Learning and Discussion of Innovative ideas about Mining Waste Management and also Mining Related News and Activities

  • Mine Waste Management Training

    Mine Waste Management Short training sponsored by Government of Japan through JICA in corporation with the Government of PNG through CEPA, MRA and DMPGM.

  • Mount Sinivit Mine

    Acid mine drainage (AMD) continues to flow from the abondoned workings (mine). It is of two types and they are Mine Drainage from underground and open-pit and the seepage water from waste dump and tailings dam.

  • Mining Warden Hearing at Ok Isai Village, Frieda River, East Sepik Province, PNG

    Landowner grievances is always a challenge for the PNG Mining Industry. However, the Regulators of the Mining Inductry facilitate Mining Warden Hearings and Development Forums to address grievances related to mining.

  • Osarizawa Underground Mine Adit

    Osarizawa Underground Mine is an abandoned mine in Akita Prefecture, Japan. Event though the mine is closed, the mine site is kept for sightseeing purposes.

  • Hidden Valley Tailings Storage Facility (TSF)

    Mine Waste refers to the waste related to mining activities such as tailings and waste rock. Management refer to how the mine derived waste is managed by the operator and or the Regulatory Body.




Showing posts with label Educational. Show all posts
Showing posts with label Educational. Show all posts

Monday, 2 July 2018

Mining Warden Hearing at Wenebele and Bawaga Villages.

The Mining Warden Hearing for Wafi-Golpu Support tenements for SML 10 application commenced on Monday the 2nd of July 2018 at Wenebele and Bawaga Villages.

The Wenebele village of Yanta clan and Bawaga Village of Hengambu comprising of Elemu Gwagu and Demagu clans and Wonkins Village of Babuaf Clans are the primary landowners of the SML area and together with the LMPs and MEs under application for the Wafi-Golpu Joint Venture.
Yanta clan have four groups while Hengambu has three Landowner groups/clans have turned up during the meeting in numbers and gave their views in regards to the tenement applications.
Firstly, the Community Affaires superintendent of Wafi-Golpu Joint Venture, Mr.David Masani introduced the warden hearing team to the landowners and opened with a word of prayer to commence hearing. Then handed over to chief Mining Warden of Papua New Guinea, Mr, and Andrew to actually proceed into the Mining Warden Hearing.
The Chief Warden , Mr, Andrew Gunua conducted the Mining Warden Hearing  for Mining Easements comprise of ME 91, Me 93, ME 94. He explained the process of application with displays of posters and explained simplified illustrations with the flow chats of how an SML and supporting tenements are granted following due processes.
He further explained the warden hearing procedure and order of hearing.
The Chief Warden then allowed the company representative to explain to the warden hearing parties of the amendment to the orginal applications and the new work plans and David Masani explained exactly the company’s intention for the three amended ME applications.
The Chief Warden after recording the company’s work plan, he allowed the landowners to give their views whether they support the company’s application or object the applications. The people raised few concerns and asked few questions for the benefit of doubts of which the Chief Warden provided response which were of satisfactory to them.
The Chief Warden then closed the meeting upon satisfying all the requirements under the Mining Act 1992 regarding wardens hearing.
Further questions regarding benefits and other agreement meetings were raised after the meeting and the landowners were advised that, there will be a Development Forum of which the Mining Minister will officially open for them to further discuss matters of this regard.
The Project Coordinator Moses Mambu briefly explained the Development Forum to the landowners and the landowners were looking forward to the Development Forum later during the month.
The same procedure was followed at the Bawaga Village and warden hearings for ME 91, me 93 and ME 94 at both venues were successfully completed. These conclude the Mining Warden Hearing for the First Day and the warden hearing team anticipates  to complete the rest of the hearings in the remaining days.

Chief Mining Warden, Mr.Andrew Gunua Explaining the Flow Chat for SML Application Process at Yanta Community Hall

Landowner Acting Deputy President of Yanta, Mr.Johnson Ruben Responded during the Warden Hearing at Wenebele(Yanta Community Hall)
 
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Thursday, 21 June 2018

Inventory Model - Mine Management Questions and Answers Series(6)

A company requires raw iron at a constant rate of 10,000 tonnes per year. Each time an order is placed, the order cost is K60.00. it cost K0.04/year to store a tonne of raw iron. Storages are not allowed. The actual cost of raw iron is K25 per tonne.
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Thursday, 14 June 2018

Queuing Theory - Mine Management Questions and Answers Series (5)

Part A

1.   What is queuing theory?
In simple terms queuing theory is the study of waiting in lines or queue. Furthermore, Queuing theory is a field of study that deals with Queuing or waiting lines that are analyzed with a set of mathematical formulas. Different queuing models and mathematical models exist to deal with different types of waiting line systems.

2.   What is finite calling population and give an example of it?
Finite calling population is when one knows exactly the population that is in a queue system. E.g. a mine engineer knows the exact number of trucks (i.e. 20) which will be served by a number of shovels (i.e. 2).
Or  finite calling population has a specific , countable number of potential customers that planning is often easy, for example fixed production equipment at mining sites, trucking terminals, trains and airplanes.

3.   What is the expected logic between the arrival and service rates? Why is this important?
The expected logic is that the Arrival Rate must be less than the Service Rate. In other words, the service rate must be greater than the arrival rate. So it is logic to assume that the rate at which services are completed must exceed the arrival rate of customers to increase productivity. This logic is very important because if this is not the case then, the waiting line will continue to grow and there will be no average solution. Hence, it is generally assumed that the service rate must exceed the arrival rate, λ<µ.

4.   What is the difference between the average number of customers in the queue system and average number of customers in the waiting line?
Average number of customers in queue system  is when customers waiting in a single or multiple line to be served in a Bank, dinner at the mess etc. not only people but trucks/machines/ships parts and products queue to be loaded/unloaded and manufacturing operation to be worked on.
Whereas
Average waiting line system comprises of arrivals, servers and waiting line structure. Waiting lines are based on averages of customers or trucks/machines/ships/plane/train arrivals and service times.

5.   Explain the cost relationships between service cost and level of service, and waiting cost and level of service. Explain with an example of each.
Generally the service cost is inversely related to waiting cost. For example, as the level of service is improve by increasing the number of servers, the cost of services will increase whereas waiting cost decreases.
Cost of providing the service is reflected in the cost of servers like loaders/shovels, bank clerk or repair crew in maintenance plant. As the number of servers is increased to reduce service time, service cost rises. The major effect of waiting cost is the loss of business because customers might be tired of waiting and leave, or loss production due to time waste (e.g. in mining) but this loss is temporary.
The curve below shows these relationships.


Part B

1.      Given the information, calculate the operating characteristics and determine if there is any improvement required, including the possibility of increasing number of   mechanics.
Number of customers in the queue is exactly known so this is a finite calling population.

Data; 
Population size N (no. of trucks) = 10 trucks
Number of servers = 1
Arrival rate (λ) per customer = 1/140 = 0.00714 per hour
Service rate (µ) is =1/4.5 = 0.2222 hour

i.     Po =  =  = 0.692162    n=0, 1, 2, 3……….10

Po is the probability of no trucks in the system.
n
0
1
2
3
4
5
6
7
8
9
10
Po
1
0.3213
0.0929
0.0239
0.0054
0.001
0.00017
2.14E-05
2.1E-06
1.3E-07
4E-09
sum(1/sum)
0.692162

ii.      Lq =   == 0.1121 waiting for maintenance.

iii.    L = Lq + (1-Po) = 0.1121 + (1 – 0.692162) = 0.419938 trucks in the queue system.

iv.    Wq  =    =1.64 hours waiting for repair.

v.      W = Wq +  = 1.64 +  = 6.14 hours’ time in the maintenance workshop.

From the results above, it is seen that the mechanics are busy for about 51% of 12 hour shift each day repairing one machine on the queue system. Out of the 10 haul trucks, an average of 0.42 machines are in the queue for maintenance or4.2 % trucks are broken down waiting for repair or under repair. Each broken down truck is idle which means in 12 hour shift, about two tucks are repaired (12hrs/6.14). Therefore there is no need to increase the number of mechanics. But if the fleet size is increased to 20 or so, then there is a need for couple of mechanics at the workshop. But otherwise, no need.

2.      Given the information, calculate the operating characteristics and recommend if there is anything to be done to improve truck – shovel productivity and reduce queue time.
Data:
λ = 4 trucks per hour (poison distribution)
µ = 3 trucks per hour
s = 2 shovels
sµ = 2*3 = 6, (>λ = 4)

Po = =   = 0.2 This is the probability that

No trucks are in the waiting line or queue.
L =  *Po +  =  *0.2 +  = 2.4, so 2 trucks arrive for loading at the pit.
                    
 =   = 0.5 hours or 30 minutes in the queue (waiting and served).

Lq = L -   = 2.4 -   = 1.0667 trucks in queue to be loaded by the two shovels.

Wq =  = 1.0667/4 = 0.267 hours or 16 minutes in the queue.

Pw =  =   = 0.533333

Now, there is 0.533 probability that trucks must wait or queue for loading which means that there is one or two trucks in the open pit queue system to be loaded by the two shovels.

Result Discussion

It can be seen from the results above which shows that the truck – shovel combination productivity is not attractive. There is more truck capacity and less number of shovels causing low truck productivity. The queuing time of 16 minutes for one or two trucks in queue is not attractive. There are 2 trucks entering the pit for loading at any time. By loading only one of the total 2, it consumes 16 minutes of production time where the truck(s) remain idle. It is now a need to reduce queuing time at all cost as it is recommended here. It is also recommended that, one shovel must be added to the existing fleet to increase productivity but keeping the fleet of trucks constant. With this recommendation the following are the results of improvements made in the queue system:

Recommended data: s = 3; λ = 4; µ = 3
Po = 0.254 probability that no trucks in the queue or loading
L = 1.48, so 1 truck enter the pit for loading.
W = 0.25 hour or 15 minutes queue time (average waiting time to load) 
Lq = 0.15 waiting to be served.
Wq = 0.0375 hour or 2.25 minutes waiting in line before one truck is being loaded.
Pw =0.18 probability that one truck must wait for loading or all 4 shovels are busy.

Upon the above recommendation, it is seen that productivity y is improved when one truck’s average queuing time is reduced to 2.25 minutes from 16 minutes. In the future the cost of adding one shovel to existing fleet will be compensated by the higher production which is objective of the mine management.

3.      Given the data, the manager wants to determine the average length of waiting line and average waiting time at counter. If the office space can only accommodate 10 clients, what do you recommend?

Data: λ = 5 rate per hour;   µ = 60/10 = 6 clients per hour.

Now, Lq =   =  = 2.08 clients waiting

Wq =  =  = 0.416 hour or 24.96 minutes waiting in line or queue.

The results above shows that the client service provided by the law firm is not effective as there is almost half an hour consumes while the clients are waiting to be served. Therefore, it is recommended that the constant service time must be reduced 8 minutes and by increasing one server to the exiting but clients arriving rate will be constant as it can’t be controlled. That means the law firm must accommodate two clients at a time with two servers serving the clients at all cost. Upon these changes, the following results are obtained;
Λ = 5 rate per hour;   µ = 60/8 = 7.5 clients per hour.

Lq = 0.667 clients waiting
Wq = 0.133 hours or 7.98 minutes waiting in line or queue.

Now this effective since the waiting line is reduced to 7.98 minutes from 24.96 minutes. It is assumed that there will be no or one client waiting to be served. The cost of the additional server will be covered the more clients paying for the business as they are being served.

Note also that the office capacity to accommodate only ten clients doesn’t mean that all of them are going to be served at the same time but, will sit comfortably waiting to be served. That doesn’t have an effect on the queuing time but queuing time depends on serving time. Therefore, I think the purpose of office capacity to accommodate ten clients is just for the firm’s good reputation and attractiveness of the business.





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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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Osarizawa Underground Mine Adit Osarizawa mine is an abandoned mine in Akita Prefecture, Japan . Event though the mine is closed, the ...

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