Decision Science.pdf

MBA PGDM Decision Science Presentation.

1. Decision Science
Aman Pandey-122
Harshada Patil-137
Mrunali Kale-147
Sagarika Bolar-155
Siddhant Bonsode –168
Shubham Surve-167

2. Assignment Model
•The assignment method is a way of allocating organizational
resources in which each resource is assigned to a particular task.
The resource could be monetary, personnel, or technological .The
assignment method is used to determine what resources are
assigned to which department, machine, or center of operation in
the production process.
•Hungarian Methos is the mostly used method of solving
assignment problems

3. Objective of the
Model
•Enhance production efficiency
•Control costs
•Maximize Profits

4. Types of
Assignment Model
•Balanced
•Unbalanced
•Minimized
•Maximized
•Prohibitive
•Non-Prohibitive

5. Unbalanced
Solve the following unbalanced assignment problem of minimizing the
total time for performing all the jobs.
Job
Workers
1 2 3 4 5
A 5 2 4 2 5
B 2 4 7 6 6
C 6 7 5 8 7
D 5 2 3 3 4
E 8 3 7 8 6
F 3 6 3 5 7

6. Solution:
In this problem, the number of jobs is less than the number of workers so we have to
introduce a dummy job with zero duration .The revised assignment problem is as
follows:-
Step 1: The cost Table.
Job
Workers
1 2 3 4 5 6
A 5 2 4 2 5 0
B 2 4 7 6 6 0
C 6 7 5 8 7 0
D 5 2 3 3 4 0
E 8 3 7 8 6 0
F 3 6 3 5 7 0

7. Step 2: Find the (FRCT)
Job
Workers
1 2 3 4 5 6
A 5 2 4 2 5 0
B 2 4 7 6 6 0
C 6 7 5 8 7 0
D 5 2 3 3 4 0
E 8 3 7 8 6 0
F 3 6 3 5 7 0

8. Step 3: Find the(SRCT)
Job
Workers
1 2 3 4 5 6
A 3 0 1 0 1 0
B 0 2 4 4 2 0
C 4 5 2 6 3 0
D 3 0 0 1 0 0
E 6 1 4 6 2 0
F 1 4 0 3 3 0

9. Step 4:Determine an Assignment
By using the Hungarian method, the assignment is made as follows:
Job
Workers
1 2 3 4 5 6
A 3 0 1 0 1 0
B 0 2 4 4 2 0
C 4 5 2 6 3 0
D 3 0 0 1 0 0
E 6 1 4 6 2 0
F 1 4 0 3 3 0

10. Step 5:The solution obtained in Step 4 is not optimal. Because we were able to make
five assignments when six were required.
Step 6:Cover all the zeros of the table shown in the Step 4 with five lines (since
already we made five assignments).
Check row E since it has no assignment. Note that row B has a zero in column 6,
therefore check column 6.
Then we check row C since it has a zero in column 6. Note that no other rows and
columns are checked. Now we may draw five lines through unchecked rows (row A, B,
D and F) and the checked column (column 6).
This is shown in the table given below:

11. 1 2 3 4 5 6
A 3 0 1 0 1 0
B 0 2 4 4 2 0
C 4 5 2 6 3 0
D 3 0 0 1 0 0
E 6 2 4 6 2 0
F 1 4 0 3 3 0
job
Workers

12. Step 7:Develop the new revised table.
Examine those elements that are not covered by a line in the table given in Step 6.Take
the smallest element, in this case the smallest element is 1. 6 Subtract this smallest
element from the uncovered cells and add 1 to elements (A6, B6, D6 and F6) that lie at
the intersection of two lines. Finally, we get the new revised cost table, which is shown
below:
1 2 3 4 5 6
A 3 0 1 0 1 1
B 0 2 4 4 2 1
C 3 4 1 5 2 0
D 3 0 0 1 0 1
E 5 0 3 5 1 0
F 1 4 0 3 3 1
Job
Workers

13. Step 8: Now, we go to Step 4 and repeat the procedure until we arrive at an optimal
solution (assignment).
Step 9: Determine an assignment.
1 2 3 4 5 6
A 3 0 1 0 1 1
B 0 2 4 4 2 1
C 3 4 1 5 2 0
D 3 0 0 1 0 1
E 5 0 3 5 1 0
F 1 4 0 3 3 1
Job
Worker

14. Since the number of assignments equal
to the number of rows (columns), the
assignment shown in the above table is
optimal. Thus, the worker A is assigned
to Job 4, worker B is assigned to job 1,
worker C is assigned to job 6, worker D is
assigned to job 5, worker E is assigned to
job 2, and worker F is assigned to job 3.
Since the Job 6 is dummy so that worker
C can’t be assigned. The total minimum
time is:
A4 + B1 + D5 + E2 + F3
2 + 2 + 4 + 3 + 3 = 14

15. Prohibtive Explanation
•say there are 4 contractors –C1, C2, C3 &
C4. And there are 4 roads to be repaired –
R1, R2, R3 & R4. But contractor C2
cannot or is not allowed to work on R3.
This is a prohibited problem. Then we
assign a very high or infinite value
(represented by M) to C2-R3 and proceed
with solution.

16. Prohibitive with Balanced
An Airline company has drawn a new flight schedule involving 5 flights to assissts in
allocating 5 pilots to this flights. It has asked them to stake a number out of 10. The
higher the number the greater is the preference. Certain of this Flights are unsuitable to
some pilots owing to some domestic reasons. This has been marked with “X”.
What should be the allocation of the pilots to flights in order to meet as many
preferences as possible.
Flight Number
Pilots1 2 3 4 5
A 8 2 X 5 4
B 10 9 2 8 4
C 5 4 9 6 X
D 3 6 2 8 7
E 5 6 10 4 3

17. •Solution
Step 1-First describe the sumtype
Min/Max -Max
Bal/Unbal -Bal
Prohibitive/ Non-prohibitive-Prohibitive
In Prohibitive Problem we will not do anything to the prohibited cell till the end.
Step 2. -As the problem is of Maximum matrix we need to convert it to minimum matrix
by deducting all the elements from the highest number (here it is 10)
So, the New matrix is as follows:
Flight Number
Pilots 1 2 3 4 5
A 2 8 X 5 6
B 0 1 8 2 6
C 5 6 1 4 X
D 7 4 8 2 3
E 5 4 0 6 7

18. •Step 3-Subtract the smallest element of each row from all the elements in the row of
the given cost matrix. See that each row contains atleast one zero. So, the New
Matrix is as follows
Flight Number
Pilots 1 2 3 4 5
A 0 6 X 3 4
B 0 1 8 2 6
C 4 5 0 3 X
D 5 2 6 0 1
E 5 4 0 6 7

19. •Step 4-Subtract the smallest element of each Column from all the elements in the
Column of the given cost matrix. See that each Column contains atleast one zero.
So, the New Matrix is as follows
Flight Number
Pilots 1 2 3 4 5
A 0 5 X 3 3
B 0 0 8 2 5
C 4 4 0 3 X
D 5 1 6 0 0
E 5 3 0 6 6

20. Step 5-Determine an Assignment
By using the Hungarian method, the assignment is made as follows:
•As the No. of lines is not equal to the No of Rows/columns, the solution is not
optimal.
So, create a New matrix by adding minimum element value of the matrix to the
intersection point and deducting it from the uncovered element value. Covered
elements other than intersection should be kept as it is.
Flight Number
Pilots 1 2 3 4 5
A 0 5 X 3 3
B 0 0 8 2 5
C 4 4 0 3 X
D 5 1 6 0 0
E 5 3 0 6 6

21. Step 6-Determine an Assignment.
By using the Hungarian method, the assignment is made as follows:
The Optimal Allocation is as under;
A -1 -8
B -2 -9
C -4 -6
D -5 -7
E -3 -10
Total score is 40
Flight Number
Pilots 1 2 3 4 5
A 0 2 X 0 0
B 3 0 11 2 5
C 4 1 0 0 X
D 8 1 9 0 0
E 5 0 0 3 3

22. Basic Sum
•Solve the following assignment
problem. Cell values represent cost
of assigning job A, B, C and D to
the machines I, II, III and IV.

23. •Solution:
•Herethenumberofrowsandcolumnsareequal.
•∴Thegivenassignmentproblemisbalanced.Nowletusfindthesolution.
•Step1:Selectasmallestelementineachrowandsubtractthisfromalltheelementsinitsrow.
•Lookforatleastonezeroineachrowandeachcolumn.Otherwisegotostep2.

24. •Step 2:Select the smallest element in each
column and subtract this from all the
elements in its column.
•Since each row and column contains at least
one zero, assignments can be made.

25. •Step 3 (Assignment):
•Examine the rows with
exactly one zero. First
three rows contain more
than one zero. Go to row
D. There is exactly one
zero. Mark that zero by
(i.e.) job D is assigned to
machine I. Mark other
zeros in its column by ×.

26. •Step 4:Now examine the columns with exactly one zero. Already there is an assignment in column I. Go to
the column II. There is exactly one zero. Mark that zero by . Mark other zeros in its row by ×.
•ColumnIIIcontainsmorethanonezero.Therefore,proceedtoColumnIV,thereisexactlyonezero.Mark
thatzeroby.Markotherzerosinitsrowby×.

27. •Step 5:Again, examine the rows.
Row B contains exactly one zero.
Mark that zero by .

28. •Thus, all the four assignments have been
made. The optimal assignment schedule and
total cost is
•The optimal assignment (minimum) cost
•= ₹ 38

29. Thank You