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Sunday, 27 May 2018

Probability-Mine Management Questions and Answers Series (4)

Question 1.


  • 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


The daily limestone production at the quarry occurs with the following probabilities:

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


16,57,17,81,70,21,73,78,86,75,27,12,76,45,64,67,34,86,70,02

34,42,59,51,76,37,56,86,87,03,73,77,67,39,07,72,47,81,71,25
90,45,25,67,55,05,98,30,50,77,35,67,85,53,91,03,20,13,40,80
If the limestone supply does not meet the mill demand, it costs the mill K10/tonne to meet that sort fall. the quarry has no stockpile and limestone production is sufficient for daily supply to the mill and any short fall is critical to the mill.
Use Monte Carlo simulation technique to determine the average daily limestone demand by the mill, daily supply by the quarry and average cost of the mill if quarry experiences that it is unable to maintain scheduled daily production rate at any one time. Use the following random numbers and do 20 trials or simulations. Answer the following questions:
1.      What is the average daily consumption of limestone? Prove it with expected value Calculation.
2.      What is the average daily production? Calculate the expected value.
3.      What is the average daily cost of production and its expected value?

Question 1: Answers
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
The average daily consumption of limestone is 117.5 tonnnes/day. However, the expected daily consumption is 116.75 tonnes/day. By comparing these values, there is a difference of 0.75 tonnes per day. This is quite ok but if we have more trials then the amount will be very close to the expected value which is 116.75. The difference is due to small number of simulations.
Therefore, the daily consumption of limestone is between 116.75 tonnes/day and 117.5 tonnes/day.
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
The average daily production of limestone is 129.25 tonnnes/day. But, the expected daily production is 126 tonnes/day. When at these values, there is a difference of 3.25 tonnes per day. This difference is due to the small number of trials analyzed during the simulation. So it is required that more trials can be used for simulations in order to achieve more accurate values which must be very close to 126 tonnes/day.
Therefore, the daily production of limestone is between 126 tonnes/day and 129.25 tonnes/day.
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
The average daily production cost of limestone is K400/t per day. But the expected daily production cost is K395/t per day. There is little difference of K5/t per day. So the average value is somewhat closer to the expected value but if number of trials used for simulation was increased then there would be more accurate value.
Therefore, the daily production cost of limestone is between K395/t per day and K400/t per day.
It is seen that the limestone production at the quarry site is between 126 tonnes and 129.25 tonnes daily and the daily consumption of limestone at the mill is between 116.75 tonnes and 117.5 tonnes daily. For effective and continuous production, the above rates must be maintained. Failure to meet these requirements, there is always costs involve at a rate of K10/t daily. As such it is estimated that the mining company spends about K395/t to K400/t daily. These constraints can be minimized by a stockpile.


Question 2
Being a campaign manager for a political candidate, you are aware of the risks and objectives to achieve. You exactly know the positions of rival candidates and how many votes they and even your candidate could get in each Council Ward (CW). You are faced with the problem of campaign budget and predict how your candidate will perform in terms of scoring votes. As an experienced person you come up with the following data to analyse.
The 10 CWs seem clear in terms of knowing how many “First Votes” you will score and respective probabilities in each CW as given in Table 1.
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
You plan to spend in each Council Ward given the assumption in Table 1.0.
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
Tell your candidate the estimated votes and average cost. What advice will you give to the candidate given the result of this analysis?
Question 2: Answers
A. First Votes.
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  1St  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):
  E(x) = 0.0x400 + 0.05x500 + 0.05x800 + 0.2x1200 + 0.1x1200 + 0.2x2500 + 0.3x3000 = 1995 votes
It is expected that a total of 1900(excluding 95 votes) 1st votes in each council wards compared to 1905 votes/CW. There is a difference of 90 votes/CW. This difference is due to small number of simulation (only 20). In order to obtain more accurate value, over 50 to 100 trials are necessary.  But it is estimated that my candidate will score between 1905 votes and 1995 votes in each council wards.
B. Cost of 1st Votes
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
As an experience person, each vote will be paid for K10/person. As such, the expected budget is K16250/CW compared to the analyzed value K14000 which gives a difference of K2250. This difference is due to less number of simulations (only 10) so it is necessary that over 20 to 100 trials are required. In council wards that we know that few votes will be scored, we will at least remove some amount of money from the budgeted amount and we might use them to buy votes at the polling booth or spend them in council wards that have huge population that we will obtain first votes.
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 1st   votes in each council wards. There may be deviations in each council wards but we just estimated that for CW with large number of 1st votes will be distributed to those that have less number of 1st 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.
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