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1. Solve the assignment problem for the optimal solution
Time taken by the workers.
workers JOB
A B C D
1 45 40 51 67
2 57 42 63 55
3 49 52 48 64
4 41 45 60 55
The solution of this problem in step wise manner..
Step 1 Row wise reduction so matrix will be
workers JOB
A B C D
1 5 0 11 27
2 15 0 21 13
3 1 4 0 16
4 0 4 19 14
Step 2 Column wise reduction
workers JOB
A B C D
1 5 0 11 14
2 15 0 21 0
3 1 4 0 3
4 0 4 19 1
Step 3: Draw the line in such a way so that minimum no. of line should cover all
zero.
So
workers JOB
A B C D
1 5 0 11 14
2 15 0 21 0
3 1 4 0 3
4 0 4 19 1
As we observed that the no. of lines and no. of columns or rows are equal, so that
we allocate the work.
workers JOB
A B C D
1 5 0 11 14
2 15 0 21 0
3 1 4 0 3
4 0 4 19 1
Assignment is made so that
1 B, 2 D, 3 C, 4 1 and time taken is (40+55+48+41) = 184 mins.
2. Using the optimal cost matrix, determine the optimal job assignment and cost of
assignment.
Machinist JOB
1 2 3 4 5
A 10 3 3 2 8
B 9 7 8 2 7
C 7 5 6 2 4
D 3 5 8 2 4
E 9 10 9 6 10
Step 1 Reduction of matrix
Machinist JOB
1 2 3 4 5
A 8 1 1 0 6
B 7 5 6 0 5
C 5 3 4 0 2
D 1 3 6 0 2
E 3 4 3 0 4
Step 2 : Column reduction
Machinist JOB
1 2 3 4 5
A 7 0 0 0 4
B 6 4 5 0 3
C 4 2 3 0 0
D 0 2 5 0 0
E 2 3 2 0 2
Draw the lines such a way so that Minimum no. of line should cover all the zeros.
Machinist JOB
1 2 3 4 5
A 7 0 0 0 4
B 6 4 5 0 3
C 4 2 3 0 0
D 0 2 5 0 0
E 2 3 2 0 2
Since the no. of lines covering all the zeros are less than the no. of columns/rows, so again
we have do the reduction by 2 as minimum value is 2 in matrix.
Machinist JOB
1 2 3 4 5
A 7 0 0 2 6
B 4 2 3 0 3
C 2 0 1 0 0
D 0 2 5 2 2
E 0 1 0 0 2
Machinist JOB
1 2 3 4 5
A 7 0 0 2 6
B 4 2 3 0 3
C 2 0 1 0 0
D 0 2 5 2 2
E 0 1 0 0 2
A 2, B 4, C5, D 1, E3
So that the cost which will occur is
A = 3
B = 2
C = 4
D = 3
E = 9
Which equal to 21.
3. Using the optimal cost matrix, determine the optimal job assignment and cost of
assignment.
Machinist JOB
1 2 3 4 5
A 10 3 3 2 8
B 9 7 8 2 7
C 7 5 6 2 4
D 3 5 8 2 4
E 9 10 9 6 10
Step 1 Reduction of matrix
Machinist JOB
1 2 3 4 5
A 8 1 1 0 6
B 7 5 6 0 5
C 5 3 4 0 2
D 1 3 6 0 2
E 3 4 3 0 4
Step 2 : Column reduction
Machinist JOB
1 2 3 4 5
A 7 0 0 0 4
B 6 4 5 0 3
C 4 2 3 0 0
D 0 2 5 0 0
E 2 3 2 0 2
Draw the lines such a way so that Minimum no. of line should cover all the zeros.
Machinist JOB
1 2 3 4 5
A 7 0 0 0 4
B 6 4 5 0 3
C 4 2 3 0 0
D 0 2 5 0 0
E 2 3 2 0 2
Since the no. of lines covering all the zeros are less than the no. of columns/rows, so again
we have do the reduction by 2 as minimum value is 2 in matrix.
Machinist JOB
1 2 3 4 5
A 7 0 0 2 6
B 4 2 3 0 3
C 2 0 1 0 0
D 0 2 5 2 2
E 0 1 0 0 2
Machinist JOB
1 2 3 4 5
A 7 0 0 2 6
B 4 2 3 0 3
C 2 0 1 0 0
D 0 2 5 2 2
E 0 1 0 0 2
A 2, B 4, C5, D 1, E3
So that the cost which will occur is
A = 3
B = 2
C = 4
D = 3
E = 9
Which equal to 21.
4.
JObs Machines
M1 M2 M3 M4
W1 40 50 60 65
W2 30 38 46 48
W3 25 33 41 43
W4 39 45 51 59
How would the task to be assigned to minimize the cost?
Answer: Step 1 Row wise reduction
Jobs Machines
M1 M2 M3 M4
W1 0 10 20 15
W2 0 8 16 18
W3 0 8 16 18
W4 0 6 12 20
Step 2 Column Wise Reduction
Jobs Machines
M1 M2 M3 M4
W1 0 4 8 7
W2 0 2 4 0
W3 0 2 4 0
W4 0 0 0 2
Jobs Machines
M1 M2 M3 M4
W1 0 4 8 7
W2 0 2 4 0
W3 0 2 4 0
W4 0 0 0 2
Step 3 In three line all zeros are covered so that we have again reduce the matrix by 2.
Jobs Machines
M1 M2 M3 M4
W1 0 2 4 7
W2 0 0 0 0
W3 0 0 0 0
W4 4 0 0 4
All zeros are covered by four lines
Jobs Machines
M1 M2 M3 M4
W1 0 2 4 7
W2 0 0 0 0
W3 0 0 0 0
W4 2 0 0 4
So work assign in this way
W1—M1, W2---M4, W3—M3, W4—M2.
40+48+41+45 = 174.
5. The college department has the problem of providing the teachers for all courses
offered by his department at the highest possible level of educational quality. He has
got the three professors and one teaching assistant (TA). Four courses must be offered
and after appropriate introspection and evaluation he has arrived at the following
relative ratings (100= best rating) regarding the ability of each instructor to teach the
four courses respectively.
Course 1 Course 2 Course 3 Course 4
Prof. 1 70 50 70 80
Prof. 2 30 70 60 80
Prof. 3 30 40 50 70
TA 40 20 40 50
How should be the assign his staff to the course to maximize educational quality of the
department.
Answer :
Step 1: We have to maximize the problem so reduce it by 70.
Course 1 Course 2 Course 3 Course 4
Prof. 1 10 30 10 0
Prof. 2 50 10 20 0
Prof. 3 50 40 30 10
TA 40 60 40 30
Step 2 : Reduce row wise
Course 1 Course 2 Course 3 Course 4
Prof. 1 10 30 10 0
Prof. 2 50 10 20 0
Prof. 3 40 30 20 0
TA 10 30 10 0
Step 3 : Reduce column wise. And cover all the zero by minimum line
Course 1 Course 2 Course 3 Course 4
Prof. 1 0 20 0 0
Prof. 2 40 0 10 0
Prof. 3 30 20 10 0
TA 0 20 0 0
So the no. of line is 4 which is equal to no. of column or rows.
Course 1 Course 2 Course 3 Course 4
Prof. 1 0 20 0 0
Prof. 2 40 0 10 0
Prof. 3 30 20 10 0
TA 0 20 0 0
So the course offer by
Prof. 1 Course 1 70
Prof. 2 Course 2 70
Prof. 3 Course 4 70
TA Course 3 40
Total 250
6. Captain MS Dhoni has to allot five Middle batting position to five batsmen. The
average run scored by each batsman at these position is as follows
BATSMAN BATTING POSITION
I II III IV V
VIRENDRA 40 40 35 25 50
GAUTAM 42 30 16 25 27
SURESH 50 48 40 60 50
YUVARAJ 20 19 20 18 25
SACHIN 58 60 59 55 53
Find out the proper batting assignement.
Answer:
Step 1 Reduction of matrix.
BATSMAN BATTING POSITION
I II III IV V
VIRENDRA 20 20 25 35 10
GAUTAM 18 30 44 35 33
SURESH 10 22 20 0 10
YUVARAJ 40 41 40 42 35
SACHIN 02 00 01 05 02
Step 2 reduction row wise
BATSMAN BATTING POSITION
I II III IV V
VIRENDRA 10 10 15 25 10
GAUTAM 0 12 26 17 15
SURESH 10 22 20 0 10
YUVARAJ 5 6 5 7 0
SACHIN 02 00 01 05 02
Step 4 Reduction Column Wise
BATSMAN BATTING POSITION
I II III IV V
VIRENDRA 10 10 14 25 10
GAUTAM 0 12 25 17 15
SURESH 10 22 19 0 10
YUVARAJ 5 6 4 7 0
SACHIN 02 00 0 05 02
Step 5 Reduce by 10
BATSMAN BATTING POSITION
I II III IV V
VIRENDRA 0 0 4 15 0
GAUTAM 0 12 25 17 15
SURESH 10 22 19 0 10
YUVARAJ 5 6 4 7 0
SACHIN 02 00 0 05 02
So the batting order is like this
Gautam—virendra—Sachin—Suresh—yuvaraj runs are 40+42+59+60+25=226
7. Darda oil mills has four plants each of which can manufacture anyone of these four
product. The manufacturing cost differs from plant to plant so the sales revenue.
Sales
Plants
Products
I II III IV
A 70 88 69 82
B 80 90 71 94
C 75 87 73 80
D 78 85 74 89
revenue
Plants
Products
I II III IV
A 59 70 55 71
B 65 73 55 79
C 62 72 59 68
D 65 74 58 76
Reduction to profit matrix
Plants
Products
I II III IV
A 11 18 14 11
B 15 17 16 15
C 13 15 14 12
D 13 11 16 13
Plants
Products
I II III IV
A 7 0 4 7
B 3 1 2 3
C 5 3 4 6
D 5 7 2 5
Column wise reduction
Plants
Products
I II III IV
A 4 0 2 4
B 0 1 0 0
C 2 3 2 3
D 2 7 0 2
repeat
Plants
Products
I II III IV
A 3 0 2 3
B 0 2 1 0
C 1 3 2 2
D 1 7 0 1
Plants
Products
I II III IV
A 2 0 2 2
B 0 3 2 0
C 0 3 2 1
D 0 7 0 0
Plants
Products
I II III IV
A 2 0 2 2
B 0 3 2 0
C 0 3 2 1
D 0 7 0 0
A= II, B=IV, C=I,D=III ==18+15+13+16 = 61
8. The planning engineering has the four jobs and four machine for the allocation
the job-wise allocation is given below.
JObs
Machines
W X Y Z
A 1200 1440 480 1080
B 720 600 360 960
C 240 360 480 600
D 720 840 360 1200
STEP : This is simple assignment problem ( minimization and unrestricted)
After reducing by row wise and column wise we get answer …2280Rs.
With Assignment A to Z, B to X, C to W and D to Y.
9. The engg. Workshop has five lathes and five operators. The workshop is involved in
the producing of standard part which is required in large quantity. Since the operator
have the varying skills and machines alos have the different efficiency the output are
varying. The wekly output has given below.
Operators
Lathes
1 2 3 4 5
O1 30 33 37 42 46
O2 29 33 39 44 50
O3 33 38 45 49 44
O4 31 34 41 57 52
O5 32 38 41 46 51
This is simple matrix of (maximization, unrestricted allocation)
So first we have to make it in minimization form by subtracting from 52.
The by row and column reduction and following the procedure.
We get the answer.209 units. O1—1, O2—4, O3—3, O4—5 AND O5—2.
10. Captain MS Dhoni has to allot five Middle batting position to five batsmen. The
average run scored by each batsman at these position is as follows
BATSMAN BATTING POSITION
I II III IV V
VIRENDRA 35 36 35 25 40
GAUTAM 52 61 14 24 27
SURESH 50 47 42 13 41
YUVARAJ 19 25 40 47 52
SACHIN 60 72 84 56 65
Find out the proper batting assignment.
11. Assign three jobs on three machines for following cost matrix:
Jobs Machines
M1 M2 M3
J1
J2
J3
Rs. 14
Rs.11
Rs.20
Rs.12
Rs.17
Rs. 8
Rs.16
Rs.21
Rs. 7
Solution
Step 1: Subtract minimum entry in each column from all the entries on that column. This is a job opportunity cost matrix.
Job-opportunity cost matrix
Step 2: Subtract minimum entry in each row of job-opportunity cost matrix from all the entries of that row. This is a total opportunity cost matrix:
Total opportunity cost matrix
Step 3: Check for Optimality
Draw minimum number of horizontal and vertical lines to cover all zeros. This can be done in 2 lines, which is one less than the number of rows (which is 3). Thus, the solution is yet not final. Hence, go
JOBS/MACHINES M1 M2 M3
J1 0 1 6
J2 0 9 14
J3 9 0 0
to step 4.
Check for optimality
Step 4: From the uncovered entries, the minimum is 1. Thus, subtract 1 from all entries, which are uncovered. Add one at junction of lines, i.e., at J3-M1. We get the following matrix now, as the received opportunity cost matrix.
JOBS/MACHINES M1 M2 M3
J1 0 0 5
J2 0 8 13
J3 10 0 0
Revised total opportunity cost matrix
Now, go to step 3 to check the optimality. We can cover all zeros of above matrix by at least three lines (which is also equal to number of rows). Hence, above solution may be used for optimal assignment.
Step 5: Assignment Scheme
Refer revised total opportunity cost matrix. Row J2 has only one zero at M1 column. Hence, assign J2 to M1. Remove row J2 and column M1.
Column M3 of remainder matrix has one zero at J3 row. Assign J3 to M3. The last assignment is remainder job J1 to M2. Thus, the final assignment is:
Job Machine Cost
J1
J2
J3
M2
M1
M3
Rs.12
Rs.11
Rs. 7
Total Cost Rs.30
12. Four persons A,B,C and D are to be assigned four jobs I, II, III and IV. The cost matrix is given as under, find the proper assignment.
A B C D
I 8 10 17 9
II 3 8 5 6
III 10 12 11 9
IV 6 13 9 7
Solution : In order to find the proper assignment we apply the Hungarian algorithm as follows: I (A) Row reduction
A B C D
I 0 2 9 1
II 0 5 2 3
III 1 3 2 0
IV 0 7 3 1
I (B) Column reduction
A B C D
I 0 0 7 1
II 0 3 0 3
III 1 1 0 0
IV 0 5 1 1
II(A) and (B) Zero assignment
A B C D
I 0 0 7 1
II 0 3 0 3
III 1 1 0 0
IV 0 5 1 1
In this way all the zero’s are either crossed out or assigned. Also total assigned zero’s = 4 (i.e., number of rows or columns). Thus, the assignment is optimal. From the table we get I to B; II to C; III to D and IV to A.
13. There are five machines and five jobs are to be assigned and the associated cost matrix is as follows. Find the proper assignment. Machines
I II III IV V
A 6 12 3 11 15
B 4 2 7 1 10
C 8 11 10 7 11
D 16 19 12 23 21
E 9 5 7 6 10
Solution: In order to find the proper assignment, we apply the Hungarian method as follows: IA (Row reduction) Machines
I II III IV V
A 3 9 0 8 12
B 3 1 6 0 9
C 1 4 3 0 4
D 4 7 0 11 9
E 4 0 2 1 5
IB (Column reduction) Machines
I II III IV V
A 2 9 0 8 8
B 2 1 6 0 5
C 0 4 3 0 0
D 3 7 0 11 5
E 3 0 2 1 1
II (Zero assignment) Machines
I II III IV V
A 0** 7 0 6 6
B 2 1 8 0** 5
C 0 4 5 0 0**
D 1 5 0** 9 3
E 3 0** 4 1 1
Thus, we have got five assignments as required by the problem. The assignment is as follows: A -- I, B-- IV, C -- V, D -- III and E -- II. This assignment holds for table given in step IV but from theorem 1 it also holds for the original cost matrix. Thus from the cost matrix the minimum cost = 6+1+11+12+5=Rs.35.
14. A firm produces four products. There are four operators who are capable of producing
any of these products. The firm records 8 hours of day and allows 30 minutes for
lunch. The processing time in minutes and the profit for each product are given below:
Operator Product
A B C D
1 15 9 10 6
2 10 6 9 6
3 25 15 15 9
4 15 9 10 10
Profit per unit 8 6 5 4
Find the optimal assignment of products to operators?
Ans: Step 1
8 hours working = 480 minutes
(-)Lunch time = 030 minutes
Available hours = 450 minutes
No. of products each operator can produce can be found out by dividing 450 by the given processing
time.
For eg: Operator 1 can produce 450/15=30 units of product A at a profit rate of Rs. 8,implies a total
profit of Rs.240
Profit Matrix
Operator Product
A B C D
1 240 300 225 300
2 360 450 250 300
3 144 180 150 200
4 240 300 225 180
To solve the problem it has to be converted into a minimisation problem
450 is subtracted from each value:
Operator Product
A B C D
1 210 150 225 150
2 90 0 200 150
3 306 270 300 250
4 210 150 225 270
We use Hungarian method to solve the Matrix, we get the following table:
Step 2 Row-wise reduction,
Operator Product
A B C D
1 60 0 75 0
2 90 0 200 150
3 56 20 50 0
4 60 0 75 120
Step 3 Column-wise reduction
Operator Product
A B C D
1 4 0 25 0
2 34 0 150 150
3 0 20 0 0
4 4 0 25 120
Step 4 Since the minimum no. of lines to cover all zeros is 4 , further reduction is done by 4
Operator Product
A B C D
1 0 0 21 (0)
2 30 (0) 146 150
3 0 20 (0) 0
4 (0) 0 21 120
Thus the optimal assignment is:
Operator Product Profit
1 D 300
2 B 450
3 C 150
4 A 240
Total 1,140
15. ABC airline, operating 7 days a week has given the following time table. The crews
must have a minimum lay-over of 5 hours between flights. Obtain the pairing of flights
that minimises lay-over time away from home. For any given pairing the crew will be
based at the city that results the smallest lay-over.
Hyderabad-Delhi Delhi-Hyderabad
Flight No. Departure Arrival Flight No. Departure Arrival
A1 6AM 8AM B1 8AM 10AM
A2 8AM 10AM B2 9AM 11AM
A3 2PM 4PM B3 2PM 4PM
A4 8PM 10PM B4 7PM 9PM
We assume that all the crew is based at Hyderabad. Using this assumption we can obtain the lay-
over times for various combinations of flights. To illustrate, the flight A1 which starts from
Hyderabad at 6AM, reaches Delhi at 8AM. If it has to return as B1, the schedule time for which is
8PM , then it can do so after 24 hrs , since it has a lay-over time of 5 hours.
Similarly, lay-over times for other flights can be obtained,
Lay-over Time-Crew at Hyderabad
Flight B1 B2 B3 B4
A1 24 25 6 11
A2 22 23 28 9
A3 16 17 22 27
A4 10 11 16 21
When crew is based at Delhi,
Lay-over Time-Crew at Delhi
Flight B1 B2 B3 B4
A1 20 19 14 9
A2 22 21 16 11
A3 28 27 22 17
A4 10 9 28 23
Now in order to obtain the minimum lay-over time we select the corresponding lower value out of
the two tables. For eg: In combining A1-B1 we select 20, which is lower than 24.
Flight B1 B2 B3 B4
A1 20* 19* 6 9*
A2 22 21* 16* 9
A3 16 17 22 17*
A4 10 9* 16 21
*when crew is based at Delhi.
Now we use the Hungarian method to solve the matrix,
Reduced Cost Table,
Flight B1 B2 B3 B4
A1 14 13 (0) 3
A2 13 12 7 (0)
A3 (0) 1 6 1
A4 (1) (0) 7 12
Here there is no need to perform column reduction, as zero is obtained in each column.
The Optimal Solution
Flight No. Flight No. Lay-over Time Crew Based At
A1 B1 6 Hyderabad
A2 B2 9 Hyderabad
A3 B3 16 Hyderabad
A4 B4 9 Delhi
Total 40
16. A company has just developed a new item for which it proposes to undertake a
national television promotional campaign. It has decided to schedule a series of one-
minute commercials during peak audience viewing hours of 1 to 5 PM. To reach the
widest possible audience, the company want to schedule one commercials on each of
the networks and to have only one commercial appear during each of the four one
hour time blocks. The exposure ratings for each hour, which represent the number of
viewers per Rs.10,000 spent, are given below.
Viewing Hours
Network A B C D
1-2PM 27.1 18.1 11.3 9.5
2-3PM 18.9 15.5 17.1 10.6
3-4PM 19.2 18.5 9.9 7.7
4-5PM 11.5 21.4 16.8 12.8
(a) Which network should be scheduled each hour to provide the maximum audience exposure?
(b) How would the schedule change if it is decided not to use network A between 1 and 3 PM?
(a) to solve this problem we multiply each exposure rating by 10 to simplify the calculation. Further
being a minimisation problem each value is subtracted from the largest value to get the opportunity
loss matrix.
Opportunity Loss Matrix
Viewing Hours
Network A B C D
1-2PM 0 90 158 176
2-3PM 82 116 100 165
3-4PM 79 86 172 194
4-5PM 156 57 103 143
Now we use the Hungarian method to solve the above matrix
Row-wise reduction,
Viewing Hours
Network A B C D
1-2PM 0 90 158 176
2-3PM 0 34 18 83
3-4PM 0 7 93 115
4-5PM 99 0 46 86
Column –wise reduction,
Viewing Hours
Network A B C D
1-2PM 0 90 140 93
2-3PM 0 34 0 0
3-4PM 0 7 75 32
4-5PM 99 0 28 3
Since the number of lines covering all zeros is 3, we improve the solution,
Viewing Hours
Network A B C D
1-2PM 0 90 137 90
2-3PM 3 37 0 0
3-4PM 0 7 72 29
4-5PM 99 0 25 0
Further reduction,
Viewing Hours
Network A B C D
1-2PM 0 83 130 83
2-3PM 10 37 0 0
3-4PM 0 0 65 22
4-5PM 106 0 25 0
The Optimal solution,
Viewing Hours
Network
1-2PM A
2-3PM C
3-4PM B
4-5PM D
(b) For the given restriction, the prohibited time-slots are replaced M , row- wise reduction are
shown in the below table,
Viewing Hours
Network A B C D
1-2PM M 0 68 86
2-3PM M 16 0 65
3-4PM 0 7 93 115
4-5PM 99 0 46 86
Column –wise Reduction,
Viewing Hours
Network A B C D
1-2PM M 0 68 86
2-3PM M 16 0 65
3-4PM 0 7 93 115
4-5PM 99 0 46 86
Further Reduction
Viewing Hours
Network A B C D
1-2PM M 0 47 0
2-3PM M 37 0 0
3-4PM 0 28 93 50
4-5PM 78 0 25 0
Ans:
Viewing Hours Network Network
1-2PM B D
2-3PM C C
3-4PM A A
4-5PM D B
17. An airline that operates seven days a week has a time table shown below. Crew must
have a minimum lay-over of 6 hours between flights.Obtain the pairing of flights that
minimises layover time away from home. For any given pairing the crew will be based
at the city thath results in the smaller layover time.
Delhi-Calcutta Calcutta-Delhi
Flight No. Departure Arrival Flight No. Departure Arrival
1 7AM 9AM 101 9AM 11AM
2 9AM 11AM 102 10AM 12AM
3 1.30PM 3.30PM 103 3.30PM 5.30PM
4 7.30PM 9.30PM 104 8PM 10PM
Layover time converted when crew is located in Delhi
Flights 101 102 103 104
1 48 50 13 22
2 44 46 57 18
3 35 37 48 57
4 23 25 36 45
*Layover time can be converted into points by assigning 3- minutes=1 point
Similarly we assign points, when crew is based in Calcutta
Flights 101 102 103 104
1 40 38 27 18
2 44 42 31 22
3 35 51 40 31
4 17 15 36 43
Now we can select the minimum layover time from both the above tables
After solving the matrix by Hungarian method we get the Following Optimal Solution:
Flights 101 102 103 104
1 (0)
2 (0)
3 (0)
4 (0)
18. An office has four workers, and four tasks have to be performed. Workers differ in efficiency and tasks differ in their intrinsic difficulty. Time each worker would take to complete each task is given in the effectiveness matrix. How the tasks should be allocated to each worker so as To minimise the total man-hour?
Tasks
Workers
1 2 3 4
18
23
12
7
A 5 23 14
B 10 25 1
C 35 16 15
D 16 23 21
Solution: I - A, II - C, III - B, IV – D.
19.You are the given information about the cost performing different jobs by different persons.
The job person marking X indicates that the individual involved cannot perform the particular job.
Using this information, state(i)the optimal assignment of the job (ii) the cost of such assignment.
person Job
J1 j2 j3 j4 j5
P1 27 18 X 20 21
P2 31 24 21 12 17
P3 20 17 20 X 16
P4 22 28 20 16 27
Balancing the problem and assigning a high cost to pairing p1-j3 and p3-j4,we have the cost table
given in table below
person Job
J1 j2 j3 j4 j5
P1 27 18 M 20 21
P2 31 24 21 12 17
P3 20 17 20 M 16
P4 22 28 20 16 27
P5(dummy) 0 0 0 0 0
Now we can drive the reduced cost table. The cell with prohibited assignment continue to be shown
with cost element M, since M is defined to be extremely large so that subtraction and addition of a
values does not practically affect it. To test optimality lines are drawn to cover all zeros. Since the
number of lines covering all zeros is less than n, we select the lowest uncovered cell, which equals
4.with this value, we can obtain the revised reduced cost table as below.
Person job
J1 j2 j3 j4 j5
P1 9 0 M 2 3
P2 19 12 9 0 5
P3 4 1 4 M 0
P4 6 12 4 0 11
P5 0 0 0 0 0
Reduced cost table
Person job
J1 j2 j3 j4 j5
P1 9 (0) M 6 3
P2 15 8 5 (0) 1
P3 4 1 4 M (0)
P4 2 8 (0) 0 7
P5 0 0 0 4 0
The number lines covering zeros is equal to 5(=n). The assignment is p1-j2, p2-j4, p3-j5, p4-j3, while
job j1 would remained unassigned. This assignment pattern would cost 18+12+16+20=66 aggregate.
20 A company plans to assigns 5 salesmen to 5 district in which it operates. Estimate of sales
revenue in thousands of rupees for each salesman in different district are given in the following
table. In your option, what should be the placement of salesmen if the objective is to maximise
the expected sales revenue?
Expected sales area
Salesman District
D1 D2 D3 D4 D5
S1 40 46 48 36 48
S2 48 32 36 29 44
S3 49 35 41 38 45
S4 30 46 49 44 44
S5 37 41 48 43 47
Since it is maximisation problem, we would first subtract each of entries in the largest one,which
equals to 49 here. The resultant data given below.
Opportunity Loss Matrix
Salesman District
D1 D2 D3 D4 D5
S1 9 3 1 13 1
S2 1 17 13 20 5
S3 0 14 8 11 4
S4 19 3 0 5 5
S5 12 8 1 6 2
Now we shall proceed as usual
STEP 1: subtract minimum value in each row from every value in the row.
Reduced Cost Table 1
Salesman District
D1 D2 D3 D4 D5
S1 8 2 0 12 0
S2 0 16 12 19 4
S3 0 14 8 11 4
S4 19 3 0 5 5
S5 11 7 0 5 1
STEP 2,3 : subtract minimum value in each column from each value of column.
Reduced Cost Table 2
Salesman District
D1 D2 D3 D4 D5
S1 8 0 0 7 0
S2 0 14 12 14 4
S3 0 12 8 6 4
S4 19 1 0 0 5
S5 11 5 0 0 1
Since the number of lines covering all zeros is fewer than n, we select the least uncovered cell value
which equals 4. With this, we can modify the table as below.
STEP 4,5,6: find improved solution. Test for optimality and make assignment.
Salesman District
D1 D2 D3 D4 D5
S1 12 (0) 0 7 0
S2 (0) 10 5 10 0
S3 0 8 4 2 0
S4 23 1 (0) 0 5
S5 15 5 0 (0) 1
There are more than one optimal assignment possible because the existence of multiple zeros in
different rows and columns. The assignments possible are:
S1-D1, S2-D5, S3-D1, S4-D3, S5-D4 ; or
S1-D2, S2-D1, S3-D5, S4-D3, S5-D4; or
S1-D2, S2-D5, S3-D1, S4-D4, S5-D3; or
S1-D2, S2-D2, S3-D5, S4-D4, S5-D3.
Each of these assignment pattern would lead to expected aggregated sales equal to 231 thousand
rupees.
21. A solicitor firm employs typist on hourly piece- rate basis for their daily work. There are 5
typist and their charges and speed are different. According to an early understanding, only one job
is given to one typist the typist is paid for a full hour even when he works for a fraction of hour.
Find the least cost allocation for the following data:
Typist Rate/hour(Rs) Number of pages Typed/hour
Job No. Of pages
A 5 12 P 199
B 6 14 Q 175
C 3 8 R 145
D 4 10 S 298
E 4 11 T 178
Using the given information we3 first obtain the cost matrix, when the different jobs are performed
by different typist.
If typist A given job P, we should require 199/12=16X7/12 hours, hence be paid for 17 hours @ Rs. 5
per hour. The result in a cost of Rs.85 for the combination.
Total Cost Matrix
Typist Job
P Q R S T
A 85 75 65 125 75
B 90 78 66 132 78
C 75 66 57 114 69
D 80 72 60 120 72
E 76 64 56 112 68
Subtracting the minimum element of each row from all its element, we obtain RCT-1, as below.
Typist Job
P Q R S T
A 20 10 0 60 10
B 24 12 0 66 12
C 18 9 0 57 12
D 20 12 0 60 12
E 20 8 0 56 12
Now subtracting the minimum element of each column from all elements, we get RCT-2, as below
Typist Job
P Q R S T
A 2 2 0 4 0
B 6 4 0 10 2
C 0 1 0 1 2
D 2 4 0 4 2
E 2 0 0 0 2
Here the minimum number of lines cover all zeros is 4, which is smaller than the order,5, of given
matrix .Accordingly the revised table is prepared by considering the least uncovered value , equal to
1, and adjusting it with uncovered cell values and those lying at the intersection of lines. table
containsRCT-3.
Reduced Cost Table 3
Typist Job
P Q R S T
A 2 1 0 3 0
B 6 3 0 9 2
C 0 0 0 0 2
D 2 3 0 3 2
E 3 0 1 0 3
4 lines can cover all the zeros, Accordingly RCT-4 is drawn in table below.
Typist Job
P Q R S T
A 2 1 2 3 (0)
B 4 1 (0) 7 0
C 0 (0) 2 0 2
D (0) 1 0 1 0
E 3 0 3 (0) 3
In this case the minimum number of lines cover all zeros equal 5, which matches with order of the
matrix. Accordingly assignment have been made as below.
Typist Job Cost
A T 75
B R 66
C Q 66
D P 80
E S 112
Total 399
The optimal solution, however is not unique.
22. Well done company has taken third floor of a multi-storeyed building for a rent with a
view to locate one of their zonal offices. There are 5 main rooms in this to be assigned to be five
managers. Each room has its own advantages and disadvantages. Some have windows, some are
closer to wash rooms or to the canteen or secretarial pool. The rooms are of all different sizes and
shapes. Each of the five managers were asked to rank their room preferences amongst the room
301,301,303,304 and 305. Their preferences are recorded as below.
M1 M2 M3 M4 M5
302 302 303 302 301
303 304 301 305 302
304 305 304 304 304
301 305 303
302
Most of the manager did not list all the five rooms since they were not satisfied with some of these
rooms and they have left these from the list. Assuming that their preferences can be quantified by
numbers, find out as to which manager should be assigned to which room so that their total
preference ranking is minimum.
In the first step, we formulate the assignment problem using preference ranks as shown in table
below. Notice that the rooms not ranked by manager are represented by M-as prohibited
assignment.
Room Manager
M1 M2 M3 M4 M5
301 M 4 2 M 1
302 1 1 5 1 2
303 2 M 1 4 M
304 3 2 3 3 3
305 M 3 4 2 M
Since greater preferences are shown by lower numbers, the optimal solution calls for minimising the
total preference ranking. To solve we obtain RCT-1 by subtracting the least value from each value of
the row for each of these rows. As table below.
Room Manager
M1 M2 M3 M4 M5
301 M 3 1 M 0
302 0 0 4 0 1
303 1 M 0 3 M
304 1 0 1 1 1
305 M 1 2 0 M
Since each of columns as well as rows has a zero, lines are drawn to cover all zeros. Further, the
number of being equal to 5, the order of given matrix, assignment can be made, as shown below.
Room Manager
M1 M2 M3 M4 M5
301 M 3 1 M 0
302 0 0 4 0 1
303 1 M 0 3 M
304 1 0 1 1 1
305 M 1 2 0 M
The optimal assignment pattern is:
Room Manager Ranking
301 M5 1
302 M1 1
303 M3 1
304 M2 2
305 M4 2
23.A company has four sales representative who are to assigned to four different sales territories.
The monthly sales increase estimated for each sales representative for different sales territories(
in lakhs or rupees) are shown in table below.
Sales representative
Sales territories
I II III IV
A 200 150 170 220
B 160 120 150 140
C 190 195 190 200
D 180 175 160 190
Suggest the optimal assignment and the total maximum sales increase per month.
If for certain reasons, sales representative B can not be assigned to sales territory III, will the optimal
assignment schedule be different? If so, find that schedule and the effect of total sales.
The given problem being maximization type, need to be converted first into minimization type as
below; wherein the various elements are obtained by subtracting each of element in the given table
from 220.
Opportunity Loss Matrix
I II III IV
A 20 70 50 0
B 60 100 70 80
C 30 25 30 20
D 40 45 60 30
Subtracting minimum value in each row from each row element. RCT-1.
Reduced Cost Table-1
I II III IV
A 20 70 50 0
B 0 40 10 20
C 10 5 10 0
D 10 15 30 0
Subtracting minimum value in each column from each column element. RCT-2.
I II III IV
A 20 65 40 0
B 0 35 0 20
C 10 0 0 0
D 10 10 20 0
Since the minimum number lines to cover all zeros(3) is smaller than matrix order(4), rct-3 IS derived
as revised table.
Reduced Cost Table-3
I II III IV
A 10 55 30 0
B 0 35 0 30
C 10 0 0 10
D 0 0 10 0
With the number of lines to cover all zeros being equal to order of given matrix, assignment can be
made as shown in table below. However the problem has an alternative optimal solution as well.
Both of these are given below:
Alternative 1
Salesman Territory Sales
A IV 220
B I 160
C III 190
D II 175
Total 745
Alternative 2
Salesman Territory Sales
A IV 220
B III 150
C II 195
D I 180
Total 745
If salesman B cannot be assigned to territory III( Alternative 2), then alternative 1 above may be
adopted without adverse on sale.
24. Four sales manager(A,B,C and D) are to be assigned to four sales zones (P,Q,R and S). The
zones differ in their sales potential and sales manager differ in their capabilities. It is estimated
that a bright sales manager operating in each sale zone would bring in the following monthly sales.
Zone P= Rs.150 lakhs
Zone Q=Rs.120 lakhs
Zone R=Rs.100 lakhs
Zone S=Rs.80 lakhs
Since the sales manager differ in their capabilities, their effectiveness on a 10 point scale( on
which a bright sales manager is rated as 10) has been assessed as follows:
Sales manager A: 7 points
Sales manager B: 5 points
Sales manager C: 9 points
Sales manager D: 4 points
Suggest the optimal allocation of sales manager to sales zones to maximize monthly business.
SOLUTION:
The 1st step to determine proportionate sale( in Rs. Lakhs) in the four zones considering
effectiveness of different sales manager. The following table gives such calculation.
Sales manager
Sales zone
P Q R S
A 7/10x150=105 7/10x120=84 7/10X100=70 7/10X80=56
B 5/10X150=75 5/10X150=60 5/10X100=50 5/10x80=40
C 9/10x150=135 9/10x120=108 9/10X100=90 9/10X80=72
D 4/10X150=60 4/10X120=48 4/10X100=40 4/10X80=32
: the optimal solution is-
A to R : Rs. 70 lakhs
B to Q : Rs. 60 lakhs
C to S : Rs.72 lakhs
D to A : Rs.60 lakhs
Rs. 262 lakhs
26. An office has four workers, and four tasks have to be performed. Workers differ in efficiency and tasks differ in their intrinsic difficulty. Time each worker would take to complete each task is given in the effectiveness matrix. How the tasks should be allocated to each worker so as To minimise the total man-hour?
Tasks
Workers
1 2 3 4
A 5 23 14 18
B 10 25 1 23
C 35 16 15 12
D 16 23 21 7
Solution: I - A, II - C, III - B, IV - D
27. A taxi hire company has one taxi at each of five depots a,b,c,d and e. A customer requires a taxi in each town, namely A,B,C,D and E. Distances (in kms) between depots (origins) and towns (Destinations) are given in the following distance matrix:
a b c d e
A 140 110 155 170 180
B 115 100 110 140 155
C 120 90 135 150 165
D 30 30 60 60 90
E 35 15 50 60 85
Solution: A -e, B - c, C - b, D - a, E - d Minimum distance (km) = 180+110+90+30+60=470 km.
28.
Jobs
Workers
1 2 3 4
A 1 19 10 4
B 6 21 7 19
C 31 12 11 1
D 12 19 17 3
Solution- A - I, B - III, C - II, D – IV
29.
Jobs
Workers
1 2 3 4
A 15 17 24 16
B 10 15 12 13
C 17 19 18 16
D 13 20 16 14
Solution- A - II, B -III, C - IV, D - I Minimum cost = 17+12+16+13=58 30.
Jobs
Workers
1 2 3 4
A 8 26 17 11
B 14 29 5 27
C 40 21 20 17
D 19 19 24 10
Solution- I -A, II - C, III - B, IV – D 31.
Find the proper assignment of the assignment problem whose cost matrix is given as under.
1 2 3 4 5
A 10 6 4 8 3
B 2 11 7 7 3
C 5 10 11 4 8
D 6 5 3 2 5
E 11 7 10 11 7
Solution- A -V, B -I, C- IV, D -III, E – II. 32. Solve the following assignment problem.
1 2 3 4
A 8 9 10 11
B 10 11 12 13
C 13 14 15 13
D 9 11 14 10
Solution- A-2, B-3, C-4, D-1 or A-3, B-2, C-4, D-1