Chapter 7 Rate of Return Analysis 1. Chapter Contents Internal Rate of Return Rate of Return...
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Chapter 7 Rate of Return Analysis 1
Chapter 7 Rate of Return Analysis 1. Chapter Contents Internal Rate of Return Rate of Return Calculations Plot of NPW versus interest rate i Fees or Discounts
Chapter Contents Internal Rate of Return Rate of Return
Calculations Plot of NPW versus interest rate i Fees or Discounts
Examples Incremental Analysis Using Spreadsheet 2Engineering
Economics
Slide 3
Rate of return analysis is the most frequently used exact
analysis technique in industry. Major advantages Rate of return is
a single figure of merit that is readily understood. Calculation of
rate of return is independent from the minimum attractive rate of
return (MARR). Rate of Return Analysis 3Engineering Economics
Slide 4
Internal Rate of Return What is the internal rate of return
(IRR)? IRR is the interest rate at which present worth or
equivalent uniform annual worth is equal to 0. In other words, the
internal rate of return is the interest rate at which the benefits
are equivalent to the costs. 4Engineering Economics
Slide 5
Internal Rate of Return Internal rate of return is commonly
used to evaluate the desirability of investments or projects. IRR
can be used to rank multiple prospective projects. Because the
internal rate of return is a rate quantity, it is an indicator of
the efficiency, quality, or yield of an investment. To decide how
to proceed, IRR will be compared to preselected minimum attractive
rate of return (Chapter 8) 5Engineering Economics
Slide 6
6 Given a cash flow stream, IRR is the interest rate i which
yields a zero NPW (i.e., the benefits are equivalent to the costs),
or a zero worth at any point in time. This can be expressed in 5
different ways as follows. NPW = 0 PW of benefits PW of costs = 0
PW of benefits = PW of costs PW of benefits/PW of costs = 1 EUAB
EUAC = 0 Internal Rate of Return (IRR) Engineering Economics
Slide 7
Example A person invests $1000 at the end of each year. If the
person would like to have $80,000 in savings at EOY 26 what
interest rate should he select? When the compound interest tables
are visited the value of i where (F/A, i%, 26)=80 is found as 8%,
so i=8% Checking the Tables 26 yrs @ 6%, F/A = 59.156 26 yrs @ 10%,
F/A = 109.182 26 yrs @ 8%, F/A = 79.954 7Engineering Economics
Slide 8
Example A person invests $1000 at the end of each year. If the
person would like to have $80,000 in savings at EOY 26 what
interest rate should he select? When the compound interest tables
are visited the value of i where (F/A, i%, 26)=80 is found as 8%,
so i=8% Checking the Tables 26 yrs @ 6%, F/A = 59.156 26 yrs @ 10%,
F/A = 109.182 26 yrs @ 8%, F/A = 79.954 8Engineering Economics
Slide 9
Example EXCEL solution RATE(n, A, P, F, type, guess) rate(26,
1000, 0, -80000) = 8% rate (26, -1000, 0, 80000) = 8% A, P, F must
have different signs (+ or )! IRR(value range, guess) value range =
the cash flow stream 9Engineering Economics
Slide 10
Example Cash flows for an investment are shown in the following
figure. What is the IRR to obtain these cash flows? YEARCASH FLOW
0($500) 1$100 2$150 3$200 4$250 10Engineering Economics
EXCEL solution IRR(C1:C5) = 12.83% C1 ~ C5 stores the stream of
the 5 cash flows: -500, 100, 150, 200, 250 17Engineering
Economics
Slide 18
Example A student, who will graduate after 4 years, borrows
$10,000 per year at 5% interest rate at the beginning of each year.
No interest is charged till graduation. If the student makes five
equal annual payments after the graduation (end-of-period
payments). a) What is each payment after the graduation? b)
Calculate IRR of loan? (hint: use cash flow from when the student
started borrowing the money to when it is all paid back) c) Is the
loan attractive to the student? 18Engineering Economics
Slide 19
EXAMPLE CONTINUES Year Cash Flow 010,000 1 2 3 40 5(9240) 6 7 8
9 a) b) 19Engineering Economics
Slide 20
INTERPOLATION: c) Since the rate is low, the loan looks like a
good choice. 20Engineering Economics
Slide 21
INTERPOLATION: c) Since the rate is low, the loan looks like a
good choice 21Engineering Economics
Slide 22
a) pmt(5%, 5, -40000) = $9,238.99 per month. b) irr(g1:g10) =
2.66%. c) Since the rate is low, the loan looks like a good choice.
22Engineering Economics EXCEL Solution
Slide 23
Plot of NPW versus Interest Rate Borrowing Cases YearCash Flow
0200 1-50 2 3 4 5 :: :: 23 p. 218 Engineering Economics
Slide 24
Plot of NPW versus Interest Rate Investment Cases YearCash Flow
0-200 150 2 3 4 5 :: :: 24Engineering Economics
Slide 25
Fees or Discounts Question: Option 1: If a property is financed
through a loan provided by a seller, its price is $200,000 with 10%
down payment and five annual payments at 10%. Option 2: If a
property is financed through the same seller in cash, the seller
will accept 10% less. However, the buyer does not have $180,000 in
cash. What is the IRR for the loan offered by seller? 25Engineering
Economics
QUESTION CONTINUES INTERPOLATION: This is a relatively high
rate of interest, so that borrowing from a bank and paying cash to
the property owner is better. 28 Engineering Economics
Slide 29
EXCEL Solution Combined cash flows (difference between options
1 & 2): At time 0:-$160,000 EOY 1-5: $47,484 IRR = rate(5,
47484, -160000) = 14.78% per year IRR = irr(a1:a6) = 14.78% per
year 29 Engineering Economics
Slide 30
Loan and Investments are Everywhere Question: A student will
decide whether to buy weekly parking permit or summer semester
parking permit from USF. The former costs $16 weekly; the latter
costs $100 due May 17 th 2010; in both cases the duration is 12
weeks. Assuming that the student pay the weekly fee on every
Monday: a) What is the rate of return for buying the weekly permit?
b) Is weekly parking attractive to student? *Total 12 weeks
30Engineering Economics
Slide 31
QUESTION CONTINUES WeekWeeklySemester May 170($16)($100) May
241($16) May 312($16) June 73($16) June 144($16) June 215($16) June
286($16) July 57($16) July 128($16) July 199($16) July 2610($16)
August 211($16) a) To find IRR%, set cash flows equal in PW terms
100 = 16 16 (P/A,i%,11) (P/A,i%,11) = (100 - 16) / 16 (P/A,i%,11) =
5.25 Looking in the table for the above value: IRR = 15% b) Nominal
interest rate for 52 weeks IRR 15%/week or 15*52 = 780%/yr Since
the rate is high, paying the semester fee looks like a good choice.
31 Effective annual interest = (1+.15)^52 1 = 1432% ! Engineering
Economics
Slide 32
EXCEL Solution Combined cash flows (difference between the 2
options): At time 0:-$84 EOM 1~11: $16 IRR = rate(11, 16, -84) =
14.92% per month IRR = irr(C1: C12) = 14.92% per month
32Engineering Economics
Slide 33
Rate Of Return Calculations Question: There are two options for
an equipment: Buy or Lease for 24 months. The equipment might be
either leased for $2000 per month or bought for $30,000. If the
plan is to buy the equipment, the salvage value of the equipment at
EOM 24 is $3,000. What is the IRR or cost of the lease?
33Engineering Economics
Slide 34
QUESTION CONTINUES MonthBuy OptionLease Option
0($30,000)($2,000) 1-23($2,000) 24$3,0000 To find IRR%, set cash
flows of Buy and Lease options equal in PW terms 34Engineering
Economics
Slide 35
35Engineering Economics Try i = 5% 28000 3000(0.3101)
20000(13.489) = 91.7 Try i = 4% 28000 3000(0.3477) 2000(14.148) =
-1339.1 i 5% per month
Slide 36
Combined cash flows (difference of buy & lease): At time
0:-$28000 EOM 1~23: $2000 EOM 24: $3000 IRR = irr(e1:e25) = 4.97%
per month EXCEL Solution 36Engineering Economics
Slide 37
When there are two alternatives, rate of return analysis is
often performed by computing the incremental rate of return, IRR,
on the difference between the two alternatives. Incremental
Analysis 37Engineering Economics
Slide 38
Incremental Analysis The cash flow for the difference between
alternatives is calculated by taking the higher initial-cost
alternative minus the lower initial-cost alternative. The following
decision path is made for incremental rate of return (IRR) on
difference between alternatives: Two -Alternative Situations
Decision IRRMARRChoose the higher-cost alternative IRR