=========================preview======================
(IELM202)[2013](f)midterm_sample~=vmzqqj^_87983.pdf
Back to IELM202 Login to download
======================================================
IELM 202: Introduction to Operations Research

Sample Mid-Term Exam: Solutions


Question 1.
The feasible region of an LP is shown as the shadow area in the following figure. Suppose the objective of the LP is to maximize the sum of x and y.



(1) Write the Formulation of the LP:

Maximize z=x+y
Subject to x-y3 (1)
6x+5y30 (2)
y4 (3)
x,y0

(2) Identify the corner point that corresponds to the optimal solution.

The optimal solution is at the corner point of line (2) and line (3), which turns out to be x=5/3, y=4, and z=17/3.

(3) Transform the LP into the standard from

Maximize z=x+y
Subject to x- y +s1 =3
6x+5y +s2 =30
y +s3=4
x,y,s1,s2,s30

(4) Initial simplex tableau:



Coefficients


Row
Basic
z
x
y
s1
s2
s3
RHS

0
Z
1
-1
-1
0
0
0
0

1
s1
0
1
-1
1
0
0
3

2
s2
0
6
5
0
1
0
30

3
s3
0
0
1
0
0
1
4



Question 2.

Zales Jewelers uses rubies and sapphires to produce two types of rings. A type-1 ring requires 2 rubies, 3 sapphires, and 1 hour of labor. A type-2 ring uses 3 rubies, 2 sapphires, and 2 hours of labor. Each type-1 ring sells for $400, and type-2 ring sells for $500. At present, Zales has 105 rubies, 120 sapphires, and 80 hours of labor. Market demand requires that the company produce at least 10 type-1 rings and at least 15 type-2 rings. The goal is to maximize profit. To make an LP formulation, we have defined the following variables:

X1 = type 1 rings produced
X2 = type 2 rings produced

(1) Formulate an LP to maximize the profit:

Max Z= 400X1 + 500X2
s.t. 2X1 + 3X2 105 (1)
3X1+2X2 120 (2)
X1+2X2 80 (3)
X1 10 (4)
X2 15 (5)
X1,X2 0

(2) Given the partial sensitivity report of Excel Solver, answer the following question:


Final
Reduced
Objective
Allowable
Allowable

Name
Value
Cost
Coefficient
Increase
Decrease

x1
30
0
400
350
66.67



Suppose the price of type 1 ring becomes $350, how will the optimal solution change in terms of X1 and the objective function value? Why?

The coefficient decrease is 400-350=50<66.67. So the current BFS is still optimal. Thus X1=30, but the objective function value will be decreased by 50*30=1500


(3) Given the partial sensitivity report of Excel Solver, answer the following question:


Final
Shadow
Constraint
Allowable
Allowable

Name
Value
Price
R.H. Side
Increase
Decrease


105
140
105
25
0


120
40
120
0
33.33


60
0
80
1E+30
20


Suppose a wor