CO 350 Assignment 1 – Winter 2010, Solutions
Due: Friday January 15 at 10:25 a.m.
Question #
Max. marks
Part marks
1
7
2
4
1
1
1
1
3
3
4
8
2
2
2
2
5
4
2
2
6
4
1
3
7(Bonus)
(2)
Total
30 + (2)
1. A widget plant makes 800 typeA widgets, 600 typeB widgets, and 400 typeC widgets every
day. Each of these products can be sold either with
or without
a wireless card. The total
number of widgets (of any of the 3 types) that can be ﬁtted with wireless cards during a
normal working day is 800. In addition, up to 220 widgets (of any of the 3 types) can be ﬁtted
with wireless cards on overtime, at a higher cost. The net proﬁts are as follows:
Without
Regular time install of
Overtime install of
wirelesscard
wirelesscard
wirelesscard
TypeA widget
$10
$20
$15
TypeB widget
$20
$30
$25
TypeC widget
$20
$50
$35
The objective is to ﬁnd a schedule that maximizes the total net proﬁt. Give a formulaton
as a linear programming problem.
Remarks:
Distribution of the points: 1 for setup, 1 for objective function, 1 for nonnegativity
constraints, and 4 for the other constraints.
Solution:
We use the following 9 nonnegative variables:
type A
type B
type C
number of widgets with no wireless card
x
1
x
2
x
3
number of widgets with cards installed on regular time
y
1
y
2
y
3
number of widgets with cards installed on overtime
z
1
z
2
z
3
We write the objective function as
10
x
1
+ 20
x
2
+ 20
x
3
+ 20
y
1
+ 30
y
2
+ 50
y
3
+ 15
z
1
+ 25
z
2
+ 35
z
3
We have three constraints for the total number of widgets produced of each type:
Type A:
x
1
+
y
1
+
z
1
= 800
Type B:
x
2
+
y
2
+
z
2
= 600
Type C:
x
3
+
y
3
+
z
3
= 400
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThere are two more constraints for the total number of widgets that can be ﬁtted with wireless
cards in normal time, and overtime:
y
1
+
y
2
+
y
3
≤
800
z
1
+
z
2
+
z
3
≤
220
.
The LP formulation is:
max
10
x
1
+ 20
x
2
+ 20
x
3
+ 20
y
1
+ 30
y
2
+ 50
y
3
+ 15
z
1
+ 25
z
2
+ 35
z
3
s.t.
x
1
+
y
1
+
z
1
= 800
x
2
+
y
2
+
z
2
= 600
x
3
+
y
3
+
z
3
= 400
y
1
+
y
2
+
y
3
≤
800
z
1
+
z
2
+
z
3
≤
220
x
1
,
x
2
,
x
3
,
y
1
,
y
2
,
y
3
,
z
1
,
z
2
,
z
3
≥
0
Note:
There are of course other correct solutions. In particular, it is possible to make a
formulation using only the variables
y
1
,y
2
,y
3
,z
1
,z
2
,z
3
, since it is possible to express the
x
j
variables in terms of them. In this way one can get the following LP formulation:
max
10
y
1
+ 10
y
2
+ 30
y
3
+ 5
z
1
+ 5
z
2
+ 15
z
3
(+28
,
000)
subject to
y
1
+
z
1
≤
800
y
2
+
z
2
≤
600
y
3
+
z
3
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 GUENIN
 Optimization, feasible solution

Click to edit the document details