• Perform Column DeductionPerform Column Deduction

0 6 11

0 3 4

0 0 0

1 2 3

1

2

3

Column DeductionColumn Deduction

• Minimum uncrossed no = 3.

• Modify Numbers

Assign Zeros Assign Zeros

Adopt Tick MarkingAdopt Tick Marking

0 3 8

0 0 1

3 0 0

1 2 3

1

2

3

• Hence optimal solution is

P1J1, P2J2, P3J3 giving Z = 19+17+13 = 49

Procedure of Assigning ZerosProcedure of Assigning Zeros Steps to be followed, after getting at-least one zero in each row &

each column.1. Start with 1st row. If there is only one uncrossed, unassigned

zero, assign it & cross other zeros in respective column (of assigned zero), if they exits, else go to next row. Repeat this for next all other rows.

2. If still uncrossed, unassigned zeros are available, start with first column. If there is only one uncrossed, unassigned zero, assign it and cross other zeros in respective row of assigned zero, if they exit, else go to next column. Repeat this for next all other columns.

3. Repeat 1 & 2 until single uncrossed, unassigned zeros are available, while going through rows & columns.

4. If still multiple uncrossed, unassigned zeros are available while going through rows & columns, it indicates that multiple (alternative) optimal solutions are possible.

Assign any one remaining uncrossed, unassigned zero & cross remaining zeros in respective row & column of assigned zero. Then go to 1.

5. If required assignments are completed then STOP, else perform “Tick Marking Procedure”. Modify numbers & go to 1.

Tick marking Procedure Tick marking Procedure

(To draw minimum number of lines through zeros.)

1. Tick marks row/rows where there is no assigned zero.2. Tick mark column/columns w.r.t. crossed zero/zeros in

marked row/rows.3. Tick mark row/rows w.r.t. assigned zero in marked

column/columns. 4. Go to to step 2 and repeat the procedure until no zero is

available for tick marking.Then

5. Draw lines through unmarked rows and marked columns (Check no. of lines = No. of assigned zeros)

Tick marking Procedure Tick marking Procedure (Continue)(Continue)

How to Modify numbers ?

Find minimum number out of uncrossed numbers.

1. Add this minimum number to crossings.

2. Deduct this number from all uncrossed numbers one by one.

3. Keep crossed numbers, on horizontal & vertical lines,

except on crossing, same

C11 C12 C13

C21 C22 C23

. . .

. . .

C51 C52 C53

Jobs (Activities)

1 2 3

1

2

3

4

5

Persons (R

esources)

• D1 and D2 Dummy Jobs are to be introduced to balance the problem

B. Unbalanced Minimization ProblemB. Unbalanced Minimization Problem

C11 C12 C13 0 0

C21 C22 C23 0 0

. . . 0 0

. . . 0 0

C51 C52 C53 0 0

Jobs (Activities)

1 2 3 D1 D2

1

2

3

4

5

Persons (R

esources)

• D1 and D2 are Dummy Jobs : Cij = 0

Prob. 2 The personnel manager of ABC Company wants to assign Mr. X, Mr. Y and Mr. Z to regional offices. But the firm also has an opening in its Chennai office and would send one of the three to that branch, if it were more economical than a move to Delhi, Mumbai or Kolkata. It will cost Rs. 2,000 to relocate Mr. X to Chennai, Rs. 1,600 to reallocate Mr. Y there, and Rs. 3,000 to move Mr. Z. What is the optimal assignment of personnel to offices ?

Office

Delhi Mumbai Kolkata

Personnel

Mr. X 1,600 2,200 2,400

Mr. Y 1,000 3,200 2,600

Mr. Z 1,000 2,000 4,600

UBMin (AP), after adding Chennai :

Delhi Mumbai Kolkata Chennai

Mr. X 1,600 2,200 2,400 2,000

Mr. Y 1,000 3,200 2,600 1,600

Mr. Z 1,000 2,000 4,600 3,000

BMin (AP), after adding Dummy Row :

Delhi Mumbai Kolkata Chennai

Mr. X 1,600 2,200 2,400 2,000

Mr. Y 1,000 3,200 2,600 1,600

Mr. Z 1,000 2,000 4,600 3,000

Dummy 0 0 0 0

After Row Deduction :

D M K Ch

X 0 600 800 400

Y 0 2,200 1,600 600

Z 0 1,000 3,600 2,000

Dm 0 0 0 0

Modified Matrix and Assignment :

D M K Ch

X 0 200 400 0

Y 0 1,800 1,200 200

Z 0 600 3,200 1,600

Dm 0 0 0 0

Modified Matrix and Assignment :

D M K Ch

X 200 200 400 0

Y 0 1,600 1,000 0

Z 0 400 3,000 1,400

Dm 400 0 0 0

Modified Matrix and Assignment :

D M K Ch

X 200 0 200 0

Y 0 1,400 800 0

Z 0 200 2,800 1,400

Dm 6 00 0 0 200

Hence, Optimal Solution is : XM, YCh, ZD

Giving Z = 2,200 + 1,600 + 1,000 = 4,800

Hence, it is economical to move Y to Chennai Office.

Jobs (Activities)

1 2 3 4 5

1

2

3

4

5

Persons (R

esources)

• Convert Profit Matrix into Relative Cost

Matrix

C. Balanced Maximization ProblemC. Balanced Maximization Problem

Pij

• How to Convert Profit Matrix into

Relative Cost Matrix ?

3. (Pij)max - Pij = RCij

1. (Pij) (-1) = RCij

2. 1/Pij = RCij

Prob. 3 Balanced Maximization Problem :Prob. 3 Balanced Maximization Problem :

C1 C2 C3 C4

P1 60 100 90 95

P2 93 60 70 60

P3 75 85 60 90

P4 95 89 65 60

Balanced RC Matrix :Balanced RC Matrix :

C1 C2 C3 C4

P1 40 0 10 5

P2 7 40 30 40

P3 25 15 40 10

P4 5 11 35 40

After Row Deduction :After Row Deduction :

C1 C2 C3 C4

P1 40 0 10 5

P2 0 33 23 33

P3 15 5 30 0

P4 0 6 30 35

After Column Deduction :After Column Deduction :

C1 C2 C3 C4

P1 40 0 0 5

P2 0 33 13 33

P3 15 5 20 0

P4 0 6 20 35

Assigning :Assigning :

C1 C2 C3 C4

P1 40 0 0 5

P2 0 33 13 33

P3 15 5 20 0

P4 0 6 20 35

Modified Matrix and Assigning :Modified Matrix and Assigning :C1 C2 C3 C4

P1 46 0 0 5

P2 0 27 7 27

P3 21 5 20 0

P4 0 0 14 29

Hence, Optimal Solution is : P1C3, P2C1, P3C4, P4C2

Giving Z = 90 + 93 + 90 + 89 = 362

Jobs (Activities)

1 2 312345

Persons (R

esources)

• Convert Unbalanced Profit Matrix into

Unbalanced Relative Cost Matrix

D. Unbalanced Maximization ProblemD. Unbalanced Maximization Problem

Pij

• Then Balance the matrix and solve

Prob. 4 Unbalanced Maximization Problem :Prob. 4 Unbalanced Maximization Problem :

J1 J2 J3 J4

P1 62 78 50 101

P2 71 84 61 73

P3 87 111 92 71

Unbalanced RC Matrix :Unbalanced RC Matrix :

J1 J2 J3 J4

P1 49 33 61 10

P2 40 27 50 38

P3 24 0 19 40

Balanced RC Matrix :Balanced RC Matrix :

J1 J2 J3 J4

P1 49 33 61 10

P2 40 27 50 38

P3 24 0 19 40

D 0 0 0 0

After Row Deduction :After Row Deduction :

J1 J2 J3 J4

P1 39 23 51 0

P2 13 0 23 11

P3 24 0 19 40

D 0 0 0 0

Assigning :Assigning :

J1 J2 J3 J4

P1 39 23 51 0

P2 13 0 23 11

P3 24 0 19 40

D 0 0 0 0

Modified Matrix and Assigning :Modified Matrix and Assigning :

J1 J2 J3 J4

P1 39 34 51 0

P2 2 0 12 0

P3 13 0 8 29

D 0 11 0 0

Modified Matrix and Assigning :Modified Matrix and Assigning :

J1 J2 J3 J4

P1 37 34 49 0

P2 0 0 10 0

P3 11 0 6 29

D 0 13 0 2

Hence, Optimal Solution is : P1J4, P2J1, P3J2

Giving Z = 101 + 71 + 111 = 283

10 5 13 15 16

3 9 18 13 6

10 7 3 2 2

7 11 9 7 12

7 9 10 4 12

Jobs (Activities)

1 2 3 4 5

1

2

3

4

5

Persons (R

esources)

Prob. 5 Practice Problem (Minimization) Prob. 5 Practice Problem (Minimization)

5 0 8 10 11

0 6 15 10 3

8 5 1 0 0

0 4 2 0 5

3 5 6 0 8

Jobs (Activities)

1 2 3 4 5

1

2

3

4

5

Persons (R

esources)

5 0 7 10 11

0 6 14 10 3

8 5 0 0 0

0 4 1 0 5

3 5 5 0 8

Jobs (Activities)

1 2 3 4 5

1

2

3

4

5

Persons (R

esources)

• Minimum uncrossed Number = 1.

6 0 7 11 11

0 5 13 10 2

9 5 0 1 0

0 3 0 0 4

3 4 4 0 7

Jobs (Activities) 1 2 3 4 5

1

2

3

4

5

Persons (R

esources)

• Hence optimal solution is

P1J2, P2J1, P3J5, P4J3, P5J4 giving Z = 5+3+2+9+4 = 23

4 7 5 6

− 8 7 4

3 − 5 3

6 6 4 2

Jobs (Activities) 1 2 3 4

1

2

3

4

Persons (R

esources)

Prob. 6Prob. 6

Problem for Alternative / Multiple Optimal Solutions :

0 3 1 2

Μ 4 3 0

0 Μ 2 0

4 4 2 0

Jobs (Activities) 1 2 3 4

1

2

3

4

Persons (R

esources)

After Row deduction

0 0 0 2

Μ 1 2 0

0 Μ 1 0

4 1 1 0

Jobs (Activities) 1 2 3 4

1

2

3

4

Persons (R

esources)

After Column deduction

Now modified matrix will be :

0 0 0 3

Μ 0 1 0

0 Μ 1 1

3 0 0 0

Jobs (Activities) 1 2 3 4

1

2

3

4

Persons (R

esources)

• Hence, this is a case of alternative optimal solutions. • Assign any one remaining zero and cross existing zeros

in respective row and column, then apply assigning procedure.

• Hence, one of the optimal solutions is P1J2, P2J4, P3J1, P4J3 giving Z = 7+4+3+4 = 18

0 0 0 3

Μ 0 1 0

0 Μ 1 1

3 0 0 0

Jobs (Activities) 1 2 3 4

1

2

3

4

Persons (R

esources)

• Hence, another optimal solution is P1J3, P2J2, P3J1, P4J4 giving Z = 5+8+3+2 = 18

To get another optimal solution, assign another remaining zero.

P1J3 can not be assigned, as it is already assigned before.

1. Restriction in Assignment.

e.g. Assignment of P3 to J4 is not possible.

Then C34 = M (Big Number)

2. Alternative Optimal Solution Possibility

-Already considered.

3. Particular assignment is prefixed.

e.g. If P3 & J4 prefixed

Then Row3 & Column4 are deleted.

Typical Cases in Assignment ProblemsTypical Cases in Assignment Problems

Typical Questions

Que. Answer each of the following questions in brief.

[ 1 ] What are the methods to solve Assignment Problems ?

Ans. : (1) Enumeration Method

(2) Integer Programming Method

(3) Transportation Method

(4) Hungarian Method

[ 2 ] How can you identify a function as Linear ?

Ans. : Power of each variable = only 1. No multiplication of variables.

[ 3 ] What is the purpose of “Tick Marking Procedure” in a method of solving Assignment Problems ?

Ans. : The purpose of “Tick Marking” procedure is to draw minimum number of lines covering zeros.

[ 4 ] What is significance of name “Hungarian Method” ?

Ans. : It is because of D. Konig of Hungary.

[ 5 ] Following is Minimization Assignment Problem.

(i) Write objective equation.

(ii) Write all possible constraints.

(iii) State “Non-Negativity” conditions for this problem.

(iv) State Optimal Solution. Is it unique optimal ?

X Y

A 3 2

B 4 5

Ans. : ( i ) Min Z = 3x11 + 2x12 + 4x21 + 5x22

( ii ) x11 + x12 = 1 x21 + x22 = 1

x11 + x21 = 1

x12 + x22 = 1

( iii ) All xij = 0 or 1 OR x11, x12, x21, x22 = 0 or 1

( iv ) Optimal solution is AY, BX giving Z = 6 (unique)

[ 6 ] Get Optimal Solution of following Minimization Assignment Problem. How many Optimal Solutions are existing for this problem ?

X Y Z

A 3 2 2

B 6 5 5

C 6 1 1

X Y Z

A 3 2 2

B 6 5 5

C 6 1 1

X Y Z

A 1 0 0

B 1 0 0

C 5 0 0

X Y Z

A 0 0 0

B 0 0 0

C 4 0 0

X Y Z

A 0 0 0

B 0 0 0

C 4 0 0

X Y Z

A 0 0 0

B 0 0 0

C 4 0 0

X Y Z

A 0 0 0

B 0 0 0

C 4 0 0

There are 4 Optimal Solutions.

RecapitulateRecapitulate• Methods to solve Assignment

Problem

• Hungarian Method

• Typical cases of Assignment

Problems

Home Assignments :

[ 1 ] Answer the following in brief :

( a ) What is the purpose of “Tick Marking” procedure in solving Assignment Problem ?

( b ) What is significance of name “Hungarian” method ?

[ 2 ] For the following Minimization Assignment Problem, answer the question given below.

J1 J2 J3

P1 2 4 3

P2 5 6 6

( a ) Formulate this problem as LPP.

( b ) Get all possible Optimal Solutions.

[ 3 ] ( i ) State possible methods of solving “Assignment Problem”.

( ii ) State possible methods of converting “Profit Matrix” into “Relative Cost Matrix”.

[ 4 ] Formulate following Assignment Problem as :

( i ) LPP

( ii ) Transportation Problem

P Q

A 2 5

B 6 4

[ 5 ] Find optimal assignments for following “Minimization Assignment Problem”.

J1 J2 J3

P1 −10 −10 −12

P2 −18 −6 −14

P3 −6 −2 −2

[ 6 ] Average time taken by operators on 4 old machines and a new machine are tabulated below. Management is considering to replace one of the old machines by a new machine. Is it advantageous to replace new machine with an old machine ? Why ?

M1 M2 M3 M4 New

O1 10 12 8 10 11

O2 9 10 8 7 10

O3 8 7 8 8 8

O4 12 13 14 14 11

[ 7 ] Consider the problem of assigning four operators to four machines. The assignment costs are given in Rupees. Operator 1 cannot be assigned to

machine 3. Also operator 3 cannot be assigned to machine 4. Find the optimal assignment.

M1 M2 M3 M4 1 5 5 − 2

2 7 4 2 3 3 9 3 5 −

4 7 2 6 7

If 5th Machine is made available and the respective costs to the four operators are Rs. 2, 1,

2 and 8. Find whether it is economical to replace any of the four existing machines. If so, which ?

[ 8 ] There are four batsman P, Q, R, & S. The batsman are to be selected for first three positions P1, P2 and P3. The expected score by these batsman at three different positions are given as below. Decide the optimal batsman for these three positions.

P1 P2 P3

P 42 16 27

Q 48 40 25

R 50 18 36

S 58 38 60

[ 9 ]

J1 J2 J3 J4

P1 20 22 28 15

P2 16 20 12 13

P3 19 23 14 25

P4 10 16 12 10

Minimization Problem :

[ 10 ]

J1 J2 J3 J4

P1 10 24 30 15

P2 16 22 28 12

P3 12 20 32 10

P4 9 26 34 16

Minimization Problem :

Thank youThank youFor any Query or suggestion :

Contact :Dr. D. B. Naik Professor & Head, Training & Placement (T&P)S. V. National Institute of Technology (SVNIT), Ichchhanath, Surat – 395 007 (Gujarat) INDIA.

Email ID : dbnaik@gmail.comdbnaik@svnit.ac.indbnaik_svr@yahoo.com

Phone No. : 0261-2201540 (D), 2255225 (O)