4.
PROJECT BENEFIT COST ANALYSIS
4.1. Least cost
This chapter deals with the comparison of alternative strategies or alternative
solutions. To illustrate this the following example is discussed.
An rather old excavator, which will be replaced after two years, needs repairing (or
overhaul) at an estimated costs of 10,000 (investment). If the overhaul is not
carried out it is anticipated that the operating expenses (M + R) for the excavator will
involve an increase of 5,600 for each of the two remaining years of its service life.
It might appear that this is a good investment because an expenditure of 10,000
leads to a total saving of 11,200 over the next two years. This is, however,
incorrect as it overlooks the time value of money.
The question is whether the present value of the anticipated savings on running
costs is greater or less than the investment costs for overhaul of 10,000. An
alternative, but equivalent, way of framing the question is: "would the owner of the
excavator earn more or less than the 5,600 per year offered by the savings on
running costs by placing the initial 10,000 into an alternative investment (for
example in stocks or bonds)". The first approach focuses upon the discounting of
future euro's into present value, whereas the second approach concentrates on the
translation of present euro's into a time stream of euro's (compounding). In principle
these approaches are equivalent. The problem can only be solved if a certain interest
rate is assumed.
Present value concept (discounting).
For an interest rate of 10 % the present value of the savings in M + R costs is:
,
5 600
600
,
5
+
= 5,091 + 4,628 = 9,719.
10
.
1
2
10
.
1
When the present values of the savings on M + R expenses is compared with the
overhaul costs of 10,000 it is obvious that the overhaul investment is not a wise
decision. If, however, the interest rate is only 5 % then the present value of the
savings on M + R expenditure is:
600
,
5
600
,
5
+
= 5,333 + 5,080 = 10,413.
05
.
1
2
05
.
1
In present euro's, which is more than the initial investment of 10,000, making the
repair at the beginning the better option.
Compounding approach.
Suppose that there is an opportunity to invest the 10,000 in a stock or a bond
offering an annual rate of return equal to 10 % (after costs etc.). The future value is
the initial investment plus the compounded interest earned. At the end of the first
year the future value will be 10,000 plus 1,000 in interest earning (10%) =
11,000. Assume (for simplicity) that the M + R extra costs are only paid once a year
(at the end of the year); the balance after paying the bill for the extra M + R costs will
be 11,000 - 5,600 = 5,400.
66
If the balance is reinvested at 10% the future value at the end of the second year is
5,400 x 1.10 = 5,940 or a surplus of 340 remains after payment of the
second extra costs for M + R. Clearly the alternative investment is a stock or a bond
is more attractive. Alternatively if the available return is 5 % rather than 10 % the
future value at the end of the first year will be 10,000 x 1.05 = 10,500 or
4,900 after payment of the extra 5,600 expenses for M + R. By the end of the
second year there would only be 4,900 x 1.05 = 5,145 available from the
original 10,000 to pay the bill of 5,600 for the extra M + R costs of the second
year or a deficit of 455. In this case the investment in the overhaul is justified.
The following comparison of options is exactly the same as the discussed example
above. An office building is to be replaced by a new building at the end of two years.
Should the insulation in this old building be improved at a cost of 10,000 if the
anticipated savings in heating and cooling cost is 5,600 for each of the remaining
years in the building's life ?
Example
The expected Nett Present Value NPV (see next paragraph) of some project is
50,000,000. Collecting more data and doing some additional studies during a period
of 3 more years are expected to increase the NPV to a value of 65,000,000
(discounted value at the end of the 3-years study period). The costs of these
additional activities is 1,000,000 at the end of each of these 3 years. Should the
extra data collection and studies be done? The discount rate is 8 %.
Answer
Original
situation : NPV = 50,000,000
i = 8 %
t = 0
t
New situation :
NPV = 65,000,000
additional
studies
t = 0 1 2 3
t
1,000,000 / year
Express new situation in Present Value (P.V.) ( t = 0 ):
discount
factor
1 / (1.08) 3 = 0.794 (for NPV of 65,000,000)
discount factor [ (1.08) 3 - 1 ] / [ 0.08 x (1.08) 3 ] = 2.577
NPV 65,000,000 x 0.794 - 1,000,000 x 2.577 = 49,023,000
As the NPV for the new situation is lower than the original schedule the conclusion is
that additional studies should not be carried out !
67
Capitalized costs
In construction works the precise life of an asset may be difficult to access with
accuracy. An initial capital investment is made in order to shape the natural ground;
the life of these earth works (for example a canal.a road cutting) may be very long or
even forever. In such cases the computation of capital recovery takes a similar form
to the computation of simple interest.
1
( + i)n
If the value of n increases , so the term
approaches to 1.
1
( + i)n -1
and the capital recovery formula becomes approximately :
A = i . P (for n = 100 years).
If the lifetime of an asset is considered to be 100 years and not in perpetuity, there
will be only a very small difference in the resulting calculation between using the
appropriate capital recovery factor itself and using the relevant interest rate.
The term capitalized costs is commonly used by engineers in cases where
comparisons of costs are made over periods of time in perpetuity and annual costs
are assumed to be incurred on a perpetual basis.
A
The Present Value of the annual costs becomes: P =
i
68
Optimum of initial cost and maintenance cost
Example
A dike is proposed for river protection. The higher the dike the greater the costs, and
the lower the risk of flooding. Estimated data are indicated in the following table:
Height
of
dike
(m) 2 3 4 5 6 7 8
Cost of dike
10 25 43 67 100 150 225
( 1,000)
Risk of flooding
4 2 1 0.5
0.1
0.05 0.01
(times per year)
If the damage by flooding is estimated at 10,000 each time it occurs, what design
height should be selected if money can be borrowed at (a) 10 %; and (b) 20 % ?
Answer:
Capital costs: the dike would be everlasting, so only annual costs need to be
considered
a. For 10 % interest rate.
Height
of
dike
(m) 2 3 4 5 6 7 8
Annual interest
1.0 2.5 4.3 6.7 10.0 15.0 22.5
( 1,000)
Annual costs of
Floods ( 1,000)
40
20
10
5.0
1
0.5
0.1
Total annual
Costs ( 1,000)
41
22.5 14.3 11.7
11
15.5 22.6
The total annual costs are minimal, at 10 % interest rate, for a design height of
6 metres.
b. For 20 % interest rate.
Height
of
dike
(m) 2 3 4 5 6 7 8
Annual interest
2.0 5.0 8.7 13.4 20.0 30.0 45.0
( 1,000)
Annual costs of
Floods ( 1,000)
40
20
10
5.0
1
0.5
0.1
Total annual
Costs ( 1,000)
42
25
18.7 18.4
21
30.5 45.1
The total annual costs are minimal, at 20 % interest rate, for a design height of
4.5 metres.
69
45.1
42
41
40,000
30.5
30,000
Annual
25
Total
costs of
annual
floods
costs for
Total
22.6
22.5
i = 20 %
21
annual
20,000
costs for
18.7
18.4
i = 10 %
15.5
14.3
11.7
11
10,000
Optimum design
height for i = 20
Annual interest i = 20 %
Optimum design
height for i = 10
Annual interest
i = 10 %
2 3
4
4.5
5
6
7 8
Height of dike in meters
Optimum design height of a river dike for flood protection, optimisation of capital
costs and the cost of flooding for different rates of interest (i = 10 % and i = 20 %).
70
4.2.
Net Present Value (NPV)
The method of appraising alternative capital investment projects by the net present
value (NPV) is long established and well tried. The net present value method is
alternative known as the present value, the present worth or the net present worth
method. The basis of this method is that all future costs and benefits concerned with
an investment project are converted (discounted) to present value, using a selected
interest rate.
B
C
NPV =
t
-
t
B = Benefits
t
1
( + i)
t
1
( + i)
C = Costs
In all cases the NPV is uniquely defined. It is widely used in the selection of projects.
If the NPV is positive, the project is considered to be profitable: it yields benefits and
exceeds investments, operating costs and taxes. It is frequently more convenient and
certainly more conventional to express all euro estimates in terms of present value.
For example consider two alternative projects, A and B, either of which would cost
10,000 today and yield benefits over a four-year period as follows:
Year
1 2 3 4
Project A
6,000
2,000
16,000
4,000
Project B
8,000
1,000
12,000
4,800
Which of these projects is preferable? From a mere comparison of the annual
benefits it is impossible to determine the answer, as project A is to be preferred for
the 2nd and 3rd year, while project B is better for the first and last year.
Once a discount rate is selected these benefits can be converted into present values
and a comparison made. Present value of the benefits for an interest rate of 5 %:
000
,
6
000
,
2
000
,
16
000
,
4
Project A :
+
+
+
05
.
1
2
05
.
1
3
05
.
1
4
05
.
1
= 5,714 + 1,814 + 13,821 + 3,291 = 24,641
000
,
8
000
,
1
000
,
12
800
,
4
Project B :
+
+
+
05
.
1
2
05
.
1
3
05
.
1
4
05
.
1
= 7,619 + 907 + 10,366 + 3,949 = 22,841
Project A is superior to project B.
Furthermore both projects have a positive NPV and are therefore economically
feasible.
Another meaning is that if 24,641 is put in a bank today at 5 % interest it would
be possible to withdraw 6,000, 2,000, 16,000 and 4,000 in the first,
second, third and fourth year respectively before the account would be depleted.
71
Example
Two different tenders have been received for works. Both quote a total price of 50
million, but they demand different payment schedules:
Tenderer A demands the following schedule:
·
initial payment (t=0): 5 million,
·
thereafter 9 equal 6-month instalments, each 5 million.
The works will be completed at the end of year 5.
Tenderer B demands the following schedule:
·
initial payment (t=0)
2.5 million
·
after
6
month 10 million
·
after
12
month 15 million
·
after
18
month 5 million
·
after 24 month
5 million
·
after
36
month 5 million
·
after
48
month 7.5 million
The works will be completed at the end of year 4.
Which tender is to be preferred if:
a.
The criterion of least cost is applied;
b.
The criterion of maximum NPV is applied; the net benefits of the project,
discounted at the moment of completion, are estimated at 80 x 106.
The discount rate is 10 %.
Answer
a. Criterion of least costs
tenderprice 50,000,000
NPV = 80.106
i = 10 %
Tender A 1
2
3
4
5
completion
end
of
year
5
5
5 5 5
5
5
5
5 5 5 x 106
NPV = 80.106
Tender B 1
2 3
4
5
completion
end
of
year
4
2.5
5
5
5
7.5 x 106
10
15
72
Simplify (not fully correct) : half-yearly interest : 5 % and bring half-yearly payments
to the beginning of the year:
Tender A
000
,
000
,
5
Total payment every year: 5,000,000 +
= 9,762,000
05
.
1
Present Value of all payments:
1
1
1
1
{ 1 +
+
+
+
} x 9,762,000 =
10
.
1
2
.
1 10
3
10
.
1
4
10
.
1
= 4.169 x 9,762,000 = 40,700,000
Tender B
Present Value of all payments:
10
15
5
5
5
5
.
7
{ 2.5 +
+
+
+
+
+
} . 106
05
.
1
10
.
1
05
.
1
x 10
.
1
2
10
.
1
3
10
.
1
4
10
.
1
= { 2.5 + 9.524 + 13.636 + 4.329 + 4.132 + 3.757 + 5.123 } . 106
=
43,001,000
Conclusion: Tender A is cheaper if criterion of least costs (cheapest) is applied.
b. Criterion of maximum NPV
Tender A
1
Present Value of net benefits of the project at t = 0: 80. 106 x
= 49.67.106
5
.
1 10
Net Present Value = 49.67 x 106 - 40.7 x 106 =
8,970,000
Tender B
1
Present Value of net benefits of the project at t = 0: 80. 106 x
= 54.64.106
4
10
.
1
Net Present Value = 54.64 x 106 - 43.0 x 106 =
11,640,000
Conclusion: Tender B is cheaper if criterion of maximum NPV is applied.
73
4. 3. Equivalent annual cost method
In using the equivalent annual cost method for the purpose of comparison, all
payments (costs) and receipts (benefits), are converted to their equivalent uniform
annual costs. Again it is necessary to make an assumption about the required rate of
return (freely interchangeable with the interest i) before it is possible to convert
varaible cash flows to an uniform series of payments over the life of an investment
proposal. The following example illustrates the application of the equivalent annual
cost technique:
Example
To cross a river, a timber bridge has been designed, at an estimated cost of
8 million. The lifetime of the bridge is estimated at 25 years and the annual costs
for maintenance at 2.5 % of the construction costs. It is believed that a concrete
bridge, with a lifetime of 50 years and annual costs for maintenance of 0.5 % of the
construction costs, could be a better alternative. What are the maximum cost of a
concrete bridge, in order to make this a viable alternative?
The discount rate is 7.5 %; the residual value is in both cases zero.
Answer
The two designs represents mutually exclusive projects with identical benefits and
constant annual costs the comparison can be made on annual costs basis.
Timber bridge
Annuity [A/P, 7.5 %, 25] =
0.0897 (say 9 %)
Depreciation & interest
9 %
Maintenance:
2.5
%
Total annual costs:
11.5 % of 8 x 106
Concrete bridge (maximum construction cost X)
Annuity [A/P, 7.5 %, 50] =
0.0771 (say 7.7 %)
Depreciation & interest:
7.7 %
Maintenance: 0.5
%
Total annual costs:
8.2 % of X . 106
Therefore:
0.082. X . 106 < 0.115 x 8 x 106
X < 11.22 . 106
Remark
If the NPV Method would have been used it has to be realised that the service life of
the timber bridhe is shorter than the concrete bridge or with other words the two
bridges do not offer the same 'service'. The timber bridge only provides a connection
for 25 years, while the concrete bridge provides the same 'service' but for 50 years.
So in order to make a correct comparison the timber bridge should be renewed after
25 years in order to provide the same duration of 'service'. In this example this is
rather simple but usually the lifetime of one alternative is not equal or a multiple value
of the other alternative. This problem is avoided by using the equivalent annual cost
method.
74
Example. Equivalent annual cost comparison
A flood control pumping station is being designed. Three possible pumping stations
are proposed and the relevant costs are shown in the table.
The cost of capital may be taken as 19 %.
Scheme
A
B
C
Cost of pumps ()
12,000
18,000
28,000
Life (years)
15
15
20
Maintenance
Per annum ()
1,000
1,500
1,500
Cost of pipes ()
22,000
18,000
12,000
Life (years)
30
30
30
Cost of pumping
( per hour)
1.20
0.90
0.80
Table: Costs of alternative schemes.
Questions
1. What is the most economic range of pumping times in hours/ year for each
scheme (demonstrate your answer by a graph).
2. What is the most economical scheme if the expected frequency of pumping is
according the following figure:
Fre-
quen-
cy
of
pum-
ping
10 % 30 %
40 % 20 %
Annual pumping hours 1000 2000
3000
4000
5000
Figure: Frequency of pumping demand
75
Answer
The solution is to plot the equivalent annual costs of each scheme for different
pumping demands and determine the range of pumping demands which are
cheapest for each scheme.
Convert cost of installation of the pumps and the pipes to an annual cost by the
annuity factor (or capital recovery factor), where i = 19 % and n = 15 or 20 years
(pumps) or n = 30 years (pipes).
annuity for i = 19 % and n = 15 years:
0.20509
annuity for i = 19 % and n = 20 years:
0.19604
annuity for i = 19 % and n = 30 years:
0.19103
The maintenance cost of the pumps is already expressed in annual costs.
Calculate the annual 'fixed' costs, which are independent of the number of hours
pumping.
Scheme A
The equivalent annual costs of installation and maintenance costs of the pumps and
pipes = 12,000 x 0.20509 + 1,000 + 22,000 x 0.19103 = 7,663.74
Scheme B
The equivalent annual costs of installation and maintenance costs of the pumps and
pipes = 18,000 x 0.20509 + 1,500 + 18,000 x 0.19103 = 8,630.16
Scheme C
The equivalent annual costs of installation and maintenance costs of the pumps and
pipes = 28,000 x 0.19604 + 1,500 + 12,000 x 0.19103 = 9,281.48
The annual 'variable' cost depending on the number of hours pumping for each
scheme are:
Scheme A
Scheme B
Scheme C
Pumping hours
0
0 0 0
1000
1,200
900
800
5000
6,000
4,500
4,000
These pumping costs vary linearly between 0 and 5000 hours.
Taking the 'fixed' equivalent annual cost and the 'variable' pumping cost the following
figure can be plotted (see next page).
Economic break-even point between Scheme A and Scheme B at
X pumping hours.
7,663.74 + X hours x 1.20 = 8,630.16 + X hours x 0.90
X hours x 0.30 = 966.42 X 3,221 pumping hours.
76
14,000 13,664
A
13,281
13,130
13,000
B
C
12,000
11,000
10,000
9,281 C
9,000
Scheme C never an economic alternative.
8,630 B
Scheme A cheapest
Scheme B cheapest
8,000
7,664
A
7,000
0
1000
2000
3000 X
4000 5000
pumping
hours
Figure: Annual costs versus pumping demand.
Question 2
For the given frequency of pumping demand the 'average' pumping hours is:
0.10 x 500 + 0.30 x 2,000 + 0.40 x 3,500 + 0.20 x 4,500 = 2,950
pumping hours; therefore Scheme A is the most economical solution.
77
4.4. Internal Rate of Return (IRR)
The IRR is defined as the discount rate at which the present value of benefits
equals the present value of costs, or at which the NPV = 0.
Whereas the determination of the NPV is straightforward, the IRR as a rule cannot be
calculated easily. Usually the IRR has to be determined by trial and error: by assu-
ming some values for i, the NPV canbe calculated and by way of interpolation the
value of i can be determined, for which the NPV = 0, thus yielding the IRR.
Nowadays, various pocket calculators are programmed to determine quickly the IRR.
The IRR is a measure for the return on the investments that the project yields. Any
project with an IRR exceeding the market rate of interest, i.e. the interest rate at
which investible funds can be obtained, is acceptable. As such it can be used with
other investment opportunities and in particular with the prevailing market rate of
interest. The underlying assumption in the calculation of the IRR is that revenues
generated by the project, can be re-invested against the same (high) rate as the IRR
itself. This may be too an optimistic assumption, particularly if the IRR is high. There
may not be other opportunities for investments which yield the same high returns.
Example
The construction of a water supply project is under construction and will be
completed on January 1, 2006. The expenditure during construction are as follows:
January 1, 2002
150,000
January 1, 2003
200,000
January 1, 2004 250,000
January 1, 2005
300,000
January 1, 2006
200,000
A final payment to the contractor will be made on January 1, 2007 of 100,000.
The useful life of the project is assumed at 20 years. The residual value of the
project at the end of this period is nil. The interest that has to be paid on the
borrowed capital is 7 %. The annual cost of operation and maintenance at the end of
every year is expected to be:
50,000
per year during the first five years,
100,000
per year during the second five years
150,000
per year during the last ten years.
It is expected that the sale of the water will be as follows:
1,000,000
m3
per year during the first ten years,
2,000,000
m3
per year during the second ten years.
Question a:
At what constant price should the water be sold in order to be able to liquidate the
project at the end of the 20 years without debt, or profit ?
Question b:
The end of years receipts are assumed to be as follows:
120,000 per year during the first five years
180,000 per years during the years 6 10
250,000 per year during the years 11 15
390,000 per year during the years 16 20
78
Determine the B/C ratio and the Net Present Value (NPV) (for 7 % interest).
Determine the maximum interest rate for which the money could be borrowed
whereby the project still is economically viable (IRR = Internal Rate of Return).
Answer
Question a
Present value at start of project (Jan. 1, 2006) of construction costs (in thousand
euro's)
2002 2003 2004 2005
2006 i = 7 %
150 200 250 300 200 100 . 103
150 .[ F/P, i, 4] + 200 .[ F/P, i, 3] + 250 . [ F/P, i, 2] + 300 .[ F/P, i, 1] + 200 + 100 .[P/F, i, 1]
= 150 . 1.3108 + 200 . 1.2250 + 250 . 1.1449 + 300 . 1.07 + 200 + 100 . 0.9346
= 196.62 + 245 + 286.23 + 321 + 200 + 93.46 = 1,342.31 . 103
2,000,000 X
1,000,000 X
X = unit rate
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
50,000
100,000
150,000
P = 1,342,310
P.V. of Benefits : 1,000,000 X [P/A, i, 10] + 2,000,000 X [P/A, i, 10] . [ P/F, i, 10]
= 1,000,000 X . 7.0236 + 2,000,000 X . 7.0236 . 0.5083
=
14,157,224 X
P.V. of Costs : 1,342,310 + 50,000 [ P/A, i, 5 ] + 100,000 [ P/A, i,5] [P/F, i, 5 ] +
150,000
[P/A, i, 10 ] [ P/F, i, 10 ] =
1,342,310 + 50,000 . 4.1002 + 100,000 . 4.1002 . 0.7130 +
150,000 . 7.0236 . 0.5083 =
1,342,310 + 205,010 + 292,338 + 535,565 = 2,375,223
Unit rate X :
2,375,223 / 14,157,224 = 0.168 per m3
79
Question b
P.V. of Benefits : 120,000 [ P/A, i, 5] + 180,000 [P/A, i, 5] . [ P/F, i, 5] +
250,000
[P/A, i, 5] . [ P/F, i, 10] + 390,000 [P/A, i, 5] . [ P/F, i, 15]
= 120,000. 4.1002 + 180,000 . 4.1002 . 0.7130 + 250,000 .
4.1002. 0.5083 + 390,000 . 4,1002 . 0.3624 =
492,024 + 526,220 + 521,033 + 579,506 = 2,118,783
P.V. of Costs : 2,375,223
B/C ratio :
0.89
NPV: 2.118,783 - 2,375,223 = (say) 256,000 (for i = 7 %) (negative !)
IRR:
try 5 %:
PV benefits:
{120,000 + 180,000 . 0.7835 + 250,000 . 0.6139 + 390,000 .
0.4810}. 4.3295 = 2.606.770
PV costs:
Construction costs: 1,300,000
Operation and maintenance:
50,000 . 4.3295 + 100,000 . 4.3295 . 0.7835 + 150,000 .
7.7217 . 0.6139 = 1,366,744
NPV (5 %)
2,606,770 - 1,300,000 - 1,266,770 = (say) + 39,000
IRR
therefore slightly above 5 % (by interpolation approximately 5.3 %).
80
4.4. Benefit-cost
ratio
The B/C - ratio has been widely used in the early stages of benefit- cost analysis.
It is defined as:
B
t
t
1
( + i)
B/C-ratio:
Present value of B = Benefits
C
t
Present Value of C = Costs
t
1
( + i)
the ratio of the present value of benefits to the present value of costs.
If the B/C ratio has a value of more than 1, then the project was considered to be
attractive; if the value was less than 1, then the project could not earn back the inputs
applied, and thus was not recommended for execution (for a certain value of i). For
nearly 60 years the B/C ratio method has been the accepted procedure for making
go / no-go decisions on independent projects and for comparing alternative projects
in the public sector, even though the other methods as discussed will lead to identical
recommendations, assuming all these procedures are properly applied.
tB
t
1
( + i)
Conventional B/C-ratio:
Present Value of B = Benefits
O & M
I +
t
Present Value of O & M =
t
1
( + i)
Operation and Maintenance Costs
Present Value of I = Initial Costs
B
O & M
t -
t
t
1
( + i)
t
1
( + i)
Modified B/C-ratio:
I
The resulting B/C-ratios will give consistent results in determining the acceptability of
a project (B/C > 1 or B/C < 1 or B/C = 0). The magnitude however of the B/C ratio
will differ between conventional and modified B/C. Therefore is the B/C factor not
used internationally anymore , for a number of reasons:
1.
Without further information, the B/C ratio is not well-defined: are the benefits
net of running costs, or are gross benefits considered?
2.
The project with the highest B/C ratio does not always yield the highest value
for other indicators (NPV, IRR), used in project appraisal.
81
Example
For the extension of the runway of an airport land needs to be purchased for
350,000. Construction cost for the runway are projected to be 600,000 and the
additional annual maintenance cost for the extension are estimated to be
22,500. If the runway is extended, a small terminal will be constructed at a cost of
350,000. The annual extra operating and maintenance cost for this terminal is
estimated at 75,000. The operational cost of the airport itself will increase by
100,000 for additional air traffic controllers to cope with the increased number of
flights. The annual benefits of this extension, consisting of extra income from airlines
leasing, airport tax, convenience benefit, additional tourism, is estimated at
490,000. Apply with a study period of 20 years and 10 % interest rate
490,000 . (P/A, 10 %, 20 years)
Conventional B/C-ratio:
=
1,200,000 + 197,500 . (P/A, 10 %, 20 years)
490,000 . 8.5136
= 1.448
1,200,000 + 197,500 . 8.5136
Modified B/C-ratio:
490,000 . (P/A, 10 %, 20 years) - 197,500 . (P/A, 10 %, 20 years)
= 2.075
1,200,000
The difference between conventional and modified B/C ratios is essentially due to
subtracting the equivalent present value of operating and maintenance from both the
numerator and the denominator of the B/C-ratio. Subtracting a constant (the present
value of O & M costs) from both numerator and denominator does not alter the
relative magnitudes of the numerator and denominator but the ratio is not the same.
An additional issue of concern is the treatment of disbenefits in benefit/cost ratio
analysis. In the example of the runway extension project the increased noise level
from commercial planes will be a serious nuisance to people living nearby the airport.
The annual disbenefits of this `noise pollution' is estimated at 100,000. Taken this
into account the conventional B/C-ratio's will change as follows:
Disbenefits considered as reduced benefits:
[
490,000 - 100,000 ] . (P/A, 10 %, 20 years)
= 1.152
1,200,000 + 197,500 . (P/A, 10 %, 20 years)
Disbenefits treated as additional costs:
[
490,000] . (P/A, 10 %, 20 years)
= 1.118
1,200,000 + [ 197,500 + 100,000 ] . (P/A,10 %, 20 years)
82
4.6. Exercises Cost Benefit Analysis
1.
At a long-term strip-mining coal site it is proposed to maintain temporary
haulage roads serving the excavation by using hand labour. The annual wage
bill is estimated to be 105,000. With other associated expenses, the total cost
of labour to the contractor will be 145,000 per year. The production of coal
on the site is expected to last for 6 years, and alternative methods of construc-
ting and maintaining haulage roads need to be investigated.
The first alternative is to buy a motor-grader for 95,000 and, as a conse-
quence, reduce the labour force. Maintenance of the grader is estimated to
average 4,000 per year for the 6 years, after which it will have a salvage
value or resale value of 20,000. The labour costs associated with the use of
the grader amount to 80,000 per year.
The second alternative is to lay more substantial roads in the first instance,
extending these after 2 years and again after 4 years. Initial costs are then
80,000 , with further invest-ments of 40,000 and 37,000 after 2 and 4
years respectively. Total labour costs in this scheme amount to 64,000 per
year.
If the return of at least 10 % is desirable on the capital invested , which is the
most economic scheme ?
Make the comparisons based on :
a. the equivalent annual cost method and
b. the present value method.
2.
The erection of a building for storage is under consideration. There are two
technical acceptable alternatives: a reinforced concrete shell roof structure
having an initial cost of 2,700,000 and a steel-framed structure with brick
cladding for an initial cost of 1,800,000. The life of the concrete building is
estimated to be 60 years and, while there will be no maintenance costs for the
building during the first 10 years, there will thereafter be an annual mainte-
nance cost of 35,000. The life of the other building is estimated to be 20
years with, an equivalent annual maintenance cost from completion of
construction of 40,000. The salvage value of the concrete building is
estimated at 80,000 and that of the steel-framed building at 27,000.
An acceptable rate of return is assessed at 10 %.
Which is the better economic proposition ?
Make the comparisons based on :
a. the equivalent annual cost method and
b. the present value method.
3.
A specialized piling rig is purchased by a contractor for one project only. The
duration of the project is two years. The economic life of the rig is 10 years, but
it is sold at the end of the project, that is after 2 years, then the contractor will
be able to get half the purchese value. If the rig costs 75,000 and the
required rate of return is 10 %, what is the annual cost of the rig to the
contractor if operating expenses are ignored ?
83
4.
A pumping scheme being developed has three different possible systems of
pumps and pipeworks. If the life of the scheme is 20 years, which scheme
should be recommended as the most economic ?
Pipe diameter (mm) Installation cost ( ) Annual running
Scheme
cost ( )
A
500
24,000
9,500
B
600
26,000
6,000
C
700
31,000
5,200
Use 10 % to represent the cost of capital. If the cost of capital was 6 %, would
the recommendation alter ?
5.
A hydroelectric project, if completely developed now, will cost 100,000,000.
Annual operation and maintenance charges will amount to 5,000,000 per
year. Alternatively, 55,000,000 may be invested in the project now and the
remainder of the work carried out in 12 years' time at a cost of 53,000,000.
In this alternative case annual operation and maintenance charges will be
3,400,000 per year for the first 12 years and 5,600,000 per year
thereafter. Both schemes are assumed to have perpetual life. Compare their
equivalent annual costs with an interest at 12 %.
6.
Water for an irrigation scheme can be supplied either by gravity (Alternative A)
or by pumping (Alternative B).
Alternative
A requires a relatively long canal with intake from a reservoir. The
total investment is estimated at 300,000. The annual costs for maintenance
and operation are estimated at 10,000. Useful service life is estimated at 30
years.
Alternative
B requires a pumping station with an intake from a nearby river. The
invest-ments are estimated at 90,000 for the civil engineering structures with
a service life of 30 years and at 25,000 for the mechanical and electrical
equipment with a service life of 15 years. The annual costs for maintenance and
operation are estimated at 20,000.
The net salvage of all investments at the end of their service life is assumed to
be zero.
a.
Determine the most economic alternative for an interest rate is 6 %
b.
Determine the most economic alternative for an interest rate is 4 %
c.
Determine the unit cost per m3 for an interest rate of 6 % if the estimated
consump-tion is 1.5 million m3 / year during the first 6 years and 2 million m3
/ year during the remaining 24 years.
7.
In an economic assessment concerned with the alignment of a new road, one of
the alternatives to be evaluated on the basis of annual cost consists of a bridge at
an estimated cost of 1,350,000, an embankment costing 215,000, and other
earthworks at an estimated cost of 38,000. Maintenance on the earthwork and
the embankment is estimated to reach an annual cost of 30,000 over the first 4
years of its service and then drop to 14,000 for every year thereafter. Mainte-
nance on the bridge is expected to remain constant throughout its life at a figure of
70,000 a year.
84
What is the total equivalent uniform annual cost of this alterna-tive if the life of the
bridge is estimated at 60 years, the life of the earthworks and the embankments is
in perpetuity and the interest rate to be used is 15 %?
8.
A proposed highway project requires an initial investment of 10 million and a
supplementary investment of 5 million at the end of the tenth year.
The project will have an useful life of 50 years, counting from the date of the initial
investment. The interest rate is 6 %. The cost of operation and maintenance is
200,000 per year. The benefits of the project has been estimated to begin
with 1.0 million per year for the first 15 years (at the end of each year),
thereafter increasing at once to 2.75 million per year and remaining
constant for the remaining 35 years. Determine the value of Benefit-cost (B/C)
ratio, Net Present Value (B-C), and Internal Rate if Return (IRR).
9.
In diverting river water for an irrigation project, two alternative schemes are
prepared, as follows:
Scheme 1. Open ditch and tunnel with a capital cost of 2,500,000 and an
annual maintenance cost of 40,000 per year.
Scheme 2. Pipework and open flume with a capital cost of 1,750,000 and a
maintenance cost of 80,000 per year, with a major replacement cost of
120,000 every 10 years.
Either of the above schemes will provide the service required. If the current
interest rate is 12 %, compare the two schemes on the basis of capitalized cost
(n is 100 = perpetuity).
10. In a remote wilderness in Africa a rich ore deposit has been discovered. It has
been estimated that all ore can be mined during a period of 20 years. The most
economical way to bring out the ore is by river. To make the river navigable
there are two alternative projects:
Plan A to regulate the river by training works, excavation and blasting of rock,
with a total initial cost of 10,000,000 and a cost of dredging of
2,000,000 per year.
Plan B to canalize the river by means of weirs and navigation locks: initial
costs 20,000,000 and cost of operation and maintenance of 400,000
per year.
Capital for both projects is available at 10 % interest.
The terminal value of the navigation works after 20 years is assumed to be nil.
Question
a: Make a cost comparison of annual costs.
The cost of dredging of Plan A is now expected to be as follows:
100,000 during the first year and then gradually increases by an amount of
200,000 per year till it would reach a cost of 3,900,000 during the
twentieth year.
Question
b: Determine which of the two projects is more economic.
85
11. A new highway of 25 m wide is in the stage of being designed. A considerable
portion of the highway has to be cut deeply (10 m) in the surrounding terrain of
sandy soils. The problem is to determine the most economic side slope of the
cut. If they are steep, they will require a lot of maintenance due to erosion
during heavy rainfall. If they are flat, they require extra excavation during the
construction of the highway. The capital cost of excavation and disposal of the
soil is 3.00 per m3.
Slope
Total excavation (m3) per km
Annual slope maintenance ( )
1 : 1 (n = 1)
250,000 + 100,000 = 350,000 80,000 per km
1 : 2 (n = 2)
250,000 + 200,000 = 450,000 50,000 per km
1 : 3 (n = 3)
250,000 + 300,000 = 550,000 34,000 per km
1 : 4 (n = 4)
250,000 + 400,000 = 650,000 24,000 per km
The capital cost of the road deck is 250,000 per km. The useful life of the
project is 50 years. Annual maintenance of the road deck costs 3,000 per
km. The interest rate is 5 %.
12. An appraisal of three alternatives, mutually exclusive projects, A, B, and C, is
being made for a company that requires a return of at least 10 % on its invested
capital. The estimated details of the investment are shown in the table below.
Which investment should be recommended and why?
Support your recommendation and reasoning by calculation.
Euro
Project A
Project B
Project C
Initial cost
100,000
160,000
280,000
Scrap value
nil
nil
40,000
Net annual
receipts
18,400
30,600
42,300
Life, years
8
8
10
13. A decision has to be made with regard to the installation of automatic control
equipment on a concrete batching plant installed at the construction site.
Quotations for the equipment show its cost to be 300,000 , but its installation
will have the effect of reducing annual labour cost from an estimated
150,000 to 45,000.
Maintenance of the automatic plant is expected to amount to 6,000 per year
more than the manually controlled plant and only this excess cost need be
considered in the analysis.
The automatic equipment, if installed, will have a salvage value of 30,000
irrespectively of the length of time it is in use. The contractor carrying out the
work state their rate of return on capital to be 10 %. Will the selection of the
automatic equipment for the contract with a duration of 3.5 years be justified,
and what is the minimum contract period that will do this?
86
14. A public agency has sufficient funds available for a number of projects. One of
these projects can be executed in four ways (A, B, C or D). The investments
and the net annual benefits of the 4 alternatives are listed in the following table:
Alternative
Investment
Net annual
benefits
A
100
20
B
200
30
C
300
50
D
500
75
All amounts are given in thousands of euros.
Assume that all alternatives have an infinitely long service life and that the net
annual benefits remain constant in the future. Questions:
a.
Which alternative has the highest rate of return ?
b.
Which alternative is to be preferred if unused funds can be invested in
other projects with a rate of return of 10 % ?
c.
Would you come to another conclusion than that given onder b, if unused
funds could be invested in projects with a rate of return of 14 % ?
15. For the installation of a pipeline connection two different payment schedules are
offered:
a. an
immediate
payment of 1,150 at the moment the connection is
made, or
b.
8 annual payments of 231.50 with the additional condition that these
payments have to be made at the beginning of each year.
Questions:
a.
Which payment proposal do you prefer if you can borrow 1,150 now
for 8 years at 12% per year under normal conditions (payment at the end
of the period).
b.
What is the effective annual interest rate in the case of 8 annual
payments?
16. The first cost of a project is 100,000. The annual equivalent operation and
maintenance costs are 15,000. The annual equivalent benefits are
26,500. The life of the investment is 25 years. Its net salvage value is zero.
Questions:
a.
Estimate the internal rate of return of the project.
b.
Could the investment be made economically if funds are available at an
interest rate of 4% per year? Explain your answer briefly.
c.
In how many years can a loan for the financing of this project be repaid, if
the loan carries an annual interest rate of 4 % and the annual surplus is
initially used for this repayment ?
87
17. Water has to be transported by gravity by means of a canal. The canal has a
useful life of 20 years and requires an investment of 1,000,000. The interest
rate is 10 % per year. The net salvage value of the canal after 20 years of
operation is assumed to be zero. The annual equivalent maintenance and
operation costs are estimated at 100,000.
Calculate the constant transportion cost (unit cost) in per m3 for the
following cases:
a. the
annual
transport is 15 million m3 throughout the 20 years' period;
b.
the annual transport is 13 million m3 during the first period of 10 years
and 17 million m3 during the second period of 10 years.
18. The following loans were taken to finance the planning, design and construction
of a project:
Loan
Annual interest rate
date
1,000,000
10 %
31st Dec.2002
2,000,000
8 %
1st Jan. 2004
5,000,000
6 %
1st Jan. 2005
5,000,000
4 %
31st Dec.2005
An additional loan will be needed for the final payment of 1,000,000 due on
the 1st of January 2007. All previous and future loans are consolidated
("refinanced") at an interest rate of 4 % per year on the 1st of January 2006, the
day the project is put into operation. The expected annual equivalent operation
and maintenance cost are 1,000,000. The expected annual revenue (gross
benefit) is 3,250,000. The net salvage value after 20 years is expected to be
1,500,000.
Questions:
a.
What is the first cost of this project and what is the total depreciation ?
b.
What is the internal rate of return ?
c.
What is the equivalent annual surplus (profit) of this project ?
d.
What is the marginal rate of return of a proposed extension which will cost
an additional 1,500,000 , which will not raise the O & M costs and net
salvage value but which will raise the annual revenue to 3,400,000 ?
e.
Will it be justified from an economic point of view to invest these
1,500,000 in the proposed extension if this money can also be invested in
another project which will have an internal rate of return of 12 % ?
19. A project according plan A requires an investment of 4,000,000. Its useful
service life is 15 years. The annual costs for maintenance and operation are
200,000. The annual benefits are estimated at 624,000. It is being
considered to extend the project by an additional investment of 1,000,000.
This plan B (the extended version of plan A) requires a total investment of
5,000,000. The total annual costs for maintenance and operation will increase
to 240,000 , whereas the total annual benefits are now estimated at
722,000.
88
Questions:
a.
Determine the Benefit-Cost ratios of the plans A and B with interest at 6
%.
b.
Determine the rate of return of the plans A and B, as well as the marginal
rate of return of plan B with respect to plan A.
c.
Will it be worthwhile to execute plan A or plan B if unused funds can be
invested in other projects having a rate of return of 5 % ?
20. The useful life of an 10 million bridge depends on how often it is repaired
and painted. Use the formula: y = x2 + 20, in which y is the useful life in years,
and x is the number of times per decade that the bridge gets a repair and paint
job at a cost of 250,000 each time. The interest rate is 5 %. Determine the
most economic frequency (in times per decade) of giving the bridge a repair and
paint job.
21. In a country a new coal mine will be put into production; the total output will be
exported. There are 2 options for the transportation of the coal to the port of
export:
a. Water
transport
The river on which the mine is situated has to be improved for navigation:
-
Length
400 km
-
Construction capacity 50 km/year
-
Start of construction 1st January 2002
-
Construction costs LC 40 x 106 per 100 km, spread evenly over the
construction period, payable at the end of each year
-
Maintenance costs LC 2 x 106 per 100 km per year
-
Transportation costs LC 0.05 per ton per 100 km
b.
Rail transport .
A new railway line has to be constructed:
-
Length
375 km
-
Construction capacity 75 km/ year
-
Start of construction 1st January 2005
-
Construction costs LC 32 x 106 per 100 km, spread evenly over the
construction period, payable at the end of each year
-
Maintenance costs LC 2.5 x 106 per 100 km per year
-
Transportation costs LC 0.07 per ton per 100 km.
Other relevant data are:
-
LC is one unit of Local Currency
-
Total production 5 x 106 ton per year
-
For water transport start of construction: 1st January 2002
-
For rail transport start of construction:
1st January 2005
-
Both options have a life time of 50 years, without any residual value.
-
All costs and benefits occur at the end of the year.
-
Discount rate 10 %.
89
Questions :
a.
If the project is financed from internal resources (local currency = L.C.),
which of the two options is to be preferred?
b.
Investments will be provided partly from external resources (foreign
currency F.C.), but all costs for maintenance and transportation will be
financed from internal resources (L.C.). The local currency (L.C.) is
overvalued by a factor 2; meaning foreign component costs (F.C.) is 2x
expressed in local currency (L.C.).
Proportion of foreign currency in total investment costs:
-
Water transport 20% F.C. (and 80 % L.C.)
- Rail
transport
80%
F.C. (and 20 % L.C.)
Which option is to be preferred now?
c.
Transportation time for the railway line is 5 hours less than for water
transport, against a value of LC 0.02 per ton per hour.
Which option is to be preferred for each of the cases 1. and 2. above?
22. The purchase price for a piece of construction equipment is 20,000 . The
operating costs based on the annual average estimated hours of operation are:
800 in the first year; 1200 in the second year, 1500 in the third
year, 1800 in the fourth year and 2100 in the fifth year.
The resale value of the plant can be assumed as follows: 15,000 after 3
years, 12,000 after 4 years and 8,000 after 5 years .
The cost of capital is 8 % per year.
Question: Calculate the optimum replacement age.
23. A reinforced concrete road pavement, including the base, is laid for 100.00
per m2.
A flexible pavement to give the same service is laid for 90.00 per m2.
The flexible pavement has major maintenance every 5 years, which costs the
equivalent of 3.25 per m2 per year. The concrete pavement has a first
lifetime of 40 years, after which it is resurfaced with asphalt costing 31.00
per m2.
Thereafter it is maintained at the same cost as a flexible pavement. In addition,
both types of road require annual maintenance estimated to amount to 0.67
per m2.
On the basis of both roads giving perpetual service, compare the capitalized
costs of 2000 m2 of road at an interest rate of 12 %.
90
4.7. Answers
exercises
Problem 1.
a. equivalent annual cost method. Scheme 1 (original situation)
0 1 2 3 4 5 6
145,000 / year
i = 10 %
Annual cost of labour = 145,000
This is the sole annual outgoing and requires no conversion to annual payments.
Scheme 2 (alternative 1)
S = 20,000
0 1 2 3 4 5 6 i = 10 %
84,000/
year
95,000
Annual maintenance costs of grader =
4,000
Annual cost of labour =
80,000
Subtotal =
84,000
Annual capital recovery cost of the motor grader (where S is the salvage value of the
grader):
( P S ) ( A/P, 10 %, 6 years) + S . i
(95,000 20,000 ) ( 0.2296) + 20,000 ( 0.10) =
19,220
Total equivalent annual cost =
103,220
Scheme 3 (alternative 2)
0 1 2 3 4 5 6
64,000
/
per
year
80,000
40,000 37,000
Annual capital recovery of initial cost:
80,000 ( A/P, 10 %, 6 years) = 80,000 ( 0.2296) = 18,368
Annual capital recovery for capital cost at end of 2 years:
40,000 ( P/F, 10 %, 2 ) (A/P, 10 %, 6) = 40,000 ( 0.8265 ) ( 0.2296 ) = 7,591
Annual capital recovery for capital cost at end of 4 years:
40,000 ( P/F, 10 %, 4 ) (A/P, 10 %, 6) = 40,000 ( 0.6830 ) ( 0.2296 ) = 5,802
Annual labour costs =
64,000
Total equivalent annual cost =
95,761
Scheme 3 is therefore the most economic on the basis of this evaluation because its
equivalent annual cost is lower than those of the other two schemes.
91
There are a number of points to be noted. The first concerns the treatment of salvage
values when computing annual capital recovery costs. The salvage value ( 20,000)
will become available from the sale of the grader at the end of 6 years. Therefore, the
part of the cost which is invested over the 6 years of the grader's useful life, and
which will not be recoverable as salvage, is the initial cost less the salvage value (
95,000 20,000 = 75,000). Since the salvage value will become available
again at the end of 6 years it is only necessary to charge to each equivalent annual
cost the interest on that amount. Treating each year separately, the salvage value
can be looked on as being locked up or loaned for the initial purpose of the grader
during each year and it is therefore not possible to earn interest or profit by investing
the money elsewhere. Account is taken of this in the calculation.
In Scheme 3, each of the payments is converted to present value before being
converted to an equivalent uniform series of payments over the 6 years of the
comparison.
Finally, the only overriding assumption is that each of the three schemes considered
will either give equally good service if put into operation and/or at least will provide
the minimum service required. In making an economic choise between the
alternatives, it is assumed that the technical merit of each alternative has been
examined and found to be satisfactory. The only considerations that may now affect
the ultimate decision are the irreducible factors.
One example of an irreducible factor might be that there is an ample supply of skilled
labour in an area where unemployment is high. It therefore becomes a social
obligation of the contractor to act beneficially as he is able towards the local
community. There may, for the contractor, be other spinoffs in doing that, which
though irreducible in themselves, create a better climate in which to work a benefit
that may well outweigh some of the other considerations.
In the above problem, the comparison between the schemes was made on the basis
that each of them represented the annual cost for 6 years. The equivalent annual
costs were therefore comparable because the lives of the alternatives were assumed
to be the same. This may not always be the case, particularly where the construction
of more permanent installations is under consideration.
b. Present value method
Scheme 1
Present value of annual labour cost over 6 years:
145,000 ( P/A, 10 %, 6 years) = 145,000 . (4.3552) =
631,504
Scheme 2
Initial cost of motor grader =
95,000
Present value of maintenance cost and labour cost
84,000 . (P/A, 10 %, 6 years) = 84,000 . (4.3552) =
365,837
Subtotal =
460,837
Less: Present value of salvage value
20,000 . (P/F, 10 %, 6 years) = 20,000 . (0.56448) =
11,290
Present value of total costs =
449,547
92
Scheme 3
Initial cost of first section of road =
80,000
Present value of second investment:
40,000 . ( P/F, 10 %, 2 years) = 40,000 . (0.82645) =
33,058
Present value of third investment:
37,000 . ( P/F, 10 %, 4 years) = 37,000 . (0.68302) =
25,272
Present value of annual labour cost over 6 years
64,000 . ( P/A, 10 %, 6 years) = 64,000 . (4.3552) =
278,733
Present value of total costs =
417,063
Therefore, on the basis of the above present value evaluation the economic appraisal
comes out in favour of Scheme 3, since, in effect, with the given interest rates, the
whole scheme can be financed with a smaller lump sum than the other two.
In the case of scheme 3, where there are several staged investments over the period
under consideration, it will be noted that one step in the computation has been
saved in considering present value rather than equivalent annual cost method for
comparison purposes. On the other hand, all the payments for labour, for example,
that are already convenient form for annual costs, need to be converted to a lump-
sum present value.
Check
Scheme Equivalent
%
Present value
%
annual cost
1
145,000
100 %
631,504
100 %
2
103,220
71.2 %
449,547
71.2 %
3
95,761
66.0 %
417,063
66.0 %
Answer problem 2a.
Reinforced concrete building
80,000
0 10 11
20
30
40
50
60
i = 10 %
35,000
/
year
2,700,000
Capital recovery (per year) = ( P S ) . ( A/P, 10 %, 60 years) + S . i =
( 2,700,000 80,000 ) . (0.1003 ) + 80,000 (0.10) = 270,786
The sum of money at the end of year 10 equivalent to 35,000 per year from years
11 to 60:
35,000 . ( P/A, 10 %, 50 years) = 35,000 . ( 9.9148 ) = 347,018
Present value of 347,018 at year 0:
347,018 ( P/F, 10 %, 10 years) = 347,018 (0.3856 ) =
133, 810
Therefore, equivalent annual cost over 60 years of 35,000 a year from years 11 to
60:
133,810 . ( A/P, 10 %, 60 years) = 133,810 (0.1003) = 13,421
Therefore, total equivalent annual cost = 270,786 + 13,431 =
284,207
93
Steel-framed building
27,000
0 1
10
20
i = 10 %
40,000 / year
1,800,000
Capital recovery = ( 1,800,000 27,000 ) . ( A/P, 10 %, 20 years) + 27,000 . i
= 1,773,000 . ( 0.1175) + 27,000 . ( 0.10 ) = 211.028 / per year
Therefore, total equivalent annual cost = 211,028 + 27,000 = 238,073
The steel-framed building is therefore cheaper when the comparison is made on
basis of annual costs.
This problem raises a number of points. A comparison has been made on the basis
of annual cost and it is therefore implicit in the calculation that after 20 years the
steel-framed building can be replaced at the same cost as the initial installation and
that the repla-cement will continue at this cost at intervals of 20 years. Rising costs
are inevitable in this context, though it is not unreasonable that such a method of
comparison should be used because, in the majority of cases, the future cost
increases, when discounted to the present time, quickly become a relatively small
proportion of present costs.
In the case of the reinforced concrete building, the capital investment is being made
now, and therefore no question of increased cost in the replacement situation arises.
If the replacement cost of the steel-framed building in 20 years' time is increased by
50 % over the present-day cost, that is, it becomes 2,700,000, then the present
value of the increase in cost under similar conditions of interest amounts to :
900,000 . ( 0.1486 ) = 133,779
If the second replacement cost in 40 years' time increases by 50 % over the first
replacement value, that is, it becomes 4,050,000, then the present value of the
total increase amounts to:
2,250,000 . ( 0.02209 ) = 49,703
The two sums produce an equivalent uniform annual cost of
( 133,779 + 49,703 ) . (0.10032 ) = 18,407
over the total life of 60 years under consideration.
The
total equivalent annual cost now becomes:
256,480
The steel-framed building remains therefore to be cheaper when the comparison is
made on basis of annual costs.
Quite apart from the financial aspects of the economical appraisal, there may be
considerable advantages within many businesses from constructing buildings with a
shorter life.
New developments in products and building materials may enable such a company
to replace the building in 20 years' time with one that gives improved performance.
94
Replacement may well take place at a cost comparable to that of the original building
or investment because of technical improvements. With the long-life building in such
a situation it may be difficult to make good use of it in changed circumstances unless
money is spent on its rehabilitation. This aspect becomes an irreducible factor in
such a situation.
Alternatively, future costs can be estimated only by the interpretation of historic
trends. Since, historically, costs have always risen continuously and steadily (with a
few exceptions), it seems likely that they will continue to do so. A longer-life
investment is clearly advantageous in this circumstance.
Answer Problem 2b.
Present value method
Reinforced concrete building
Initial cost of building =
2,700,000
Less: Present value of salvage value +
80,000 . (P/F, 10%, 60 years) = 80,000 . (0.0033) =
264
Subtotal =
2,699,736
Equivalent capital value at the end of year 10 of annual maintenance of 35,000
per year from years 11 to 60:
35,000 . ( P/A, 10 %, 50 years) = 35,000 . ( 9.9148 ) = 347,018
Present value of 347,018 at year 0:
347,018 . ( P/F, 10 %, 10 years) = 347,018 . (0.3856 ) = 133, 810
Present value of total payments over 60 years = 2,833,546
Steel-framed building
Initial cost of building =
1,800,000
Present value of maintenance cost:
40,000 . (P/A, 10 %, 60 years) = 40,000 . (9.9671) = 398,684
Present value of renewal cost less salvage cost at the end of 20 years:
( 1,800,000 27,000 ) . (P/F, 10 %, 20 years) = 1,773,000 . (0.14865)
= 263,556
Present value of renewal cost less salvage cost at the end of 40 years:
( 1,800,000 27,000 ) . (P/F, 10 %, 40 years) = 1,773,000 . (0.02210)
= 39,183
Subtotal =
2,501,423
Less:
Present value of 27,000 . (P/F, 10 %, 60 years):
27,000 . (0.00328) =
89
Present value of total payments over 60 years = 2,500,334
This confirms the result of the analysis made by the equivalent uniform annual cost
method .
In the above problem, using the present value method where the buildings have
different lives, it should be noted that the comparison has to be made over a period
of time that is the lowest common multiplier of the lives of the alternatives.
95
It is therefore necessary in the case of the steel building to consider the replacement
costs at the end of 20 and 40 years, together with salvage values at the end of 20,
40, and 60 years.
The present value of the series of maintenance payments for the concrete building
could have been calculated in a different way. The payments did not commence until
year 11 and they continue until the end of year 60. If the factor for conversion of an
annual payment to present value for the first 10 years is subtracted from the similar
factor over a 60-year period and is then multiplied by the annual amount, the same
result will be obtained (note small arithmetical error due to the rounding of the
factors).
Present value of payments for years 11 60:
35,000 . [ (P/A, 10 %, 60 years) - (P/A, 10 %, 10 years) ] =
35,000 . ( 9.9671 - 6.1445 ) = 35,000 . ( 3.8226 ) =
133,791
Having obtained either total equivalent annual costs or total present values, then
either of these amounts can readily be converted into the other. For example, the
total payments at total present value of the concrete building can be converted to
total annual costs as follows:
Equivalent annual cost:
2,833,546 . (A/P, 10 %, 60 years ) = 2,833,546 . (0.1003) = 284,205
Answer problem 3
Annual capital recovery cost of the piling rig
( 75,000 - 37,500 ) . (A/P, 10 %, 2 years) + 37,500 . i =
= 37,500 . (0.5762) + 3,750 = 21,608 + 3,750 = 25,358
Answer problem 4
Calculate the present value of each scheme using 10 %
Scheme A
Present value of installation cost =
24,000
Present value of maintenance costs:
9,500 . (P/A, 10 %, 20 years) = 9,500 . ( 8.5135 ) =
80,878
total present value =
104,878
Scheme B
Present value of installation cost =
26,000
Present value of maintenance costs:
6,000 . (P/A, 10 %, 20 years) = 6,000 . ( 8.5135 ) =
51,081
total present value =
77,081
Scheme C
Present value of installation cost =
31,000
Present value of maintenance costs:
6,000 . (P/A, 10 %, 20 years) = 5,200 . ( 8.5135 ) =
44,270
total present value =
75,270
At 10 % Scheme C is the most economical because it has the smallest present
value.
96
Repeating the calculations at 6 %
Scheme A
Present value of installation cost =
4,000
Present value of maintenance costs:
9,500 . (P/A, 6 %, 20 years) = 9,500 . ( 11.4679 ) =
108,945
total present value =
132,945
Scheme B
Present value of installation cost =
26,000
Present value of maintenance costs:
6,000 . (P/A, 6 %, 20 years) = 6,000 . ( 11.4679 ) =
68,807
total present value =
94,807
Scheme C
Present value of installation cost =
31,000
Present value of maintenance costs:
6,000 . (P/A, 6 %, 20 years) = 5,200 . ( 11.4679 ) =
59,633
total present value =
90,633
Scheme C is at 6 % the most economical; the difference has become larger due to
the lower interest rate . Only for a certain interest rate higher then 10 % there will be
a certain interest rate whereby Scheme B becomes more economical as the
difference in maintenance costs has less weight.
Answer problem 5
Alternative 1
i = 12 %
n =
O & M: 5 . 106
100 . 106
Annuity A = P . i (for n = ) = 100 . 106 . 0.12 =
12 . 106
Operation and Maintenance (O & M)
5 . 106
Total equivalent annual costs:
17 . 106
.
5 106
Or capitalized costs: P +
= ( 100 + 41.67) . 106 =
141.67
.106
.
0 12
97
Alternative 2
12
n =
i = 12 %
A1 = O & M: 3.4 . 106
A2 = O & M: 5.6 . 106
P1 = 55 . 106
P2 = 53 . 106
Present value (Capitalized costs):
A
A
P
1
2
1 +
+ [ P2 +
] . ( P/F, i, 12 ) =
i
i
4
.
3
2
.
2
1
{
55
+
+ [ 53 +
] } . 106 .
=
12
.
0
12
.
0
12
12
.
1
{ 55 + 28.33 + 71.33 . 0.2567 } . 106 =
101.64
.106
The second alternative is much cheaper.
The total equivalent annual costs of alternative 2: 101.64 . 106 . 0.12 = 12.2 . 106
Answer problem 6
Question a (i = 6 %)
Alternative A
i = 6 %
p x 2.0 x 106
p x 1.5 x 106
0
6 15
n = 30
10,000
10,000
300,000
Alternative B
i = 6 %
p x 2.0 x 106
p . 1.5 x 106
0
6
15
n = 30
20,000
20,000
90,000
25,000
25,000
Equivalent annual costs
Alternative A
300,000 . [ A/P, 6 %, 30 ] + 10,000 = 300,000 . 0.07265 + 10,000
= 31,795 ( P.V. = 437.6 x 103 )
98
Alternative B
{ 115,000 + 25,000 . [P/ F, 6 %, 15 ] } . [ A/ P, 6 %, 30 ] + 20,000 =
{ 115,000 + 25,000 x 0.4172 } . 0.07265 + 20,000 =
125,430 x 0.07265 + 20,000 = 29,112
( P.V. = 400.7 x 103 )
Conclusion:
Alternative B is the most economic alternative.
Or: 90,000 . [ A/ P, 6 %, 30] + 25,000 [ A/ P, 6 %, 15] + 20,000 =
90,000 . 0.0727 + 25,000 . 0.103 + 20,000 =
6,543 + 2,575 + 20,000 = 29,118
Answer Question 6b.
( i = 4 % )
Equivalent annual costs
Alternative A
300,000 . [ A/P, 4 %, 30 ] + 10,000 = 300,000 . 0.05783 + 10,000
=
27,349 ( P.V. = 472.9 x 103 )
Alternative B
[ 115,000 + 25,000 . (P/F, 4 %, 15)] . ( A/P, 4 %, 30 ) + 20,000 =
[ 115,000 + 25,000 . 0.5552 ] . 0.05783 + 20,000 =
128,881 . 0.05783 + 20,000 = 27,453
( P.V. = 474.7 x 103 )
Conclusion:
Alternative A is the most economic alternative (just).
Question c.
P.V. annual benefit (for both alternatives) for p = unit cost per m3 :
(p . 1.5 . 106 ) . ( P/A, 6 %, 6) + ( p . 2.0 .106 ) . ( P/A, 6 %, 24) . ( P/F, 6%, 6)
[ (1.5 . 4.9164 ) + ( 2.0 . 12.5502 . 0.7050 ) ] . p. 106 =
[ 7.3746 + 17.6958 ] . p . 106 = 25.07 . p . 106
Cost per m3 (alternative B):
400.7 x 103 = 25.07 x p x 106
p = 0.016 / m3
Answer problem 7
i = 15 %
Investment
Maintenance
Lifetime
(per year)
Bridge
1,350,000
70.000
n = 60 years
Embankment
215,000
30,000 for the
Other earthworks
38,000
first 4 years ;
n =
14,000 thereafter
99
Embankment & other earthworks:
000
,
14
Present value: : 215,000 + 38,000 + 30,000 . (P/A, i, 4) +
. (P/F, i, 4) =
i
4
15
.
1
-1
000
,
14
1
215,000 + 38,000 + 30,000 .
+
.
=
4
15
.
0
1
x 15
.
0 15
.
4
15
.
1
263,000 + 30,000 . 2.855 + 14,000 . 3.8117 =
263,000 + 85,649 + 53,364 = 402,013
Equivalent annual costs: P . i = 402,013 . 0.15 = 60,302
Bridge:
Annuity (Capital recovery): 1,350,000 . ( A/P, i, 60 ) =
.
0 15 1
x 15
. 60
1.350,000 .
= 1,350,000 . 0.15003 = 202,546
15
.
1
60 -1
Annual maintenance :
70,000
Total equivalent uniform annual costs
332,848
Summary:
Capital recovery for the bridge: 1,350,000 . 0,15003 =
202,546
Annual maintenance of bridge:
70,000
Interest for embankment and earthworks: 263,000 . 0.15 =
39,450
Basic annual maintenance on embankment and earthworks
14,000
Equivalent annual cost of extra maintenance during first 4 years:
16,000 . (P/A, i, 4) . 0.15 =
6,852
Total equivalent uniform annual costs:
332,848
Answer problem 8
2.75 . 106
1.0 . 106
10 15
n = 50
O & M = 0.2 . 106
10.106
5.106
i = 6 %
Costs :
10. 106 + 5. 106 (P/F, i, 10) + 0.2 . 106 . (P/A, i, 50) =
1
50
06
.
1
-1
10. 106 + 5. 106 .
+ 0.2 . 106 .
=
10
06
.
1
50
0 06
.
1
. 06
.
10. 106 + 5. 106 . 0.5584 + 0.2 . 106 . 15.76 =
(10 + 2.792 + 3.152 ) . 106 = 15.944 . 106
100
Benefits : 1.0. 106 . (P/A, i, 15) + 2.75. 106 (P/A, i, 35) . (P/F, i, 15) =
15
.
1 06
-1
35
1 06
.
-1
1
( 1.0 .
+ 2.75. 106 .
.
) . 106 =
15
.
0 06 .
1
. 06
35
0 06
.
1
. 06
.
15
.
1 06
( 1.0 . 9.7122 + 2.75 . 14.4925 . 0.417 ) . 106 =
( 9.7122 + 16.6498 ) . 106 = 26.342 . 106
B/C ratio:
Benefits / Costs = 26.342 / 15.944 = 1.65
NPV
Benefits - Costs = ( 26.342 - 15.944 ) . 106 =
10.399 . 106
IRR
Try i = 10 % (as 6 % gives a positive NPV)
Costs :
10. 106 + 5. 106 (P/F, i, 10) + 0.2 . 106 (P/A, i, 50) =
1
50
10
.
1
-1
10. 106 + 5. 106 .
+ 0.2 . 106 .
=
10
10
.
1
50
0 10
.
1
x 10
.
10. 106 + 5. 106 . 0.3855 + 0.2 . 106 . 9.92 =
( 10 + 1.93 + 1.98 ) . 106 = 13.91 . 106
Benefits : 1.0. 106 . (P/A, i, 15) + 2.75. 106 (P/A, i, 35) . (P/F, i, 15) =
15
10
.
1
-1
35
10
.
1
-1
1
( 1.0 .
+ 2.75. 106 .
.
) . 106 =
15
.
0 10x .
1 10
35
0 10
.
1
x 10
.
15
06
.
1
( 1.0 . 7.606 + 2.75 . 9.644 . 0.239 ) . 106 =
( 7.606 + 6.35 ) . 106 = 13.95 . 106
The Internal rate of return is 10 %.
Answer problem 9
Scheme 1
Capitalized cost:
2,500,000 + 40,000 / i = 2,500,000 + 333,333 =
2.833
million
Scheme 2
Capitalized cost
1,750,000 + 80,000 / i + 120,000 (P/F, 12 %, 10 years) + 120,000
.
(P/F, 12 %, 20 years) + 120,000 . (P/F, 12 %, 30 years) + 120,000 .
(P/F, 12 %, 40 years) + etc. =
1,750,000 + 666,667 + 38,637 + 12,440 + 4,005 + 1,290 +
...
= 2.474 million
Remark: The replacement cost of 120,000 every 10 years can be considered as
an equivalent 'annual' cost, whereby annual is now 10 years and the compounded
interest rate for 10 years is 1.1210 = 3.10585 - 1 = 2.10585
101
Scheme 2
Capitalized cost
1,750,000 + 80,000 / 0.12 + 120,000 / 2.10585 =
1,750,000 + 666,667 + 56,984 = 2.474 million
Answer problem 10
Question a
Annual costs
Plan A Capital recovery cost: 10,000,000 . [ A/P, i, 20 ] =
1,175,000
Operation and maintenance :
2,000,000
total
3,175,000
Plan B Capital recovery cost: 20,000,000 . [ A/P, i, 20 ] = 2,350,000
Operation and maintenance :
400,000
total
2,750,000
Plan B is less costly then Plan A.
Question b
Present value
Plan A Initial costs
10,000,000
Dredging 100,000 . [P/A, i, 20 ] + 200,000 [ P/C, i, 20 ]
= 100,000 . 8.5136 + 200,000 . 55.41 =
11,953,000
total
21,953,000
or 2,580,000 / year
Plan B Initial costs
20,000,00
Operation and maintenance:
400,000 . [P/A, i, 20 ] = 400,000 . 8.5136 = 3,405,000
total
23,405,000
or 2,750,000 / year
Plan A is less costly then Plan B.
Answer problem 11
i = 5 %, n = 50, annuity factor = 0.0548
Slope Capital
recovery
Annual cost
Total annual
costs
slope
cost ( )
of excavation ()
maintenance
1 : 1 (n = 1)
350.000 . 3.00 .
0.0548 = 57,540
80,000
137,540
1 : 2 (n = 2)
450.000 . 3.00 .
0.0548 = 73,980
50,000
123,980
1 : 3 (n = 3)
550.000 . 3.00 .
0.0548 = 90,420
34,000
124,000
1 : 4 (n = 4)
650.000 . 3.00 .
0.0548 = 106,860
24,000
130,860
The most economical slope will be around 1 : 2.5 ( n = 2.5 ).
102
Answer problem 12
i = 10 %
Project A
Project B
Project C
n (life)
8 years
8 years
10 years
[ P / A]
5.335
5.335
6.145
P.V. benefits
18,400 . 5.335 =
30,600 . 5.335 =
42,300 . 6.145 =
98,163
163,251
259,915
Initial cost
100,000
160,000
280,000
P.V. of scrap
40,000 . 0.3856 =
value
nil
nil
15,422
NPV
- 1.837
+ 3,251
- 4,663
Recommen-
return < 10 %
return > 10 %
return < 10 %
Dation
rejected
acceptable
rejected
Answer problem 13
i = 10 %, n = 3,5 years
Costs:
Capital recovery cost: ( 300,000 30,000) . (A/P, i, n) =
270,000 . (A/P, 10 %, n)
Annuity for 3.5 years: 0.353; capital recovery:
270,000 . 0.353 =
95,200
Interest: 30,000 . 0.10 =
3,000
Extra maintenance costs:
6,000
Total costs
104,200
Benefits: Nett annual cost savings: ( 150,000 45,000) =
105,000
As benefits exceeds costs (just) the investment is justified (for i = 10 %)
103
Answer problem 14
n =
Alternative A
Alternative B
Alternative C
Alternative D
Investment
100
200
300
500
Annual
20
30
50
75
benefits
20 / i
30 / i
50 / i
75 / i
PV benefits
20 / i - 100
30 / i - 200
50 / i - 300
75 / i - 500
NPV
20 / 100 = 0.2
30 / 200 =
50 / 300 =
75 / 500 = 0.15
IRR
(20 %)
0.15 (15 %)
0.167 (16.7 %) ( 15 %)
Conclusion:
Highest rate
Of return
Unused funds: (i = 10 %)
NPV
20 / 0.10 100 30 / 0.10 200 50 / 0.10 300 75 / 0.10 500
= 200 100 =
= 300 200 =
= 500 300 =
= 750 500 =
or
100
100
200
250
Unused funds 500 100 =
500 200 =
500 300 =
nil
400
300
200
Annual
400 . 0.10 + 20 300 . 0.10 + 30 200 . 0.10 + 50 0 + 75 = 75
benefits
= 40 + 20 = 60
= 30 + 30 = 60 = 20 + 50 = 70
Preferred
Conclusion:
Alternative
Unused funds: (i = 14 %)
NPV
400 . 0.14 + 20 300 . 0.14 + 30 200 . 0.14 + 50
= 56 + 20 = 76 = 42 + 30 = 72 = 28 + 50 = 78 0 + 75 = 75
Conclusion:
Preferred
Alternative
104
Answer problem 15
1,150
Payment schedule a.
Annual payment for i = 12 %: 1150 . (A/ P, 12 %, 8) = 1150 . 0.2013 =
231.50. This payment is done at the end of the year !
231.5 231.5 231.5 231.5
231.5 231.5 231.5 231.5
Payment schedule b.
Present Value (PV) of payment schedule B (i = 12 %):
231.5 + 231.5 . ( P / A, 12 %, 7 years) = 231.5 + 231.5 . 4.564 =
231.5 ( 1 + 4.564 ) = 231.5 . 5.564 = 1,288
The same payment of 231.50 is done at the beginning of the year.
Schedule A is the better schedule for the party that is paying; schedule B is the better
schedule for the receiving part; the difference is 231.50 . 0.12 = 27.78 per
year. For payment schedule b one only needs to borrow
1,150 - 231,5 = 918.50.
n = 7 years:
918.50 . (A / P, i, 7 ) = 231, 50
(A / P, i, 7 ) = 231,50 / 918.50 = 0.2520
i = 16 %: Annuity = 0.2476; i = 18 %: Annuity = 0.2624
Interpolation gives an effective annual interest of 16.6 %.
Answer problem 16
Question a
Compare annual costs against annual benefits (NPV = 0)
Capital recovery: 100,000 . (A/P, i, 25)
Annuity factors:
i = 10 % : 0.1102; i = 12 % : 0.1275
NPV = 0:
100,000 . (A/P, i, 25) + 15,000 = 26,500
(A/P, i, 25) = (26,500 15,000 ) / 100,000 = 0.1150
By interpolation one finds the IRR = approx. 10.5 %
Question b : i = 4 %
Yes, investments can be made economically because the cost of money at 4 % will
result in a positive NPV (NPV = 0 for i = 10.5 %).
Question c
Annual surplus: 26,500 - 15,000 = 11,500
This surplus is being used to repay the loan, which carries an annual interest of 4 %.
So 100,000 . (A / P, 4 %, n ) = 11,500 (n = ?)
(A / P, 4 %, n ) = 0.1150
n = 10 years : annuity = 0.1233
n = 12 years: annuity = 0.1066
By interpolation one finds n = 11 years.
105
Answer problem 17
Question a , i = 10 %
Annual capital recovery cost:
1,000,000 . (A / P, I, 20) = 1,000,000 . 0.1175 =
117,500
Annual equivalent maintenance & operation costs:
100,000
Total annual costs:
217,500
Annual benefit:
15 . 106 x unit cost
Transportation cost (unit cost) per m3: 217,000 / 15,000,000 = 0.0145
Question b
Present value of all costs: PV = 217,500 . (P / A, 10%, 20 years) = 1,851,000
Present value of benefits, whereby X = unit cost:
13 . 106 . X . (P/ A, 10%, 10) + 17 . 106 . X . (P/ A, 10%, 10) . (P/ F, 10%, 10)
=
13 . 106 . X . 6.1446 + 17 . 106 . X . 6.1446 . 0.3855 =
(79.88 + 40.27 ). 106 . X = 120.15 . 106 . X
Transportation cost X (unit cost) per m3: 1,851,000 / 120,150,000 = 0.0154
Answer problem 18
Start project
1,500,000
1
2
20
2002 2003 2004 2005 2006 2007
2025
1,000,000 2,000,000 5,000,000 5,000,000 1,000,000
(10 % ) ( 8 % )
(6 %)
(6 % )
(4 % )
a. First
costs
Is compunded costs on the day the project is put into operation (1-1-2006) =
1,000,000 . (F / P, 10 %, 3) + 2,000,000 . (F / P, 8 %, 2) + 5,000,000 .
(F / P, 6 %, 1) + 5,000,000 + 1,000,000 . (P / F, 4 %, 1) =
1,000,000 . 1.331 + 2,000,000 . 1.1411 + 5,000,000 . 1.06 + 5,000,000 +
1,000,000 / 1.04 =
1,331,000 + 2,282,200 + 5,300,000 + 5,000,000 + 961,500 =
14,874,700 Say 14.9 million.
b. Total
depreciation
First cost salvage value = 14.9 million - 1.5 million =
13.4 million
c.
Internal rate of return
NPV = 0 or all costs = all benefits
13,374,000 . (A / P, i %, 20) + 1,500,000 . i + 1,000,000 =
3,250,000
Find i by trail and error.
106
For i = 10 %, annuity factor = 0.1175
13,374,000 . 0.1175 + 1,500,000 . 0.10 + 1,000,000 = 2,725,000
For i = 12 %, annuity factor = 0.1339
13,374,000 . 0.1339 + 1,500,000 . 0.12 + 1,000,000 = 2,975,000
For i = 14 %, annuity factor = 0.1510
13,374,000 . 0.1510 + 1,500,000 . 0.14 + 1,000,000 = 3,235,000
IRR = 14 % (slightly more).
d. Equivalent
annual
surplus
Actual `costs' of money is 4 %.
Annual surplus = annual revenue - annual costs =
3,250,000 - 1,000,000 - 1,500,000 . 0.04 - 13,374,000 . (A / P, 4
%, 20) = 3,250,000 - 1,000,000 - 60,000 - 13,374,000 .
0.0736
= 3,250,000 - 1,000,000 - 60,000 - 986,000 =
1,204,000
e.
Marginal rate of return
Annual capital recovery cost of additional initial cost of 1,500,000 =
1,500,000 . (A / P, i, 20 )
Additional revenues: 3,400,000 - 3,250,000 = 150,000 .
(A / P, i, 20 ) = 150,000 / 1,500,000 = 0.100
for i = 6 % annuity = 0.0872; for i = 8 % annuity = 0.1019; so i = 7.9 %
(approx.)
f. Justification
The investment of 1,500,000 is not justified because these amount can yield
12 % in another project against about 8 % in this project.
Answer problem 19
Question a ( i = 6 %, n = 15 years )
Plan A
Present Value Benefits: 624,000 . ( P / A, 6 %, 15) = 624,000 . 9.7122 = 6,060,500
Present Value All Costs:
4,000,000 + 200,000 . (P / A, 6 %, 15) =
4,000,000 + 200,000 . 9.7122 = 5,942,000
B / C factor :
6,060,500 / 5,942,000 = 1.02
Plan B
Present Value Benefits: 722,000 . ( P / A, 6 %, 15) = 722,000 . 9.7122
= 7,012,000
Present Value All Costs:
5,000,000 + 240,000 . (P / A, 6 %, 15) = 5,000,000 + 240,000 . 9.7122 =
7,330,900
B / C factor :
7,012,000 / 7,330,900 = 0.96
107
Question b
Plan A
The IRR > 6 % as B / C- factor > 1 for i = 6 %.
Try 7 %:
discounting factor : 9.1079 and B / C = 0.98
IRR = 6.5 %
Plan B
The IRR < 6 % as B / C factor < 1 for i = 6 %.
Try 5 %:
discounting factor : 10.38 and B / C = 1.02
Try 5.5 %: discounting factor : 10.0376 and B / C = 0.98
IRR = 5.25 %
Marginal rate of return of Plan B with respect to Plan A:
Plan B Plan A (= actual extension)
Present Value Benefits:
( 722,000 - 624,000) . ( P / A, i, 15) = 98.000 . (P / A, i, 15)
Present Value All Costs:
( 5,000,000 - 4,000,000) + ( 240,000 - 200,000) . (P / A, i, 15)
For i = 5 % B / C factor = (98,000 . 10.38) / (1,000,000 + 40,000 . 10.38) = 0.72
For i = 2 % B / C factor = (98,000 . 12.85) / (1,000,000 + 40,000 . 12.85) = 0.83
For i = 0 % B / C factor = (98,000 . 15 ) / (1,000,000 + 40,000 . 15 ) =
0.92
the marginal rate of return of the extension is negative !
Question c
Unused funds are defined as the difference in investment of Plan A and Plan B:
1,000,000 (the additional investment). As the marginal rate of return of the
additional investment is lower than 5 % (and even negative) the unused funds should
be invested in other projects (with a rate of return of 5 %).
Answer problem 20 i = 5 %
Frequency x
Useful life y
Annual capital
Annual repair &
Total
(times/
y = x2 + 20
recovery cost
maintenance cost
annual
decade)
(in years)
(depreciation)
at
costs
= 10. 106 .
250,000 / time
annuity
x = 0
y = 20 years
10. 106. 0.0802
(no painting)
= 820,000
0
820,000
x = 5 / decade y = 45 years
10. 106. 0.0563 5 x 250,000 / 10
(every 2
= 563,000
= 125,000
688,000
years)
y = 56 years
10. 106. 0.0535 6 x 250,000 / 10
x = 6 / decade
= 535,000
= 150,000
685,000
y = 69 years
10. 106. 0.0518 7 x 250,000 / 10
x = 7 / decade
= 518,000
= 175,000
693,000
y = 120 years
10. 106. 0.05
x = 10
= 500,000
= 250,000
750,000
(every year)
Most economic frequency: 6 times / decade
108
Answer problem 21
Question a
Water transport
400
Construction time:
= 8 years
8
Start of construction:
1st January 2002
(40 106
x
)
400
Construction cost/ year:
x
= LC 20 . 106 / year
8
100
End of construction:
31st December 2009
2002
2003
2004
2005
2006
2007
2008 2009
20
20
20
20
20
20
20
20 x 106 LC
F
1
( +
)
10
.
0
8 -1
Compounding factor:
( F/ A, 10 %, 8 years ) =
= 11.44
0 10
.
Construction cost at the end of the project: F = 11.44 . (20 . 106 ) = LC 228.8 .106
10
20 30 40 50
Annual cost of construction: 228.8 x 106 . (A/P, 10%,50 years)
23.11 x 106
Annual maintenance cost: 400 km/ 100 x 2 x 106
8 x 106
1 x 106
Annual transportation cost: 5 m. ton x 0.05 x 400/ 100
LC 228.8 .106
10
.
0
x 1
( +
)
10
.
0
50 -1
Annuity (10 %, 50 years):
= 0.101
1
( +
)
10
.
0
50 -1
Annual cost of construction costs:
LC (228.8 .106 ) . 0.101 = LC 23.11 .106
400
Annual maintenance costs:
LC (2 .106 ) .
=
LC 8.00 .106
100
400
Transportation costs:
LC 0.05.
. 5 .106 =
LC 1.00 .106
100
Total annual costs
LC 32.11 .106
109
Rail transport
375
Construction time:
= 5 years
5
Start of construction:
1st January 2005
32
(
106
x
)
375
Construction cost/ year:
x
= LC 24 . 106 / year
5
100
End of construction:
31st December 2009
2002 2003 2004
2005
2006
2007
2008 2009
24
24
24
24
24 x 106 LC
F
1
( + 0 10
. )5 - 1
Compounding factor:
( F/ A, 10 %, 5 years ) =
= 6.11
10
.
0
Construction cost at the end of the project: F = 6.11 . (24 . 106 ) = LC 146.64 .106
10
20 30 40 50
Annual cost of construction: 146.64 x 106 . (A/P, 10%,50 years)
14.81 x 106
Annual maintenance cost: 375 km / 100 x 2.5 x 106
9.38 x 106
1.31 x 106
Annual transportation cost: 5 m. ton x 0.07 x 375/ 100
LC 146.64 .106
10
.
0
x 1
( +
)
10
.
0
50 -1
Annuity (10 %, 50 years):
= 0.101
1
( +
)
10
.
0
50 -1
Annual cost of construction costs: LC (146.64 .106 ) . 0.101 =
LC 14.81 .106
375
Annual maintenance costs:
LC (2.5 .106 ) .
=
LC 9.38 .106
100
375
Transportation costs:
LC (0.07.
) . 5 .106 =
LC
1.31 .106
100
Total annual costs
LC 25.50 .106
110
So, Project B is preferred.
Question b
Water transport (20 % FC)
Construction cost / year: + 20 %
Annual cost of construction costs: + 20 % = 1.2 . LC 23.11 .106 =
LC 27.73 .106
Annual maintenance costs (same as question a):
LC 8.00 .106
Transportation costs (same as question a):
LC 1.00 .106
Total annual costs
LC 36.73 .106
Rail transport (80 % FC)
Construction cost / year: + 80 %
Annual cost of construction costs: + 80 % = 1.8 . LC 14.81 .106 =
LC 26.66 106
Annual maintenance costs costs (same as question a):
LC 9.38 .106
Transportation costs (same as question a):
LC 1.31 .106
Total annual costs
LC 37.35 .106
In this case, Project A is prefererred.
Question c
Compare the difference between the two projects.
The 'costs' of the transportation time are relative:
the railway time is faster and therefore cheaper, by
5 hours x LC 0.02 / ton x 5 x 106 ton per year = LC 0.5 . 106
For the first case the difference between the two alternatives becomes larger.
For the second case the difference between the two alternatives becomes smaller
and the two alternatives are about the same.
111
Answer problem 22
15,000
i = 8 %
1 2 3
3 years
800
1,200
1,500
20,000
12,000
1 2 3 4
4 years
800
1,200 1,500 1,800
20,000
8,000
1 2 3 4
5
5 years
800
1,200
1,500 1,800 2,100
20,000
Depreciation
Interest (8
Equivalent annual cost
Total
(annual)
%) on resale
of maintenance costs
annual
value
costs
3
years 5,000 . ( A/P, i, 3) = 15,000 . 0.08 ( 800/ 1.08 + 1200/ 1.082 +
5,000 . 0.3880 =
= 1,200 1500 / 1.083 ) . ( A/P, i, 3 )
4,289
1,940
= 2,960 . 0.3880 = 1,149
4
years 8,000 . ( A/P, i, 4) = 12,000 . 0.08 ( 800/ 1.08 + 1200/ 1.082 +
8,000 . 0.3019 =
= 960 1500 / 1.083 + 1800/
4,668
2,415
1.084) . ( A/P, i, 4) =
4,283 . 0.3019 = 1,293
5
years 12,000 . ( A/P, i, 5)
8,000 . 0.08 { 800/ 1.08 + 1200/ 1.082 +
= 12,000 . 0.2505 =
= 640 1500 / 1.083 + 1800/ 1.084
5,142
3,006
+ 2100/1.085 } . [ A/P, i, 5]
= 5,712 . 0.2505 = 1,496
Sell the equipment after 3 years !
112
Answer problem 23
i = 12 %, n =
Reinforced concrete road pavement per m2
Present value costs:
100 + 0.67 / 0.12 + 31 . (P/ F, 12%, 40 years) +
3.25 / 0.12 . (P/ F, 12%, 40 years) =
100 + 5.583 + 31 . 0.0107 + 27.08 . 0.0107 =
100 + 5.583 + 0.624 = 106.207 / m2
per 2,000 m2 : 2,000 . 106.207 / m2 =
212,414
Flexible pavement per m2
Present value costs:
90 + 3.25 / 0.12 + 0.67 / 0.12 =
90 + 3.92 / 0.12 = 122.667 / m2
per 2,000 m2 : 2,000 . 122.667 / m2 =
245,333
113
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