 A mineral processing plant requires excessive limestone and its consumption is critical where the supply must be consistent daily. The limestone is supplied by a satellite quarry owned by the mining company. From Past experience, the daily consumption of limestone by the mill is given by the following discrete probabilities:
Limestone Consumption

100

110

115

120

125

130

135

Probability, P(X)

0.2

0.150

0.2

0.1

0.15

0.10

0.10

Daily Limestone Production, x (t/day)

90

100

120

125

130

140

140

Probability, P(X)

0.05

0.1

0.10

0.20

0.25

0.15

0.15

Production Cost/Day($/t)

25

30

35

40

45

50

55

Probability, P(X)

0.1

0.05

0.25

0.2

0.3

0.05

0.05

A. Limestone consumption/day
 
Probability P(x)

Cumulative Probability

Limestone consumption(tons/day)

0.2

0.2

100

0.15

0.35

110

0.2

0.55

115

0.1

0.65

120

0.15

0.8

125

0.1

0.9

130

0.1

1

135

1

No.Trials

Random #

Simulated value

Cumulative consumption
 
1

0.16

100

100
 
2

0.57

120

220
 
3

0.17

100

320
 
4

0.81

130

450
 
5

0.7

125

575
 
6

0.21

110

685
 
7

0.73

125

810
 
8

0.78

125

935
 
9

0.86

130

1065
 
10

0.75

125

1190
 
11

0.27

110

1300
 
12

0.12

100

1400
 
13

0.76

125

1525
 
14

0.45

115

1640
 
15

0.64

120

1760
 
16

0.67

125

1885
 
17

0.34

110

1995
 
18

0.86

130

2125
 
19

0.7

125

2250
 
20

0.02

100

2350
 
Total

2350
 
Average

117.5
 
Expected value E(x):
 
E(x)= 0.2x100 + 0.15x110 + 0.2x115 + 0.1x120 + 0.15x125 + 0.1x130 + 0.1x135 = 116.75 tonnes

B. Limestone Production/day
 
Probability P(x)

Cumulative Probability

Limestone consumption(tons/day)
 
0.05

0.05

90
 
0.1

0.15

100
 
0.1

0.25

120
 
0.2

0.45

125
 
0.25

0.7

130
 
0.15

0.85

140
 
0.15

1

140
 
1

No.Trials

Random #

Simulated value

Cumulative consumption
 
1

0.34

125

125
 
2

0.42

125

250
 
3

0.59

130

380
 
4

0.51

130

510
 
5

0.76

140

650
 
6

0.37

125

775
 
7

0.56

130

905
 
8

0.86

140

1045
 
9

0.87

140

1185
 
10

0.03

90

1275
 
11

0.73

140

1415
 
12

0.77

140

1555
 
13

0.67

130

1685
 
14

0.39

125

1810
 
15

0.07

100

1910
 
16

0.72

140

2050
 
17

0.47

130

2180
 
18

0.81

140

2320
 
19

0.71

140

2460
 
20

0.25

125

2585
 
Total

2585
 
Average

129.25
 
Expected value E(x):
 
E(x) = 0.05x90 + 0.1x100 + 0.1x120 + 0.2x125 + 0.25x130 + 0.15x140 + 0.15x140 =126 tonnes

C. Production cost /day
 
Probability P(x)

Cumulative Probability

Limestone Consumption(tons/day)

0.1

0.1

25

0.05

0.15

30

0.25

0.4

35

0.2

0.6

40

0.3

0.9

45

0.05

0.95

50

0.05

1

55

1

No.Trials

Random #

Simulated value

Cumulative consumption
 
1

0.9

50

50
 
2

0.45

40

90
 
3

0.25

35

125
 
4

0.67

45

170
 
5

0.55

40

210
 
6

0.05

25

235
 
7

0.98

55

290
 
8

0.3

35

325
 
9

0.5

40

365
 
10

0.77

45

410
 
11

0.35

35

445
 
12

0.67

45

490
 
13

0.85

45

535
 
14

0.53

40

575
 
15

0.91

50

625
 
16

0.03

25

650
 
17

0.2

35

685
 
18

0.13

30

715
 
19

0.4

40

755
 
20

0.8

45

800
 
Total

800
 
Average

40
 
Expected value E(x):
 
E(x) = 0.1x25 + 0.05x30 + 0.25x35 + 0.2x40 + 0.3x45 + 0.05x50 + 0.05x55 = K39.5/t per day

Rest Houses

A

B

C

D

E

F

G

H

I

Est.Votes

0

500

1200

800

2500

1700

3000

1200

400

Probability (Px)

0.0

0.05

0.20

0.05

0.20

0.10

0.30

0.10

0.00

Rest Houses

A

B

C

D

E

F

G

H

I

Est.Votes

0

1000

500

1000

3000

2000

3000

1000

500

Probability (Px)

0.0

0.05

0.10

0.15

0.20

0.15

0.10

0.10

0.05

Probability P(x)

Cumulative Prob.

First Votes

0

0

400

0.05

0.05

500

0.05

0.1

800

0.2

0.3

1200

0.1

0.4

1200

0.1

0.5

1700

0.2

0.7

2500

0.3

1

3000

1

No.of trials

Random #

Simulated value

Cumulative 1^{St} votes
 
1

0.45

1700

1700
 
2

0.22

1200

2900
 
3

0.98

3000

5900
 
4

0.28

1200

7100
 
5

0.94

3000

10100
 
6

0.36

1200

11300
 
7

0.57

2500

13800
 
8

0.16

1200

15000
 
9

0.21

1200

16200
 
10

0.12

1200

17400
 
11

0.92

3000

20400
 
12

0.81

3000

23400
 
13

0.98

3000

26400
 
14

0.88

3000

29400
 
15

0.07

800

30200
 
16

0.33

1200

31400
 
17

0.48

1700

33100
 
18

0.21

1200

34300
 
19

0.88

3000

37300
 
20

0.05

800

38100
 
Total

38100
 
Average

1905
 
Expected value E(x):

Probability P(x)

Cumulative Prob.

First Votes

0

0

0

0.1

0.1

500

0.15

0.25

500

0.05

0.3

1000

0.15

0.45

1000

0.1

0.55

1000

0.15

0.7

2000

0.2

0.9

3000

0.1

1

3000

1

No.of trials

random#

simulated value

cummulative 1St votes
 
1

0.04

500

500
 
2

0.23

500

1000
 
3

0.33

1000

2000
 
4

0.64

2000

4000
 
5

0.06

500

4500
 
6

0.16

500

5000
 
7

0.59

2000

7000
 
8

0.43

1000

8000
 
9

0.81

3000

11000
 
10

0.94

3000

14000
 
Total

14000
 
Average

1400

Per CW
 
Expected Value E(x):
 
E(x) = 0.1x500+0.15x500+0.05x1000+0.15x1000+0.1x1000+0.15x2000+0.2x3000+0.1x3000=K1625/CW

To sum up, it is predicted that the candidate should plan ahead and must be prepared to allocate a budget between K14000 and K16250 to each council ward in order to avoid inconvenience. In doing this, the candidate is expecting about 1905 to 1995 1^{st} votes in each council wards. There may be deviations in each council wards but we just estimated that for CW with large number of 1^{st} votes will be distributed to those that have less number of 1^{st} votes so that we have a fair distribution of our estimates. Having done all these, we are confident that we will spend the above amount of money to score the ex amount of 1st votes.