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FACULTY OF SCIENCE AND TECHNOLOGY

SEMESTER SEPTEMBER TAHUN 2012

EBTM 4203

OPERATIONAL MANAGEMENT

NO. MATRIKULASI : 681104035475001

NO. KAD PENGNEALAN : 681104-03-5475NO. TELEFON : 019-9575476

E-MEL : nordin_yusof@oum.edu.my

PUSAT PEMBELAJARAN : PP KUALA TERENGGANU

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Page 1 of11

Table of Contents

QUESTION 1 ( i ) ............................................................................................................... 2

QUESTION 1 ( ii ) .............................................................................................................. 7

QUESTION 2.................................................................................................................... 10

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QUESTION 1 ( i )

Variables

Objective

Total Expected Profit,

Therefore,

Constant

Daily hours for painting,

Daily hours for finishing,

Daily hours for framing,

Maximum hours allocated ;

for painting,

for finishing,

for framing,

Therefore (non-negativity conditions),

Problem Statement;

Maximize,

Subject to :-

, ,

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For Value ;

a) Painting,

When, and

b) Finishing,

When, and

.00

c) Framing,

When, and

For Value ;

a) Painting,

When, and

.00

b) Finishing,

When, and

.00

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c) Framing,

When, and

.00

For Value ;

a) Painting,

When, and

.00

b) Finishing,

When, and

.00

c) Framing,

When, and

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The optimum solution

Equation 1 ;

Equation 2 ;

Equation 3 ;

Multiply Equation 3 by -2 for eliminate X1 using equation 2 and equation 3

Therefore,

Equation 4 ;

Multiply Equation 1 by 3 and Equation 2 by -2 for eliminate X1

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Therefore,

Now, substitute into equation 4 to find.

Now, substitute and into equation 1 to find.

Therefore, the optimal solution is for producing of each paintings type to obtain the

maximum profit is:-

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QUESTION 1 ( ii )

To determine the optimal step-by-step solution in simplex method.

Problem Statement;

Maximize,

Subject to :-

By introducing slack variablesMaximize,

Subject to :-

Using the basic feasible solution:

.

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Iteration 0 ; New Pivot Row = Curent Pivot Row / Pivot No.

= Minimum Ratio = 16

Basic

VariableEq.

Coefficient ofSolution Ratio

(0)

(1)

(2)

(3)

IterationBasic

VariableEq.

Coefficient ofSolution

0

(0)

(1)

(2)

(3)

Pivot column =

IterationBasic

VariableEq.

Coefficient ofSolution

1

(0) 116

(1) 4.4 0

(2)

(3) 0.6

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New Pivot column =

IterationBasic

VariableEq.

Coefficient ofSolution

2

(0) 860

(1) 2 0

(2)

(3) 0

New Pivot column =

IterationBasic

VariableEq.

Coefficient ofSolution

3

(0) 580

(1) 0 0.3

(2)

(3) 0

MAXIMIZED PROFIT = 4079

This solution is optimal because none of the coefficients in row Eq(0) is negative.

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QUESTION 21) Unbalanced problem ; Create a dummy Construction Site

PlantConstruction Site Supply

(tons )A B C Dummy

19 12

12010

600

180

2 1306

205 9 0

150

37

1004

708 0

170

Demands

( tons )130 120 190 60 500

2) Evaluate the unoccupied cells using 170

System of equations: Setting R1 = 0

R1 + K3 = 10 K3 = 100 = 10

R1 + K4 = 0 K4 = 00 = 0

R2 + K1 = 6 K1 = 6(1) = 7

R2 + K2 = 5 R2 = 56 =1

R3 + K2 = 4 K2 = 4(2) = 6

R3 + K3 = 8 R3 = 810 =2

K1=7 K2=6 K3=10 K4=0

PlantConstruction Site Supply

(tons )A B C Dummy

R1= 01

9 12120

1060

0180

R2= -12 130

620

5 9 0150

R3= -23

7100

470

8 0170

Demands

( tons )130 120 190 60 500

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3) Improvement indices for all the unoccupied cells,

Unoccupied Cell

Improvement Index

(lij = cijRK)

1A 907 = +2

1B 1206 = +6

2C 9(1)10 = 0

2Dummy 0(1)0 = +1

3A 7(2)7 = +2

3Dummy 0(2)0 = +2

All of improvement index are nonnegative and therefore this solution is optimal.

4) Total minimized the transportation cost.

Route

Units Supply

(tons)

Cost per unit

(RM)

Total Cost

(RM)From

(Plant)

To(Construction

Site)

1 C 120 10 1,200

1 Dummy 60 0 0

2 A 130 6 780

2 B 20 5 100

3 B 100 4 400

3 C 70 8 560

Total minimised transportation cost 3,040

TOTAL MINIMISED TRANSPORTATION COST IS RM3,040.00