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Variance and Standard Daviation Part 4-Basic Mathematics-Assignment Solution, Exercises of Mathematics

Institute of Mathematical SciencesMathematics

This is solution to assignment of Basic Mathematics course. This was submitted to Karunashankar Sidhu at Institute of Mathematical Sciences. It includes: Phase, Method, Variance, Standard, Lower, Bound, Artificial, Simplex

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

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Assignment: # 04 (Spring-2012)
Mth601 (Operations Research)
Lecture: 24– 30
Total Marks = 30
Due date: 14th June, 2012
INSTRUCTIONS
Please read the following instructions before attempting to solve this assignment
1. In order to attempt this assignment you should have full command on
Lecture # 24 to Lecture # 30
2. Try to get the concepts, consolidate your concepts and ideas from these questions
which you learn in Lecture # 24 to Lecture # 30
3. You should concern recommended books for clarify your concepts as handouts
are not sufficient.
4. Try to make solution by yourself and protect your work from other students. If we
found the solution files of some students are same then we will reward zero marks
to all those students.
5. You are supposed to submit your assignment in Word format any other formats
like scan images, PDF format etc will not be accepted and we will give zero
marks to these assignments.
6. Assignments through e-mail will be acceptable after (before uploading the
solution file) due date but with deduction of 30% of obtained marks.
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Download Variance and Standard Daviation Part 4-Basic Mathematics-Assignment Solution and more Exercises Mathematics in PDF only on Docsity!

Assignment: # 04 (Spring-2012) Mth601 (Operations Research) Lecture: 24– 30 Total Marks = 30

Due date: 14

th

June, 2012

INSTRUCTIONS

Please read the following instructions before attempting to solve this assignment

1. In order to attempt this assignment you should have full command on Lecture # 24 to Lecture # 30 2. Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in Lecture # 24 to Lecture # 30 3. You should concern recommended books for clarify your concepts as handouts are not sufficient. 4. Try to make solution by yourself and protect your work from other students. If we found the solution files of some students are same then we will reward zero marks to all those students. 5. You are supposed to submit your assignment in Word format any other formats like scan images, PDF format etc will not be accepted and we will give zero marks to these assignments. 6. Assignments through e-mail will be acceptable after (before uploading the solution file) due date but with deduction of 30% of obtained marks.

Max: z =2 x 1 – x 2 + x 3 by using TWO phase method, Subject to: x 1 + x 2 –3 x 3 ≤ 8 4 x 1 – x 2 + x 3 ≥ 2 2 x 1 +3 x 2 – x 3 ≥ 4 x 1 , x 2 , x 3 ≥ 0

Solution: In first phase we minimize the objective function in terms of artificial variables but to express this in terms of non-basic variables.

Since both artificial variables leaves the basis and objective is minimized to zero. So we proceed to 2nd^ phase.

2 x 2 – x 3 ≥4, x 1 , x 2 ≥0 and x 3 is unrestricted. Solution : Since x 3 is unrestricted so it can be replace by difference of two supposed decision variables t 1 and t 2 such that x 3 = t 1 – t 2 So the complete LP problem becomes; Min z = x 1 + x 2 + t 1 – t 2 Subject to x 1 – x 3 +4( t 1 – t 2 )= x 1 –2 x 2 ≤ 3 2 x 2 –( t 1 – t 2 )≥ 4 x 1 , x 2 , t 1 , t 2 ≥ 0 Since the modified LP problem contains constraints of ‘ = ’ and ‘ ≥ ’ type, so big-M method can be applied and for computational purpose put M =100.