1-) Write down the principles of engineering economy and explain them shortly.
Answer 1
a) Develope the alternatives: The final choice is among alternative. The alternatives need to be
idientified and then defined for subsequent analysis.
b) Focus on differences : Only differences in expected future outcomes among the alternatives
should be considered in the decision
c) Use consistent viewpoint: The prospective outcomes of the alternatives should be developed
from a defined view point.
d) Use a common unit of view point : This will make easier the analysis and compertion of
alternatives.
e) Consider all relevant criteria : Selection of a preferred alternative requires the are of a criterion
( or several criteria)
f) Make uncertainty explicit : Uncertainity is available in estimating the future outcomes of the
alternatives and should be recognized in compersion.
g) Revisit your decision : initial projected outcomes of the selected alternative should be
compered with actual results achieved.
2-) indicated whether each of the following statment is true (T) or false (F).
T F Shaft maintenance can be classifed into fixed cost.
T F Interest is money paid for the use of equity capital.
T F IRR is the rate of return at which NPV is maximum.
T F Sink fund function spreads a future value evenly over time.
T F The costs for blasting material can be classified into variable costs.
3-) It is estimatet that a copper mine will produce 10000 tons of ore during the coming year.
Production is expected to increase by 5 % per year thereafter in each of the following four years.
Profit per ton will be $14 for years one through five.
a) Draw a cash-flow diagram for this copper mine operation from the company’s viewpoint.
b) If the company can earn 15 % per year on its capital, what is the net future value of the
copper mine’s cash flows at the end of year five ?
F=P(1+i)N
Answer 3
Years
Production
Profit , $
1
10000
140000
2
10500
147000
3
11025
154350
4
11576.2
162067
5
12155
170170
NFV = 140000*[1+ 0.15]1 +147000*[1+0.15]2 + 154350*[1+0.15]3 + 162067*[1+0.15]4 +
170170*[1+0.15]5
= 161 000 + 194407 + 234 747 + 283 456 + 342 273 = $ 215 883
4-) The lignite produced from Söke field is used for domestic purposes. The selling price of lignite is
p=150-0.02*D TL. The fixed cost is 44 000 TL/year and the variable cost is 18 TL /ton.
a) What if the optimal amount of the yearly lignite production that maximizes profit for the
mine?
b) Determine breakeven points for profitable production of lignite.
a  cv  a  cv 2  4 b  C F 
Db 
2 b 
1/ 2
Answer 4
A=150 , Cv = 18 , b=0.02 , CF = 44 000
D* =
a  cv 
2b
=
10 = is (a-Cv ) >0
150 ytl / ton  18 ytl / ton
= 3300 ton / year
2 * 0.02
150-18 =132 ytl/ton , yes
20 = is [ TR – Ct ] >0 for D* =3300 ton/year
18*3300]
{150*330-0.02 [3300]2 } – [44000 +
277200 – 130400 =173800 > 0, yes

 =  132  18 = 6250 ton/year
 150  18  150  18  4 0.02 *  44000
D1 =
2 *  0.02
2
D2 = (-132+118) /-0.04 =350 ton/year
1/ 2
 0.04
5-) A capital investment of $ 79 000 000 can be made in a gold mining project that will produce an
annual revenue of $85 000 000 an annual operating cost of $32 000 000 for six years. The company is
willing to accept any project that will earn 15% per year on all invested capital. Draw a cash flow
diagram and show whether this is a desirable investment by using NPV method.
P = F(1+i)-N
Answer 5
79
85
85
85 85 85 85
32
32 32 32 32 32 $ *1000 000
NPV = PV of cash flows – PV of cash outflows
PV of cash inflows =$ 85 M *(P/A, 15 % , 6)
= 85 M * 3.7845 = $ 321 682 500
PV of cash outflows = $ 32 M *( P/A,15 % , 6)+$79M
= 32 M *3.7845 + $ 79 M
= $ 200 104 000
NPV = 321 682 500 – 200 104 000 ≈ $ 121 000 000
6-) Consider a mine where production rate is 1 000 000 tons/year, selling price of ore is $20/ton and
total cost of production is $8/ton. The mine has three million tons of reserve.
a) When the discount rate is 15 % , prepare a cash flow table, calculate net present value and also
profit/ton.
b) At the end of exploration works, the staff has found another three millions of tons of reserve,
making total amount of reserve six million tons. The mine has recommended two alternatives
to produce the ore. The first alternative suggests that the life of the mine can be extend over
six years [ in other words, three years are added to first three years while production rate
remains the same( 1 000 000 tons/year) ]. The second alternative suggests that the life of the
mine remains the same (three years) but the production rate is increased to 2 000 000
tons/year. İf you were the person who is in charge of selecting the best alternative , which
alternative would you select ?
( Use NPV criterion and profit /ton in your choice ). What is the main conclusion that you draw
from this result ?
N
 1  İ N  1
 1 
P  F
 , P  A
N 
1 İ 
 İ 1  İ  
Answer 6
6/a
1 2
Revenue ( $/ton) 20 20
Cost ($/ton )
8 8
Profit ( $/ton)
12 12
3
20
8
12
NPV 1 = $ 12 M (P/A,15,3)
= $ 12 M *2.2283 = $27.4 M
Profit/ton = 27.4 M/3M = 9.13 $/ton
6/b
NPV1 = 12 M (P/A, 15 ,6) =12 M * 3.784 = 45.4 M 45.4/6 = 7.57 $/ton
NPV2 = 24 M *2.283 = 54.8 M = 54.8/3 = 18.26 $/ton
Because profit /ton in (2) is greater than profit in (1) ,(2) is prefered. The main conclusion drawn this
reesult is that making production with high capacity is always advantageous over production with
longer life.
7-) Three mutually exclusive altenatives for mine investment are being evaluated, and their costs and
revenues are itemized in the following table.
Capital investment, $
Annual profit, $
Market Value
Useful life ( years)
A
4 620 000
704 000
0
20
Mutually Exclusive Alternatives
B
C
2 420 000
6 380 000
440 000
825 000
0
0
20
20
İf the MARR is 5 % per year, determine which alternative should be selected by using incremental
analysis procedure.
Answer 7
a) Arrange the feasible alternatives based on increasing capital investment
Capital investment
Profit
Project life, year
NPV
IRR
B
2420
440
20
3062
17.5
A
4620
704
20
4152
14.2
C
6380
825
20
3900
11.5
b) Calculate IRR and NPV values
NPVB = -2420 +440*( P/A,5,20)
= -2420 + 440*12.46 = 3062
NPVA = -4620 +704 * (P/A,5,20)
= -4620 + 704*12.46 = 4152
NPVc = -6380 + 825*12.46
NPVB it is a base alternative
c ) Calculate the incremantal vaues
Capital İnvestment
Profit
IRR
A-B
2200
264
10.4
9
NPV IRR
A  B = -2200 + 264*(P/A,9,20)
= -2200 + 264*9.135 = 211.6
11
NPV IRR
A B = -2200 + 264*(P/A,11,20)
= -2200 + 264*7.962 = -98.0
11  9
IRR  9

 IRR=10.4 %
211 .6  (98 .0)
211 .6
3
NPV CIRR
 A = -1760 + 121*(P/A,3,20)
= -1760 + 121*14.926 = 46
4
NPV CIRR
 A = -1760 + 121*(P/A,4,20)
C-A
1760
121
3.3
= -1760 + 121*13.596 = -114.9
43
IRR  3

 IRR =3.3 %
46  (114 .9)
46
IRR (C-A) = 3.3 < MARR = 5
Alternative C is discarded. Therefore A is prefered alternative because invested incremantal value
(2200) is justified at profit 264 with IRR > MARR
8-) A dozer costs $ 500 000 and has a useful life of 5 years. Its estimated market value at the end of
the 5 year is $20 000.
a) Determine depreciations for years between 1 and 5 using i) the SL method and ii) the 200% DB
method.
b) Compute the net present value of depreciation deduction for each year of the two methods. The
MARR is 10% per year. If a large NPV is desirable which method is preferred ?
dk = B(1-R)k-1 R , dk = ( B-SVN )/Q, dk =(B-SVN ) [2(N-k+1)/N(N+1)], dk = (B-SVN/N)
Answer 8
B=$500 000, N=5, SV=$20 000
a) SL method , dk = ( B-SVN )/N
d1 =(500 000-20 000)/5 = 96 000 , d2= d3= d4= d5= 96 000
DB Method , R=2/N = 2/5=0.4 , dk = B(1-R) k-1 * R
d1 = 500 000 (1 – 0.4)1-1 *0.4 = 200 000 , d2 = 500 000*(0.6)2-1 * 0.4 =120 000
d3 = 500 000*(0.6)3-1 * 0.4 = 72 000 , d4= 500 000*(0.6)4-1 * 0.4 = 43 200 ,d5 = 500 000*(0.6)5-1 * 0.4
= 25 920
b) NPV SL = 96 000 (P/A, 10 ,5) =96 000 * 3.791 = $ 363 936
NPV DB = 200 000 [1/(1+0.1)]1 + 120 000 [1/(1+0.1)]2 + 72 000 [1/(1+0.1)]3 + 43 200 [1/(1+0.1)]4
+ 25 920 [1/(1+0.1)]5
= 181818 + 99 173 + 54 095+ 25 506 + 16 094 =$ 380 686
NPV DB > NPV SL so DB method is prefered.
9-) A marble quarry is estimated to contain 900 000 tons of stone, and the X mining Company just
purchased this quarry for $ 1 800 000. If 100 000 tons of marble can be produced each year and the
average selling price per ton is $ 8.60 , calculate the first year’s depletion allowance for the cost
method.
Answer 9
Depletion unit = basis/ reserve = $ 1 800 000 / 400 000 tons = 2 $/ton
Depletion allowance for the first year = 100 000 tons * 2 $/ton = $ 200 000
10-) Consider a power plant that uses lignite as fuel. In this plant, the cost of producing electricity is
calculated by TC= 0.27 D2 +0.3D + 12 where D is the amount of electricity produced ( in megawatts).
If the revenue obtained from the sale of electricity is TR = -0.2 D2 +15D,
a) Determine the value of D that maximizes profit,
b) Show that profit has been maximized rather than minimized,
c) What is the range of profitable electricity production ?
Answer 10
a) Profit = TR – TC = -0.2 D2 +15D – [ 0.27 D2 + 0.3D+12] = -0.2 D2 -0.3D -12
= -0.47 D2 + 14.7 D -12
ɗP / ɗD = ɗ( -0.47D2 +14.7D -12)/ ɗD = -0.94 D +14.7 = 0 → D = 15.64 megawatt.
b) ɗ2 P/ ɗD2 =-0.94 < 0 profit maximized.
c) a= -0.47 , b= 14.7, c= - 12
=
b ^ 2  4ac = 4.7^2  4 * (0.47 * 12) = 13.91
D1 = (-b-∆^½)/2a = (-14.7-13.91)/(-0.94)=30.4 megawatt
D2 = (-b+∆)/2
11-) A certain coal stripping operation presently uses three dozers for reclamation work. To reduce
costs, three alternatives are being considered for this job in the future : rebuild present equipment (A),
purchase new dozers (B), and employ a concrator (C). Details for the alternatives are given in the
following table. If the MARR is 8 % per year, which alternative should be selected? useNPV criterion.
Number of dozers
Capital investmen, $
Annual costs, $:
Maintenance
Labor
Supplies
Useful liffe (years)
Salvage value,$
A
3
360 000
B
2
920 000
C
Not applicable
0
140 000
240 000
58 000
8
0
85 000
460 000
42 000
8
120 000
0
525 000
0
8
0
A
360
438
8
0
B
920
287
8
120
C
0
525
8
0
Answer 11
CI
AC
Life
SV
NPVA = -360 -438*(P/A,8,8)
= -360 -438*5.747 = -2 877
NPVB = -920 -287*(P/A,8,8)+120*(P/F,8,8)
= -920 -287*5.747 + 120 *0.54 = -2 504
NPVC = -525*(P/A,8,8) = -525 * 5.747 = -3 017
Select the alternative with the least cost , select B.
12-) The following table shows annual cash flows for two mutually exclusive alternatives. If the
MARR is 15% than which alternative should be chosen? Use incrementl analysis.
Year
0
1
2
3
4
Project A
-$10 000
+$1000
+$5000
+$6000
+$7000
-Project B
-$10 000
+$7000
+$5000
+$3000
+$1000
Answer 12
A
B
A-B
IRR 24.7 30.5 16.54
NPV 2.60 2.41 0
Select A
13-) Two drilling machines can perform drilling equally well. The capital investment and annual
expenditures for these machines are given in the following table.
Capital investment, $
Annual expenditure, $
Life, years
Salvage value, $
Drilling Machine A
150 000
18 000
5
2 000
Drilling Machine B
200 000
10 000
10
0
If the MARR is 10 % , which machine should be selected ? Use repeatability assumption.
Answer 13
NPVA = -150 -150(P/F,10%,5) -18(P/A,10%,10) + 2(P/F,10%,5) + 2(P/F,10,10)
= -150 – 150*0.6209 – 18*6.145 + 2*0.6145 + 2*0386 = -351.7
NPVB = -200 -10(P/A,10%,10) = -200 -10*6.145 = -261.4
Select the alternative with the least cost , Select B.
14-) Write down unique characteristics of mining and explain them shortly.
Answer 14
Capital insensity – mining ventures are extremely capital intensive - $500 M - $10 billion
Long production period-4.8 years even longer times ( 10-12) for enviromental considerations
High risks ( geological, engineering, economical and political risks)
Non-renewable resources. ( ore resources depletes new resource have to be fpund. –closing mine has
on local communities)
15-) It is estimatet that a copper mine will produce 10000 tons of ore during the coming year.
Production is expected to increase by 5 % per year thereafter in each of the following four years.
Profit per ton will be $14 for years one through five.
a) Draw a cash-flow diagram for this copper mine operation from the company’s viewpoint.
b) If the company can earn 20 % per year on its capital, what is the net future value of the
copper mine’s cash flows at the end of year five ?
F=P(1+i)N
Answer 15
Years
Production
Profit , $
1
10000
200000
2
10500
210000
3
11025
220500
4
11576.2
234525
5
12155
243100
NFV = 200 000*[1+ 0.15]1 +210 000*[1+0.15]2 + 220500*[1+0.15]3 + 231525*[1+0.15]4 +
243100*[1+0.15]5
=230 000 + 277725+ 335336 + 404937 + 488 947 = $ 1 736 945 $
16-) The lignite produced from Söke field is used for domestic purposes. The selling price of lignite is
p=150-0.02*D YTL. The fixed cost is 45000YTL/year and the variable cost is 20 YTL /ton.
a) What if the optimal amount of the yearly lignite production that maximizes profit for the
mine?
b) Determine breakeven points for profitable production of lignite.
a  cv  a  cv 2  4 b  C F 
Db 
2 b 
1/ 2
Answer 16
A=150 , Cv = 20 , b=0.02 , CF = 45 000
D* =
a  cv 
2b
=
10 = is (a-Cv ) >0
150 ytl / ton  20 ytl / ton
= 3250 ton / year
2 * 0.02
150-20 =130 ytl/ton , yes
20 = is [ TR – Ct ] >0 for D* = 3250 ton/year
18*3250]
{150*3250 - 0.02 [3250]2 } – [4500 +
277200 – 130400 =166 250 > 0, yes

 =  130  115.32 = 367 ton/year
 150  20  150  20  4 0.02 *  45000
D1 =
2 *  0.02
2
1/ 2
 0.04
D2 = (-130 – 115.32) /-0.04 = 6125 ton/year
17-) A capital investment of $ 80 000 000 can be made in a gold mining project that will produce an
annual revenue of $90 000 000 an annual operating cost of $30 000 000 for six years. The company is
willing to accept any project that will earn 15% per year on all invested capital. Draw a cash flow
diagram and show whether this is a desirable investment by using NPV method.
 (1  i) N  1
N 
 i(1  i) 
P = F(1+i)-N or P  A
Answer 17
80
90
90
90 90 90
90 90
30
30 30 30 30 30 30$ *1000 000
NPV = PV of cash flows – PV of cash outflows
PV of cash inflows =$ 90 M *(P/A, 15 % , 6)
= 90 M * 3.7845 = $ 340 605 000
PV of cash outflows = $ 30 M *( P/A,15 % , 6)+$79M
= 32 M *3.7845 + $ 80 M
= $ 193 535 000
NPV = 321 682 500 – 200 104 000 ≈ $ 147 070 000
18- ) Indicate whether each of the following statements is true(T) or false (F).
T F as the discount rate increases, theNPV increases.
T F The greater the NPV, the richer the invstors become by undertaking a project.
T F The mutuallyt exclusive alternative with the highest IRR is chosen in incrimental analysis.
T F Book value is the estimated value of property at the end of its useful life.
T F Land is not a depriciable property.
19-) A capital investment of $ 100 000 can be made in a mining project that will produce an annual
revenue of $53 100 for five years and then a salvage value of $20 000. Annual expenses will be $30
000. If the MARR is 10 % per year, determine whether the project is acceptable by using the IRR
method.
 (1  i) N  1
P = F(1+i) or P  A
N 
 i(1  i) 
-N
Answer 19
3
2
15680
A
B
d
1
0
-1
-2
5
10
e
15
C
-12620
NPVA = 0= - $100 000 +( $53 100 - $ 30 000)(P/A,i%,5)+$20 000(P/F,i%,5); i% =?
At i=5% : NPV = - $ 100 000 + $ 23 100 (4.3295) + $ 20 000 (0.7835) = $ 15 680
At i=15% : NPV = - $ 100 000 + $ 23 100 (3.3522) + $ 20 000 (0.4972) = - $ 12 620
BA dA

BC de
15 %  5%
İ %  5%

$15680  $12620 $15680  $0
İ%=5%+
$15680
10%
$28300
İ%=5% +5.5% =10.5%
20-) Consider the following two mutually exclusive alternatives related to a mining project, and
recommend which one should be chosen using NPV method. The MARR is 15% per year, and the
study period is 10 years. Assume repeatability is applicable.
Capital investment
Annual revenue
Market value
Useful life ( years)
A
$ 200 000
56 000
40 000
5
B
$ 300 000
54 000
0
10
Answer 20
NPVA = - 200 000 -200 000( P/F,15% , 5) + 56 000 [P/A,15%,10] + 40 000[P/F,15%,5] + 40
000[P/F,15%,10]
= - 200 000 – 99 400 + 281 053 + 19 888 + 9 888
=11 389
NPVB = -300 000 + 54 000 [P/A,15% , 10 ] = -28 947
A is chosen.
21-) A dozer costs $ 500 000 and has a useful life of 5 years. Its estimated market value at the end of
year 5 is $20 000. Determine depreciations for years between 1 and 5 using i) the SL method and ii)
DB method.
dk = B(1-R)k-1 R , dk = ( B-SVN )/Q, dk =(B-SVN ) [2(N-k+1)/N(N+1)], dk = (B-SVN/N)
Answer 21
i)
SL method , dk (B-SVn) / N
d1 = (500 000 – 20 000)/5 = 96 000, d2 = d3 = d4 = d5 = 96 000
ii) DB Method , R2/N =2/5=0.4 , dk = B ( 1-R) k-1 *R
d1 = 500 000 (1-0.4)1-1 *0.4 = 200 000
d2 = 500 000 (0.6)2-1 * 0.4 = 120 000
d3 = 500 000 (0.6)3-1 * 0.4 =72 000
d4 = 500 000 (0.6)4-1 * 0.4 = 43 200
d5 = 500 000 (0.6)5-1 * 0.4 = 25 920
22-) Indicated whether each of the following statements is true (T) or false ( F) .
T F Insurance and tax are available costs
T F Profit maximization occurs when marginal cost is equal to marginal revenue
T F Breakeven points occur when total cost
T F NPV increase with increasing discount rate
T F IRR is internal rate that makes NPV zero
23-) Amining company produces nickel from a latheritic nickel deposit. The price of nickel is
$10 000/ton. The company manager thinks that the total cost function is Ct = 100 000 + 9 500 D +
0.5D2 where D is the amount of metal proceduced. ( Total revenue is product of price and the amount
of nickel produced)
a) Derive profit function,
b) Find an optimal value of D making profit maximum,
c) Determine breakeven points for profitable production of nickel.
Answer 23
a) P= TR-Ct = 10 000*D –[ 100 000 +9500 D + 0.5D2 ]
= 10 000*D -100 000 -9500D-0.5D2 =0.5D2 +500 D – 100 000
b) ɗP / ɗD = -D + 500 = 0
D=500 ton ɗ2 P/ ɗD2 = -1 (max attained)
c) TR =Ct = 10 000 D= 100 000 + 9500 D +0.5 D2
0.5D2 -500D+100 000= 0
D1.2 =
 (500)  (500) 2  4(0.5)(100000)  500  223.6
=
; D1 = 276 ton , D2 = 724 ton
1
2 * 0.5
24-) A mining firm has just purchased a new drilling machine that has a useful life of five years. The
engineer estimates that maintenancce cost for the machine during the five year will be $10 000. As the
machine will ages, maintenance costs are expected to increase at a rate of $ 1 000 per yaar over the
remaining life. Assume that maintance costs occur at the end of each year. The machine will at the end
of five years with a salvage value of $ 10 000. Draw cash flow diagram for the drilling machine.
Answer 24
$10000
$11 000
$ 10 000
$12 000
$13 000
$14 000
25-) Consider a mining project outlined in the following table( all dollars are in thousands).
Years
Capital
investment, $
Revenue
Operating
cost, $
0
1
2
3
4
5
41 848
21 965
76 467
31 379
64 726
31 379
66 242
31 379
33 486
6 168
103 581
When a discount rate of 16% is chosen , determine whether the project is feasable or not. Use NPV
method.
P = F(1+i)-N
Answer 25
Year
profit
0
-103 581
1
19 883
2
45 088
3
33 347
4
34 863
5
27 318
NPV = PV of cash[ınflows-outflows]
NPV  
103581
19883
45088
33347
27318




0
1
2
4
(1  0.16 )
(1  0.16 )
(1  0.16 )
(1  0.16 )
(1  0.16 ) 5
= -103 581 + 17140.5 + 21364 + 19254.5 + 13006.4 + 33507.7 = $692 100 project is feasable
26-) A capital investment of $ 80 000 000 can be made in a gold mining project that will produce an
annual revenue of $90 000 000 an annual operating cost of $30 000 000 for six years. What is the
projected IRR on this investment ? And if MARR is known to be 13%, is this investment justifiable ?
 (1  i) N  1
P  A
N 
 i(1  i) 
Answer 26
Choose IRR1 =70 % and IRR2 80 %
NPV at 70 % → NPV = $ 90 M (P/A,70 % , 6) -$ 30 M (P/A,70 % , 6) -$ 80 M (P/F,70%,6)
(P/A,70 % , 6) = [(1+0.7)6 -1]/[0.70(1+0.8)6 ] = 23.13/16.89 =1.37
NPV at 70% = 90*1.37 – 30*1.37 -80 = $ 2.2 M
NPV at 80% →(P/A,80%,6) = [ (1+0.8)6 -1]/[0.80(1+0.80)6 ] = 33.012/27.209 = 1.213
NPV at 80% → 90*1.213 -30*1.213 – 80 = -$ 26.4 M
A=2.2 M
d
B
IRR
70 %
e
80 %
C= -26.4 M
BA dA

BC de
BA= 80% -70 % =10%
BC = 2.2 M –(-26.4 M) =28.6 M
dA = IRR % -70%
de = 2.2 M
10% IRR%  70%
10 * 2.2

 IRR% 
 70  IRR%  70.8%
$28.6
$2.2M
28.6
IRR%=70% > MARR = 13% project is feasable
27-) Given the following three mining project alternatives with MARR = 15 %, choose the best
alternative by using the incremental IRR procedure.
Capital investment ,$*
Annual net cash flow,$
Salvage value , $
Useful life( year)
*) All dollars are in millions
A
28
5.5
1.5
10
B
16
3.3
0
10
C
23.5
4.8
0.5
10
Order the feasable alternatives based on increasing capital investment
B
CI
16
ACF 3.3
SV
0
C
23.5
4.8
0.5
A
28
5.5
1.5
→ compute IRR orNPV
NPVB = -16 + 3.3*(P/A,10,15%) + 0(.) = 0.563
NPVC = -23.5 + 4.8( P/A,10 15%) + 0.5(P/F,10,15%) =0.715
NPVA = -28 + 5.5(P/A,10,15%) + 1.5(P/F, 10 ,15%) = -0.02
NPVB = 0 , B is a base alternative , Drop A from consideration due to NPVA < 0 calculate incremental
values.
C-B
CI
7.5
ACF 1.5
SV
0.5
NPV C-B = 0 = -7.5 +1.5 ( p/A,10 , i%) + 0.5(p/F , 10 ,i%) →i=15.5%
iC-B = MARR , keep the alternative C as current best alternative , drop alternative B
from further consideration. Recommend thealternative C for investment.
28-) A mining company is considering to purchase a newforklift truck. There are two alternatives :
Alternative A costs $11 000, requires $ 2 000 per year in operating expense , and is expected to have a
3 year economic life. Its salvage value at the time is projected to be $ 4 000. Alternative B costs $15
000 but requires $1500 per year to operate. It is expected that its economic life wil be 5 years, with
salvage value of $4000. If MARR is 12% which alternative is superior? Use coterminated assumption.
 (1  i) N  1
P = F(1+i) or P  A
N 
 i(1  i) 
-N
Answer 28
3
A
5
B
NPVA= [-11000 – 2 000(P/A,12%,3)+4000(P/F,12% ,3)] *(P/F,12%,2) = -10 332
NPVB = -15 000 -1 5 000 ( P/A,12% ,5)+ 4 000 (P/F,12%,5) = -18 130
Choose alternative A as a minimum cost alternative.
29-) A mining company purchased a drilling machine at the cost of $ 140 000. The machine is
estimated to have 5 year useful life. At the end of the 5th year , the salvage value will be $ 20 000.
Calculate the depreciation expenses for the first ,second and third yeras using a) straight line method
and b) the 150% DB method.
dk = B(1-R)k-1 R , dk = ( B-SVN )/Q, dk =(B-SVN ) [2(N-k+1)/N(N+1)], dk = (B-SVN/N)
Answer 29
B = 140 000 , N = 5 , SV = 20 000
a) dk = (B-sv)/N = (140 000 – 20 000)/5 = 24 000 for 2010,2011 and 2012
b) dk = B(1-R) k-1 *R
R = 1.5/N =1.5/5 = 0.3
k=1 d1 = 140 000 (1 – 0.3) 1-1 *0.3 = 42 000
k=2 d2 = 140 000 (1 – 0.3) 2-1 *0.3 = 29 400
k=3 d2 = 140 000 (1 – 0.3) 3-1 *0.3 = 20 580
30-) A mining company operating a polymetalic deposit produces lead and Zinc which are sold at
$0.99/lb and $1.25/lb respectively. The mine manager thinks that the ttotal cost function is:
TC = 2 + 0.1Dpb + 0.15 DZn + 0.05 D2Zn + 0.05 DPbDZn
Where DPb and DZn are outputs of lead and Zinc recpectively ( total revenue is product of price and
output of ore produced )
a) Derive the profit function for both products
b) Find the optimal value od DPb and DZn along with maximum profit
c) Make sure that profit is maximized at this point
Answer 30
TRPb = 0.99*DPb
TRZn = 1.25*DZn
P=TR-TC
PZn = TRZn - TCZn
PPb = 0.99*DPb –[ 2 + 0.1DPb +0.15....+ 0.025DPb DZn ]
PZn =1.25*DZn –[...]
ɗP Pb/ ɗD Pb = 0.89 – 0.1DPb – 0.025 D Zn = 0 ɗPZn / ɗD Zn = 1.10 – 0.24DZn 0.025 DPb =0
DZn = 3.74 lb , DPb = 7.96 Lb
ɗ2 PPb/ ɗD2 Pb = - 0.1
ɗ2 PZn/ ɗD2 Zn = - 0.24
31-) consider a mine with the following data :
Capital investment = $30 000 000,
Reserve = 6 000 000 ton
The selling price of the ore = $20/ton
Total cost of production = $8/ton and
The discount rate = 15%
For ore production the mine management suggests two alternatives on the production rate : the first
one with 1 000 000 ton/year and the second ne with 2 000 000 ton/year. If you were person who is in
charge of selecting the best alternative which alternative would you select? Assume that repeatability
is applicable.
Answer 31
N= 6 000 000 / 1 000 000
= 6 years
Annual profit = $ 12/ton*1 000 000 ton
=$ 12M/year
0
A
N = 6 000 000 / 2 000 000
= 3 years
Annual profit : $12/ton * 2 000 000 ton
= $ 24M / year
6 0
B
3
6
NPVA = -$30 M +$ 12M ( P/A,15%,6)
NPVA = -$30 M +$ 12M ( P/A,15%,6)
=-$30 M + $45.4 = $15.4
= -$30M - $19.7 M +$ 90.8 M = $41.5
Choose the alternative B
32-) An energy company wants to invest on a new hydro-electricity central ( HEC) that produces
electricity. The initial cost of the HEC is $10.5 million. The HEC will produce 40 million kwh with
81% efficiency per year. The goverment will buy the electricity at a price of 5 cent per kwh. After 30
years, the company is 12 % , determine the payback period of this project.
Answer 32
CI= $ 10.5 M P = 40 M *0.81 / 100 = $ 1.62 M
Year
0
1
PV of $ 1.62 M
At i=12%/yr
-10.5
1.45
7
0.73
8
0.65
-9.05 -7.76 -6.61
-5.58 -4.66 -3.84 -3.11
-2.46
10
0.52
12
0.42
13
0.37
14
0.33
15
0.30
-1.88 -1.36 -0.89 -0.47
-0.1
0.23
0.53
Cumulative Pv at i=12%/yr
Year
PV of $ 1.62 M
At i=12%/yr
Cumulative Pv at i=12%/yr
9
0.58
2
1.29
11
0.47
-$10.5M + $1.62 M + (P/A,12%,PB) = 0
3
1.15
4
1.03
5
0.92
6
0.82
(P/A,12%,PB) = 10.5/1.62 = 6.4815 PB=13.05 ≈13 years
33-) Consider a hypotetical gold mine with 5 year life. Assume that the initial capital cost is $30 000
000, the annual revenue is $25 000 000 and the annual operating cost is $15 000 000 over 5 years. The
salvage value of the equipment at the end of five year is $30 000 000. The company will open up the
mine if it will earn at least 15% per year. Determine whether it is accepable by using Rate of Return
method.
Answer 33
At IRR = 10%
NPV = -$30 M + $ 25 M ( P/A ,10%,5) + 3M( P/F, 10%,5)
= $ 9.8 M
At IRR = 25%
NPV= -$30 M + $25 M (P/A , 25%,5) - $15(P/A,25%,5) + $3 M ( P/F, 25% 5)
= -$4.09 M
BA dA

BC de
A=9.8 M
15%
IRR  10%

IRR = 20.6%
$13.9M
$9.8M
d
B
IRR
e 20%
25 %
C= -4.09 M