Operating system

1. TYPE 3 ASSIGNMENT PROBLEM
PREPARED BY :SOORAJ. R

2. Type 3 problem
1. Step1 : Subtract the row minima
2. Step2 : Subtract the column minima

3.  Cross out all zeros In the Corrsponding this assigned now
 Proceeds in the same way until the last row is examined.
 Cross out all the zeros in the corresponding row of this assigned column.
 If any zeros are still left out or needs to be treated then we use trial and error
method.
Step 3

4. Step4:COVER ALL ZEROS WITH A MINIMUM
NUMBER OF LINE’S
(a)Mark all the row for which assignments have not been made.
(b)Mark corresponding columns which have zeros of marked row.
(c) Mark rows which have assignment of marked columns.
(d)Repeat the above steps.
(e)Draw lines through unmarked rows and marked columns

5. Step 5
Select the smallest of the elements that is not covered by lines.
Subtract it from all the elements that do not have a line through them.
Add it to every elements that line at the intersection of two lines.
Leave the rest or remaining elements of the matrix unchanged.

6. Step 6
 Re-apply the steps 3-5to get the modified matrix and also
the optimum solution

7. Example
Question:
A project work consist of three major jobs for which three contractor have
submitted tenders. The tender amounts quoted in lakhs of rupees are given in
the matrix below.
Job
a b c
1 17 25 31
Contractors 2 10 25 16
3 12 14 11

8. a b c
1 0 5 14
2 0 12 6
3 1 0 0
Subtracting the least element of every row from all elements of
that row and then subtracting the least element of every column
from all elements of that column. Then making zero
assignments

9. Now all the rows and columns do not have assignment.
So mark in the 2 nd row as it has no assignment. Then
mark in the first column as it has zeros of ticked row.
Then mark in the first row as it has assigned zero of
ticked column. There is only one zero for this ticked row
and under it there is mark already made draw lines in
the unticked rows and ticked columns

10. a b c
1 0 5 14
2 0 12 6
3 1 0 0
a b c
1 0 0 9
2 0 7 1
3 6 0 0

11. The smallest element not covered by the lines is 5. i) subtract 5 from all elements not
covered by the lines. Ii) Add 5 to the elements lying at the intersection of two lines.
Then make zero assignment. Now every row and every column has assignment. The
solution is 1to b, 2to a, 3toc.The total cost =25+10+11=46 lakh rupees.

12. Thank you