Catalog
Time Value Money········································································································ 1
Present Worth Comparison························································································· 12
Equivalent Annual Worth Comparisons (EAWC)························································ 19
Rate of return Method of Comparison········································································· 26
Payback Comparison Method····················································································· 33
Break Even Analysis··································································································· 36
Replacement Analysis·································································································44
Analysis of Public Projects B_C ratio·········································································· 55
Depreciation················································································································ 66
ENGINEERING ECONOMY
INTEREST & ECONOMIC ANALYSIS
I-Time value of money
1- Simple interest:
• When a simple interest rate is quoted, the interest earned is directly proportional to the capital
involved in the loan.
• Expressed as a formula:
𝐼 = 𝑃𝑖𝑁
Where
I = amount of interest to be paid
P= Present amount
i = interest rate / period
N= Number of interest periods
• So, the total amount to be paid after a period N is
F= P + I = P + PiN
F = P(1+iN)
Where
•
F = Future sum of money
Remark: ‘N’ is generally taken as a period of I year if:
1 year = 12 months → it gives ordinary simple interest.
1 year = 365 days → it gives exact simple interest.
Example (1) :
A loan of 1000 L.E is taken for 2 months at 10%, so the Future sum is
- With ordinary simple interest, the amount to be repaid is
2
𝐹 = 1000 [1 + 0.10 (12)] = 1016.67 L.E
-
With exact simple interest when the two months are January and February in a non-leap year
(February is 28 days):
𝐹 = 1000 [1 + 0.10 (
31+28
365
)] = 1016.16 L.E
Example (2) :
A loan of 1000 L.E is taken for 2 years at 10%, so the Future sum is
- With ordinary simple interest, the amount to be repaid is F=P(1+iN)
𝐹 = 1000[1 + 0.10(2)] = 1200 L.E
Note that the principal 1000 L.E has earned 100 L.E of interest for the first year and 100 L.E of
interest for the second year.
2- Compound interest:
•
•
Again assume a loan of 1000 L.E., for 2 years at an interest rate of 10% compounded annually.
The pattern of interest compounding is shown in the following table:
Amount owed at
Interest on
Amount owed
Year
Beginning of year
Amount owed
At end of year
1
1000 L.E
1000 x 0.10 = 100
1000+100 = 1100 L.E
L.E
2
1100 L.E
1100 x 0.10 = 110
1100+110 = 1210 L.E
L.E
F2 = P + Pi + (P + Pi)i
1st year interest 2nd year interest
Amount borrowed
F2 = P(1+i+i+i2)
= P(1+2i+i2)
= P(1+i)2
=
Fn = P(1+i)
n
1000(1+0.10)2 = 1210 L.E
In general
3- Nominal interest rates: ( compound interest )
• Interest rates are normally quoted on an annual basis.
• In some cases, interest is compounded several times per year: monthly, quarterly, semiannually,
etc… in this case, the interest rates are called Nominal Interest rates.
• For example: a year divided into four quarters with interest at 2% per quarter is typically quoted as
“8% compounded quarterly”.
• The future value at the end of 1 year for 200 L.E earning interest at 8% compounded quarterly is
developed as follows:
F3 (end of 1st quarter) = P(1+i) = 200(1+0.08/4) = 204 L.E
F6 (end of 2nd quarter) = 204(1+0.08/4) = 208.08 L.E
F9 (end of 3rd quarter) = 208.08(1+0.08/4) = 212.24 L.E
F12 (end of 4th quarter) = 212.24(1+0.08/4) = 216.48 L.E
• In general,
𝒓
𝑭 = 𝑷(𝟏 + )𝒎
𝒏
Where: m = number of compounding periods per year.
r = nominal interest.
•
For the previous example:
𝑟
𝐹 = 𝑃(1 + )𝑚
𝑛
0.08 4
𝐹 = 200(1 +
) = 216.48 L.E
4
Effective interest rate:
𝑟
𝑖𝑒𝑓𝑓 = (1 + )𝑚 − 1
𝑛
If Nominal 12% compounded monthly, therefore
𝑖𝑒𝑓𝑓
𝑟
𝑖𝑒𝑓𝑓 = (1 + )𝑚 − 1
𝑛
0.12 12
= (1 +
) − 1 = 0.127
12
II•
•
•
Time value equivalence
The value of money varies with time.
For example:
If I have today 500 L.E and I want to talk about them after 3 years from now, they will be
more than 500 L.E. In some way, If I want to talk about them 3 years ago, they would be less than
500 L.E.
Time value of money is very important when studying the economical feasibility of projects.
Seven factors are used to find the equivalent value of the money at the past, present, or future time.
1- Compound – Amount Factor ( Single payment):
P
F?
i%
N
•
( Find F, given P, at i% and after N periods)
We know from the compound interest formula that:
F = P(1+i)N
Factor for equivalence
The factor (1+i)N = Compound-Amount Factor = (F/P,i,N)
Example:
P=1000 L.E
i=10% N=5 years
F=1000(1+0.10)5
F=?
F=1000(F/P,10%,5)
2- Present – Worth Factor : ( Single payment):
P?
F
i%
N
( Find P, given F, at i% and after N periods)
•
We know that F=P(1+i)N
•
Then, 𝑃 = 𝐹 (1+𝑖)𝑁
1
Factor for equivalence
F=1000 x 1.6105 = 1610.5 L.E
•
1
The factor (1+𝑖)𝑁 = Present-Worth Factor = (P/F,i,N)
Example:
F=5000 L.E
1
𝑃 = 5000 (1+0.08)7
i=8% N=7 years
P=?
𝑃=𝐹
1
(1 + 𝑖)𝑁
P=5000(P/F,8%,7)
= 5000 x 1/1.7138
= 2917.49 L.E
P= 5000 x 0.58349
= 2917.49 L.E
Note that
(F/P,i,N) = 1/(P/F,i,N)
3- Sinking – Fund Factor :
A?
i%
0
F
A
A
A
A
1
2
3
4
A
5
N
( Find A, given F, at i% and after N periods)
•
•
•
•
•
I know F and I want to know the equivalent of F in the form of a uniform series of payments, each
payment has a constant value ‘A’ made at the end of interest period.
F=A(1+i)N-1 + A(1+i)N-2 + A(1+i)N-3 + A(1+i)N-4 + A(1+i)N-5 + …..
(1)
Multiplying both sides by (1+i):
F(1+i) = A(1+i)N + A(1+i)N-1 + A(1+i)N-2 + A(1+i)N-3 + A(1+i)N-4 +……. (2)
Subtracting eq.(1) from eq.(2) :
F(1+i)-F = A(1+i)N - A(1+i)N-5
F + Fi - F = A(1+i)N – A(1+i)0
Fi
= A(1+i)N – A
= A[(1+i)N – 1]
𝑖
𝐴 = 𝐹 ((1+𝑖)𝑁 −1)
Factor for equivalence
𝑖
The factor ((1+𝑖)𝑁 −1)= sinking-Fund Factor = (A/F,i,N)
Example:
F=5000 L.E
i=8% N=7 years
A=?
𝑖
𝐴=𝐹 (
)
(1 + 𝑖)𝑁 − 1
0.08
𝐴 = 5000 (1+0.08)7 −1
A=5000(A/F,8%,7)
= 5000 x 0.11207
= 560.35 L.E
A= 5000 x 0.11207
= 560.35 L.E
4- Series – Compound Amount Factor :
F?
i%
0
A
A
A
A
1
2
3
4
A
5
N
( Find F, given A, at i% and after N periods)
•
𝑖
𝐴 = 𝐹 ((1+𝑖)𝑁 −1) then,
We know that
(1 + 𝑖)𝑁 − 1
𝐹 = 𝐴[
]
𝑖
Factor for equivalence
The factor [
(1+𝑖)𝑁 −1
𝑖
] = Series Compound – Amount Factor = (F/A, i, N)
Example:
A=1000 L.E
i=8% N=5 years
F=?
(1 + 𝑖)𝑁 − 1
𝐹 = 𝐴[
]
𝑖
𝐹 = 1000 [
(1+0.08)5 −1
0.08
]
F=A(F/A,8%,5)
= 1000 x 5.8666
= 5866.6 L.E
F= 1000 x 5.8665
= 5866.5 L.E
5- Capital - Recovery Factor :
0
A?
i%
P
A
A
A
A
1
2
3
4
A
5
N
( Find A, given P, at i% and after N periods)
P=A(1+i)-1 + A(1+i)-2 + A(1+i)-3 + A(1+i)-4 + A(1+i)-5 + ……. (1)
Multiplying both sides by (1+i)-1
P(1+i)-1=A(1+i)-2 + A(1+i)-3 + A(1+i)-4 + A(1+i)-5 + A(1+i)-N-1 + ……. (2)
Eq. (2) – Eq. (1)
P(1+i)-1 - P=A(1+i)-N-1 - A(1+i)-1
P[(1+i)-1 -1] = A[(1+i)-N-1 – (1+i)-1 ]
𝑃[
1
1
1
− 1] = 𝐴 [
−
]
(1 + 𝑖)
(1 + 𝑖)𝑁+1 (1 + 𝑖)
Multiplying both sides by –(1+i)
𝑃[
−(1 + 𝑖)
−(1 + 𝑖)
(1 + 𝑖)
+ (1 + 𝑖)] = 𝐴 [
+
]
(1 + 𝑖)
(1 + 𝑖)𝑁+1 (1 + 𝑖)
−1
𝑃[−1 + 1 + 𝑖] = 𝐴 [
+ 1]
(1 + 𝑖)𝑁
1
]
(1 + 𝑖)𝑁
(1 + 𝑖)𝑁 − 1
𝑃𝑖 = 𝐴 [
]
(1 + 𝑖)𝑁
𝑖(1 + 𝑖)𝑁
𝐴 = 𝑃[
]
(1 + 𝑖)𝑁 − 1
𝑃𝑖 = 𝐴 [1 −
𝑖(1+𝑖)𝑁
The factor [(1+𝑖)𝑁 −1] is the capital recovery factor. From the table: (A/P,i%,N).
Example:
P=3933 L.E
i=8% N=5 years
A=?
𝑖(1 + 𝑖)𝑁
𝐴 = 𝑃[
]
(1 + 𝑖)𝑁 − 1
0.08(1+0.08)5
𝐴 = 3933 [ (1+0.08)5 −1 ]= 985.044 L.E
A=3933(A/P,8%,5)
A= 3933 x 0.25046
= 985.059 L.E
6- Series – Present Worth Factor :
i%
P?
0
A
A
A
A
1
2
3
4
A
N
( Find P, given A, at i% and after N periods)
We know that :
Therefore
𝑖(1+𝑖)𝑁
𝐴 = 𝑃 [(1+𝑖)𝑁 −1]
5
(1 + 𝑖)𝑁 − 1
𝑃 = 𝐴[
]
𝑖(1 + 𝑖)𝑁
The factor :
[
(1+𝑖)𝑁 −1
𝑖(1+𝑖)𝑁
] is the Series Present-Worth Factor. From the table: (P/A,i%,N).
Example:
A=1000 L.E
i=8% N=5 years
P=?
(1 + 𝑖)𝑁 − 1
𝑃 = 𝐴[
]
𝑖(1 + 𝑖)𝑁
(1+0.08)5 −1
𝑃 = 1000 [0.08(1+0.08)5 ] =3992.71 L.E
P=1000(P/A,8%,5)
P= 1000 x 3.9926
= 3992.6 L.E
7- Arithmetic – Gradient Conversion Factor :
F?
i%
A1
2G
G
A
3G
4G
A’
0
1
2
3
4
5
N
(Find F, given G, at i% and after N periods)
•
•
•
We must first convert the arithmetic-gradient into uniform series A1
F=G(1+i)3 + 2G(1+i)2 + 3G(1+i) + 4G …..
(1)
Multiplying both sides by (1+i)
F(1+i)=G(1+i)4 + 2G(1+i)3 + 3G(1+i)2 + 4G(1+i) + …… (2)
Eq.(2) – Eq.(1)
F(1+i) - F=G(1+i)4 + G(1+i)3 + G(1+i)2 + G(1+i) - 4G ……
Fi=G(1+i)4 + G(1+i)3 + G(1+i)2 + G(1+i) – (N-1)G
=G(1+i)4 + G(1+i)3 + G(1+i)2 + G(1+i) – NG + G
=G[(1+i)4 + (1+i)3 + (1+i)2 + (1+i) + 1]– NG
So:
Fi=G (F/A, i, N) – NG
F= {G [(F/A, i, N) – N]}/i
Series-Compound Amount Factor (F/A, i%, N)
To convert F into A1: from sinking fund factor
A1 = F (A/F, i, N)
[(𝐹/𝐴, 𝑖, 𝑁) − 𝑁]
𝐴1 = 𝐺 [
] (𝐴/𝐹, 𝑖, 𝑁)
𝑖
[(𝐹/𝐴, 𝑖, 𝑁)(𝐴/𝐹, 𝑖, 𝑁) − 𝐺𝑁(𝐴/𝐹, 𝑖, 𝑁)]
𝐴1 = [
]
𝑖
But
(F/A, i, N) = 1/ (A/F, i, N)
Therefore
𝐺 𝐺𝑁(𝐴/𝐹, 𝑖, 𝑁)
𝐴1 = −
𝑖
𝑖
𝐺
1 𝑁(𝐴/𝐹, 𝑖, 𝑁)
𝐴1 = [1 − 𝑁(𝐴/𝐹, 𝑖, 𝑁)] = 𝐺 [ −
]
𝑖
𝑖
𝑖
The factor
1 𝑁(𝐴/𝐹, 𝑖, 𝑁)
[ −
]
𝑖
𝑖
is the Arithmetic-Gradient Conversion Factor (A/G, i, N)
Now: A=A1 + A’
Example:
F?
i =8%
A1
2G
G
A
3G
4G
A=769.26 L.E
A’
0
0
1
2
3
4
5
1
2
3
N
If we have a series of 5 payments beginning with 400 L.E and increasing by 200 L.E for each
successive year therefore A’ = 400 L.E
A1 = 200[A/G, i, N] = 200[A/G,8%,5] = 200(1.8463) = 369.26 L.E
A = A’ + A1 = 400 + 369.26 = 769.26 L.E
4
5
2-Present Worth Comparison
•
Many economist prefer a Present-Worth analysis because it reveals the sum in today’s L.E.
that is equivalent to a future cash flow system.
•
•
Present Worth (PW) models are less subject to misinterpretation.
Today is usually represented as time zero in a cash flow
Basic Present-Worth Comparison Patterns:
•
Two general patterns are apparent in Present-Worth calculations:
ab-
Present-Worth Equivalence (PWE)
Net Present Worth (NPW)
(A) Present-Worth Equivalence (PWE):
-
PWE pattern determines the present-worth equivalence of a series of future transactions to
secure one figure that represents all the transactions.
This figure can then be compared to a corresponding figure that represents transactions from
a competing alternative.
Example on PWE:
An investor can make 3 annual end-of-year payments of 15000 L.E which will generate receipts
of 10000L.E at the end of year 4 and will increase annually by 2500 L.E. for the following 4
years. If the investor can earn a rate of return of 10%,[on alternative 8-year investments], is
this [alternative] attractive?
Solution:i=10%
G=2500
A=15000 L.E.
A’=10000 L.E.
0
1
2
3
4
5
6
7
8
PW= -15000(P/A,10,3)+{[10000 + 2500(A/G,10,5)](P/A,10,5)}(P/F,10,3)
= -15000(2.4868) + [10000+2500(1.8100)](3.7907)(0.75132)
= 4066 L.E
(The transactions provide a return of 10% on the investment plus a sum of 4066 L.E. This
investment is therefore preferable to one that would return exactly 10% over 8-years)
(B) Net Present Worth (NPW):
Net Present Worth = PW(Benefits) – PW(Costs)
In choosing between alternatives, the criterion is to select the one that maximizes net present
worth, or simply, the one that yields the larger positive PW.
A negative PW means that the alternative does not satisfy the rate-of-return requirement.
Example on NPW:
Two devices are available to perform a necessary function for 3 years. The initial cost for each
device at time zero and subsequent annual savings produced by the device are shown below.
The required interest rate is 8 %.
Year
Devise A
Devise B
0
-9000 L.E.
-14000 L.E.
1
4500 L.E.
6000 L.E.
2
4500 L.E.
6000 L.E.
3
4500 L.E.
8000 L.E.
Solution:9000
i=8%
4500 / year
1
0
2
3
NPW (device A) = -9000 + 4500(P/A,8,3) = -9000 + 4500(2.577) = 2597 L.E.
14000
0
i=8%
6000
6000
1
2
8000
3
NPW(device B) = -14000 + 6000(P/A,8,2) + 8000(P/F,8,3)
= -14000 + 6000(1.7832) + 8000(0.79383) = 3049.84 L.E.
-
Both alternatives meet the minimum acceptable rate of return, because both are positive
Device B is preferred because its net present worth is higher than that of device A.
Comparison of Assets that have unequal lives:
•
•
The utilization of present-worth comparisons implies that the lives involved have a common
endpoint (unequal lives)
Two methods are used to find common lives for comparison.
a- Common-Multiple method
b- Study-period method
(a) The Common Multiple Method:
-
The common live is the least common multiple of the lives of involved assets.
-
If assets had lives of 2,3,4 and 6 years. The least common multiple is 12 years
This means:
a- The asset with a life of 2 years would be replaced 6 times during the analysis.
b- The assets with the 3,4 and 6 years lives would be replaced 4,3, and 2 times
respectively.
-
The use of the least-common multiple of lives depends on the validity of the
assumption that assets will be repeatedly replaced by successors having identical cost
characteristics. This assumption is more often reasonable when the least common
multiple is small.
(b) Study-Period Method:
-
The common life is a specified duration that corresponds to the length of a project or
the period of time the assets are expected to be in service.
-
Some of the possibilities are to set the study period as the length of the:
i-
The shortest life of all competing alternatives. (This gives a protection against
technological obsolescence).
ii-
The known duration of a required service. (if we compare two machines of 4
and 7 years as economical lives, to be used in a project for 3 years. Then the 3
years is the common life.
-
A study-period comparison presumes that all assets will be disposed of at the end of
the time period. Therefore it is usually necessary to estimate the income that can be
realized from the sale of an asset which can still provide useful service.
Example (1) PW Comparisons of alternatives with unequal economic lives:
Assets A1 and A2 have the capability of satisfactorily performing the required function. A2 has an
initial cost of 3200 L.E. and an expected salvage value of 400 L.E. at the end of its 4-year
economic life. Asset A1 costs 900 L.E. less initially, with an economic life 1 year shorter than
that of A2, but it has no salvage value and its annual operating costs exceed those of A2 by 250
L.E. When the required rate of return is 15%, which alternative is preferred when compared by:
a- The least common-multiple method?
b- A 2-year study-period (assuming the assets are needed for only 2 years)?
Solution:a- The least common multiple of 3 and 4 is 12 years
For alternative A1:
2300 L.E
2300 L.E
2300 L.E
0
2300 L.E
i=15%
1
2
3
4
5
6
7
250 L.E
8
9
10
11
PA1 = -[2300+2300(P/F,15,3)+2300(P/F,15,6)+2300(P/F,15,9)+250(P/A,15,12)]
= -[2300+2300(0.65752)+2300(0.43233)+2300(0.28426)+250(5.4206)]
= -6815.603 L.E.
12
For alternative A2:
3200 L.E
3200 L.E
3200 L.E
i=15%
1
0
2
3
5
4
6
400 L.E
400 L.E
400 L.E
7
8
9
10
11
12
PA2 = -[3200-2800(P/F,15,4)-2800(P/F,15,8)+400(P/F,15,12)]
= -[3200-2800(0.57175)-2800(0.32690)+400(0.18691)]
= -5641.45 L.E.
Conclusion:
•
•
The minus sign means that we are talking about costs.
Alternative A2 is preferred because it has less present worth of costs than alternative A1.
b- Solution by the study-period method:
• The study period is 2 years as given in the problem.
• If the asset A1 is sold after 2 years, its sale value is
Initial cost = 2300 L.E.
Salvage value = 0
Economical life = 3 years.
The value (2300 – 0) should be depreciated in 3-years.
The depreciation value/year = (2300 – 0)/3 = 766.67 L.E.
Therefore the value of the asset at the end of 2 years is equal to the uncovered
depreciation which is 2300 – 2x766.67 = 766.67 L.E
•
The same logic is applied to Asset A2:
Depreciation/year = (3200 – 400)/4 = 700 L.E
Value of A2 at the end of year 2: = 3200 – 2x700 = 1800 L.E
For alternative A1:
2300
0
250
766.66
6
250
1
2
PA1 = -2300 – 250(P/A,15,2) + 766.66(P/F,15,2)
= -2300 – 250(1.6257) + 766.66(0.75614) = -2126.725 L.E
i=15%
For alternative A2:
1800 L.E
3200
i=15%
0
1
2
PA2 = -3200 + 1800(P/F,15,2) = -3200 + 1800(0.75614) = -1838.948 L.E
Conclusion:
Alternative A2 is preferred because it has the lowest Present Worth of Costs.
Comparison of assets assumed to have infinite life
•
•
Long-lived assets: dams, railway, Tunnels,
N in this case is considered infinity.
The sum of the first cost plus the present worth of disbursements is called a capitalized
cost.
Capitalized Cost = P + A(P/A,i,N) = P + A(P/A,i,∞) = 𝑃 + 𝐴 [
(1+𝑖)∞
= 𝑃 + 𝐴 [𝑖(1+𝑖)∞ ] = 𝑃 +
(1+𝑖)𝑁 −1
(1+𝑖)∞ −1
𝑖(1+𝑖)𝑁
𝑖(1+𝑖)∞
] = 𝑃 +𝐴[
]
𝐴
𝑖
Capitalized cost = P+A/i
•
The capitalized cost is equivalent to the present worth of long-lived assets.
Example: An Asset that last forever:
A 500000 L.E gift was given to a city for the construction and continued upkeep of a music
hall. Annual maintenance for a hall is estimated at 15000 L.E. In addition, 25000 L.E will be
needed every 10 years for painting and major repairs. How much will be left for the initial
construction costs after funds are allocated for perpetual upkeep? Deposited funds can earn
6% annual interest, and the returns are not subject to taxes.
Solution:Capitalized cost = P+A/i
Capitalized cost = P+A/i = 500000 L.E
A=A1+ A2
Where A1 = maintenance cost/year = 15000 L.E
A2 = Equivalent painting/year = 25000(A/F,6,10) = 25000(0.07587)
P=First Cost = Capitalized Cost – A/i = 500000 – (15000 + 25000x0.07587)/0.06
= 218387.5 L.E
The sum left for the initial construction costs after funds are allocated for perpetual upkeep
= 218387.5 L.E
3-Equivalent Annual-Worth Comparisons
•
In an EAWC method, all the receipts and disbursements occurring over a period of time are
converted to an equivalent uniform yearly amount.
•
•
It is a popular method especially because cost-accounting procedures, depreciation expenses,
tax calculations, and other summary reports are annual in nature.
EAWC produce results compatible with present-worth comparisons.
Example (1):
Equivalent net Worth of Cash Flows:
A consulting firm proposes to provide “self-inspection” training for clerks who work with
insurance claims. The program lasts 1 year, costs 2000 L.E. per month, and professes to improve
quality while reducing clerical time. A potential user of the program estimates that savings in the
first month should amount to 800 L.E. and increase by 400 L.E. per month for the rest of year.
However, operation confusion and work interference are expected to boost clerical costs by 1200
L.E. the first month, but this amount should decline in equal increments at the rate of 100 L.E. per
month. If the required return on money is 12% compounded monthly and there is a stipulation that
the program must pay for itself within 1 year, should the consultants be hired?
i=12/12 = 1% per month
Solution:-
G=400 L.E.
A’=800 L.E.
A=2000 L.E.
0
1
2
3
4
5
6
7
8
A’=1200 L.E.
G= -100 L.E.
EAW= 800 + 400(A/G,1%,12) – 2000 - [1200 - 100(A/G,1%,12)]
= 800 + 400 x 5.3682 – 2000 - [1200 – 100 x 5.3682]
= 284 L.E.
Conclusion:
Therefore the program looks promising (+ve EAW)
9
10
11
12
The above diagram is equivalent to the following diagram:
A=284 L.E.
0
2
1
3
4
5
6
7
8
9
10
11
12
Example (2):
Net Annual Worth of Single Project:
The purchase of a truck will reduce labor costs by 10,000 L.E per year. The price of the truck is
57,000 L.E and its operating costs will exceed those of the present equipment by 100 L.E. per
month. The resale value is expected to be 6000 L.E. in 12 years. Should the truck be purchased
when the prevailing interest rate is 12%?
Solution:6000 L.E.
57000 L.E.
A=10000 L.E./year
0
1
2
3
5 6
7 8 9 10
A=100 x 12 = 1200 L.E./year
4
11
12
i = 12%
A=10000 – 1200 -57000(A/P,12,12) + 6000(A/F,12,12)
= 10000 – 1200 -57000(0.16144) + 6000(0.04144)
= -153 L.E.
Equivalent annual-worth calculations indicate that the purchase and use of the truck will cause a
loss equivalent to 153 L.E. per year for 12 years.
Conclusion:
Therefore it is not recommended to purchase the truck.
Example (3):
Comparison of Net Annual Worth
A supplier of laboratory equipment estimates that profit from sales should increase by 23000
L.E./year if a mobile demonstration unit is built. A large unit with sleeping accommodations for
the driver will cost 71000 L.E., while a smaller unit without sleeping quarters will be 55000 L.E.
Salvage values for the large and small units after 5 years of use will be respectively 8000 L.E. and
3500 L.E. Lodging costs saved by the larger unit should amount to 3000 L.E. annually, but its
yearly transportation costs will exceed those of the smaller unit by 1300 L.E. With money at 15%,
should a mobile demonstration unit be built, and if so, which size is preferable.
Solution:For the large unit:
8000 L.E
71000 L.E
A2 = 3000 L.E
A1 = 23000 L.E
0
1
5
A3 = 1300 L.E
i =15%
Net Annual Worth large unit = 23000 + 3000 – 1300 – 71000(A/P,15,5) + 8000(A/F,15,5)
= 23000 + 3000 – 1300 – 71000(0.29832) + 8000(0.14832)
= 4706 L.E
For the small unit:
3500 L.E
55000 L.E
A1 = 23000 L.E
0
1
i = 15%
5
Net Annual Worth small unit = 23000 – 55000(A/P,15,5) + 3500(A/F,15,5)
= 23000 – 55000(0.29832) + 3500(0.14832)
= 7111.48 L.E
Conclusion:
• As both the net AW are positive, therefore the two alternatives are attractive.
• Alternative II (small unit) is the best as it shows a high value of the net AW.
Example (4):
Comparison of assets with unequal lives
Two models of machines can be purchased to perform the same function. Type I has a low initial
cost of 3300 L.E high operating costs of 900 L.E per year, and a short life of 4 years. The more
expensive Type II costs 9100 L.E, has annual operating expenses of 400 L.E., and can be kept in
service economically for 8 years. The scrap value of each machine is zero. Which is preferred
when the minimum attractive rate of return is 8 %?
Solution:
As the assets have unequal lives, try to find a common life using the LCM (least common multiple)
method
Common life = 8 years
For alternative I (with N=8 years)
3300 L.E
i = 8%
3300 L.E
A1 = 900 L.E
0
1
4
AI = - 900 – 3300(A/P,8,8) – 3300(P/F,8,4)(A/P,8,8)
= - 900 – 3300(0.17402) – 3300(0.73503)(0.17402)
= -1896.3 L.E
8
For alternative I (with N=4 years)
i = 8%
3300 L.E
A1 = 900 L.E
0
4
1
AI = - 900 – 3300(A/P,8,4)
= - 900 – 3300(0.30192)
= -1896.3 L.E
As we see, using the 4 years or the 8 years give exactly the same result. Therefore in the AW
comparison, it is not necessary to calculate a common life.
i = 8%
9100 L.E
A1 = 400 L.E
0
1
4
8
AII = - 400 – 9100(A/P,8,8)
= - 400 – 9100(0.17402)
= -1983.59 L.E
Conclusion:
Therefore alternative I is the best as it gives the lowest of the Annual Worth of Costs.
Perpetual Life:
• The factors used to calculate A are:
𝑖(1+𝑖)𝑁
The Capital-Recovery Factor (A/P,i,N)= (1+𝑖)𝑁 −1
•
•
•
For the factor (A/P,I,N):
𝑖(1+𝑖)𝑁
(1+𝑖)𝑁 −1
𝑖(1+𝑖)∞
⇒ (1+𝑖)∞ −1 = 𝑖
So in an economic comparison involving an asset with an infinite life, such as land, the
capital recovery factor is replaced by the interest rate.
It is difficult in this case to use the sinking fund factor, so the annual worth (A) should be
calculated by the capital-recovery factor for the assets that have perpetual lives.
Assets that have perpetual lives are: Dams, Tunnels, Canals, Monuments…They are public
projects
Example (5):
Annual Worth of an asset that have perpetual life
A short concrete canal can be constructed as part of a flood-control project; the placement of a
large galvanized culvert will serve the same function. The cost of the canal, which will last
indefinitely, is 75000 L.E, and maintenance costs will average 400 L.E/year. A culvert, which will
have to be replaced every 30 years, will cost 40000 L.E. and have annual maintenance costs of 700
L.E. Salvage values are negligible for both alternatives, and the government interest rate is 6%.
Which alternative has the lowest equivalent annual cost?
Solution:
This is a comparison between a canal with perpetual life and a culvert with an economic life of 30
years.
Canal:
i = 6%
75000 L.E
A1 = 400 L.E
0
∞
1
(AW) canal = – 400 – 75000(A/P,i,N)
for N=∞
(A/P,i,N) = i
= – 400 – 75000(i) = – 400 – 75000(0.06) = – 4900 L.E.
Galvanized Culvert:
i = 6%
40000 L.E
A1 = 700 L.E
0
30
1
(AW) Galv, Culvert = – 700 – 40000(A/P,6,30)
= – 700 – 40000(0.07265) = – 3606 L.E.
Conclusion:
The Galvanized Culvert has the advantage of a lower equivalent annual cost.
Rate of return Method of Comparison
The Internal Rate of Return (IRR)
PW
PW
IRR
IRR
i%
i%
PW (+) – PW (-) = 0
IRR is the value of i for which
Example:
A parcel of land adjacent to a proposed freeway exit is deemed likely to increase in value. It
can be purchased now for $80,000 and is expected to be worth $150,000 within 5 years.
During that period, it can be rented for pasture at $1500 per year. Annual taxes are presently
$850 and will likely remain constant. What rate of return will be earned on the investment if
the estimates are accurate?
Solution: Piece of land $80,000 Purchasing price, $1500 Rent, $850 Taxes, $150,000 Selling price.
Therefore, 1500(P/A, i,5) + 150,000(P/F, i,5) = 80,000 + 850(P/A, i,5)
150000
By trial and error,
850
i = 10%
+ve PW
i = 12%
+ve PW but less
i = 14%
+136.95
i = 15%
-3244.14
1500
0
1
i %=?
80000
IRR is between 14% and 15%, by interpolation IRR ≈ 14%
The 72-rule
A sum doubles in value every
Therefore 𝑖 =
𝟕𝟐
𝟓
𝟕𝟐
𝒊
years and since 80000 almost doubled in 5 years
= 14.4%
so, we check i% = 14%
i % = 15% to start
5
Minimum Acceptable Rate of Return (MARR):
MARR is a lower limit for investment acceptability set by organizations or individuals.
If
IRR > MARR → Project is acceptable (attractive)
IRR < MARR → Project is not acceptable (not attractive)
Single source of MARR
Multiple Source of MARR
MARR = 10%
1x106 Loan
0.5x106 → i = 18%
Personal Money
0.3x106 → i = 10%
A friend Money
0.2x106 → i = 12%
MARR = weighted average
= 0.5 x 18% + 0.3x10% + 0.2x12%
= 9% + 3% + 2.4%
= 14.4%
Comparison of Two Projects:
An old hotel was damaged by fire, it can be renovated to either offices or Apartments.
First cost of renovation
Increase in salvage value from renovation
Annual receipts
Annual disbursements
Present value of fire-damaged building
Expected salvage value of the fire-damaged
building after 30 years
120000
Offices
Offices
$340000
$120000
$212000
$59100
$485000
$266000
88000
251000
212000
IRR = 18.5%?
340000
IRR = 44.97%?
MARR = 12%
190000
Apartments
59100
0 1
Apartments
$490000
$190000
$251000
$88000
$485000
$266000
30
0 1
IRR = 16.7%?
490000
IRR = 33.26%?
30
Incremental Analysis
Apartment – Offices: 70000
28900
39000
0
1
IRR = 6.1%?
30
150000
PW Ap-Offices = -15000 + 10100(P/A,i,30) + 70000(P/F,i,30) = 0
By trial and error: IRR Ap-Offices = 6.1% < MARR
Therefore, Offices is better
Classification of alternatives:
- Independent → selection of a project is not affected by the selection of another one.
- Mutually exclusive → selection of one eliminates the opportunity to accept any of the others.
Example: A
B
C
Initial Investment
$170000 $260000 $300000
Annual Receipts
$114000 $120000 $130000
Annual disbursements $70000 $71000 $64000
IRR
22½ %
13.6%
17.7%
For 10 years
B – A → IRR B-A < MARR
⇒ A better than B
C – A → IRR C-A > MARR
⇒ C better than A (10.9%)
D – C → IRR D-C < MARR
⇒ C better than D
Therefore, C is the best.
D
$330000
$147000
$79000
15.9%
MARR = 10%
If independent, all
projects that have
IRR>MARR → selected if
sufficient budget
Example: - Cost-Reduction Proposal
Subassemblies for a model IV scope are purchased for $71 apiece. The annual demand is 350 units,
and it is expected to continue for 3 years, at which time the model V scope now under development
should be ready for manufacturing. With equipment purchased and installed for $21,000 the
production costs to internally produce the subassemblies should be $18,500 for the first year and
$12,250 each of the last 2 years. The equipment will have no salvage value. Should the company
make or buy the subassemblies?
Solution: Present annual cost = 350 x $71 = $24,850
Net savings year 1 = $24,850 - $18,500 = $6,350
Net savings (years 2,3) = $24,850 - $12,250 = $12,600
PW = -$21,000 + $6,350(P/F, i,1) + $12,600(P/F, i,2) + $12,600(P/F, i,3) = 0
$12,600
$12,600
$6,350
i%
PW
0%
→
10550
10%
→
4652.62
15%
→
2333.89
20%
→
333.78
25%
→
-1404.80
0
1
2
3
i=?
$21,000
By interpolation, the internal rate of return (IRR) = 21%
PW
$10,000
21%
$6,000
$2,000
i%
10%
20%
30%
←Accepted Proposal→ ←Rejected Proposal→
Multiple Rates of Return
Example: - Two Solutions for an IRR Evaluation
One of the alternatives for improving an operation is to do nothing to it for 2 years and then spend
$10,000 on improvements. If this course of action is followed, the immediate gain is $3000
followed by two years of breakdown operations. Thereafter, annual income should be $2,000 per
year for 4 years. What rate of return can be expected from following this course of delayed action?
Solution: The cash-flow diagram shown below suggests that there might be a sign reversal in the flow pattern.
This is confirmed by tabulating the cumulative cash flow, where it is clear that the total transactions
reverse from positive to negative at year 2 and again reverse signs at year 6.
End of year
0
1
2
3
4
5
6
Cumulative cash flow
+$3000
+$3000
-$7000
-$5000
-$3000
-$1000
+$1000
$3000
$2000/year
2
0
1
3
4
5
6
$10,000
PW = 3000 – 10,000(P/F, i,2) + 2,000(P/A, i,4) (P/F, i,2) = 0
i%
PW
0%
→
1000
10%
→
-25
Which indicates by the sign reversal from PW at i=0 that one IRR is slightly less than 10%
At i=51%, PW = 2 where the second sign reversal confirms there is a second IRR.
The profile of present worth over the discounting range of 0 to 51% is shown
The equivocal answer for the proposal’s IRR is that either 9.4% or 51% when returns can be
assumed to be reinvested at either rate.
PW
$1000
IRR=9.4%
IRR=51%
$750
$500
$250
i%
10%
50%
Example: - Ranking Reversal
Project
0
1
2
X
Y
-$1000
-1000
$100
100
3
4
$600
200
$850
200
End-of-Year Cash Flow
$350
200
The two projects are first compared by their present worth when the minimum required rate of
return is 10%
PW(X) = -1000 + [100+250(A/G,10%,4)] (P/A,10%,4)
= -1000 + [100+250(1.3810)] (3.1698) = $411.56
PW(Y) = -1000 + [1000+200(P/A,10%,3)] (P/F,10%,1)
= -1000 + [100+250(2.4868)] (0.90909) = $361.27
This ranks project X higher than project Y
When an IRR comparison is made, the rankings switch, as shown by the following calculations.
For project X;
PW = -1000 + [100+250(A/G, i%,4)] (P/A, i%,4) = 0
⇒ IRR = 23.4%
For project Y;
PW = -1000 + [1000+200(P/A, i%,3)] (P/F, i %,1) = 0
⇒ IRR = 34.5%
NPV
i=10%
IRR = i = 34.5%
361
Project Y
i%
i=10%
i=20%
External Rate of Return
The occasional occurrence of multiple rates of return is avoided by using the external rate-of-return
(ERR) method. The main appeal of the ERR method, however, is its pragmatic assumption that
receipts are actually reinvested at a generally available interest rate. This rate is typically taken to
be the MARR. Another advantage of the ERR method is that it usually can be computed directly
rather than by trial and error.
An external rate of return is calculated by equating the future worth of receipts (positive cash flows)
compounded at an explicit interest rate to the future worth of disbursements (negative cash flows)
compounded at ERR. If the explicit interest rate is represented by i%, then FW (receipts
compounded at i%) = FW (disbursements compounded at i) and ERR is the value of i that conforms
to the equality. When i% is the MARR and ERR exceeds i%, the investment is attractive because
it promises a yield greater than the lower limit of acceptability. The ERR method is consistent with
the PW method.
That is, ERR > MARR if and only if PW > 0 and
ERR < MARR if and only if PW < 0
Example: Evaluate the cash flow described in example of “Two solutions for IRR Evaluation” above, when
receipts are reinvested at the MARR of 15%.
Solution: Given i = 15%, the FW of receipts and disbursements are equated as
3000(F/P, i %,6) + 2000(F/A, i %,4) = 10,000(F/P, i,4)
From which
(F/P, i,4) =
3000(2.3130)+2000(4.9933)
10,000
= 1.69256
Which, by interpolation, produces an external rate of return of 14.8%.
Financial Analysis
Payback Comparison Method:
•
Payback Period is the period of time required by the proposal to return its original investment
from the savings it generates.
•
•
The formula for obtaining a rough measure of the time an investment takes to pay for itself is
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
𝑃𝑎𝑦𝑏𝑎𝑐𝑘 𝑝𝑒𝑟𝑖𝑜𝑑 =
𝐴𝑛𝑛𝑢𝑎𝑙 𝑅𝑒𝑐𝑖𝑒𝑝𝑡𝑠 − 𝐴𝑛𝑛𝑢𝑎𝑙 𝑑𝑖𝑠𝑏𝑢𝑟𝑠𝑒𝑚𝑒𝑛𝑡𝑠
𝐹𝑖𝑟𝑠𝑡 𝐶𝑜𝑠𝑡
=
𝐴𝑛𝑛𝑢𝑎𝑙 𝑛𝑒𝑡 𝑐𝑎𝑠ℎ 𝑓𝑙𝑜𝑤
Data utilized in applying the formula are direct, not discounted, cash-flow amounts, and no
salvage values are included.
Example (1):
Initial investment for purchasing a machine is = 30000 L.E. The reduction in operating costs =
10000 L.E./year for the next 5 years.
Payback = 30000/10000 = 3 years.
If the annual net cash flow is not constant, the following procedure is used to determine the
payback period:
Example:Consider an investment of 9000 L.E which will result in the following receipts:
Year Receipts
The 9000 L.E. will be returned by 5000
1st year
Therefore period = 2½ years
1
5000 L.E.
2
3000 L.E.
3
2000 L.E.
4
1000 L.E.
5
500 L.E.
+ 3000
2nd year
+ 2000/2
Half the third year
Criterion for choosing between projects:
1- The management establishes a maximum period for returning the capital invested. (ingeneral 2,3,4 or 5 years)
2- Projects that have a payback period greater than the maximum permissible period set by
the management are eliminated from the comparison.
3- Among the remaining projects, select the one with the lowest payback period.
Advantages of the payback method:
1- Simple, rapid calculations and decision.
2- It is recommended when:
a- The enterprise considers the rapid recovery of its invested capital is the criterion the
most important
b- When it is needed to make a preliminary selection between various projects. So,
applying the Payback period criteria will eliminate a number of alternatives, then other
methods can be applied to compare between the rest of alternatives.
Disadvantages of the payback method:
1- It does not take into consideration the cash flow further to the payback period.
Example:
If you have to decide between two machines:
Machine (1) : has a first cost of 4500 L.E. Annual income of 1000 L.E for 10 years.
Machine (2) : has a first cost of 3000 L.E. Annual income of 1000 L.E for 3 years.
Payback of M/C 1 = 4500/1000 4.5 years
Payback of M/C 2 = 3000/1000 3 years
According to the payback criterion, M/C 2 is the best. But no further income will be
realized by M/C 2 while M/C 1 will realize an income of 1000 L.E for 5½ more years.
This means that the payback period method does not take into consideration the benefits
realized further to the recovery period.
Conclusion:
The method should be applied only when comparing projects with equal lives
2- The method does not take into consideration the time value of money.
Example:
If you have to decide between the following two projects:
Project A
Project B
Initial cost
10000 L.E.
10000 L.E
Annual income year 1
5000 L.E
2000 L.E
year 2
4000 L.E
3000 L.E
year 3
1000 L.E
5000 L.E
year 4
2000 L.E
2000 L.E
Both projects have a payback period of 3 years.
But the first project will earn 3000 L.E more than the second one for the first, 1000 L.E for
the second year. Which means that project A recovers its invested capital faster than project
B and this money can be reinvested in another profitable project.
Conclusion:
a- The Payback method is useful for comparing projects having the same pattern of cash
flow, and having equal lives.
b- If these two conditions are not realized, the study should begin by the Payback Period
to eliminate undesirable projects, than continue by other methods of evaluation.
Break Even Analysis:
•
Most organizations look for profits
Selling price = Total Cost + Profit
•
Total Cost include all items of cost which can be divided into two categories:
Total Cost = Fixed Cost + Variable Cost
-
Fixed Costs : are those costs which remain relatively constant regardless of the
level of activity. They are also known as the Indirect Costs.
This description implies that the fixed level is maintained whether output is
zero or at 100% capacity.
In some cases this assumption is not valid, fixed costs may tend to increase as
output increases, and they can vary with time.
However, the change is usually not significant for short-run studies.
Example of F.C. : Rent – Interest – Research – Insurance – Depreciation –
Taxes – Advertising budget – Technical services – Executive salaries
-
Variable Costs : are those costs which are generally proportional to output.
They are also called “Direct Costs”, because they are directly associated with
a specific product or service. When there is no output, variable costs are zero.
Example of V.C. : Direct Labor – Direct Material (sometimes : Packaging,
Maintenance, Direct supplies, Direct supervision, sales commissions)
Graphical Representation of B.E. Charts :
Revenue and Costs
T.R (Total Revenue)
Selling Price (SP)
Profit
T.C
TR=TC at BEP
Total Variable
Cost at ‘n’
output
BEP
Total Cost
F.C
Fixed Cost
BEP
•
•
Total Cost (TC) = VC + FC
Total Revenue (TR) = TC + Profit
n
Units of output
Algebraic Relationships:
•
•
•
The graphic representation of the BEA is convenient for clarify or present economic
relationships. It is possible to obtain quantities for particular conditions by scaling
values from the chart.
The same conditions can be easily quantified by formulas. Calculations generally
provide accuracy.
Using the symbols already defined, we have:
Total Revenue at n output = TR = n* (SP)
Total Cost at n output
= TC = n* (VC) + FC
Gross Profit at n output = Z = TR – TC = n*(SP – VC) – FC
At BEP, TC = TR or Z = 0
Therefore:
n(SP-VC)-FC = 0
n(SP) = n(VC) + FC
where n = BEP
Therefore:
BEP = FC/(SP-VC)
•
The term (SP-VC) is called contribution. It indicates the portion of the selling price
that contributes to paying off the fixed cost
At n=BEP, the sum of contributions from BEP units equals the total fixed cost. The
contribution of each unit sold beyond n=BEP is an increment of profit.
Linear Breakeven Analysis:
•
A breakeven chart displays “Cost-Volume-Profit” relationships that hold only over
a short run.
Assumptions:
- Analysis is valid in the short run.
- Selling Price/unit remains constant during the period of analysis
- Variable Cost/unit remains constant
Example (1): BEA of a single product
An enterprise expects to sell 2750 units at a selling price/unit of 60 L.E during the next
year. Estimation of variable cost/unit includes the following:
Raw material
9.75 L.E.
Direct Labor
14.50 L.E.
Manufacturing expenses
6.25 L.E.
Sales commissions
3.00 L.E.
33.50 L.E.
The expected fixed costs are as follows:
Rent
10,000 L.E.
Salaries
25,000 L.E.
Depreciation
7,500 L.E.
Advertising
13,000 L.E.
Other fixed expenses
2,800 L.E.
58,300 L.E.
Check whether the company will realize profits next year or not.
Solution:At BEP
volume of production = n
Therefore Total Revenue = Total Cost
n (60)
= 58300 + (33.5)n
60n – 33.5n = 58300
n = 2200 units
Conclusion: Since 2750>2200 therefore the company will realize profit.
Example (2): Comparison between two or more alternatives using BEA
An enterprise needs to purchase a motor for operating a water pump used for raising water
from a tunnel. The economic life of the pump is 4 years. Two options are available:
Option (1):
Construction of a line of electricity and purchasing an electric motor at a total cost of 5500
L.E. The salvage value of the electric motor is 700 L.E. at the end of its 4-year economic
life. The cost of electricity is estimated to be 3 L.E./hr. Maintenance costs are estimated to
be 450 L.E./year.
Option (2):
Purchasing a gasoline motor at a cost of 2100 with zero salvage value at the end of its 4year economic life. The cost of oil and gas necessary for the operation of the motor is
estimated to be 1.34 L.E./hr. Maintenance cost/hr is estimated to be 0.45 L.E. Labor cost
is evaluated at 5.70 L.E./hr.
Determine the BEP in terms of number of operating hours and specify which one should
be used.
Solution:First identify the type of cost:
Option (1): First cost/year:
Depreciation (straight line method) = (5500 - 700)/4
= 1200 L.E.
Maintenance
= 450 L.E.
Total fixed cost
= 1650 L.E.
Variable cost:
Cost of electricity/hour
= 3 L.E./hr
Option (2): First cost/year:
Depreciation (straight line method) = (2100 - 0)/4
= 525 L.E.
Variable cost:
Labor cost/hr
= 5.70 L.E.
Maintenance cost/hr
= 0.45 L.E.
Gaz and oil cost / hr
= 1.35 L.E.
Total variable cost/hr
= 7.50 L.E./hr
Assume n → number of working hours
Therefore
(Total Cost)option 1 = 1650 + 3n
(Total Cost)option 2 = 525 + 7.5n
At the BEP:
(Total Cost)option 1 = (Total Cost)option 2
1650 + 3n
= 525 + 7.5n
n = 250 hours
Therefore two options are equivalent for 250 working hours per year.
The previous result can be shown on the following chart:
cost
Total Cost (2)
BEP
1650
Total Cost (1)
Fixed Cost (1)
525
Fixed Cost (2)
250
Number of working hours
Conclusion:
•
•
•
If the number of working hours of the motor is less than 250 hrs per year, the
gasoline motor (option 2) is better (because of its least total cost).
If the number of working hours of the motor is greater than 250 hrs per year, the
electric motor (option 1) is better (because of its least total cost).
So the choice of either one of the two options will depend on the expected number
of working hours.
Non-Linear Breakeven Analysis:
•
•
Cost and Revenue functions do not always follow convenient linear patterns.
More often, realistic cost relationships develop a nonlinear pattern.
TC
TC
Revenue
and
costs
TR
Revenue
and
costs
BEP2
TR
BEP2
BEP1
BEP1
Units of output
Units of output
TC
Revenue
and
costs
TR
BEP2
BEP1
Units of output
Example:If we have the case where:
Selling price of finished units varies according to: SP = (100-0.001n) L.E/unit
Fixed costs are considered reasonable at 200000 L.E./month
Variable costs varies according to : VC = (0.005n + 4) L.E./unit and the maximum capacity
of production of the plant is 12000 units/month.
-
To find the BEP or BEPs and the point of maximum profit algebraically, we
do the following:
Equation of the profit function Z;
Z = TR – TC
= n(SP) – (FC + n(VC))
= n(100 – 0.001n) –[200000 + n(0.005n +4)]
= 100n – 0.001n2 – 200000 – 0.005n2 – 4n
= - 0.006n2 + 96n – 200000
At the BEP, profit = 0
0 = - 0.006n2 + 96n – 200000 → This is an equation of the second order
Therefore n1 = 2467 units and n2 = 13533 units
-
To find the point of maximum profit, differentiate the Z-function with respect
to n and equate to zero.
𝑑𝑍 𝑑(−0.006𝑛2 + 96𝑛 − 200000)
=
𝑑𝑛
𝑑𝑛
0 = -2(0.006n)+96
n = 8000 units → volume of production that gives max profit
Therefore Z8000 = -0.006(8000)2 + 96(8000) – 200000 = 184000 L.E
Remark:Of course to realize profit, the volume of production should lay in the region between the two
BEPs.
Replacement Analysis
•
A replacement analysis is conducted to determine “if” and “when” an asset currently in
service (the defender) should be replaced by a more economical alternative (a challenger).
•
Analysis is usually done using the EAW analysis (PW and IRR can also be used for the
analysis).
Existing assets may be replaced because of:
1- Deteriorating Performance or
2- Obsolescence or
3- Inadequate capacity
•
But before talking about the techniques used in each case, we must calculate the economic
life for cyclic replacements.
Economic Life for cyclic Replacements
•
Many mechanical items used in service agencies and manufacturing are replaced by
essentially the same machine when the original wears out.
•
Equipment become less efficient and accumulate higher and higher repair (maintenance) cost
as they age.
•
Conversely, the longer they are kept in operation, the lower will be their average annual
capital cost because the purchase price is spread over more years.
•
The sum of these two costs is the total cost of providing the machines services.
•
The objective of a cyclic replacement study is to determine the pattern of replacement that
will minimize the average annual cost.
•
Data for future costs must be estimated or available from internal or suppliers’ records.
Example:
A company has a fleet of small trucks for store-to-factory deliveries. The purchase price per
truck is 1000 L.E. , and the anticipated schedule of future operating costs and Truck values is
Year
1
2
3
4
5
6
7
Operating costs, L.E 100 300 500 700 900 1100 1300
Truck Value, L.E
600 500 400 300 200 100 0
Determine the economical life for cyclic replacement.
Solution:
We have 7 policies for the cyclic replacement
1- Replacing each year
2- Replacing every 2 years.
3- Replacing every 3 years.
4- Replacing every 4 years.
5- Replacing every 5 years.
6- Replacing every 6 years.
7- Replacing every 7 years.
Cyclic Replacement:Policy (1):
$600
$100
$1000
1
0
A1 = -1000(A/P,20,1)+500(A/F,20,1)
= -1000(1.20)+500(1)
= -$700
Policy (2):
$500
$1000
$100
0
1
$300
2
A2=-1000(A/P,20,2)-[100+200(A/G,20,2)]+500(A/F,20,2)
= -1000(0.65455) -[100+200(0.4545)]+500(0.45455)
= -$618.18
Policy (3):
$1000
$100
1
0
$400
$300
2
$500
A3=-1000(A/P,20,3)-[100+200(A/G,20,3)]+400(A/F,20,3)
= -1000(0.47473) -[100+200(0.8791)]+400(0.27473)
= -$640.66
3
Policy (4):
$1000
$100
1
0
$300
$300 $500
3
2
$700
A4=-1000(A/P,20,4)-[100+200(A/G,20,4)]+300(A/F,20,4)
= -1000(0.38629) -[100+200(1.2742)]+300(0.18629)
= -$685.24
4
Policy (5):
$1000
$300
$500 $700
$100
1
0
2
3
4
$200
$900
A5=-1000(A/P,20,5)-[100+200(A/G,20,5)]+200(A/F,20,5)
= -1000(0.33438) -[100+200(1.6405)]+200(0.13438)
= -$735.60
5
Policy (6):
$100
$900
$300
$500 $700
$100
$1000
0
1
2
3
4
5
$1100
6
A6=-1000(A/P,20,6)-[100+200(A/G,20,6)]+100(A/F,20,6)
= -1000(0.30071) -[100+200(1.9788)]+200(0.10071)
= -$786.39
Policy (7):
$900
$1000
$1100
$1300
$300
$500 $700
$100
0
1 12
3
4
5
56
A7=-1000(A/P,20,7)-[100+200(A/G,20,7)]
= -1000(0.27742) -[100+200(2.2901)]
= -$835.44
7
EAC
(Equivalent Annual
Cost)
A*
Minimum Equivalent
Annual Cost
N*
Economic Life for cyclic replacement
Economic life Curve for any Asset
years
Approaches of solving a replacement problem:
Example: Consider the following problem
A defender has a current market value of 5000 L.E. and a challenger that can be
purchased for 7500 L.E. Both have a service life of 3 years with no salvage
value expected at the end of that time. Their operating costs are shown in the
following table.
Year
0
1
2
3
Defender (D)
P=5000 L.E.
1700
2000
2500
Challenger (C)
P=7500 L.E.
500
1100
1300
i=12%
D-C
-2500 L.E.
1200
900
1200
Solution:
1- First approach:
To consider the (salvage value) of the old asset to be the cost of keeping the
defender in service.
Therefore
EAC (D)=-[5000+1700(P/F,12%,1)+2000(P/F,12%,2)+2500(P/F,12%,3)](A/P,12%,3)
= -4119 L.E.
EAC (C)=-[7500+500(P/F,12%,1)+1100(P/F,12%,2)+1300(P/F,12%,3)](A/P,12%,3)
= -4059 L.E.
Which indicates that the Defender should be replaced.
Savings = 4119-4059 = 60 L.E.
2- Second approach:
To consider the Defender’s market value as a receipt (positive cash flow) that
offsets part of the purchase price (negative cash flow) of each challenger. Then
the net differences between the cash flows of the defender and each challenger
are compared.
Therefore
EAW (D-C)=[-2500+1200(P/F,12%,1)+900(P/F,12%,2)+1200(P/F,12%,3)](A/P,12%,3)
= -60 L.E.
Which means that the challenger is preferred and the Defender should be replaced.
Inflated Salvage value of a Defender:
Example: Assume that a second challenger competes with the Defender and
has a purchase price of 9000 L.E., but 6000 L.E. is offered for the Defender as a Tradein and the seller guarantees that operating costs will be no more than 800 L.E./year.
Should the offer be accepted when the required rate of return is 12% and no salvage
value is expected at the end of challenger 3-year life?
Solution:
We consider that the 1000 L.E above the market value (6000-5000=1000) is decreased
from the purchasing price of the second challenger.
Applying the first approach:
EAC(C2) = -[(9000-1000)(A/P,12%,3)+800] = -4131 L.E.
Comparing this value with those of the Defender and the first challenger, we conclude
that the new challenger is rejected.
Defender and Challenger with Different lives:
• Two ways to solve the problem if PW method is used:
1- Study period: when the asset is needed for service for a known period.
2- Least Common Multiple method: when there is a continuing need for its
service.
• When comparing with the AW method, there is no need of calculating a
common life.
Example: Consider the same Defender described above, and a challenger with
a first cost of 12000 L.E, salvage value of 2000 L.E at the end of its 5-year economic
life, and annual operating costs of 700 L.E. Service provided by the equipment will be
needed indefinitely. When MARR=12%
EAC (Challenger) = -[12000(A/P,12%,5) - 2000(A/F,12%,5) +700] = - 3714 L.E.
Which means that this challenger is better than the Defender.
Replacement Analysis
Replacement due to deterioration:
•
Deterioration is manifested through:
- Excessive operating costs
- Increased maintenance costs
- Higher reject rates
Example (1):
An existing machine is worth 2500 L.E. today and will loose 1000 L.E. in value by next year plus 500
L.E./year thereafter. Its 8000 L.E. operating cost for this year is predicted to increase by 1000 L.E.
annually, owing to deterioration. It will be retired in 4 years when its salvage value will be zero. A
new improved machine that satisfactory performs the same function as the existing machine can be
purchased for 6000 L.E. and is expected to have relatively constant annual operating costs of 6000
L.E. to the end of its 7-year economic life, at which time the salvage value will be 1500 L.E. No major
improvements are expected in designs for machines of this type within 7 years. If MARR = 12%,
should the existing machine be replaced? If so, when?
Solution:
1- If we decided to take the challenger, we assume that it will be kept for all its economic life.
2- For the Defender, we will consider to keep it for one more year and compare its EAC with the
EAC of the challenger.
Challenger:
16000
i=12%
1500
6000 L.E/year
0
1
EAC (Challenger) = -16000 (A/P,12%,7) + 1500 (A/F,12%,7) – 6000 = -9357 L.E.
7
Defender:
2500
i=12%
G=1000
0
1
A’=8000
2L.E/year
1500
1000
3
500
4
Annual cost of keeping the Defender one more year:
= –2500(A/P,12%,1) – 6500(A/F,12%,1) = -9300 L.E.
8000
2500
1500
0
1
Since the Defender has a lower annual cost for next year, it should be retained.
Annual cost of keeping the Defender a second year:
= –1500(A/P,12%,1) – 8000(A/F,12%,1) = -9680 L.E.
9000
1500
1
The EAC of keeping the Defender a second year is higher than that of the challenger.
So, the purchasing of the challenger should be anticipated 1 year from now.
1000
2
Replacement due to obsolescence:
•
•
Dramatic operating-cost reductions or impressive quality gains provided by technologically
advanced challengers are the typical causes of obsolescence.
Progressive design enrichments can outdate an existing asset well before its economic life has
expired.
Example:
A low-volume office copying machine was purchased 2 years ago for 700 L.E. At the time of
purchase it was believed that the machine would have an economic life of 5-years and a salvage
value of 100 L.E. Operating costs over the first 2 years for material, labor, and maintenance
have averaged 4200 L.E./year and are expected to continue at the same level. The same
company that manufactured the presently used copying machine has a new model which costs
1000 L.E. but will perform the current workload with operating costs of 3500 L.E./year. They
are offering 500 L.E. for the old model as a trade-in on the new machine. The expected salvage
value for the new model is 200 L.E. at the end of 10 years.
Another company has a different type of a copier which is available only on a lease basis. The
company claims that leasing their copier at 750 L.E. per year will reduce the operating expense
for the present amount of work to 2750 L.E. Since they do not accept trade-ins, the machine
now in use would have to be sold in the open market, where it is expected to bring only 250
L.E. If MARR = 10%, should the defending copier be replaced by one of the challengers?
Solution:
In order to compare both challengers to the Defender under equitable conditions, the apparent
market value (250 L.E.) is used, and the price for purchasing a copier is discounted
accordingly.
100
250
4200 L.E/year
0
1
3
(EAC)defender = - 250(A/P,10%,3) + 100(A/F,10%,3) – 4200 = -4270 L.E
200
1000-250
3500 L.E/year
0
1
10
(EAC)challenger1 = - (1000-(500-250))(A/P,10%,10) + 200(A/F,10%,10) – 3500 = -3610 L.E
2750 L.E/year
750 L.E/year
0
1
(EAC)challenger2 = - 2750 – 750 = -3500 L.E
The EAC analysis indicates that both challengers are preferred to the defender. Leasing is more
attractive than buying, if it can be assumed that lease charges will not increase.
Replacement due to Inadequacy:
•
•
When current operating conditions change, an older asset occasionally lacks the capacity to
meet new requirements.
The decision of changing the existing equipment is taking, and the problem is to find which
alternative to choose.
Example:
Machine A was installed 6 years ago at a total cost of 8400 L.E. At that time it was
estimated to have a life of 12 years and a salvage value of 1200 L.E. Annual operating costs
have held relatively constant at 2100 L.E. The successful marketing of a new product has
doubled the demand for parts made by machine A. The new demand can be met by purchasing
an identical machine which now costs 9600 L.E. installed. The economic life and operating
costs for the two machines will be the same. The salvage value for the second A-type machine
will be 1600 L.E.
Machine B, a different type, costs 17000 L.E. installed but has twice the capacity of
machine A. Its annual operating costs will be about 3100 L.E., which should be relatively
constant throughout its 10-year economic life. Salvage is expected to be 4000 L.E. The present
machine can be used as a trade-in on the new machine B. It is worth 3000 L.E. If MARR =
10%, compare the two alternatives.
Solution:
(Alternative I) Use two type A machines (old + new)
1200
MARR = 10%
3000
0
2100 L.E/year
1
(EAC) old = - 3000(A/P,10%,6) + 1200(A/F,10%,6) – 2100 = -4270 L.E
6
MARR = 10%
9600
1600
2100 L.E/year
0
1
12
(EAC)new = - (9600)(A/P,10%,12) – 2100 + 1600(A/F,10%,12)= -6067 L.E
(Alternative II): Purchase Type B machine (Twice capacity of A)
4000
MARR = 10%
17000
3100 L.E/year
0
1
10
(EAC)II = - (17000)(A/P,10%,10) + 4000(A/F,10%,10) – 3100 = -5616 L.E
Therefore: Alternative II is preferable.
Analysis of Public Projects
-
Example of public projects: schools, hospitals, ways, damps, etc.…
The government is the investor of the public projects.
The public projects are all the projects that offer services to the public sector
The method used to analyze the public projects is the “Benefit-Cost” analysis (B/C)
method
Benefit Cost Analysis:
•
•
The B/C analysis method is based on the ratio between Benefits and costs associated to a
project.
The first step in a B/C analysis is the determination of the elements considered as
Benefits and those considered as Costs.
-
In general:
Benefits: represent the advantages, expressed in L.E, for the public.
Losses (dis-benefits): represent disadvantages, expressed in L.E, for the public.
Costs: represent expenses paid by the government for the construction, operation,
maintenance, etc. of the project.
The Basic comparison formulas in a B/C analysis are:
𝐵/𝐶 =
𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑊𝑜𝑟𝑡ℎ 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠−𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑊𝑜𝑟𝑡ℎ 𝑜𝑓 𝑙𝑜𝑠𝑠𝑒𝑠
Also 𝐵/𝐶 =
𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 𝐶𝑜𝑠𝑡𝑠
𝐴𝑛𝑛𝑢𝑎𝑙 𝑊𝑜𝑟𝑡ℎ 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠−𝐴𝑛𝑛𝑢𝑎𝑙 𝑊𝑜𝑟𝑡ℎ 𝑜𝑓 𝑙𝑜𝑠𝑠𝑒𝑠
𝐴𝑛𝑛𝑢𝑎𝑙 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 𝐶𝑜𝑠𝑡𝑠
A Benefit-Cost analysis may also be expressed as a (B-C) analysis where:
Present Value of net Benefits (B-C) = PW (Benefits) – PW (Costs) or
Annual Value of net Benefits (B-C) = AW (Benefits) – AW (Costs)
Determination of the feasibility of a public project
For a public project to be feasible (acceptable): B/C ratio ≥ 1.0
Or (B-C) ≥ 0
Example (1): Basic Application on the (B/C) method
The government studies the economic feasibility of constructing a bridge. The project needs a
first investment of 10 million L.E. Operation and maintenance costs are estimated to be 250,000
L.E/year for its 20 years economic life. Savings in time and distances due to the use of this
bridge are estimated to be 1,950,000 L.E/year. Is the project acceptable if i=7%?
Solution: 10,000,000
0
1,950,000 L.E/year
250,000 L.E/year
1
i = 7%
Using the Present Worth method:
PW (Benefits) = 1,950,000(P/A,7%,20) = 1,950,000x10.593
B
= 20,656,350 L.E
PW (Costs) = 10,000,000 + 250,000(P/A,7%,20) = 10,000,000 + 250,000x10.593
C
B/C
= 12,648,250 L.E
= 20656350/12648250 = 1.63
As B/C > 1.0 the project is acceptable.
Example (2): (B/C) Analysis
The government is studying two sites for the construction of a highway.
For the first site A:
Construction Cost = 4,000,000 L.E
Maintenance cost/year = 200,000 L.E
Annual Benefits for the residents = 125,000 L.E
For the second site B:
Construction Cost = 6,000,000 L.E
Maintenance cost/year = 120,000 L.E
Annual Benefits for the residents = 100,000 L.E
If the economic life of the highway is 20 years, and i=8%, which site is preferable.
20
Solution: Using the Annual Worth method:Alternative A:
4,000,000
0
125,000 L.E/year
200,000 L.E/year
1
20
i = 8%
C = 4,000,000 (A/P,8%,20) + 200,000 = 4,000,000x0.10185 + 200,000 = 507,400 L.E
B = 125,000 L.E
(B/C)A = 125000/507400 = 0.21 which is < 1.0
Alternative A is rejected.
Alternative B:
6,000,000
0
100,000 L.E/year
120,000 L.E/year
1
20
i = 8%
C = 6,000,000 (A/P,8%,20) + 120,000 = 6,000,000x0.10185 + 120,000 = 731,000 L.E
B = 100,000 L.E
(B/C)A = 100000/731000 = 0.136 which is < 1.0
Alternative B is rejected.
Conclusion:
The two alternatives are not profitable (since B/C < 1.0), but if it is necessary to choose one of
them, then we choose alternative “A” because its B/C ratio is higher.
Example (3): Projects that contain only Costs
Consider the case of having two sites for the construction of a highway, with the following data:
Site N
Initial Cost
10,000,000 L.E.
Maintenance Cost/year
35,000 L.E.
Costs paid by the users / year 450,000 L.E.
Site S
15,000,000 L.E.
55,000 L.E.
200,000 L.E.
If the economic life of the Highway is 30 years, and i=5%, which site should be selected?
Solution: -
To apply the B/C method, we must have benefit items in the data of the problem
analyzed.
In the given problem, we don’t have any benefits, so to overcome this difficulty, we take
the difference between the two alternatives.
Let us consider the advantage of S over N or calculate S-N:
(Initial Cost)S-N = 15,000,000 – 10,000,000 = 5,000,000 L.E.
{
Cost (Government)
(Maintenance Cost/year)S-N = 55,000 – 35,000 = 20,000 L.E
Benefit (Public) (Savings for the users) N-S = 450,000 – 200,000 = 250,000 L.E.
Using the Annual Worth method:
C = 5,000,000(A/P,5%,30) + 20,000 = 5,000,000 x 0.06505 + 20,000 = 345,250 L.E.
B = 250,000 L.E.
⸫ (B/C)S-N = 250000/345250 = 0.724 < 1.0
This means that N is better than S
Analysis B/C for several alternatives:
If the alternatives are independent:
Projects are said to be “independent” when the choice of one does not prevent the choice of
another. In this case, just find the ratio for each alternative. All the alternatives are acceptable if
we have enough budget to do all the acceptable projects, we do them all. If the budget is limited,
we choose the most attractive projects.
Example:
Suppose we have to evaluate the following 3 projects: A, B, C .These projects are independent
{A→ Construction of a small road; B→ Extension of an elementary school; C→ Repair of a
saffle in a museum}. The expected costs and benefits of these projects has an economic life of 15
years and i = 6%
Project
Initial Cost
Benefit
A
185,000 $
20,000 $
B
220,000 $
30,000 $
C
310,000 $
50,000 $
Which projects should be chosen if the budget is limited to 550,000?
Solution: For Project A:
Annual Equivalent (Costs) = 185000(A/P, 6%, 15) = 185,000x0.10296 = 19,047.6$
(B/C)A = 20000/19047.6 = 1.05
For Project B:
Annual Equivalent (Costs) = 220000(A/P, 6%, 15) = 185,000x0.10296 = 22,651.2$
(B/C)B = 30000/22651.2 = 1.32
For Project C:
Annual Equivalent (Costs) = 310000(A/P, 6%, 15) = 185,000x0.10296 = 31,917.6$
(B/C)C = 50000/31917.6 = 1.56
Conclusion: 1) As long as the report for all projects, so they are all feasible.
2) But as long as the budget is limited to 550,000 so we limit ourselves to
choosing projects B which require a budget of 220,000+310,000 = 530,000$
Analysis B/C for several alternatives: In the case of several mutually exclusive alternatives, a multiple evaluation must
be made by an analysis of the increase in profits and costs.
If the alternatives are mutually exclusive
Mutually exclusives “means that the choice of one alternative prevents the choice
of another the goal is to choose one of several alternatives.
In this case you have to do a multiple analysis as shown in the following example.
Example:
To reduce the effects of flooding from a certain river, a number of small dams can
be built. The dams will decrease the production of floods, and will also cause
decreases in the fires, and an increase in the activities of the place. The costs and
benefits resulting from the different combinations of dams are shown in the
following table
The Dams Constructional
Costs
1
1,200,000 $
1,2
1,500,000
1,2,3
2,700,000
1,2,3,4
3,500,000
N=40 years, i=4%,
Maintenance
Costs
20,000 $
35,000
50,000
60,000
Benefit
Benefit
Benefit
200,000$
190,000
280,000
300,000
20,000 $
40,000
60,000
70,000
30,000 $
30,000
60,000
70,000
Which of the 4 alternatives will you choose?
Solution: Using annual equivalence:
CI = 1,200,000(A/P, 4%, 40) + 20,000 = 1,200,000x0.0505 + 20,000 = 80,624 $
CII = 1,500,000(A/P, 4%, 40) + 35,000 = 1,500,000x0.0505 + 20,000 = 110,780 $
CIII = 2,700,000(A/P, 4%, 40) + 50,000 = 2,700,000x0.0505 + 50,000 = 186,404 $
CIV = 3,500,000(A/P, 4%, 40) + 60,000 = 3,500,000x0.0505 + 60,000 = 236,820 $
BI = 200,000+20,000+30,000 = 250,000 $
BII = 190,000+40,000+30,000 = 260,000 $
BIII = 280,000+60,000+60,000 = 400,000 $
BIV = 300,000+70,000+70,000 = 440,000 $
These data give the following table
Alternative
Do Nothing
I 1
II 1,2
III 1,2,3
IV 1,2,3,4
Steps to follow:
B ($)
0
250,000
260,000
400,000
440,000
C ($)
0
80,624
110,780
186,404
236,820
B/C
0
3.1
2.34
2.14
1.85
∆B ($)
∆C ($)
∆B/∆C
250,000
10,000
150,000
40,000
80,624
30,156
105,780
50,416
3.1
0.33
1.42
0.79
1) We arrange the alternatives in ascending order of their initial investment cost
2) We eliminate the alternatives which have B/C < 1.0
3) Any remaining alternatives are acceptable, and therefore should be
compared by ∆B and ∆C. (∆B/∆C).
- We start with the alternatives that has the lowest investment cost: here it is
the “Do Nothing” alternative which seems to be an acceptable alternative.
- Now we take another alternative, which constitutes the minimum investment
cost among the remaining alternatives. Here is the alternative “I”
• Comparing “I” with “Do Nothing” and calculating ∆B, ∆C, and ∆B/∆C:
∆B = 250000 – 0 = 250000
∆C = 80624 – 0 = 80624
∆B/∆C = 250000/80624 = 3.1
• Since ∆B/∆C > 1.0, So alternative “I” is preferred over the alternative “Do
Nothing”. Now the alternative “Do Nothing” is completely eliminated from
the comparison. Now we take another alternative which constitutes the
minimum investment cost among the remaining alternatives.
• The remaining alternatives are II, III, IV
• The alternative with minimum cost is II.
• Comparing II with I (the last one accepted), and calculating ∆B, ∆C and
∆B/∆C.
∆B = 260000 – 250000 = 10000 $
∆C = 110780 – 80624 = 30156 $
∆B/∆C = 10000/30156 = 0.33
• Since ∆B/∆C < 1.0 → alternative II is completely eliminated from the
comparison and is rejected.
• The remaining alternatives are III, IV
• We now take III because it is the alternative which gives the minimum
investment cost. We compare III with I (which is always the last accepted),
by calculating ∆B, ∆C and ∆B/∆C.
∆B = 400000 – 250000 = 150000 $
∆C = 186404 – 80624 = 105780 $
∆B/∆C = 1.42
• Since ∆B/∆C > 1.0, so alternative ‘III’ is acceptable. So alternative ‘I’ is
completely eliminated from the comparison and is rejected.
• The remaining alternatives are IV. We compare IV with III (which is always
the last accepted), by calculating ∆B, ∆C and ∆B/∆C.
∆B = 440000 – 400000 = 40000 $
∆C = 236820 – 186404 = 50416 $
∆B/∆C = 0.76
• Since ∆B/∆C < 1.0, so alternative IV is rejected and accept alternative III.
Conclusion: the best alternative is III, i.e. build 3 dams at locations 1, 2, 3
Other method of Benefit-Cost analysis (B-C):
B-C > 0 → The project is attractive
(∆B - ∆C) X→Y > 0 → Project X is better than Y
Example:
Consider the simplified data in the following table that describe the alternatives for a small
flood-control project. Three feasible options are available to reduce the damages of floods. Each
larger investment of public funds provides greater protection.
Alternative
Equivalent Annual Cost of project
A: No flood control 0
B: Construct levees 40,000 L.E.
C: Small reservoir 120,000 L.E.
D: Large reservoir 160,000 L.E.
(Alternatives are mutually exclusive)
Annual Benefit
0
70,000 L.E.
160,000 L.E.
190,000 L.E
Solution: In this problem, costs and benefits are directly given as equivalent annual values, so no
calculations are to be made.
Alternative
B
x103
C
x103
Table
B/C
B-C
x103
A
0
0
0
0
B
70
40
1.75
30
C
160
120
1.33
40
D
190
160
1.19
30
C is the best according to ∆B/∆C or ∆B - ∆C analysis
∆B
Incremental
∆C
∆B/∆C
∆B-∆C
70
90
30
40
80
40
30
10
-10
1.75
1.125
0.75
Of course we can set other criteria for comparing between projects:
a)
b)
c)
d)
e)
f)
g)
h)
Minimum Investment:
Maximum Benefit (B):
Maximum advantage of benefits over costs (B-C):
Highest benefit-to-cost ratio (B/C):
Largest investment that has a B/C ratio greater than 1.0:
Maximum incremental advantage of Benefit over Cost (∆B-∆C):
Maximum incremental Benefit-to-Cost ratio (∆B/∆C):
Largest investment that has an incremental B/C ratio greater than 1.0:
choose A
choose D
choose C
choose B
choose D
choose B
choose B
choose C
Summary:
Benefit/Cost Ratio Evaluation
(Public Projects)
Benefits (B): advantages expressed in terms of $(L.E.) which happen to the owner.
Dis-benefits (D): disadvantages to the owner.
Cost (C): expenses for construction, operation, maintenance…
B/C for a single project: (All values calculated as AW or PW)
1) 𝐵/𝐶 =
𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠−𝐷𝑖𝑠𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠
=
𝐵−𝐷
> 1.0
𝐶𝑜𝑠𝑡𝑠
𝐶
𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠−𝐷𝑖𝑠𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠−𝑂&𝑀 𝐶𝑜𝑠𝑡𝑠
2) Modified B/C =
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
3) B-C = Benefits – Costs > 0
4) B-D-C = Benefits – Dis-benefits – Costs > 0
=
𝐵−𝐷−𝑂&𝑀
𝐼
> 1.0
(Condition shown for attractiveness)
1
𝐵/𝐶 =
𝑃𝑊(𝐵) − 𝑃𝑊(𝐷)
𝐴𝑊(𝐵) − 𝐴𝑊(𝐷)
=
> 1.0
𝑃𝑊(𝐶)
𝐴𝑊(𝐶)
𝑃𝑊(𝐵)−𝑃𝑊(𝐷)−𝑃𝑊(𝑂&𝑀)
2
Modified B/C =
3
B-C = PW(B) – PW(C) > 0
= AW(B) – AW(C) > 0
4
B-D-C = PW(B) – PW(D) – PW(C) > 0
= AW(B) – AW(D) – AW(C) > 0
𝑃𝑊(𝐶)
=
𝐴𝑊(𝐵)−𝐴𝑊(𝐷)−𝐴𝑊(𝑂&𝑀)
𝐴𝑊(𝐶)
> 1.0
Example: Values are calculated by AW
B = 500,000 $/year
I = 1.5x106 (A/P, 6%, 10) = 203,805 $/year
D = 200,000 $/year
O&M = 50,000 $/year
⸫
𝐵/𝐶 =
Modified B/C =
𝐵 − 𝐷 500,000 − 200,000
=
= 1.18 > 1.0
𝐶
203,805 + 50,000
500,000−200,000−50,000
203,805
= 1.23 > 1.0
B-D-C = 500,000 – 200,000 – (203,805 + 50,000) = 46,195$ > 0
Remark:
These ratios use the interpretation B/C >1.0 to accept
Change in Benefits Change in Costs B/C
Accept/Reject
+100$ (gain)
+200$(cost)
+0.5 <1.0 Reject
+100$ (gain)
+50$(cost)
+2
>1.0 Accept
These ratios use the interpretation B/C <1.0 to accept
Change in Benefits
+100$ (gain)
+100$ (gain)
-100$ (loss)
-100$ (loss)
Change in Costs
-200$(savings)
-50$(savings)
-200$(savings)
-50$(savings)
B/C
-0.5
-2
+0.5
+2
<1.0
<1.0
<1.0
>1.0
Accept/Reject
Accept
Accept
Accept
Reject
Depreciation
-
-
Depreciation means a decrease in worth. Most assets are worth less as they get older.
The objectives for calculating a depreciation cost:
1- To recover capital invested in purchasing production assets by including the
cost of depreciation in operating expenses
2- To maintain a book value (asset value at the end of each year of its
economical life)
Instead of charging the full purchase price of a new asset as a one-time expense,
the outlay is spread over the life of the asset in the accounting records.
Courses of declining value:
1- Physical depreciation:
• Everyday wear and tear of operation
• Accidental physical damage
2- Functional depreciation:
• Demands made on an asset may increase beyond its capacity to produce
• Demand for services may cease to exist (e.g. a product no longer in demand).
3- Technological depreciation:
• Newly developed means of accomplishing a function may make the present
means uneconomical (e.g. new materials, improved safety, better quality and
lower cost from new developments make old designs obsolete.
4- Depletion:
• Depletion is the consumption of an exhaustible natural resource to produce
products or services. (e.g. removal of oil, rock, or minerals from a site
decreases the value of the site).
Depreciation Methods:
-
There are many depreciation methods, we are going to study 3 of them:
1- The straight line method
2- The declining balance method
3- The sum of the years method
-
The three methods are based on time.
Symbols used in the development of the formulae are:
P → Purchase price of the asset (at time zero).
S → Salvage value at the end of asset’s useful life.
N → Useful life of asset
n → Number of years of depreciation or use from time of purchase.
d → Annual charge for depreciation.
Dn → Cumulative depreciation for n years
BVn → Book value shown on accounting records.
1- The Straight Line Method: “SL”
- It is the simplest and most widely used method.
- The annual depreciation is constant
𝑑=
-
𝑃−𝑆
𝑁
The book value is the difference between the purchase price and the product of
the number of years of use times the annual depreciation charge:
𝐵𝑉(𝑒𝑛𝑑 𝑜𝑓 𝑦𝑒𝑎𝑟 𝑛) = 𝑃 − 𝑛
(𝑃−𝑆)
𝑁
Example:
Trucks purchased by a delivery company cost 7000$ each. Past records indicate the
trucks should have a useful life of 5-years. They can be sold for an average of 1000$ each
after 5 years of use. The company currently receives 7% interest on invested funds.
Determine:
a- the depreciation charge during year 1.
b- the depreciation charge during year 2.
c- the accumulated depreciation by the end of year 3.
d- the book value at the end of year 3.
Solution:abcd-
d (end of year 1) = (P-S)/N = (7000 – 1000)/5 = 1200$
d (end of year 2) = (P-S)/N = (7000 – 1000)/5 = 1200$
Accumulated depreciation by the end of year 3 = 1200 x 3 = 3600$
BV (at the end of year 3) = 7000 – 3 x 1200 = 3400$
3
Or 𝐵𝑉(3) = 7000 − 5 (7000 − 1000) = 3400$
2- The declining Balance Method: “DB”
- The depreciation allowed at the end of each year is a constant fraction (k) of the
unrecovered investment at the end of the previous year. (k→ depreciation rate)
-
d1 = kP0
d1 → depreciation of year 1
P0→ unrecovered capital at the end of previous year (year 0)
-
d2 = kP1
-
dn = kPn-1
d2 → depreciation of year 2
P1→ unrecovered capital at the end of previous year 1
P1 = P0 – d1
P2 = P1 – d2
dn → depreciation of year n where 1 ≤ n ≤ N
Pn-1→ unrecovered capital at the end of previous year n-1
Also we have:
P1 = BV1 = P -d1 = P – KP = P(1-K) Book Value at the end of year 1
P2 = BV2 = P1 -d2 = P1 – KP1 = (P-KP) – K(P-KP) = P-KP-KP+K2P = P(1 -2k +K2)
= P(1-K)2
Pn = BVn = P(1-K)n
PN = P(1-K)N > 0
1
But PN = S = salvage value
S = P(1-K)N
2
From 1 and 2 we conclude that
1
1
𝑆 𝑁
𝐵𝑉𝑛 𝑛
)
𝐾 =1− ( ) = 1− (
𝑃
𝑃
Advantages of the declining – Balance Method:
1- Easy
2- It permits a rapid recovery of the capital investment in the first years.
Example:
P = 120 L.E
S=20 L.E
N = 10 years
1
1
𝑆 𝑁
20 10
) = 0.1641
𝐾 =1− ( ) = 1− (
𝑃
120
Book Value at the end of each year:
P1 = BV1 = P(1-K)
= 120(1-0.1641) = 100.308 L.E
P2 = BV2 = P(1-K)2
= 83.847 L.E
P3 = BV3 = P(1-K)3
= 70.088 L.E
P4 = BV4 = P(1-K)4
= 58.587 L.E
P5 = BV5 = P(1-K)5
= 48.97 L.E
P6 = BV6 = P(1-K)6
= 40.94 L.E
P7 = BV7 = P(1-K)7
= 34.22 L.E
P8 = BV8 = P(1-K)8
= 28.60 L.E
P9 = BV9 = P(1-K)9
= 23.91 L.E
P10 = BV10 = P(1-K)10 = 19.986 L.E
Annual Depreciation:
d1 = KP = 19.692 L.E.
KP
d2 = KP1 = 16.46 L.E.
K(BV1) = KP(1-K)
d3 = KP2 = 13.76 L.E.
K(BV2) = KP(1-K)2
d4 = KP3 = 11.5 L.E.
K(BV3) = KP(1-K)3
d5 = KP4 = 9.614 L.E.
K(BV4) = KP(1-K)4
d6 = KP5 = 8.036 L.E.
K(BV5) = KP(1-K)5
d7 = KP6 = 6.718 L.E.
K(BV6) = KP(1-K)6
d8 = KP7 = 5.615 L.E.
K(BV7) = KP(1-K)7
d9 = KP8 = 4.69 L.E.
K(BV8) = KP(1-K)8
d10 = KP9 = 3.92 L.E.
K(BV9) = KP(1-K)9
dn = KP(1-K)n-1
Cumulative depreciation at the end of year 6:
Dn = P – BVn = D6 = P – BV6 = 120 – 40.94 = 79.06 L.E.
Book Value at the end of year 6:
BV6 = P(1-K)6 = 40.94 L.E.
Summary of the formulae:
K=1−
1
1
S N
(P)
BV n
( P n)
=1−
S = P(1-K)n
dn = KP(1-K)n-1
(Annual charge)
Pn = BVn = P(1-K)n
(Book Value)
Dn = P – BVn
(Cumulative depreciation for year n)
3- Sum of the Years Method:”SYD”
- Also known as
1- Digits Method
2- SYD Method (Sum of the years’ digits method)
-
-
In this method, “The depreciation allowed at the end of each year is a variable
fraction (K) of a constant amount (P-S)”
The depreciation factor “K” for a certain year corresponds to the number of the
inverse series of the numbers for this year divided by the sum of the years
number.
The sum of the years Number of the economical life N is calculated from
N(N+1)/2
Example:
If N = 5 years (1 2 3 4 5)
therefore sum of the years Number of (N = 5) is = 1+2+3+4+5=15
or using the formula: 5(5+1)/2 = 15
To calculate K:
K1 = depreciation factor of the first year = 5/15
K2 = depreciation factor of the 2nd year = 4/15
K3 = depreciation factor of the 3rd year = 3/15
K4 = depreciation factor of the 4th year = 2/15
K5 = depreciation factor of the 5th year = 1/15
𝑁−𝑛+1
In general : 𝐾𝑛 = 2 [𝑁(𝑁+1)]
5−5+1
For example 𝐾5 = 2 [5(5+1)] =
1
15
Summary of the Sum-of-the-Years Method formulae
dn = Kn (P-S)
𝑁−𝑛+1
𝑑𝑛 = (𝑃 − 𝑆) 𝑥 2 [𝑁(𝑁+1)]
𝐷𝑛 = ∑𝑛1 𝑑𝑛
BVn = P - Dn
1 ≤𝑛 ≤𝑁
Example:
P = 120 L.E.
S = 20 L.E.
N = 10 years
Sum of the years Number = 1+2+3+4+5+6+7+8+9+10=55
or 10(11)/2 = 55
Annual Depreciation:
d1 = (120 – 20) x 10/55 = 18.18 L.E.
d2 = (120 – 20) x 9/55 = 16.36 L.E.
d3 = (120 – 20) x 8/55 = 14.54 L.E.
d4 = (120 – 20) x 7/55 = 12.73 L.E.
d5 = (120 – 20) x 6/55 = 10.91 L.E.
d6 = (120 – 20) x 5/55 = 9.09 L.E.
d7 = (120 – 20) x 4/55 = 7.27 L.E.
d8 = (120 – 20) x 3/55 = 5.45 L.E.
d9 = (120 – 20) x 2/55 = 3.64 L.E.
d10 = (120 – 20) x 1/55 = 1.818 L.E.
Cumulative depreciation at the end of year 6:
𝐷𝑛 = ∑𝑛1 𝑑𝑛 = ∑61 𝑑𝑛 = 18.18 + 16.36 + 14.54 + 12.73 + 10.91 + 9.09 = 81.81 𝐿. 𝐸
BV6:
BV6 = P – D6 = 120 – 81.81 = 38.19 L.E