DECISION MATHEMATICS 2010

8/7/2019 DECISION MATHEMATICS 2010

1/21

TEACHER EDUCATION INSTITUTE OF MALAYSIA

SULTAN ABDUL HALIM CAMPUS

08000 SUNGAI PETANI,KEDAH

BACHELOR EDUCATION DEGREE (MATHEMATICS)

JANUARY 2010 INTAKE

NAME : MOHAMMAD FALAKHUDDIN BIN HARON

I/C NO. : 901225-08-6131

GROUP 1P2D


2/21


3/21

CONTENTS

No. Title Page Number

1. Preface i

2. Contents ii

3. TAS

1

Q1

Q2 & Q3

Q 4

1-2

3

4-18


4/21

1. Using Pto represent the total net benefits and xithe number of ringgit (in millions)

awarded to project i, formulate the problem as a linear programming (LP) problem.

Solution :

Problem involving constraint:

The simplex method involves iterations which move from a feasible vertex to an

improved vertex. For example, in the case: x 1 330. This means that the origin has always

been a feasible solution and so could serve as a starting point for the process. However, if

there is a constraint of the form, for example in this case x 1 + x2 400, then the origin will

not be feasible and you will need some other feasible point from which to begin the process.

With more or three variables you would not be able to do the simplex method that we usually

use. You need to able to modify the simplex algorithm so that it can either find an initial

feasible solution or do without one is some way. You will examine two ways of doing this:

y Two-stage simplex.

y The big M method.

For this case, I used two-stage simplex to find the solution.

In producing a tableau for the extended problem the extra constraint, x 5 330 and x1 + x2

400, creates a problem. To convert it into equality I need surplus variables s 8 and s9, not

slack variables. Then, I use another variable that called an artificial variable:

x5 - s8 + a1 = 300

x1 + x2 - s9 + a2 = 400


5/21

This is the table of inequalities and equation for the simplex method:

Inequalities Equation

P =3.6x1 + 3.8x2 + 3.9x3 + 3.7x4 + 4.1x5 P - 3.6 x1 - 3.8x2 3.9 x3 - 3.7 x4 - 4.1 x5=0

x1 + x2 + x3 + x4 + x5 1400 x1 + x2 + x3 + x4 + x5 + s1 = 1400

X1 330 x1 + s2 = 330

x2 380 x2 + s2 = 380

x3 350 x3 + s4 = 350

x4 400 x4 + s5 = 400

x5 400x

5+ s

6= 400

x4 240 x4 s7 + a1 = 240

x1 + x2 420 x1 + x2 s8 + a2 = 420

** In cases in which there are more than constraint this new objective will be of the

form A = a1 + a2.

So, A + x1 + x2 + x4 s7 s8 = 660


6/21

2. Explain why solving the problem using a graphical method is not feasible?

This is because the five projects are not linked to each other. So, there is n o

intersection between these five lines in the graph. Therefore, using a graphical method is not

feasible.

3. Construct an Excel worksheet to represent the model and let the Solver add -in to

find a solution to the problem. Interpret fully the solution.

PROJECT CLASSIFICATION NO. OF RINGGIT AWARDED (millions) NET BENEFIT PER RINGGIT INVESTED REQUESTED FUNDING (millions)

Solar 1,u 420 3.6 330

Solar 2,v 420 3.8 380

Coal,e 350 3.9 350

Nuclear,b 240 3.7 400

Geothemal,n 0 4.1 400

Total ringgit awarded (million) 1400

budget 1400

total net benefit, P 5289

Net benefits per ringgit invested 1 2 3 4 6 7 8

Solar 1 3.6

Solar 2 3.8

Coal 3.5

Nuclear 5.1

Geothemal 3.2

Availabalities 350 380 350 400 400 240 400

Net benefits 1260 1444 1225 2040 1280 240 400


7/21

Solution:

From the inequalities that have done in the Microsoft Excel, I changed all the

inequalities to equations which are:


8/21

THE INEQUALITIES

P = 3.6x1 + 3.8x2 + 3.9x3 + 3.7x4

+ 4.1x5

x1 + x2 + x3 + x4 + x5 1400

X1 330

x2 380

x3 350

x4 400x5 400

x4 240

x1 + x2 420

A = a1+ a2

THE EQUATIONS

P - 3.6 x1 - 3.8x2 3.9 x3 - 3.7 x4 -

4.1 x5=0

x1 + x2 + x3 + x4 + x5 + s1 = 1400

x1 + s2 = 330

x2 + s2 = 380

x3 + s4 = 350

x4 + s5 = 400

x5 + s6 = 400

x4 s7 + a1 = 240

x1 + x2 s8 + a2 = 420

A + x1 + x2 + x4 s7 s8 =660


9/21

Convert all the equations that have found form it into a tabular form.

This come a new thing, which I must modify the simplex algorithm into two-stage simplex because there aremore thanthree variables that have in the table below.

Row A represents the new objective, which I will use it in the first stage. It gives the sum of the artificial variables. Thi s is

becausethe variables are not allowed to take negatives values.

Row T shows the original objective, which I will use in the second stage. The remaining rows are the constraints.

Thus in the first iteration we can choose to increase variables x, since some of them have positive coefficients in the first

objective row. Make the arbitrary choice of increasing x1.

A T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 a1 a2 RHS

1 0 1 1 0 0 1 0 0 0 0 0 0 -1 -1 0 0 660 R1

0 1 -3.6 -3.8 -3.9 -3.7 -4.1 0 0 0 0 0 0 0 0 0 0 0 R2

0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1400 R3

0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 330 R4

0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 380 R5

0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 350 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 400 R7

0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 400 R8

0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 1 0 240 R9

0 0 1 1 0 0 0 0 0 0 0 0 0 0 -1 0 1 420 R10


10/21

FIRST STAGE

PIVOT 1

y Choose pivot column that show the most positive number which is column x1.

y Find the pivot element by using the ratio test. 1400/1= 1400, 330/1= 330, and 240/1= 240. So, the pivot element is the

number 1 in the row 4.

y Then, continue solve the row by manipulate the table so that the pivot element becomes 1, and the remaining elements of

the pivot column all become 0.

A T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 a1 a2 RHSRemarks

1 0 1 1 0 0 1 0 0 0 0 0 0 -1 -1 0 0 660 R1 R1 R4

0 1 -3.6 -3.8 -3.9 -3.7 -4.1 0 0 0 0 0 0 0 0 0 0 0 R2 3.6R4 +R2

0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1400 R3 R3 - R4

0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 330 R4 Pivot row

0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 380 R5

0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 350 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 400 R7

0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 400 R8

0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 1 0 240 R9

0 0 1 1 0 0 0 0 0 0 0 0 0 0 -1 0 1 420 R10 R10 - R4


11/21

PIVOT 2

y After that, choose the next most positive pivot column which is column x2

y Find the pivot element by using the ratio tes t. This compares 1070/1= 1070, 380/1= 380, and 90/1= 90. So, the pivot element is

the 1 in the last row.

y Also, I must manipulate the table until the pivot element becomes 1, and the remaining elements of the pivot column all becom e

0.


1 0 0 1 0 0 1 0 -1 0 0 0 0 -1 -1 0 0 330 R1 R1 R10

0 1 0 -3.8 -3.9 -3.7 -4.1 0 3.6 0 0 0 0 0 0 0 0 1188 R2 R2 + 3.8R10

0 0 0 1 1 1 1 1 -1 0 0 0 0 0 0 0 0 1070 R3 R3 R10

0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 330 R4

0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 380 R5 R5 R10

0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 350 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 400 R7

0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 400 R8

0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 1 0 240 R9

0 0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 0 1 90 R10 Pivot row


12/21

PIVOT 3

y Then, continue by choosing the next pivot column which is column x4

y Find the pivot element by using the ratio test. 980/1= 980, 400/1= 400, and 240/1= 240. So, the pivot element is the in the row

9.

y And once again manipulate the table.


1 0 0 0 0 0 1 0 0 0 0 0 0 -1 0 0 -1 240 R1 R1 + R9

0 1 0 0 -3.9 -3.7 -4.1 0 -0.2 0 0 0 0 0 -3.8 0 3.8 1530 R2 R2 + 3.7R9

0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 0 -1 980 R3 R3 R9

0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 330 R4

0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 -1 290 R5

0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 350 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 400 R7 R7-R9

0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 400 R8

0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 1 0 240 R9 Pivot row

0 0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 0 1 90 R10


13/21

With the answer of A = 0, the working have arrived at a feasible solution to the original problem.

Interpreting the tableau, the obvious solution is:

x3 = x6 =s1 = s2 = s7 = s8 = a1 = a2 = 0

A= 0

T= 2418

x1 = 330

x2 = 90

x4 = 240

s1 = 740

s3 = 290

s4 = 350

s5 = 160

s6 = 400

A T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 a1 a2 RHS

1 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 -1 -1 0 R1

0 1 0 0 -3.9 0 -4.1 0 -0.2 0 0 0 0 -3.7 -3.8 3.7 3.8 2418 R2

0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 -1 -1 740 R3

0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 330 R4

0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 -1 290 R5

0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 350 R6

0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 -1 0 160 R7

0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 400 R8

0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 1 0 240 R9

0 0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 0 1 90 R10


14/21

Then, the equations that I got from the final table for the stage 1 are:

A = 0 - a1 a2

T= 2418 3.9 x3 - 4.1 x5 0.2 s2 3.8s8 - 3.7a1+3.8 a2

x3 + x6 +s1 + s7 + s8 - a1 - a2

x1 + s2 = 330

x2 + s3 = 90

x3 + s4 = 350

x4 + s5 = 240

x5 s8 + a1 = 240

x6 + s7 = 400

s6 + s8 a1 = 240

x2 s2 s8 + a2 = 90

s2 + s3 + s8 a2 = 29


15/21

STAGE 2

y For stage 2, there is a different with the stage 1. If the stage 1 I choose the column that show the most positive number. But for this

stage, I must choose the most negative number and continue manipulate the table to get the answer.

y So, now I choose the s8 column.

y Find the pivot element by using the ratio test. 740/1= 740 and 400/1= 4 00. So, the pivot element is the number 1 in the row 7 .

y Then, continue solve the row by manipulate the table so that the pivot element becomes 1, and the remaining elements of the pivot

column all become 0.

T x1 x2 x3 x4x5 s1 s2 s3 s4 s5 s6 s7 s8 RHS

Remarks

1 0 0 -3.9 0 -4.1 0 -0.2 0 0 0 0 -3.7 -3.8 2418 R1 R1 + 4.1R8

0 0 0 1 0 1 1 0 0 0 0 0 1 1 740 R2 R2 R7

0 1 0 0 0 0 0 1 0 0 0 0 0 0 330 R30 0 0 0 0 0 0 0 1 0 0 0 0 1 290 R4

0 0 0 1 0 0 0 0 0 1 0 0 0 0 350 R5

0 0 0 0 0 0 0 1 0 0 1 0 1 0 160 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 400 R7 Pivot row

0 0 0 0 1 0 0 0 0 0 0 0 -1 0 240 R8

0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 90 R9


16/21

PIVOT 1

y Then, I choose the most negative column which is x3.

y Find the pivot element by using the ratio test. 300/1= 300 and 350/1= 350. So, the pivot element is the number 1 in the row

2.

y Then, continue solve the row by manipulate the table so that the pivot element becomes 1, and the remaining elements of

the pivot column all become 0 as usual.

T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 RHSRemarks

1 0 0 -3.9 0 0 0 -0.2 0 0 0 4.1 -3.7 -3.8 4058 R1 R1 + 3.9R2

0 0 0 1 0 0 1 0 0 0 0 -1 1 1 300 R2 Pivot row

0 1 0 0 0 0 0 1 0 0 0 0 0 0 330 R3

0 0 0 0 0 0 0 0 1 0 0 0 0 1 290 R4

0 0 0 1 0 0 0 0 0 1 0 0 0 0 350 R5 R5 R2

0 0 0 0 0 0 0 1 0 0 1 0 1 0 160 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 400 R7

0 0 0 0 1 0 0 0 0 0 0 0 -1 0 240 R8

0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 90 R9


17/21

PIVOT 2

y Then, continue by choosing the next pivot column which is column x9.

y Find the pivot element by using the ratio test. 330/1= 330 and 160/1= 160. So, the pivot element is the in the row 6.

y And once again manipulate the table.

T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 RHS Remarks

1 0 0 0 0 0 3.9 -0.2 0 0 0 0.2 -0.2 -0.1 5228 R1 R1 + 0.2 R6

0 0 0 1 0 0 1 0 0 0 0 -1 1 1 300 R2 R2 R6

0 1 0 0 0 0 0 1 0 0 0 0 0 0 330 R3

0 0 0 0 0 0 0 0 1 0 0 0 0 1 290 R4

0 0 0 0 0 0 -1 0 0 1 0 1 -1 -1 50 R5 R5 + R6

0 0 0 0 0 0 0 1 0 0 1 0 1 0 160 R6 Pivot row

0 0 0 0 0 1 0 0 0 0 0 1 0 0 400 R7

0 0 0 0 1 0 0 0 0 0 0 0 -1 0 240 R8 R8 + R6

0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 90 R9


18/21

PIVOT 3

y Then, continue by choosing the next pivot column which is column s8.

y Find the pivot element by using the ratio test. 140/1= 1400 and 290/1= 290. So, the pivot element is the in the row 2 .

y And once again manipulate the table like below.


1 0 0 0 0 0 3.9 0 0 0 0.2 0.2 0 -0.1 5260 R1 R1 + 0.1 R2

0 0 0 1 0 0 1 -1 0 0 -1 -1 0 1 140 R2 Pivot row

0 1 0 0 0 0 0 1 0 0 0 0 0 0 330 R3

0 0 0 0 0 0 0 0 1 0 0 0 0 1 290 R4 R4 R2

0 0 0 0 0 0 -1 1 0 1 1 1 0 -1 210 R5 R5 + R2

0 0 0 0 0 0 0 1 0 0 1 0 1 0 160 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 400 R7

0 0 0 0 1 0 0 1 0 0 1 0 0 0 400 R8

0 0 1 0 0 0 0 -1 0 0 0 0 0 -1 90 R9 R9 + R2


19/21

PIVOT 4

y Then, continue by choosing the next pivot column which is column s2.

y Find the pivot element by using the ratio test. 330/1= 330, 150/1 = 150 and 160/1= 160. So, the pivot element is the in the row

4.

And once again manipulate the table as usual


1 0 0 0.1 0 0 4 -0.1 0 0 0.1 0.1 0 0 5274 R1 R1 + 0.1 R4

0 0 0 1 0 0 1 -1 0 0 -1 -1 0 1 140 R2 R2 + R4

0 1 0 0 0 0 0 1 0 0 0 0 0 0 330 R3 R3 R4

0 0 0 -1 0 0 -1 1 1 0 1 1 0 0 150 R4 Pivot row

0 0 0 1 0 0 0 0 0 1 0 0 0 0 350 R5

0 0 0 0 0 0 0 1 0 0 1 0 1 0 160 R6 R6 R4

0 0 0 0 0 1 0 0 0 0 0 1 0 0 400 R7

0 0 0 0 1 0 0 1 0 0 1 0 0 0 400 R8 R8 R4

0 0 1 1 0 0 1 0 0 0 -1 -1 0 0 230 R9


20/21

PIVOT 5

y Table below is the final pivot because all numbers in row 1 are positive.

From the table that have been manipulating, finally I got the answers which are:

T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 RHS

1 0 0 0 0 0 3.9 0 0.1 0 0.2 0.2 0 0 5289 R1

0 0 0 0 0 0 0 0 1 0 0 0 0 1 290 R2

0 1 0 1 0 0 1 0 -1 0 -1 -1 0 0 180 R3

0 0 0 -1 0 0 -1 1 1 0 1 1 0 0 150 R4

0 0 0 1 0 0 0 0 0 1 0 0 0 0 350 R5

0 0 0 1 0 0 1 0 -1 0 0 -1 1 0 10 R6

0 0 0 0 0 1 0 0 0 0 0 1 0 0 400 R7

0 0 0 1 1 0 1 0 -1 0 0 -1 0 0 250 R8

0 0 1 1 0 0 1 0 0 0 -1 -1 0 0 230 R9


21/21

T = 5289

x1 = 180

x2 = 230

x3 = 0

x4 = 250

x5 = 400

s1 = 0

s2 = 150

s3 = 0

s4 = 350

s5 = 0

s6 = 0

s7 = 10

s8 = 290

DECISION MATHEMATICS 2010

Documents

Transcript of DECISION MATHEMATICS 2010