MGMT 406 Operations Research
Answer set 3
(Reference chapters 3-B.W. Taylor III-2007)
1. Southern Sporting Goods Company makes basketballs and footballs. Each product is produced
from two resources—rubber and leather. The resource requirements for each product and the total
resources available are as follows: Each basketball produced results in a profit of $12, and each
football earns $16 in profit.
Product
Resource Requirements per Unit
Rubber(lb)
Leather (ft.2)
Basketball
3
4
Football
2
5
Total resources available
500 lb.
800 ft.2
a. Formulate a linear programming model to determine the number of basketballs and footballs to
produce to maximize profit.
b. Transform this model into standard form.
c. Solve the model formulated in the Problem for Southern Sporting Goods Company graphically.
d. Identify the amount of unused resources (Le., slack) at each of the graphical extreme points.
e. What would be the effect on the optimal solution if the profit for a basketball changed from
$12 to $13? What would be the effect if the profit for a football changed from $16 to$15?
f. What would be the effect on the optimal solution if 500 additional pounds of rubber could be
obtained? What would be the effect if 500 additional square feet of leather could be obtained?
For the linear programming model for Southern Sporting Goods Company, formulated in section
a and solved graphically in section b:
g. Determine the sensitivity ranges for the objective function coeffıcients and constraint quantity
values, using graphical analysis.
h. Determine the shadow prices for the resources and explain their meaning.
1
ANS:
2
3
4
(h) Maximize Z = $12x + $16x
1
2
subject to:
3x + 2x  500 (rubber)
1
2
1
2
4x + 5x  800 (leather)
x ,x 0
1
2
A profit-max primal problem has a cost-min dual problem and vice-versa. The solution of
a dual problem yields the shadow prices. They give the change in the value of the obj.
Function per unit change in each constraint in the primal problem.
Min C= $500v1 + $800v2
subject to:
3v1 + 4v2  12
2v1 + 5v2  16
v1, v2  0
3v1 + 4v2 = 12
v2 = 3-(3/4) v1
2v1 + 5(3-(3/4) v1=16
v1=-4/7, v1 = 0 so v1=0
v2 = 16/5=3.20
Briefly, this means that every additional foot square increases in leather, profit will be
expected to increase by the amount of 3.20 dollar.
5
2. An Aluminium Company produces three grades (high, medium, and low) of aluminium at two
mills. Each mill has a different production capacity (in tons per day) for each grade, as follows:
Mill
Aluminium Grade 1 2
High
Medium
Low
6 2
2 2
4 10
The company has contracted with a manufacturing firm to supply at least 12 tons of high-grade
aluminium, 8 tons of medium-grade aluminium, and 5 tons of low-grade aluminium. It costs
United $6,000 per day to operate mill 1 and $7,000 per day to operate mill 2. The company wants
to know the number of days to operate each miii to meet the contract at the minimum cost.
a. Formulate a linear programming model for this problem.
b. Solve the linear programming model formulated in section a for United Aluminium Company
graphically.
c. How much extra (i.e., surplus) high-, medium-, and low-grade aluminium does the company
produce at the optimal solution?
d. What would be the effect on the optimal solution if the cost of operating mill 1 increased from
$6,000 to $7,500 per day?
e. What would be the effect on the optimal solution if the company could supply only 10 tons of
high-grade aluminium?
f. Identify and explain the shadow prices for each of aluminium grade contract requirements
g. Determine the sensitivity ranges for the objective function coefficients and for the constraint
quantity values.
ANS:
6
(f) Min C= $6000x + $7000x
1
2
subject to:
6x + 2x  12 (high)
1
2
1
2
2x + 2x  8 (medium)
4x + 10x  5 (low)
1
2
x,x 0
1
2
A cost-min primal problem has a profit-max dual problem and vice-versa. The soln. Of a
dual problem yields the shadow prices. They give the change in the value of the obj.
Function per unit change in each constraint in the primal problem.
7
Max Z= $12v + $8v + $5v
1
2
3
subject to:
6v + 2v + 4v  6000
1
2
3
2v + 2v + 10v  7000
1
2
3
v,v , v0
1
2 ,
3
Since v is a slack variable, we set v =0
3
3
6v + 2v =6000
1
2
v = 3000-3 v
2
1
2v +6000-(3- 6 v =16
1
1
v =- 250, v  0 so v =0
1
1
1
v = 3000
2
Briefly, this means that every additional aluminium ton decrease in medium grade,
cost will be expected to decrease by the amount of 3000 dollars.
3. A company produces two types of cotton cloth—denim and corduroy. Corduroy is a heavier
grade of cotton cloth and, as such, requires 7.5 pounds of raw cotton per yard, whereas denim
requires 5 pounds of raw cotton per yard. A yard of corduroy requires 3.2 hours of processing
time; a yard of denim requires 3.0 hours. Although the demand for denim is practically unlimited,
the maximum demand for corduroy is 510 yards per month. The manufacturer has 6,500 pounds
of cotton and 3,000 hours of processing time available each month. The manufacturer makes a
profit of $2.25 per yard of denim and $3.10 per yard of corduroy. The manufacturer wants to
know how many yards of each type of cloth to produce to maximize profit.
a. Formulate a linear programming model for this problem.
b. Transform this model into standard form.
c. Solve the model graphically and calculate How much extra cotton and processing time are left
over at the optimal solution? Is the demand for corduroy met?
d. What is the effect on the optimal solution if the profit per yard of denim is increased from
$2.25 to $3.00? What is the effect if the profit per yard of corduroy is increased from $3.10 to
$4.00?
e. What would be the effect on the optimal solution if the company could obtain only 6,000
pounds of cotton per month?
f. Determine the sensitivity ranges for the objective function coefficients and for the constraint
quantity values.
g. Determine the shadow prices for additional cotton or processing time. Explain your answer.
8
ANS:
9
Max Z= $2.25x + $3.10x
1
2
subject to:
5x + 7.5x  6500 (Cotton)
1
2
3x + 3.2x +  3000 (Labour)
1
2
x  510 (demand)
2
x , x , 0
1
2
x =510
2
5x + 7.5x  6500 (Cotton)
1
2
1
2
5x + 7.5x  6000 (Cotton) decrease to 6000 so
5x + 7.5 (510) = 6000
1
x =435.
1
(g) Min C= $6500v + $3000v + $510v
1
2
3
subject to:
5v + 3v =2.25
1
2
7.5v + 3.2v + v  3.10
1
2
v,v , v0
1
2 ,
3
3
Since v is a slack variable, we set v =0
3
3
10
v =0.32
1
v =0.20
2
Briefly, this means that every additional pound of cotton increases whereas every additional
processing hours increases, profit will be expected to increase by the amount of 0.32 and
0.20 dollars respectively.
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