Internal Rate of Return (IRR) - Industrial Engineering and

Report
Applications of MoneyTime Relationships
MARR by PW Method
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All cash inflows and outflows are discounted to
the present point in time at an interest rate that
is generally the MARR.
PW(i) = 0; i = MARR
If PW(i) > 0 => The investment is “good”.
If the return rate of an investment is greater
than MARR then the investment is
“profitable”.
MARR

MARR(hurdle rate) - maximize the economic
well-being of a company.
Example 1
An investment of $10,000 can be made in a
project that will produce a uniform annual
revenue of $5,310 for five years and then
have a market (salvage) value of $2,000.
Annual expenses will be $3,000 each year.
The company is willing to accept any
project that will earn 10% per year or
more, before income taxes, on all invested
capital. Show whether this is a desirable
investment by using the PW method.
 PW(10%) = -10,000 + 5,310(P/A, 0.1, 5) +
2,000 (P/F, 0.1, 5) - 3,000(P/A, 0.1, 5) = 0
=> This investment is marginally acceptable.

Example 2

A piece of new equipment has been proposed by
engineers to increase the productivity of a certain
manual welding operation. The investment cost is
$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 to $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 this proposal a sound one?
PW(20%) = $8,000(P/A,20%,5) +$5,000(P/F,20%,5)
- $25,000 = $934.29
Because PW(20%) > 0 => “Investment justifiable”
Bond Value Determination
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Z = face, or par, value
C = redemption or disposal price (usually
equal to Z)
r = bond rate (nominal interest) per interest
period
N = number of periods before redemption
i = bond yield rate per period
VN = value (price) of the bond N interest
periods prior to redemption
VN = C(P/F, i %, N) + rZ(P/A, i %, N)
Example 1
Find the current price (PW) of a 10-year bond
paying 6% per year (payable semiannually)
that is redeemable at par value, if bought by a
purchaser to yield 10% per year. The face
value of the bond is $1,000.
Sol: i6mon = (1.1)1/2 - 1 = 0.049
VN = 1,000 (P/F, 4.9%, 20) +
1,000 (0.03) (P/A, 4.9%, 20)
= 384.10 + 377.06 = 761.16

Example 2
A bond with a face value of $5,000 pays interest of
8% per year. This bond will be redeemed at par
value at the end of its 20-year life, and the first
interest payment is due one year from now.
(a) How much should be paid now for this bond in
order to receive a yield of 10% per year on the
investment?
(b) If this bond is purchased now for $4,600, what
annual yield would the buyer receive?
(a) VN= 5,000(P/F,10%,20) + 5,000(0.08)(P/A,10%,20)
= 743 + 3,405.44 = $4,148.44
(b) 4,600 = 5,000(P/F, i, 20) + 5,000(0.08)(P/A, i, 20)
i = 8.9% per year.
Example 3

A bond that matures in 8 years has a face value of
$10,000. The bond stipulates a fixed nominal interest rate
of 8% per year, but interest payments are made to the
bondholder every three months. A prospective buyer of
this bond would like to earn 10% nominal interest per
year on his or her investment because interest rates in the
economy have risen since the bond was issued. How
much should this buyer be willing to pay for the bond?
i3mon = (1.1)1/4 - 1 = 0.024 = 2.4%
VN = 10,000(0.02) (P/A, 2.4%, 32) + 10,000(P/F, 2.4%, 32)
= 4,369.84 + 4,537.71 = 8,907.55
Thus, the buyer should pay no more than $8,907.55
when 10% nominal interest per year is desired
MARR by FW Method
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All cash inflows and outflows are calculated to
the future point in time at an interest rate that is
generally the MARR.
FW(i) = 0; i = MARR
If FW(i) > 0 => The investment is “good”.
Example

A piece of new equipment has been proposed by
engineers to increase the productivity of a certain
manual welding operation. The investment cost is
$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 to $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 this proposal a sound one?
FW(20%) = $8,000(F/A,20%,5) + $5,000
- $25,000(F/P,0.2,5) = $2324.8 > 0
Because FW(20%) > 0, i.e., “Investment justifiable”
The Annual Worth Method (AW)
Capital Recovery (CR) - the minimum annual
profit to recover the initial investment
 The capital recovery (CR) amount for a project
is the equivalent uniform annual cost of the
capital invested. It is an annual amount that
covers the following two items:
1. Loss in value of the asset
2. Interest on invested capital (i.e., at the MARR)
CR(i) = I(A/P, i, n) - S(A/F, i, n)
I = initial investment
S = market (salvage) value at the end of project
n = project duration
The Annual Worth Method (AW)
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AW(i) = R - E - CR(i)
R - Annual Revenue
E - Annual Expense
If AW(i) > 0 => The project is “good”
If i = MARR then
PW = FW = AW = 0
Consistency among PW, FW, and AW methods
Example 1

A piece of new equipment has been proposed by
engineers to increase the productivity of a certain
manual welding operation. The investment cost is
$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 to $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 this proposal a sound one?
AW = R - E - CR(i), R - E = 8,000
AW(20%) = 8,000 - [25,000 (A/P,20%,5) - 5,000
(A/F,20%,5)] = 312.4 > 0 => The project is good.
Example

An investment company is considering building a 25unit apartment complex in a growing town. Because of
the long-term growth potential of the town, it is felt that
the company could average 90% of full occupancy for
the complex each year. If the following items are
reasonably accurate estimates, what is the minimum
monthly rent that should be charged if a 12% MARR
(per year) is desired?

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Land investment cost
$ 50,000
Building investment cost
$225,000
Study period, N
20 years
Rent per unit per month
R
Upkeep expense per unit per month $35
Property taxes and insurance per year 10% of total initial
investment
Initial investment cost = 50,000 + 225,000 = 275,000
Taxes and insurance/year = 0.1 (275,000) = 27,500
Upkeep/year = 35(12 x 25)(0.9) = 9,450
CR cost/year = 275,000(A/P,12%,20) - 50,000(A/F,12%,20)
= $36,123
Revenue = Rx12x25x0.9 = 270R
Equivalent AW (of costs) = R - E - CR(i)
= 270R - (9450 +27,500) - 36,123 = 0 => R = 270.64
this is the minimum annual rental required equals 73,073
and with annual compounding (M = 1) the monthly rental
amount
If M = 12, 73,073 = 0.9R(F/A, 0.01, 12)(25) = 260
Internal Rate of Return
(IRR)
IRR Calculation
N
PW =  Rk(P/F, i'%, k) - Ek(P/F, i'%, k) = 0
k=0
where Rk = net revenues or savings for the kth
year

0
Ek = net expenditures including any
investment costs for the kth year
N = project life (or study period)
OR

N
FW =  Rk(P/F, i'%, N-k) - Ek(P/F, i'%, N-k) = 0
k=0
Negative IRR

The IRR will be positive if satisfies two
conditions:
(1) both receipts and expenses are present in
the cash flow pattern and
(2) the sum of receipts exceeds the sum of all
cash outflows.

Be sure to check both of these conditions in
order to avoid the unnecessary work
involved with finding that the IRR is
negative
Example

A capital investment of $10,000 can be made in a
project that will produce a uniform annual
revenue of $5,310 for five years and then have a
salvage value of $2,000. Annual expenses will be
$3,000. The company is willing to accept any
project that will earn at least 10% per year, before
income taxes, on all invested capital. Determine
whether it is acceptable by using the IRR method.
SOLUTION
Sum of positive cash flows ($13,550) exceeds the
sum of negative cash flows ($10,000) => IRR >0.
PW= 0 = - 10,000 + (5,310 -3,000)(P/A,i'%,5)
+ 2,000(P/F, i'%, 5)
At i' = 5%: PW = - 10,000 + 2,310(4.3295)
+ 2,000(0.7835) = 1,568
At i' = 15%: PW = -10,000 + 2,310(3.3522)
+ 2,000(0.4972) = - 1,262
1568 - 0
= 0.05 - i’
=> i’ = 10.5%
1568 - (-1262)
0.05 - 0.15
Due to approximation error, the true i’ = 10%
For more accuracy: PW(10.5%) < 0 ,PW (8.5%) >0 =>
Linear Interpolate 10.5% & 8.5%
Example

A piece of new equipment has been proposed by
engineers to increase the productivity of a certain
manual welding operation. The investment cost is
$25,000, and the equipment will have a salvage
value of $5,000 at the end of its expected life of five
years. Increased productivity attributable to the
equipment will amount to $8,000 per year after
extra operating costs have been subtracted from the
value of the additional production. Evaluate the IRR
of the proposed equipment. Is the investment a good
one? Recall that the MARR is 20%.
Solution
PW = 8,000 (P/A, i',5) + 5,000 (P/F, i', 5) - 25,000
PW(MARR) = PW(0.2) = 934.3 > 0
=> IRR > MARR => This investment is a good one.
To calculate IRR
PW(0.25) = -1847.1
Linear Interpolation
943 - 0
= 0.2 - i’
=> i’ = 21.58%
943 - (-18147.1)
0.2 - 0.25
i’ > MARR = 20% => This equipment is
economically attractive
Example
In 1915 Albert Epstein allegedly borrowed $7,000 from a
bank on the condition that he would repay 7% of the
loan every three months, until a total of 50 payments
had been made. At the time of the 50th payment, the
$7,000 loan would be completely repaid. Albert
computed his interest rate to be
[0.07(7,000) x 4] / 7,000 = 0.28 (28%).
(a)What true effective annual interest rate did Albert pay?
(b) What, if anything, was wrong with his calculation?
Example
In 1915 Albert Epstein allegedly borrowed $7,000 from a
bank on the condition that he would repay 7% of the loan
every three months, until a total of 50 payments had been
made. At the time of the 50th payment, the $7,000 loan
would be completely repaid. Albert computed his interest
rate to be
[0.07(7,000) x 4] / 7,000 = 0.28 (28%).
(a)What true effective annual interest rate did Albert pay?
(b) What, if anything, was wrong with his calculation?
PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter
0.07(P/A, i'%, 50) = 1
Example
In 1915 Albert Epstein allegedly borrowed $7,000 from a
bank on the condition that he would repay 7% of the
loan every three months, until a total of 50 payments
had been made. At the time of the 50th payment, the
$7,000 loan would be completely repaid. Albert
computed his interest rate to be
[0.07(7,000) x 4] / 7,000 = 0.28 (28%).
(a)What true effective annual interest rate did Albert pay?
(b) What, if anything, was wrong with his calculation?
PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter
0.07(P/A, i'%, 50) = 1 => (P/A, i’%, 50) = 14.28
(P/A, 6%, 50) = 15.77 (P/A,7%,50) = 13.8
15.77 - 14.28 = 0.06 - i’
=> i’ = 6.73%
15.77 - (13.8) 0.06 - 0.07
Example
In 1915 Albert Epstein allegedly borrowed $7,000 from a
bank on the condition that he would repay 7% of the
loan every three months, until a total of 50 payments
had been made. At the time of the 50th payment, the
$7,000 loan would be completely repaid. Albert
computed his interest rate to be
[0.07(7,000) x 4] / 7,000 = 0.28 (28%).
(a)What true effective annual interest rate did Albert pay?
(b) What, if anything, was wrong with his calculation?
PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter
0.07(P/A, i'%, 50) = 1 => (P/A, i’%, 50) = 14.28
(P/A, 6%, 50) = 15.77 (P/A,7%,50) = 13.8
15.77 - 14.28 = 0.06 - i’
=> i’ = 6.73%
15.77 - (13.8) 0.06 - 0.07
4
Effective Rate = (1 + 0.673) - 1 = 30%
Example

A finance company advertises a "bargain 6%
plan" for financing the purchase of automobiles.
To the amount of the loan being financed, 6% is
added for each year money is owed. This total is
then divided by the number of months over which
the payments are to be made, and the result is the
amount of the monthly payments. For example, a
woman purchases a $10,000 automobile under
this plan and makes an initial cash payment of
$2,500. She wishes to pay the balance in 24
monthly payments
Example

A finance company advertises a "bargain 6% plan" for
financing the purchase of automobiles. To the amount of
the loan being financed, 6% is added for each year money
is owed. This total is then divided by the number of
months over which the payments are to be made, and the
result is the amount of the monthly payments. For
example, a woman purchases a $10,000 automobile under
this plan and makes an initial cash payment of $2,500. She
wishes to pay the balance in 24 monthly payments
Purchase price = $10,000 - 2,500 = 7,500 = P0
6% finance charge = 0.06 x 2 years x $7,500 = 900
=>Total to be paid = 8,400
Monthly payments = $8,400/24 = $ 350
What effective annual rate of interest does she actually pay?
In Reality …..
P0 = A(P/A, i'%, N) => 7,500 = 350(P/A, i'%,24)
(P/A, i'%, 24) = 7,500 / 350 = 21.43
(P/A, 0.75%, 24) = 21.89 and
(P/A, 1%, 24) = 21.24
Linear Interpolation
21.43 - 21.24 = i’ - 1%
=> i’ = 0.93%
21.89 - 21.24 0.75% - 1%
12
Effective Rate = (1 + 0.0093) -1 = 12%
Difficulties with IRR Method
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The PW, AW, and FW methods assume that net receipts
less expenses (positive recovered funds) each time
period are reinvested at the MARR during the study
period, N.
IRR method assuming that recovered funds, if not
consumed in each time period, are reinvested at IRR
(i'%) rather than at the MARR. This assumption may
not mirror reality in some problems, thus making IRR
an unacceptable method for analyzing engineering
alternatives.
Multiple IRR
External Rate of Return (ERR)

The reinvestment assumption of the IRR method noted
previously may not be valid in an engineering economy
study. For instance, if a firm's MARR is 20% per year
and the IRR for a project is 42.4%, it may not be possible
for the firm to reinvest net cash proceeds from the project
at much more than 20%.
N
 Ek(P/F,  %, k)(F/P, i'%, N) =
k=0
N
 Rk(F/P,  %, N - k)
k=0
where Rk = excess of receipts over expenses in
period k
Ek = excess of expenditures over receipts in
period k
N = project life or number of periods for the study
 = external reinvestment rate per period
Example

The equipment investment cost is $25,000, and the
equipment will have a salvage value of $5,000 at the
end of its expected life of five years. The equipment
will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the
additional production. Suppose that
 = MARR = 20%

What is the alternative's external rate of return, and is
the alternative acceptable?
Example

The equipment investment cost is $25,000, and the
equipment will have a salvage value of $5,000 at the
end of its expected life of five years. The equipment
will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the
additional production. Suppose that
 = MARR = 20%
What is the alternative's external rate of return, and is
the alternative acceptable?
25,000(F/P, i'%,5) = 8,000(F/A,20%,5) + 5,000

(F/P,i'%, 5) = 2.5813, i' = 20.88%
Because i' > MARR, the alternative is barely justified.
Example

When  = 15% and MARR = 20%, determine whether the
project whose total cash flow diagram appear below is
acceptable. Notice in this example that the use of an  is
different from MARR. This might occur if, for some
reason, part or all of the funds related to a project are
"handled" outside the firm's normal capital structure.
Eo = $10,000 (k = 0), E1 = $5,000 (k = 1)
Rk = $5,000 for k = 2,3, . . .,6
| - 10,000 - 5,000(P/F,15%, 1)|(F/P, i'%, 6)
= 5,000(F/A,15%,5); i'% = 15.3%
Advantages of ERR over IRR
It can usually be solved directly rather than
by trial and error.
 ERR is unique.

The Payback Period Method
Suppose that a project where all capital
investment occurs at time 0.
The Payback Period Method
Suppose that a project where all capital investment
occurs at time 0.
 Simple Payback period ( N ) is the smallest
integer such that
N
 ( Rk - Ek )  I (Initial Investment)
k=0
The Payback Period Method
Suppose that a project where all capital investment
occurs at time 0.
 Simple Payback period ( N ) is the smallest
integer such that
N
(
R
E
)

I
(Initial
Investment)
k
k

k=0

Discounted Payback period ( N ) is the smallest
integer such that
N
 ( Rk - Ek )(P/F, i, k)  I ; i = MARR
k=0
Example

The equipment investment cost is $25,000, and the
equipment will have a salvage value of $5,000 at the
end of its expected life of five years. The equipment
will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the
additional production.
Example

The equipment investment cost is $25,000, and the
equipment will have a salvage value of $5,000 at the
end of its expected life of five years. The equipment
will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the
additional production.
Example

4

The equipment investment cost is $25,000, and the
equipment will have a salvage value of $5,000 at the
end of its expected life of five years. The equipment
will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the
additional production.
(8,000)  25,000 => Simple Payback Period = 4
k=0
Example

4

The equipment investment cost is $25,000, and the
equipment will have a salvage value of $5,000 at the
end of its expected life of five years. The equipment
will amount to $8,000 per year after extra operating
costs have been subtracted from the value of the
additional production.
(8,000)  25,000 => Simple Payback Period = 4
k=0
5
 8,000(P/A,20%,5) + 5,000(P/F,20%,5) > 25,000
k=0
=> Discounted Payback Period = 5
Homework
Page 177
 # 2, 3, 5, 9, 12, 13, 19, 24, 25, 30,
33, 37, 41, 49, 51
 Due Date: Oct. 27, 1998


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