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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
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CONTENTS
No. Title Page Number
1. Preface i
2. Contents ii
3. TAS
1
Q1
Q2 & Q3
Q 4
1-2
3
4-18
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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
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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
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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
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Solution:
From the inequalities that have done in the Microsoft Excel, I changed all the
inequalities to equations which are:
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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
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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
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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
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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.
A T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 a1 a2 RHSRemarks
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
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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.
A T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 a1 a2 RHSRemarks
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
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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
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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
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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
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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
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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
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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.
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 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
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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
T x1 x2 x3 x4 x5 s1 s2 s3 s4 s5 s6 s7 s8 RHS Remarks
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
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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
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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