3360 Unit 07.1 2014-I-01

ENGR 3360U Winter 2014Unit 7

Internal Rate of Return Analysis

Dr. J. Michael Bennett, P. Eng., PMP, UOIT,

Version 2014-II-01

Unit 7 Rate of Return

Change Record

• 2014-I-01 Initial Creation

2014-I-017-2 Dr. M. Bennett, P.Eng., PMP


Course Outline1. Engineering Economics2. General Economics

1. Microeconomics2. Macroeconomics3. Money and the Bank of

Canada3. Engineering Estimation4. Interest and Equivalence5. Present Worth Analysis6. Annual Cash Flow7. Rate of Return Analysis8. Picking the Best Choice9. Other Choosing Techniques

10. Uncertainty and Risk11. Income and Depreciation12. After-tax Cash Flows13. Replacement Analysis14. Inflation15. MARR Selection16. Public Sector Issues17. What Engineering should know

about Accounting18. Personal Economics for the

Engineer

2014-I-01 Dr. M. Bennett, P.Eng., PMP 7-3



LEARNING OBJECTIVES

1. Definition of ROR/IRR2. IRR using PW and AW3. Calculations about IRR4. Multiple IRRs5. ROR of bonds

ROR = Rate of Return



7. 1 Rate of Return - Introduction Referred to as ROR or IRR (Internal Rate of Return) method

It is one of the popular measures of investment worth DEFINITION -- ROR is either the interest rate paid on the unpaid balance of a loan, or the interest rate earned on the unrecovered investment balance of an investment such that the final payment or receipt brings the terminal value to exactly equal “0”

The ROR of found using a PW or AW relation. The rate determined is called i*



Unrecovered Investment Balance ROR is the interest rate earned/charged on the unrecovered balance of a loan or investment project ROR is not the interest rate earned on the original loan amount or investment amount (P) The i* value is compared to the MARR --

If i* > MARR, investment is justified If i* = MARR, investment is justified (indifferent

decision) If i* < MARR, investment is not justified



Valid Ranges for usable i* ratesMathematically, i* rates must be:

*100% i 1. An i* = -100% signals total and complete loss

of capital2. One can have a negative i* value (feasible)

but not less than –100%3. All values above i* = 0 indicate a positive

return on the investment


Internal Rate of Return (IRR)

Is the interest rate at which the PW or AW = 0.

2014-I-01 Dr. M. Bennett, P.Eng., PMP 0.1-8



7.2 Calculation of i* using PW or AW Relations

Set up an IRR equation using either PW or AW relations and equate to zero

0 = - PW of disbursements + PW receipts

= - PWD + PWR

0 = - AW of disbursements + AW receipts

= - AWD + AWR



i* by Trial and Error by Hand Using a PW Relation

1. Draw a cash flow diagram2. Set up the appropriate PW equivalence equation

and set equal to 03. Select values of i* and solve the PW equation4. Repeat for values of i until “0” is bracketed, i.e.,

the equation is balanced5. May have to interpolate to find the approximate

i* value



ROR using Present Worth

0 1 2 3 4 5

-$1,000

+$500

+$1,500

1000 = 500(P/F, i*,3) +1500(P/F, i*,5)

Assume you invest $1,000 at t = 0; receive $500 @ t = 3 and $1500 at t = 5. What is the IRR of this project?

The above PW expression must be solved by trial and error

•Guess at a rate and try it•Adjust accordingly•Bracket•Interpolate•i* approximately 16.9% per year on the unrecovered investment balances


Trial and Error


1000 = 500(P/F, i*,3) +1500(P/F, i*,5)2 = (P/F, i*,3) +3(P/F, i*,5)Try 15% 2= .6575 + 3(.4972) = 2.149Try 18% 2 = .6086 + 3(4371) = 1.919i* = (2.149-2)/(2.149-1.919)*3 +15 = 16.9



Spreadsheet MethodsExcel supports ROR analysis with 2 functions:

=RATE(n,A,P,F) when A series is present

=IRR(first_cell:last_cell, guess) when cash flows vary

RATE is used when an investment (P) is made followed by “n” equal, end of period cash flows’



The IRR Excel Function

When cash flows vary from period to period Entries: Enter the cash flow values into contiguous cells (including any $0 amounts) Enter the IRR function

=IRR(first_cell:last_cell,guess) “guess” is an optional starting value the user feels is in the “vicinity” of the true i* value If omitted, Excel assumes a starting value of 10%



Example 7.2

The Moscow City Tower was expected to have been the world’s tallest building when finished in 2010. The HVAC engineer for a company involved in its construction has requested that $500,000 be spent now during construction on software and hardware to improve the efficiency of the environmental control systems. This is expected to save $10,000 per year for 10 years in energy costs and $700,000 at the end of 10 years in equipment refurbishment costs. Find the internal rate of return by hand.



Cash Flow

0 1 2 3 4 5 6 7 8 9 10

$500K

$10K

$700K



Use the trial-and-error procedure based on a PW equation.1. Previous figure shows the cash flow diagram.2. Set up the PW=0 equation. 0 = -500,000 + 10,000(P/A, i*,10) + 700,000(P/F, i*,10) 3. Use the estimation procedure to determine i* for the first trial.

All income will be regarded as a single F in year 10 so that the P/F factor can be used. The P/F factor is selected because most of the cash flow ($700,000) already fits this factor and errors created by neglecting the time value of the remaining money will be minimized. Only for the first estimate of i* define

P = $500,000, n = 10, and F = 10(10,000) + 700,000 = $800,000. Now we can state that

500,000 = 800,000(P/F,i,10) (P/F,i,10) = 0.625



The roughly estimated i* is between 4% and 5%. Use 5% as the first trial because this approximate rate for the P/F factor is lower than the true value when the time value of money is considered. At i* = 5%, the IRR equation is

0 = -500,000 + 10,000(P/A,5%,10) + 700,000(P/F,5%,10) 0 < $6946The result is positive, indicating that the return is more than 5%. Try i* = 6%. 0 = -500,000 + 10,000(P/A,6%,10) + 700,000(P/F,6%,10) 0 > $-35,519Since the interest rate of 6% is too high, linearly interpolate between 5% and

6%.i* = 5.00 + 6946/(6946 + 35519) = 5.16%


Example 3

An engineer invests $5,000 at the end of every year during a 40-year career. If she wants $1 million in savings at retirement, what interest must the investment earn?PW = 0 = -5000(F/A, i, 40)+ 1000000(F/A, i, 40) = 200(F/A, 7, 40) = 199.636 so 7%.



Example 4

A bond is sold to an investor for $1,000 paying $40/6 months for 10 years. He sells the bond a year later for $950.a) what rate of return did the first buyer get?b) what ror can the new buyer expect if she holds it to maturiy?



Solution


1000

40 40

950

1 2 Six-month

PWc = PWb; 1000= 40(P/A,i,2) + 950(P/F,i,2)

At 1.5%, 1000 = 1000.41NIR = 3%; EIR = (1+0.015)2 -1 = 3.02%


Solution B


950 = 40(P/A, i, 18) + 1000(P/F, i, 18)4% = 999.96 too high5% = 883.10 too lowIRR – 4% + 1%(999.66-950)/(999.96-883.10)IRR = 4.43% NIR = 8.86% EIR = 9.05%


Example 7.7

An engineering student is deciding whether to buy two one-term parking permits or a single annual one. The annual one costs $180 and the semi-annual ones, $130 (assume Aug 2012 and Jan 2013). What is the ROR for buying the annual one?



solution

-180 = -130(1+(P/F,imon, 5))

(P/F,imon, 5) = -50/-130 = 0.3846

[using the formula]1/(1+imon )5 = 0.3846

(1+imon )5 = 2.6

1+imon = 1.2106

imon = 21.06%

Annual (1.2106 12 -1) = 891%



Example 7-8

An engineering firm can buy liability insurance either quarterly or annually. The quarterly cost is $10,000 and the annual, $35,000. What is the ROR for buying annually?



Solution

35000 =10000(1+(P/A, iqrt,3))

(P/A, iqrt,3) = 2.5

(P/A, 9%,3) = 2.531(P/A, 10%,3) = 2.487iqrt = 0.09 + (0.10 – 0.09)(2.531-2.5)/(2.531-2.487)

= 9.7%


3360 Unit 07.1 2014-I-01

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