Module 4 Present Worth Analysis

Module 4: Present Worth Analysis

SI-4251 Ekonomi Teknik

Muhamad Abduh, Ph.D.


Outline Module 4 Proposals for Investment Alternatives Selection of Alternatives Present-worth comparison Future Worth Analysis Capitalized-cost calculation

Muhamad Abduh, Ph.D.4-2


Proposals for Investment Alternatives An investment is usually established from

(engineering) proposal. Every proposal can be considered as an investment

alternatives, but An investment alternative can consist of a group or

set of proposals, which in turn may include option to “do nothing” Independent Proposal

The condition at which the acceptance of a proposal from a set of proposals has not effect on the acceptance of any of the other proposals in the set.

Dependent Proposal The condition at when the acceptance of a proposal from the set will influence the

acceptance of the others Mutually exclusive the acceptance of a proposal from the set precludes the acceptance

of any others



Selection of Alternatives Decision criteria in an economic analysis

can be done comparing mutually exclusive alternatives: The differences between alternatives

highest inflow or lowest outflow The minimum attractive rate of return

highest rate of return Payback period

shortest payback period Do nothing



Selection of Alternatives Selection of two or more (investment)

alternatives by comparing their economic values. Method for comparison of alternatives:

present values, future worth, capitalized cost, annual values, rate of returns, or payback period

Conditions of comparison: equal lives different lives


SI-4251 Ekonomi Teknik6

MARR Minimum Acceptable Rate of Return

Minimum amount the investor is willing to accept for the use of money

Different from lending rates More akin to Opportunity Cost of the money Determined by Company Policy

Rate of Return

Expected Rate of Return on New Proposal

Range for ROROn accepted proposals

MARR

ROR on Safe investment

All proposals must offer at Least MARR

to be considered



Present-worth comparison

The comparison of alternatives is made by transforming all future receipts and expenditures into equivalent today’s rupiah

Example: Two types of production systems are being considered based on

MARR of 12% per year and the following characteristics:


system A system B

Initial cost Rp. 625.000.000,- Rp. 570.000.000,-

Monthly expenses

Rp. 45.750.000,- Rp. 55.750.000,-

Receipts Rp. 152.000.000,- / quarter

Rp. 140.000.000,- / 4 months

Salvage value Rp. 225.000.000,- Rp. 190.000.000,-

Life 3 tahun 3 tahun



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System A:

I

0 1 2 3

R R R R R R R R R R R R

E

SV

?PA = -I - E(P/A, iA, 36) + R(P/F, iA, 3) + R(P/F, iA, 6) + … + R(P/F, iA, 36) +

SV(P/F, iA, 36)

Effective monthly interest rate, iA = i / 12 = 1%




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System B:

I

0 1 2 3

E

SV

R R R R R RR R R


PB = -I - E(P/A, iB, 36) + R(P/F, iB, 4) + R(P/F, iB, 8) + … + R(P/F, iB, 36) + SV(P/F, iB, 36)

Select System A if PA > PB


Present-worth comparisonTo have a fair comparison of alternatives with different lives, the time span over must made equal:

a)The time period of comparison is made equal to the least common multiple (LCM) for their lives.Cash flows of the shorter period will be extended up to the remaining time period of comparison

b)At any time span to be considered (the study period approach or planning horizon approach), when LCM is impossible to perform. Only cash flows up to the time span is to be considered;




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Example:Two types of production systems are being considered based on MARR of 12% per year and the following characteristics:

system A system B

Initial cost Rp. 625.000.000,- Rp. 770.000.000,-

Monthly expenses

Rp. 45.750.000,- Rp. 55.750.000,-

Monthly receipts

Rp. 32.000.000,- Rp. 40.000.000,-


Life 2 tahun 4 tahun




4-12

System A:

1 2

I

0

E

SVR

1 2

I

0

E

R

3 4

SV

System B:



Present-worth comparison (LCM)

4-13

System A:

1 2

I

0

E1

SVR1

1 2

I

0

E

R

3 4

SV

System B:

3 4

I2E2

SV2R2



Present-worth comparison (Study Periods: 2 years)

4-14

System A:

1 2

I

0

E

SVR

1 2

I

0

E

R

3 4

SV

System B:


Estimated


Example 1: Three Alternatives

Assume i = 10% per year

A1Electric Power

First Cost: -2500Ann. Op. Cost: -900Sal. Value: +200Life: 5 years

A2Gas Power


A3Solar Power


Which Alternative – if any, should be selected based upon a present worth analysis?



Example 1: Cash Flow Diagrams

0 1 2 3 4 5

0 1 2 3 4 5

0 1 2 3 4 5

-2500

-3500

-6000

A = -900/Yr.

A = -700/Yr.

A = -50/Yr.

FSV = 200

FSV = 350

FSV = 100

A1:

Electric

A2: Gas

A3:Solar

i = 10%/yr and n = 5Muhamad Abduh, Ph.D.4-16


Calculate the Present Worth's

Present Worth's are:

1. PWElec. = -2500 - 900(P/A,10%,5) +

200(P/F,10%,5) = $-5788

2. PWGas = -3500 - 700(P/A,10%,5) +

350(P/F,10%,5) = $-5936

3. PWSolar = -6000 - 50(P/A,10%,5) +

100(P/F,10%,5) = $-6127Select “Electric” which has the min. PW Cost!



Example 2:

Two Location Alternatives, A and B where one can lease one of two locations.

Which option is preferred if the interest rate is 15%/year?

Location ALocation BFirst cost, $ -15,000 -18,000Annual lease cost, $ per year -3,500 -3,100Deposit return,$ 1,000 2,000Lease term, years 6 9



Use LCM, where “n” = 18 yrs.

The Cash Flow Diagrams are:



Unequal Lives: 2 Alternatives

i = 15% per year

A

LCM(6,9) = 18 year study period will apply for present worth

Cycle 1 for A Cycle 2 for A Cycle 3 for A

Cycle 1 for B Cycle 2 for B

18 years

6 years

6 years

6 years

9 years 9 yearsB



LCM Present Worth's

Since the leases have different terms (lives), compare them over the LCM of 18 years.

For life cycles after the first, the first cost is repeated in year 0 of the new cycle, which is the last year of the previous cycle.

These are years 6 and 12 for location A and year 9 for B.

Calculate PW at 15% over 18 years.


SI-4251 Ekonomi Teknik22

PW Calculation for A and B -18 yrs

PWA = -15,000 - 15,000(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 = -18,000 - 18,000(P/F,15%,9) +

2000(P/F,15%,9) + 2000(P/F,15%,18) - 3100(P/A,15 %,18)

= $-41,384 Select “B”: Lowest PW Cost @ 15%



Use The Study Period Approach

An alternative method; Impress a study period (SP) on all of the alternatives;

A time horizon is selected in advance; Only the cash flows occurring within

that time span are considered relevant; May require assumptions concerning

some of the cash flows. Common approach and simplifies the

analysis somewhat.



Example Problem with a 5-yr SP

Assume a 5- year Study Period for both options:

For a 5-year study period no cycle repeats are necessary. PWA = -15,000 - 3500(P/A,15%,5) +

1000(P/F,15%,5)= $-26,236

PWB = -18,000- 3100(P/A,15%,5) +

2000(P/F,15%,5) = $-27,397Location A is now the better choice.Note: The assumptions made for the A and B

alternatives! Do not expect the same result with a study period approach vs. the LCM approach!



Future Worth Analysis

In some applications, management may prefer a future worth analysis;

Analysis is straight forward: Find P0 of each alternative:

Then compute Fn at the same interest rate

used to find P0 of each alternative.

For a study period approach, use the appropriate value of “n” to take forward.



Future Worth Approach (FW)

Applications for the FW approach: Wealth maximization approaches; Projects that do not come on line

until the end of the investment (construction) period: Power Generation Facilities Toll Roads Large building projects Etc.



Capitalized Cost Calculations

CAPITALIZED COST- the present worth of a project which lasts forever. Government Projects; Roads, Dams, Bridges, project that possess

perpetual life; Infinite analysis period; “n” in the problem is either very long,

indefinite, or infinity.



Derivation of Capitalized Cost

We start with the relationship: P = A[P/A,i%,n] Next, what happens to the P/A factor

when we let n approach infinity. Some “math” follows.



P/A where “n” goes to infinity

The P/A factor is:

(1 ) 1

(1 )

n

n

iP A

i i

On the right hand side, divide both numerator and denominator by (1+i)n 1

1(1 )ni

P Ai



CC Derivation…

Repeating: 11(1 )ni

P Ai

If “n” approaches the above reduces to:

AP

i



CC Explained

For this class of problems, we can use the term “CC” in place of P.

Restate:

Or,

AW: annual worth

ACC

i

AWCC

i



CC Problem: Public Works Example

Problem Parameters

The suspension bridge will cost $50 million with annual inspection and maintenance - costs of $35,000. In addition, the concrete deck would have to be resurfaced every 10 years at a cost of $100,000. The truss bridge and 'approach roads' are expected to cost $25 million and have annual maintenance costs of $20,000. The bridge would have to be painted every 3 years at a cost of $40,000. In addition, the bridge would have to be sandblasted every 10 years at a cost of $190,000. The cost of purchasing right-of-way is expected to be $2 million for the suspension bridge and $15 million for the truss bridge. Compare the alternatives on the basis of their capitalized cost if the interest rate is 6% per, year.


Two, Mutually Exclusive Alternatives: Select the best alternative based upon a CC analysis


Bridge Alternatives: Suspension

Cash Flow Diagrams

0 1 2 3 4 . . . . . 9 10 11 ……..

$50 Million

$35,000/yr

$100,000$2 Million

Suspension Bridge Alternative

i = 6%/year


Suspension Bridge Analysis

CC1= -52 million at t = 0.

1

2

1 22

A $35,000

A 100,000( / ,6%,10) $7,587

35,000 ( 7,587)$709,783.

0.06

A F

A ACC

i

Total CC – suspension bridge is:

-52 million + (-709,783) = -$52.71 million

Muhamad Abduh, Ph.D.4-34 SI-4251 Ekonomi Teknik


Truss Bridge Alternative

For the Truss Bridge Alternative: Cash Flow Diagram:

/ / / / / /0 1 2 3 4 5 6 7 8 9 10

11 ….. A. Maint. = $20,000/yr

-25M +(-15M)

Paint: -40,000

Paint: -40,000

Paint: -40,000

Sandblast: -190,000

Truss Design:

i = 6%/year

n =




1. CC1 Initial Cost:

-$25M + (-15M) = -$40M

/ / / / / /0 1 2 3 4 5 6 7 8 9 10

11 ….. A. Maint. = $20,000/yr

-25M +(-15M)

Paint: -40,000

Paint: -40,000

Paint: -40,000

Sandblast: -190,000

Truss Design:

i = 6%/year

n =




2. Annual Maintenance is already an “A” amount so: A1 = -$20,000/year

/ / / / / /0 1 2 3 4 5 6 7 8 9 10

11 ….. A. Maint. = $20,000/yr

-25M +(-15M)

Paint: -40,000

Paint: -40,000

Paint: -40,000

Sandblast: -190,000

Truss Design:

i = 6%/year

n =




3, A2: Annual Cost of Painting

/ / / / / /0 1 2 3 4 5 6 7 8 9 10

11 ….. A. Maint. = $20,000/yr

-25M +(-15M)

Paint: -40,000

Paint: -40,000

Paint: -40,000

Sandblast: -190,000

Truss Design: i = 6%/year

n =

For any given cycle of painting compute:

A2 = -$40,000(A/F,6%,3) = -$12,564/year

Use A/F,6%,3




3, A3 Annual Cost of Sandblasting

/ / / / / /0 1 2 3 4 5 6 7 8 9 10

11 ….. A. Maint. = $20,000/yr

-25M +(-15M)

Paint: -40,000

Paint: -40,000

Paint: -40,000

Sandblast: -190,000

Truss Design: i = 6%/year

n =

For any given cycle of Sandblasting Compute

A3 = -$190,000(A/F,6%,10) =-$14,421

Use The A/F,6%,10 to convert to an equivalent $/year amount



Bridge Summary for CC(6%)

CC2 = (A1+A2+A3)/i

CC2 = -(20,000+12,564+14,421)/0.06

CC2 – $783,083

CCTotal = CC1 + CC2 =-40.783 million•CCSuspension = -$52.71 million

•CCTruss - -40.783 million

•Select the Truss Design!



Exercise:


The municipal government of Bandung is considering two proposals to build new toll highway system. The first alternative calls for upgrading the existing

system, which would cost Rp. 225,75 B for construction and additional Rp 210 M and Rp 250 M per year for maintenance and operation cost

The other option is to build a new elevated highway that is estimated to cost Rp. 885 B for construction and annual maintenance cost of Rp 325 M

If either alternatives would yield Rp 815 M revenue per year, and the interest rate is set at 8% p.a., which alternative should the government go about realizing the toll highway system?


Homework #41. A manufacturing company is trying to decide among three different pieces of equipment that have the

following characteristics:

useful life = 6 years and interest rate of 12% per year.

2. Which of these two machines that have the following costs is to be selected for a continuous production process, if the i = 15% p.a:

3. Ganesha consulting firm is considering to build or lease an office space. For interest rate of 6% compounded semiannualy compare and select alternative.

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equipment A equipment B equipment C

First cost Rp. 975.000.000,- Rp. 854.500.000,- Rp. 1.025.000.000,-

Annual M&O cost Rp. 89.700.000,- Rp. 95.000.000,- Rp. 75.000.000,-

Salvage value Rp. 161.000.000,- Rp. 205.000.000,- Rp. 321.000.000,-

Overhaul cost Rp. 175.000.000,- / 2 years Rp. 135.000.000,- / 3 years Rp. 175.000.000,- / 3 years

machine X machine Y

First cost Rp. 3.800.000.000,- Rp. 1.675.000.000,-

Annual operating cost Rp. 289.700.000,- Rp. 315.000.000,-


Life 5 years 3 years

build own lease

Construction cost Rp. 8.750.000.000,- -

Lease cost - Rp. 415.000.000,- / 3 years

Maintenance cost Rp. 110.000.000,- / year Rp. 75.000.000,- / year

Period ∞ 3 years


Documents

Module 4 Present Worth Analysis