Operations research: An Introduction to Management Science, Quantative Approaches to Decision Making.

1. halNah-ALAbdelhamidFatenDr.
AbuMandeelAhmedAmani
)(EBUS3305researchOperations
2017Sep26,
Science,ManagementtontroductionIAn:IIAssignment
Making.DecisiontoApproachesQuantative

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Given data:
We have two models, regular and catcher's model.
No. of hrs.Dep.
900 hrs.Cutting
&sewing
300 hrs.Finishing
100 hrs.Packaging&
ship.
ProfitPackaging
shippingand
inishingFandCutting
sewing
Model
5$1/81/21Regular
model
$81/41/33/2Catcher`s
model

3. SOLUTION:
( A ) Let: A: No. of units of regular model.
B: No. of units of catcher's model.
Max. ( Z ) = 5 A + 8 B
Objected to:
Cutting & sewing: A+ 3/2 B ≤ 900
Finishing: 1/2 A + 1/3 B ≤ 300
Packaging & shipping: 1/8 A + 1/4 B ≤ 100
A, B ≥ 0
( B )
According to graphical solution, we have FOUR points that shape Feasible region:
Point ( W ): ( 0 ,400 ) Max ( profit ) = 5 A + 8 B
= 5 ( 0 ) + 8 ( 400 )
= 3200
Point ( X ): ( 500, 150 ) Max ( profit ) = 5 ( 500 ) + 8 ( 150 )
=3700
the maximum profit, 500 units of regular model and 150 units of catcher's model.
Point ( Y ): ( 600, 0 ) Max ( profit ) = 5 ( 600 ) + 8 ( 0 )
=3000
Point ( Z ): ( 0 ,0 ) Max ( profit ) = 5 (0 ) + 8 ( 0 )
= 0
( C ) Max ( profit ) = 5 ( 500 ) + 8 ( 150 )
= 3700

4. ( D ) In cutting and sewing :
A + 3/2 B = ( 500 ) + 3/2 ( 150 ) = 725 hrs.
In finishing:
1/2 A + 1/3 B = 1/2 ( 500 )+ 1/3 ( 150 ) = 300 hrs.
Packaging and shipping:
1/8 A + 1/4 B = 1/8 ( 500 ) + 1/4 (150 ) =100 hrs.
( E )
SLACK TIME = CAPACITY – USAGE TIME
In cutting and sewing department:
900 – 725 = 175 hrs.
In finishing department:
300 – 300 = 0 hrs.
In Packaging and Shipping department:
100 – 100 = 0 hrs.

5. small manufacturer of plastic products used in theJackson Hole Manufacturing is a
automotive and computer industries. One of its major contracts is with a large
computer company and involves the production of plastic printer cases for the
r cases are produced ontwocomputer company’s portable printers. The printe
100 machine has a production capacity of 25-molding machines. The M-injection
200 machine has a productioncapacity of 40 cases-printer cases per hour, and the M
the printer cases;per hour. Both machines use the same chemical material to produce
200 uses 50-100 uses 40 pounds of the raw material per hour and the M-the M
pounds per hour. The computer company asked Jackson Hole to produceas many of
sonthe cases during the upcoming week as possible; it will pay $18 for each caseJack
Hole can deliver. However, next week is a regularly scheduled vacation period for
most of Jackson Hole’s production employees; during this time, annual maintenance
is performed for all equipment in the plant. Because of the downtime for
200-100 will be available for no more than 15 hours, and the M-e Mmaintenance, th
will be available for no more than 10 hours. However, becauseof the high setup cost
involved with both machines, management requires that, if production is scheduled
machine must be operated for at least 5 hours. The supplier ofon either machine, the
the chemical material used in the production process informed Jackson Hole that a
maximum of 1000 pounds of the chemical material will be available for next week’s
s raw material is $6 per pound. In addition to the rawproduction; the costfor thi
100 and-material cost, JacksonHole estimates that the hourly costof operating the M
200 are $50 and $75, respectively.-the M
a. Formulate a linear programming model that can be used to maximize the
contribution to profit.
b. Find the optimal solution.
Given data:
Capacity of each machine:
M-100 = 25 printer cases per hour.
M-200 = 40 printer cases per hour.
Usage of raw material:
M-100 = 40 pounds of R.M per hour.
M-200 = 50 pounds of R.M per hour.
The costof raw material is $6 per pound.

6. SOLUTION:
( A )
We can assume that:
X = No. of hrs spent on the M-100 machine
Y = No. of hrs spent on the M-200 machine
T.C= costofR.M × No. of pounds
6(40) X + 6(50) Y + 50 X + 75Y = 290 X + 375 Y
T.R =
25(18) X + 40(18) Y = 450 X + 720 Y
Profit = T.R – T.C
(450 X + 720 Y ) – ( 290 X + 375 ) = 160 X + 345 Y
So,
Max ( Z ) = 160 X + 345 Y
Subject to:
At the case when M-100 max. X
At the case when M-200 max. Y
At the case when M-100 mini. X
At the case when M-200 mini. Y
40 X + 50 YThe available amount of R.M
X, Y ≥ 0
( B ) The optimal solution:
When y = 10
So x will equal :
40 X + 50 Y = 1000
40 X + 50 ( 10 ) = 1000
40 X = 500
X = 12.5
( 12.5, 10 ) the optimal point.

7. Max ( Z ) = 160 X + 345 Y
= 160 ( 12.5) + 345 ( 10 )
= 5450
And there is a slack time for each constrains:
X ≤ 15 Slack time= 15 - 12.5 = 2.5
Y ≤ 10 Slack time = 10 – 10 = 0 Dual price = 145
X ≥ 5 slack time = 12.5 – 5 = 7.5
Y ≥ 5 slack time = 10 – 5 = 5
40 x + 50 y ≤ 1000 slack time = 0 Dual price = 4
NOTE:Dual price: is a quantative technique to analyze the improvement in
contribution or costs by having one additional unit of a resource.
The optimal Solution is to schedule 12.5 hours on the M-100 and 10 hours on the
M-200.