Wednesday, January 6, 2016

OT-2015 Question 3 Answer-- LP

Answer available here:
https://drive.google.com/file/d/0B6PHuk6I5SxbRzdxdVRTeGg3LWM/view?usp=sharing

You can post your approach as comment here.

For ready reference, question is copy-pasted here:


Below Simplex Table is for a typical product mix problem with three products and 3 resources.  It is an optimal table. Why?
It also has alternate optimal.  Find that.                                               6 marks


p
x1
x2
x3
s1
s2
s3
R.H.S.
P
1
0
0
0
10
15
0
7500
x3
0
0
10
1
8
-10
0
305
X1
0
1
25
0
-5
8
0
670
s3
0
0
15
0
12
5
1
320

a) Let us assume that one unit of these products or resources require one unit of storage space each. Which of the two optimum solutions obtained by you will require lesser storage than other?                                            6 marks                                                                 
b) Find one more optimal solution of the same problem.                6 marks
c) What will happen to the objective function value (increase, decrease or remain unchanged) If we choose s1 as entering variable and do the iteration in the table given in this question.                                                    2 marks

1 comment:

  1. Common Mistakes and marks:
    Students have mixed the feasibility and optimality. Wrong reason of optimality, wrong reason of having alternate solution but correct iteration has been given 3-4 marks. X2 taken as entering variable will give alternate solution not because it is a non-basic decision variable but because it is a non-basic variable with 0 coefficient in the objective function line. Even a slack variable meeting this criteria will give alternate solution. Single calculation mistake has been ignored while awarding marks.

    For finding the storage space, even s3 will occupy that because it is giving the left-out resource. Those who have not made this clear but did rest of the things right have got 3-4 marks.

    If an LP has feasible solutions then there will be either unique optimum solution or infinite number of alternate optimum solutions. Even though you have answered question related to this on Edmodo right, here nobody found the third optimum.

    For finding the impact on objective function value if s1 is taken as entering variable, iteration was not required. Students wrote that s1 cannot be taken as entering variable or some said nothing will happen to objective function etc. Such answers got 0.

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