CH 5Present Worth Analysis –Part1

04/13/232

Chapter 5 Subjects

Measures of Profitability PW method

The planning horizon (study period);Equal lives using PW & AW;Different lives using PW & AW;

FW method AW method IRR method & MARR (Next Chapter) ERR method & MARR (is not covered in text)

Measures of Liquidity Payback period Investment balance diagram

PW of Bonds

Measure of Profitability

1)PW, FW, & AW methods2)IRR method & MARR3)ERR method & MARR

04/13/234

PW, AW, & FW

For a single alternative, compute either PW, AW, FW. If PW, AW, or FW is equal or greater than zero then the project is viable, otherwise it should be rejected.

04/13/23Dr. Ayman Abu Hammad5

SELECT THE EQUIVALENT WORTH ALTERNATIVE WITH THE GREATER WORTH

If : PWA (i) < PWB (i)

thenPWA (i) ( A / P,i,N ) < PWB (i) ( A / P,i,N )

andAWA (i) < AWB (i)

similarlyPWA (i) ( F / P, i, N ) < PWB (i) ( F / P, i, N )

andFWA (i) < FWB (i)

Select alternative B

04/13/23 6

COMPARING COST ALTERNATIVES• For cost alternatives that are compared using the PW

method, the alternative that has the least negative PW is most economically desirable.

• For cost alternatives that are compared using the AW method, the alternative that has the least negative AW is most economically desirable.

• For cost alternatives that are compared using the FW method, the alternative that has the least negative FW is most economically desirable.

04/13/237

THREE GROUPS OF MAJOR INVESTMENT ALTERNATIVES

1. Mutually exclusive : At most one project out of the group can be chosen

2. Independent : The choice of a project is independent of the choice of any

other project in the group, so that all or none of the projects may be selected or some number in between

3. Contingent : The choice of the project is conditional on the choice of one or

more other projects

04/13/238

CASH FLOW ANALYSIS METHODS

The cash-flow analysis methods (previously described in Chapter 4) for a single project are:

Present Worth ( PW )Annual Worth ( AW )Future Worth ( FW )Internal Rate of Return ( IRR )External Rate of Return ( ERR )

04/13/239

COTERMINATED ASSUMPTIONYou cannot compare alternatives using PW method if

life is not equal. A finite and identical study period is used for all

alternatives. Cash flows are repeated per the least common multiple of lives.

A planning horizon is proposed when AW method is used. Truncate the remaining useful life by substituting a salvage value. This planning horizon, combined with appropriate adjustments to the estimated cash flows, puts the alternatives on a common comparable basis

04/13/2310

PLANNING HORIZONThe selected time period over which mutually

exclusive alternatives are compared -- study periodMay be influenced by factors including:

service period required useful life of the shorter-lived alternative useful life of the longer-lived alternative company policy

It is key that the study period be appropriate for the decision situation under investigation

04/13/2311

COTERMINATED ASSUMPTION

Guidelines when useful life(s) different in length than study period

Useful life < study perioda. Cost alternatives -- each cost alternative must provide same

level of service as study period : 1) contract for service or lease equipment for remaining time; 2) repeat part of useful life of original alternative until study period ends

b. investment alternatives -- assume all cash flows reinvested in other opportunities at MARR to end of study period

04/13/2312

Comparing Projects of Equal Lives

04/13/2313

PW of equal live alternatives

Perform a PW analysis of three machines. MARR= 10%. Revenues are common for the three optionsExample 5.1 Electric

PoweredGas- Powered

Solar- Powered

First cost -2500 -3500 -6000

Annual Operating Cost (AOC), $ -900 -700 -50

Salvage Value SV, $ 200 350 100

Life (n), years 5 5 5

• PWE=-2500-900(P/A,10%,5)+200(P/F,10%,5)=$-5788

• PWP=-3500-700(P/A,10%,5)+350(P/F,10%,5)=$-5936

• PWS=-6000-50(P/A,10%,5)+100(P/F,10%,5)=$-6127

04/13/2314

Comparing Projects of Different Lives

04/13/2315

PW of different live alternatives Perform a PW analysis of following two lease options. MARR= 15%. If a

study period of 5 years and returns would not change, which option should be selected

Location A Location B

First cost, $ -15,000 -18,000

Annual Operating Cost (AOC), $ -3,500 -3,100

Salvage Value SV, $ 1,000 2,000

Life (n), years 6 9

• Compare using LCM of lives= 18 years• For life cycles after the first year, the first cost is repeated in year 0 of each new

cycle, which is the last year of the previous cycle. These are years 6 and 12 for A, and 9 for B.

• Draw cash flow of repeated cycles for the two options over 18 years.• Calculate the PW at MARR=15%

04/13/23Dr. Ayman Abu Hammad16

PWA= -15000-15000(P/F,15%,6)+1000(P/F,15%,6)-15,000(P/F,15%,12)+1000(P/F,15%,12)+1000(P/F,15%,18)-3500(P/A,15%,18)=$-45,036

PWB= -18000-18000(P/F,15%,9)+2000(P/F,15%,9)+2000(P/F,15%,18)-3100(P/A,15%,18)=$-41,384

PW of different live alternatives

0 1 2 3 4

A=$3,1005 6 7 8 9 10 11 12 13 14 15 16 17 18

PWB?=

A=$3,5007 8 9 10 11 12 13 14 15 16 17 180 1 2 3 4 5 6

PWA?=

$15,000

$1,000 $1,000

$15,000

$1,000

$15,000

$1,000

$2,000

$18,000 $18,000

$2,000

04/13/2317

PW of alternatives using a Planning Horizon

For a 5 year planning horizon:PWA= -15000 -3500(P/A,15%,5) +1000(P/F,15%,5)

= $-26,236PWB = -18000 -3100(P/A,15%,5) +2000(P/F,15%,5)

= $-41,384

04/13/2318

COMPARING ALTERNATIVES USING THE CAPITALIZED WORTH METHOD

Capitalized Worth (CW) method -- Determining the present worth of all revenues and / or expenses over an infinite length of time

Capitalized cost -- Determining the present worth of expenses only over an infinite length of time

Capitalized worth or capitalized cost is a convenient basis for comparing mutually exclusive alternatives when a period of needed services is indefinitely long and the repeatability assumption is applicable

04/13/2319

CAPITALIZED WORTH METHOD

Capitalized worth of a perpetual series of end-of-period uniform payments, A, with interest i% per period:

A ( P /A, i%, )

CW = PWN --> = A ( P / A, i%, )

( 1+i )N - 1

= A lim ------------- = A ( 1 / i )

N -->

i ( 1 + i )N8

8 8

8

04/13/2320

Capitalized Cost CalculationPW Computations when N

These cases are also called "perpetuities" - e.g. scholarships, endowments, etc.

There are many engineering projects where the life of the project is so long as to be considered ‘forever’. Usually 50 years as a rule of thumb, but assume cash flows continue indefinitely.

There are other projects where a sum of money is provided at the beginning of the project and it is then invested to yield the amount used for the project (annuities for 50 years)

For both, the PW of the infinitely long uniform series of cash flows becomes the Capitalized Value

04/13/2321

PW when N

Until now, we assumed that the horizon N is a fixed number of years.

The present worth of a very long and uniform series of cash flows is calculated as

lim ( / , , )

(1 ) 1lim

(1 )

11

(1 )lim

N

N

NN

N

N

P A P A i N

iP A

i i

iP Ai

AP

i

Example 5.4The system has an installed cost of $150,000 and

an additional cost of $50,000 after 10 years. The annual software maintenance contract cost is $5000 for the first 4 years and $8000 thereafter. In addition, there is expected to be a recurring major upgrade cost of $15,000 every 13 years. Assume that i = 5% per year for county funds.

Example – cont. Find the present worth of the nonrecurring costs of $150,000 now

and $50,000 in year 10 at i = 5%. Label this CCI‘ CCI = - 150,000 - 50,000(P/F,5%,10) = $-180,695

Convert the recurring cost of $15,000 every 13 years into an annual worth AI for the first 13 years. AI = - 15 ,000(A/ F,5 %, 13) = $-847

The same value, AI = $- 847, applies to all the other 13-year periods as well.

The capitalized cost for the two annual maintenance cost series may be determined in either of two ways: (I) consider a series of $- 5000 from now to infinity and find the

present worth of -$8000 - ($-5000) = $-3000 from year 5 on; or (2) find the CC of $-5000 for 4 years and the present worth of $-

8000 from year 5 to infinity. Using the first method, the annual cost (A2) is $- 5000 forever. The capitalized cost CC2 of $- 3000 from year 5 to infinity is found using Equation [5.1] times the P / F factor.

CC2=-3000/0.05(P/F,5%,4)=$-49,362

Example –cont.

The two annual cost series are converted into a capitalized cost CC3. CC3=(A1+A2)/I = (-847+ -5000) / 0.05 = -116,940

The total capitalized cost CCT is obtained by adding the three CC values. CCT = - 180,695 - 49,362 - 116,940 = $- 346,997

determines the A value forever.A = Pi = CCT(i) = $346,997(0.05) = $17,350Correctly interpreted, this means Marin County officials

have committed the equivalent of $17,350 forever to operate and maintain the property appraisal software.

Problem 2Determine the capitalized cost of an expenditure

of $200,000 at time 0, $25,000 in years 2 through 5, and $40,000 per year from year 6 on. Use an interest rate of 12% per year.

Soln:CC = -200,000 – 25,000(P/A,12%,4)(P/F,12%,1)

– [40,000/0.12])P/F,12%,5)= -200,000 – 25,000(3.0373)(0.8929) –

[40,000/0.12])(0.5674)= $-456,933

Problem 3What is the capitalized cost (absolute value) of

the difference between the following two plans at an interest rate of 10% per year? Plan A will require an expenditure of $50,000 every 5 years forever (beginning in year 5). Plan B will require an expenditure of $100,000 every 10 years forever (beginning in year 10).

Soln:

Problem 3- Cont.

Find AW of each plan, then take difference, and divide by i.

AWA = -50,000(A/F,10%,5)= -50,000(0.16380)= $-8190AWB = -100,000(A/F,10%,10)= -100,000(0.06275)= $-6275CC of difference = (8190 - 6275)/0.10= $19,150

Problem 4Two processes can be used for producing a

polymer that reduces friction loss in engines.Process K will have a first cost of $160,000, an

operating cost of $7000 per quarter, and a salvage value of $40,000 after its 2-year life. Process L will have a first cost of $210,000, an operating cost of $5000 per quarter, and a $26,000 salvage value after its 4-year life. Which process should be selected on the basis of a present worth an alysis at an interest rate of 8% per year, compounded quarterly?

Soln:

Problem 4 – Cont. PWK = -160,000 – 7000(P/A,2%,16) –

120,000(P/F,2%,8) + 40,000(P/F,2%,16)= -160,000 – 7000(13.5777) –120,000(0.8535) +

40,000(0.7284)= $-328,328PWL = -210,000 – 5000(P/A,2%,16) +

26,000(P/F,2%,16)= -210,000 – 5000(13.5777) + 26,000(0.7284)= $-258,950Select process L