Mine Waste Management

Mine Waste Management

Mine Management Questions and Answers Series (2)

Mine Management Questions and Answers Series (1)

Tuesday, 8 May 2018

May 08, 2018Educational, Mine Management, Mining Engineering No comments

Question 1

Solve the following using graphical method.

i). Maximize Z = 3x1 + 4x2 soln:

Subject to: x1 ≤ 350 when x1=0, x2 ≤ 400

x2 <=250

x1 + x2 <=400

x1, x2>=0

When plotting these inequalities on graph paper, we obtained the following solutions. As the objective function was drawn and moved away from the origin, it approached x1=150 and x2=250. Therefore, the optimal solution is (150,250).

so the maximum value is:

Z = 3*150 + 4*250 = 1450

ii). Maximize Z = 4x1 + x2` 1) let x1 = 0, x2 <=2.5

subject to : x1 + 2x2 <= 5 x2 = 0, x1 <= 5

3x1 + 2x2 <= 9 2). Let x1 =0, x2 <= 4.5

x1,x2 >= 0 x2 = 0, x1 <= 3

By plotting the above values on the graph and drawing objective function and by moving it away from the origin, the optimal solution that this linear curve approached is: (3,0)

X1 = 3, X2 = 0.

So the maximum value is, Z = 4*3 + 0 = 12

iii). Minimize Z = x1 - x2 Working:

Subject to : x1 - 2x2 ≤ 3 1) Let x1 = 0, x2 >= -1.5

x1 - 4x2 <= 5 x2 = 0, x1 <= 3

x1,x2 >= 0 2). Let x1 = 0, x2 >= - 1.25

x2 = 0, x1 <= 5

These inequalities were plotted and as the objective linear curve was drawn and moved towards the origin, the optimal point it approached was (0,0). Actually this is the origin. Therefore, the minimum value is: Z = 0

iv). Maximize Z = 400x1 + 200x2 working:

Subject to: x1 + x2 = 30 1). Let x1 = 0, x2 = 30

2x1 + 8x2 >=80 x2 = 0, x1 = 30

x1 <= 20 2). Let x1 = 0 , x2 >=10

x1,x2 >= 0 x2=0, x1 >=40

Upon plotting the inequalities on the graph paper, there were two possible solutions to this problem but the most acceptable optimal solution is: (20,10).

x1 = 20, x2 = 10

Thus, the maximum value is; Z = 400*20 + 200*10 = 10 000

v). Minimize Z = 400x1 + 200x2 working:

Subject to: x1 + x2 = 30 1). Let x1 = 0, x2 = 0

2x1 + 8x2 >=80 x2 = 0, x1 = 30

x1 <= 20 2). Let x1 = 0 , x2 >=10

X1, x2 >= 0 x2=0, x1 >=40

From the outer space and drawing objective linear curve towards the origin, the point which the curve reached was (0,10), thus x1 = 0 & x2 = 10

Thus, the minimum value is:

Z = 200*10 = 2000

Question 2


Product	BoP	HE	Constraints
Labor(hrs)	4	3	30
Timber(m³)	2	8	80

1. Objective function:

profit from Bop + profit from HE

Z = K400/BoP + K200/HE Let x1 = BoP, x2 = HE

Z = 400x1 + 200x2

2. The key resources are time and special timber.

a). Time: 4hrs/BoP + 3hrs/HE <= (restricted to 30 hrs/month)

i.e. 4x1 + 3x2 <= 30hrs

b). Timber: 2m3/BoP + 8m3/HE <= (restricted to 80m3/month)

i.e. 2x1 + 8x2 <= 80

LP Model: working

Maximize Z = 400x1 + 200x2 1). Let x1 = 0 , x2 <= 10

Subject to: 4x1 + 3x2 <= 30 x2=0, x1 <= 7.5

2x1 + 8x2 <= 80 2). Let x1 =0, x2 <= 10

x1 <= 20 x2=0, x1 <= 40

x1,x2 >= 0

In order to satisfy the above conditions to reach the optimum solution, there was a linear objective curve drawn and moved away from the origin. As such, the curve approached (7.5, 0). So this is the

optimum solution, x1 = 7.5 & x2 = 0

Hence, the maximum profit is: Z = 400*7.5 = K3000

Using Excel Solver.

Question 2.

	BoP x₁	HE x₂	Profit		Limit
Decision variables	7.5	0
Objective function	400	200	3000
A	4	3	30	<=	30
B	2	8	15	<=	80

Sensitivity Report

	Adjustable Cells
		Final	Reduced	Objective	Allowable	Allowable
Cell	Name	Value	Cost	Coefficient	Increase	Decrease
$B$2	Decision variables x1	7.5	0	400	1E+30	133.3333333
$C$2	Decision variables x2	0	-100	200	100	1E+30

	Constraints
		Final	Shadow	Constraint	Allowable	Allowable
Cell	Name	Value	Price	R.H. Side	Increase	Decrease
$D$4	A Profit	30	100	30	130	30
$D$5	B Profit	15	0	80	1E+30	65

Answer Report

	Target Cell (Max)
Cell	Name	Original Value	Final Value
$D$3	Objective function Profit	0	3000

	Adjustable Cells
Cell	Name	Original Value	Final Value
$B$2	Decision variables x1	0	7.5
$C$2	Decision variables x2	0	0
	Constraints
Cell	Name	Cell Value	Formula	Status	Slack
$D$4	A Profit	30	$D$4<=$F$4	Binding	0
$D$5	B Profit	15	$D$5<=$F$5	Not Binding	65

There are three parts to the answer report. (1). Target cell, 2). Adjustable cell and constraint

cell. The optimal value is given in the final value column of the target cell.

The amount of slack for each constraints which means there is a flexibility existing when

removing one labor will affect the production. the sensitivity report shown above contain

relevant information concerning the effect of changes to objective function coefficient.

adjustable cells section include the reduced cost and ranges of optimality for the objective

function coefficient. The constraint section gives the shadow prices and ranges of feasibility

for right hand side values(allowable increases or decreases)

Question3

Product	Copper	Gold	Constraints
Labor (hrs)	3	4	240
Ore (tons)	2	1	1000*T

1. objective function:
	profit from copper + profit from gold
	Z = $1000/ton Cu + $1500/ ton Au	Let x₁ = copper
	Z = $1000x₁ + $1500x₂	x₂=gold

2. Key resources are tonnes of ore and time.
a). Time:	3hrs/Cu + 4hrs/Au <= (restricted to 240hrs/week)
	i.e. 2x₁ + 4x₂ <=240 hrs
b). Ore:	2ton/Cu + 1ton/Au <= (restricted to 1000*T tons/week)
	i.e. 2x₁ + x₂ <=1000*T tons

3. Market condition, production at two given conditions.
a). X₁ + x₂ <= 700
b). X₁ - x₂ < = 350

LP Models
Maximize Z = 1000x₁+1500x₂		Working.
subject to:	3x₁ + 4x₂ <= 240	1) x₁ <= 80, x₂<=60
	2x₁ + x₂ <= 1000*T	2) x₁ <=500T, x₂ <= 1000T
	x₁ + x₂ = 700	3)x₁ <=700, x₂ <= 700
	x₁ - x₂ <= 350	4) x₁ <= 350, x₂ >= -350
	x₁,x₂>=0

By plotting these inequalities on the graph paper and drawing the linear objective curve and moving

towards the origin, the optimal solution that this curve approached was such that it satisfy the above

given conditions. So the optimal solution is: (80, 0).

x1 = 80 & x2 = 0

Thus, the maximum profit is: Z = 1000*80 = $80 000.

Using the Excel Solver:

	copper x₁	gold x₂	Profit		Limit
Decision variables	80	0
Objective function	1,000	1500	80000
Labor (hrs)	3	4	240	<=	240
Ore (tonnes)	2	1	160	<=	1,000,000
total production	1	1	80	<=	700
Mix production	1	-1	80	<=	350

Answer Report

Solver Options

Max Time 100 sec, Iterations 100, Precision 0.000001

Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%.

Objective Cell (Max)
Cell	Name	Original Value	Final Value
$D$3		0	240

		Variable Cells
Cell	Name	Original Value	Final Value	Integer
$B$1		0	80	Contin
$C$1		0	0	Contin
Constraints
Cell	Name	Cell Value	Formula	Status	Slack
$D$3		240	$D$3<=$F$3	Binding	0
$D$4		160	$D$4<=$F$4	Not Binding	840
$D$5		80	$D$5<=$F$5	Not Binding	620
$D$6		80	$D$6<=$F$6	Not Binding	270

Sensitivity Report

		Variable Cells
		Final	Reduced	Objective	Allowable	Allowable
Cell	Name	Value	Cost	Coefficient	Increase	Decrease
$B$1		80	0	3	0	3
$C$1		0	0	4	1E+30	0

		Constraints
		Final	Shadow	Constraint	Allowable	Allowable
Cell	Name	Value	Price	R.H. Side	Increase	Decrease
$D$3		240	1	240	810	1E+30
$D$4		160	0	1000	1E+30	840
$D$5		80	0	700	1E+30	620
$D$6		80	0	350	1E+30	270

The optimal value is given in the final value column of the target cell.

The amount of slack for each constraint which means there is a flexibility existing for example,

Removing 1 labor will affect the production. The sensitivity report shown above contain

Relevant information concerning the effect of changes to objective function coefficient.

Adjustable cells sections include the reduced cost and ranges of optimality for the objective

function coefficient. The constraint section gives the shadow prices and ranges of feasibility

for right hand side values(allowable increase or decrease)

2. answer the following questions:

i). What are the decision variables?

The decision variables are: x1 = 80 &, x2 = 0.

ii). What is the maximum profit? Z = $80000

iii). If the objective function curve equals the curve for production function of copper and gold., what other decision variables are available?

If this situation arises, there the will be multiple solutions to this problem. As such,

we would have an additional optimal solutions, and they are:

x1 = 0, & x2=60

This gives us a maximum value of: Z = 60*1500 = $90 000

a).

	Copper x₁	Gold x₂	Profit		Limit
Decision variables	80.00	0.00
Objective function	1,000	4500	80000
Labor (hrs)	3	4	240	<=	240
Ore (tonnes)	2	1	160	<=	1,000,000
Total production	1	1	80	<=	700
Mix production	1	-1	80	<=	350

b).

	Copper x₁	Gold x₂	Profit		Limit
Decision variables	234.29	-115.71
Objective function	1,000	-4500	755000
Labor (hrs)	3	4	240	<=	240
Ore (tonnes)	2	1	352.8571	<=	1,000,000
Total production	1	1	118.5714	<=	700
Mix production	1	-1	350	<=	350

c).

	Copper x₁	Gold x₂	Profit		Limit
Decision variables	234.29	-115.71
Objective function	1,000	500	176428.6
Labor (hrs)	3	4	240	<=	240
Ore (tonnes)	2	1	352.8571	<=	1,000,000
Total production	1	1	118.5714	<=	700
Mix production	1	-1	350	<=	350

There were no separate graphs drawn for the above objective functions because the inequalities are the same ones and only the objective functions changed. But they were solve using excel solver only. As such there were some negative value generated by the excel solver. These were resulted when in the options section of solver, the “assume non-negative” option was not clicked. If it was click then it would give zero as a substitute of negative values. After all, the maximum values or maximum profits were given at the last row of objective function under the profit column for the above tables.

It was seen that, for different objective functions there were different decision variables being generated and consequently various maximum values were also generated. However, when solved graphically, the solution obtained cannot be changed even with the various objective functions because it must satisfy the given conditions, i.e. inequalities or the constraints. Therefore, no matter what, the graphical solution remains constant so only one graph was drawn. In order to satisfy this condition; x1, x2 ≥ 0, the negative values can be ignored and only the values from zero to positive values can be accepted.

Sunday, 6 May 2018

May 06, 2018Learning Material, Mine Management, Mining Engineering No comments

Question 1.

Figure 1. Represents all activities, starting from activity 1 to 12. Event 1 is the beginning and event 12 is the end of my task assigned (thick line represent critical path)

1. Describe PERT/CPM network analysis, applications, advantages and disadvantages.

Project evaluation and review technique (PERT) is a project – scheduling tool applied to ensure sequence of different activities eventually lead to a desired completion of a project in consideration. PERT techniques in network analysis is a method of minimizing trouble spots, production bottlenecks, delays associated costs, interruptions and reduce mis-allocations.

Advantages:

PERT is a planning and control technique that uses a network for scheduling and budgeting (time & finance) to accomplish a task.

Disadvantage:

The disadvantage in PERT network is that, an event cannot be accomplished until and unless all activities in a network have been completed. So in a PERT network, we are actually estimating the total project time. The path of the longest duration is called critical path and activities lying this path are critical activities. A delay in any of the activities in the critical path will result in delay, extra costs and resources, redesign, and re – routing of the entire project.

12. What is the difference between an event an activity in a PERT/CPM network?

Activity is the task allocated to be accomplished towards the total completion oif the project on time. It is represented by an arrow. The tail of the arrow represents the beginning of the activity and its head represents its completion. Whereas,

Event is designation the beginning of one activity and ending of another activity. So the starting and ending points of activities are events.

23. Describe how you would calculate earliest event (T_E) and latest event (T_L) times and define what they represent.

The earliest event time (T_E) is the earliest time at which the activities originating from an event can be started.

The latest event time (T_L) is the latest event time which activities terminating an event can be competed on scheduled.

So the earliest event time is calculated by adding earliest event time at the tail of activity arrow with the duration or earliest time of activity. Note that when there is more than one activity flowing into an event; choose the maximum value of T_Ejcalculated for that event.

The latest event time is calculated by subtracting the T_Lof the very last event by the duration/time for activity ij. Subsequently doing the same for rest of the events. When there are more than activities lead to an event, calculate and get the minimum value of T_L

14. Determine the critical path of the exploration project schedule.

The critical path starts from event 1 to event 3 then to event 5 to 6, from event 6 to event 11 and ends at event 12. At these events the T_E value is equal to T_L at each event given above.

15. What is a float and slack in PERT/CPM network and state why is having these advantageous in project scheduling?

Float in PERT network is time available in the non – critical path that could be reallocated towards an activity in the critical path. It is advantageous having float because it represents underutilized time or underutilization of resources. Not only that but also float represents flexibility of an activity and disappearance of float signifies a loss of flexibility for non – critical path activities.

Slack in PERT network is the difference between the earliest and latest times of any event. Slack is important as it help us to see whether it is a critical or non – critical path by analyzing the T_E and T_L values.

16. Calculate head slack, tail slack, float, free float and independent float for one non – critical path activity (ij).

Tail slack Ts = T_si=T_Li - T_Ei = 15– 11 = 4 weeks (event 7)

Head slack HS = Tj = T_Lj - T_Ej = 29 – 18 = 11 weeks (event 7)

Float F = (TLj - TEi) – teij = (29 – 11) – 7 = 11 weeks

Free float FF = (TEj – Tei) - teij = (18 – 11) – 7 = 0

Independent float IF = FF – TS = 0 – 4 = - 4 weeks

Note: a negative independent float is taken as zero for all practical purposes.

17. Briefly commend on your understandings of three extremes of expected times : (a) optimistic, (b) pessimistic and (c) most likely time (m) as represented by normal probability distribution curves.

· Optimistic – is the distribution that satisfy most circumstances at shortest times if execution goes very well.

· Pessimistic – is the distribution that satisfy most circumstances at longest times if everything goes bad.

· Most likely - is the distribution that satisfies most circumstances at normal times or at middle grounds if execution is normal. In other words, it is the mean of the distribution.

18. Complete the following table by calculating the variance (σ²).

Predecessor	Successor	a	M	b	te	σ²
1	2	3	2.5	5	3	0.11
1*	3	5	4.5	13	6	1.78
2	4	7	2	9	4	0.11
2	6	10	10.75	7	10	0.25
4	6	8	1	12	4	0.44
3*	5	5	6.5	11	7	1.00
5*	6	6	7.5	6	7	0.00
3	7	9	2.75	10	5	0.03
7	8	7	2	9	4	0.11
8	9	4	5	12	6	1.78
6*	11	8	12.5	14	12	1.00
9	11	2	5.75	17	7	6.25
7	10	11	5.75	8	7	0.25
10	12	3	14.25	12	12	2.25
11*	12	6	8	6	9	0.00

*critical path.

29. Calculate Z (number of standard divisions), find corresponding probability value in Z table a conclusion on the probability of implementing your schedule as planned.

Sum of critical path σ² is = σ² _(1-3) + σ²_(3-5) + σ²_(5-6)+ σ²_(6-11) + σ²_(11-12)

1.78 + 1 + 0 + 1 + 0

σ² = 3.78

σ = (3.78)^0.5 = 1.9442

Assume X = 44 weeks, and µ = 41 weeks

Z = X - µ = (44 – 41)/1.9442 = 1.5431

Using the table (given), probability of success is: Z value of 1.5431 corresponds to probability of 0.4382 from the table. This means there is 0.9382 (0.4382+0.5) probability or 94% chance of completing the project in 45 weeks or less.

Basing on the probability analysis, I am too optimistic that the project will be completed in 45 weeks or less.

110. Calculate the total float for the non – critical path.

Total float F_T = (T_Lj – T_Ei) – tij = (29 – 11) – 7 = 11 weeks (from event 7 to event 10)

Note that here, only one total float for only one non – critical path.

iv). Find optimal solutions if the objective function changes by:

a). Maximize Z = 1000x₁ + 4500x₂

solutions:

b). Maximize Z = 1000x₁ - 4500x₂

x₁ = 80, & x₂ = 0

Z = $80000

c). Maximize Z = 1000x₁ + 500x₂

x₁ = 234.29, & x₂ =-115.71

Z = $755000

x₁ = 234.29,& x₂ =-115.71

Z = $176428.6