Engineering Economy
Engineering Economy

Topic 7 – Annuity

Topic 8 – Gradient

Topic 9 – Application of Money Time Relationship

Topic 9.1 – Methods to Evaluate Profitability

Topic 9.2 – Methods to Evaluate Liquidity
Engineering Economy
MINIMUM ATTRACTIVE RATE OF
RETURN ( MARR )
Minimum Attractive Rate of Return (MARR) is
usually organization-specific and determined based
on the following:
1. Cost of money available for investment.
2. Number
of good projects
available
for
investment.
3. Risks involved in investment opportunities.
Engineering Economy
How to Use MARR
Use it as an interest rate to convert cash flows
into equivalent worth at some point in time. The
proposed problem solution (project or alternative)
is profitable if it generates sufficient cash flow to
recover the initial investment and earn an interest
rate that is at least as high as MARR.
Engineering Economy
Methods to Evaluate Profitability
1. Present Worth (PW)
2. Future Worth (FW)
3. Annual Worth (AW)
4. Internal Rate of Return (IRR)
5. External Rate of Return (ERR)
Engineering Economy
Methods to Evaluate Liquidity
1. Simple Payback Period
2. Discounted Payback Period
Engineering Economy
Present Worth (PW) Method
Based on concept of equivalent worth of all cash
flows relative to the present as a base.
All cash inflows and outflows discounted to present
at interest -- generally MARR.
PW is a measure of how much money
afforded for investment in excess of cost.
can be
PW is positive if dollar amount received for
investment exceeds minimum required by investors.
Engineering Economy
Present Worth (PW) Method
Discount future amounts to the present by using the
interest rate over the appropriate study period
𝑘
PW = 𝑁
0 𝐹𝑘 ( 1 + 𝑖) −
i = effective interest rate, or MARR per com. period
k = index for each compounding period
Fk = future cash flow at the end of period k
N = number of compounding periods in study period
interest rate is assumed constant through project
The higher the interest rate and further into future a
cash flow occurs, the lower its PWa
Engineering Economy
Present Worth (PW) Method
1.
Compute the present equivalent of the
estimated cash flows using the MARR as the
interest rate.
2.
If PW(MARR) ≥ 0, then the project is profitable.
If PW(MARR) < 0, then the project is not
profitable.
Engineering Economy
Sample Problem - Present Worth
Consider a project that has the following cash flows
over a study period of 5 years:
Initial investment: $100,000
Annual revenues: $40,000
Annual expenses: $5,000
Salvage value: $20,000
MARR: 20%.
PW(20%)= -100,000 + (40,000-5,000)(P/A,20%,5) +
20,000(P/F,20%,5) =$12,709.
Is this a profitable project? Yes! Because: We recovered $100,000
(investment). We earned an interest rate of 20% which was desired.
We even made a profit that has a present equivalent value of
$12,709.
Engineering Economy
Future Worth (FW) Method
FW is based on the equivalent worth of all cash inflows
and outflows at the end of the planning horizon at an
interest rate that is generally MARR.
The FW of a project is equivalent to PW
FW = PW ( F / P, i%, N )
If FW > 0, it is economically justified
𝑁
FW ( i % ) = FW = 𝑁
0 𝐹𝑘 ( 1 + 𝑖)
Engineering Economy
Future Worth (FW) Method
1.
Compute the future equivalent of the estimated
cash flows using the MARR as the interest rate.
2.
If FW (MARR) ≥ 0, then the project is profitable.
If FW (MARR) < 0, then the project is not
profitable.
Engineering Economy
Sample Problem - Future Worth
Consider a project that has the following cash flows
over a study period of 5 years:
Initial investment: $100,000
Annual revenues: $40,000
Annual expenses: $5,000
Salvage value: $20,000
MARR: 20%.
FW(20%)= -100,000(F/P,20%,5)+(40,000-5,000)(F/A,20%,5)
+20,000 = $31,624.
Is this a profitable project? Yes! Because FW ≥ 0: We earned an
interest rate of 20% which was desired. We even made a profit
that has a future equivalent value of $31,624.
Engineering Economy
Annual Worth (AW) Method
◘ AW is an equal annual series of dollar amounts,
over a stated period ( N ), equivalent to the cash
inflows and outflows at interest rate that is
generally MARR
◘ AW is annual equivalent revenues ( R ) minus
annual equivalent expenses ( E ), less the annual
equivalent capital recovery (CR)
AW ( i % ) = R - E - CR ( i % )
AW = PW ( A / P, i %, N )
AW = FW ( A / F, i %, N )
◘ If AW > 0, project is economically attractive
◘ AW = 0 : annual return = MARR earned
Engineering Economy
Capital Recovery (CR)
CR is the equivalent uniform annual cost of the
capital invested
CR is an annual amount that covers:
- Loss in value of the asset
- Interest on invested capital ( i.e., at the MARR )
CR ( i % ) = I ( A / P, i %, N ) - S ( A / F, i %, N )
I = initial investment for the project
S = salvage ( market ) value at the end of the study
period
N = project study period
Engineering Economy
Capital Recovery (CR)
CR is also calculated by adding sinking fund amount
(i.e., deposit) to interest on original investment
CR ( i % ) = ( I - S ) ( A / F, i %, N ) + I ( i % )
CR is also calculated by adding the equivalent annual
cost of the uniform loss in value of the investment to
the interest on the salvage value
CR ( i % ) = ( I - S ) ( A / P, i %, N ) + S ( i % )
Engineering Economy
Sample Problem - Annual Worth
Consider a project that has the following cash flows
over a study period of 5 years:
Initial investment: $100,000
Annual revenues: $40,000
Annual expenses: $5,000
Salvage value: $20,000
MARR: 20%.
AW(i%)= R – E - CR(i%)
AW(20%) = 40,000 - 5,000 - 30,752 = $4,249.62
Since AW(20%)≥ 0, the project is profitable.
Is this a profitable project? Yes! Because AW ≥ 0: We earned an
interest rate of 20% which was desired. We even made a profit
that has an annual equivalent value of $4,249.62.
Engineering Economy
Practice Problem
Uncle Tom’s ranch is now for sale for $30,000.
Annual property taxes, maintenance and supplies,
and so on are estimated to continue to be $3,000
per year. Revenues from the ranch are expected to
be $10,000 next year then to decline by $400 per
year thereafter through the 10th year. If you bought
the ranch, you would plan to keep it for only 5 years
and at that time to sell it for the value of the land ,
which is $15,000. If your desires MARR is 12%,
should you become a rancher? Use PW, AW and
FW
Engineering Economy
Uncle Tom’s ranch is now for sale for $30,000. Annual property taxes,
maintenance and supplies, and so on are estimated to continue to be
$3,000 per year. Revenues from the ranch are expected to be
$10,000 next year then to decline by $400 per year thereafter through
the 10th year. If you bought the ranch, you would plan to keep it for
only 5 years and at that time to sell it for the value of the land , which
is$15,000. If your desires MARR is 12%, should you become a
rancher? Use PW,AW and FW
Engineering Economy
Practice Problem
A piece of new equipment has been proposed by
engineers to increase the productivity of a certain
manual welding operation. The investment cost
$25,000, and the equipment will have a market value
of $5,000 at the end of a study period of five years.
Increased productivity attributable to the equipment
will amount $8,000 per year after extra operating
costs have been subtracted from the revenue
generated by the additional production. If the firm’s
MARR is 20% per year, is the proposal a sound one?
Use the (1) PW (2) FW and (3) AW Method. Draw the
cash flow diagram.
Engineering Economy
The investment cost $25,000, and the equipment will have a
market value of $5,000 at the end of a study period of five years.
Increased productivity attributable to the equipment will amount
$8,000 per year after extra operating costs have been subtracted
from the revenue generated by the additional production. If the
firm’s MARR is 20% per year, is the proposal a sound one? Use
the (1) PW (2) FW and (3) AW Method.
Engineering Economy
Internal Rate of Return (IRR)
IRR solves for the interest rate that equates the
equivalent worth of an alternative’s cash inflows
(receipts or savings) to the equivalent worth of cash
outflows (expenditures)
Also referred to as:
• investor’s method
• discounted cash flow method
• profitability index
IRR is positive for a single alternative only if:
• both receipts and expenses are present in the cash
flow pattern
• the sum of receipts exceeds sum of cash outflows
Engineering Economy
Internal Rate of Return (IRR)
The IRR method assumes recovered funds, if not
consumed each time period, are reinvested at i ‘ %,
rather than at MARR
The computation of IRR may be unmanageable
Multiple IRR’s may be calculated for the same
problem
The IRR method must be carefully applied and
interpreted in the analysis of two or more alternatives,
where only one is acceptable
Engineering Economy
Internal Rate of Return (IRR)
IRR is i’ %, using the following PW formula:
 R k ( P / F, i’ %, k ) =  E k ( P / F, i’ %, k )
R k = net revenues or savings for the kth year
E k = net expenditures including investment costs for the kth
year
N = project life ( or study period )
If i’ > MARR, the alternative is acceptable
To compute IRR for alternative, set net PW = 0
PW =  R k ( P / F, i’ %, k ) -  E k ( P / F, i’ %, k ) = 0
i’ is calculated on the beginning-of-year unrecovered investment
through the life of a project
Engineering Economy
Sample Problem - IRR
Consider a project that has the following cash flows
over a study period of 5 years:
Initial investment: $100,000
Annual revenues: $40,000
Annual expenses: $5,000
Salvage value: $20,000
MARR: 20%.
Solution : We will use trial-and-error and linear interpolation.
PW (il %) = -100, PW (il %) = -100, 000 + (40, 000 - 5, 000)(P/A, il%, 5)
+ 20, 000(P/F, il%, 5) = 0.
Engineering Economy
Cont
Start trial-and-error
PW(20%)=12,709. Therefore, iI > 20%. (Remember
the inverse relation between PW and i)
PW(25%)=679.5. Therefore, iI > 25%.
PW(30%)=-9,368.48. Therefore, iI < 30%.
Use linear interpolation to find iI: 25% < iI < 30%.
i%
PW
25%
679.50
i’
0
30%
-9368.48
i’ = 25.31%
Calcu-shift solve
i’ = 25.38%
MARR(20%), therefore the project is profitable
If i’ > MARR, the alternative is acceptable
Engineering Economy
External Rate of Return (ERR)
ERR directly takes into account the interest rate (  )
external to a project at which net cash flows
generated over the project life can be reinvested (or
borrowed ).
If the external reinvestment rate, usually the firm’s
MARR, equals the IRR, then ERR method produces
same results as IRR method
Engineering Economy
External Rate of Return (ERR)
1. All net cash outflows are discounted to the present
(time 0) at  % per compounding period.
2. All net cash inflows are compounded to period N
at  %.
3. ERR -- the equivalence between the discounted
cash inflows and cash outflows -- is determined.
The absolute value of the present equivalent worth
of the net cash outflows at  % is used in step 3.
A project is acceptable when i ‘ % of the ERR
method is greater than or equal to the firm’s
MARR
Engineering Economy
Advantages of ERR over IRR
The
1 ERR can be solved directly, without needing to
resort
2 to trial and error. The ERR is not subject to
the possibility of multiple rates of return.
If the external investment rate (E) happens to equal
the project’s IRR, then the ERR method produces
results identical to those of the IRR method.
Engineering Economy
Sample Problem - ERR
Consider a project that has the following cash flows
over a study period of 5 years:
Initial investment: $100,000
Annual revenues: $40,000
Annual expenses: $5,000
Salvage value: $20,000
 = MARR: 20%.
100, 000(F/P, i’%, 5) = 35, 000(F/A, 20%, 5) + 20, 000
(F/P, i’%, 5) = 2.8046
(1 + i’)5 = 2.8046 ⇒ i’ = 22.9%.
Therefor the project is profitable.
Engineering Economy
Sample Problem - ERR
For the cash flows given below, find the ERR
when the external reinvestment rate is ε = 12%
(equal to the MARR).
Year
Cash Flow
0
1
2
3
4
-$15,000
-$7,000
$10,000
$10,000
$10,000
Expenses
Revenue
Solving, we find
Engineering Economy
Practice Problem
a. Determine the IRR of the project. Is it acceptable
b. What is the ERR for this project? Assume that  = 15% per
year.
c. What is the simple payback period for this proposal?
Proposal A
Investment cost
Expected life
Market (salvage) value a
Annual receipts
Annual expense
Engineering Economy
$10,000
5 years
–$1,000
$8,000
$4,000
Practice Problem
A project is estimated to cost $100,000, last 8 years,
and have a salvage value of $10,000. The annual
gross income is expected to average $24,000 and
annual expenses excluding depreciation is $6,000. If
capital is earning 10% before income taxes,
determine if this is desirable investment using PW,
AW and IRR
Engineering Economy
Payback Period
The simple payback period is the number of years
required for cash inflows to just equal cash outflows.
It is a measure of liquidity rather than a measure of
profitability.
Engineering Economy
Payback Period
Payback is simple to calculate.
The payback period is the smallest value of θ (θ ≤ N)
for which the relationship below is satisfied.
For discounted payback future cash flows are
discounted back to the present, so the relationship
to satisfy becomes
Engineering Economy
Problems with the payback period
method.
It doesn’t reflect any cash flows occurring after θ, or
θ'.
It doesn’t indicate anything about project desirability
except the speed with which the initial investment is
recovered.
Recommendation: use the payback period only as
supplemental information in conjunction with one or
more of the other methods in this chapter.
Engineering Economy
Sample – Payback Period
Finding the simple and
discounted
payback
period for a set of cash
flows.
The cumulative cash
flows in the table were
calculated using the
formulas for simple and
discounted payback.
From the calculations θ
= 4 years and θ' = 5
years.
Engineering Economy
End of
Year
Net Cash
Flow
Cumulative
PW at 0%
Cumulative
PW at 6%
0
-$42,000
-$42,000
-$42,000
1
$12,000
-$30,000
-$30,679
2
$11,000
-$19,000
-$20,889
3
$10,000
-$9,000
-$12,493
4
$10,000
$1,000
-$4,572
5
$9,000
$2,153