Module 5 Annual Worth Analysis

Module 5: Annual Worth Analysis

SI-4251 Ekonomi Teknik

Muhamad Abduh, Ph.D.


Outline Module 5 Selection of Alternatives with Different Lives Salvage Sinking-Fund Method Salvage Present Worth Method Capital-Recovery-Plus-Interest Method AW of a Perpetual Investment

Muhamad Abduh, Ph.D.5-2


Selection of Alternatives with Different Lives1. Selection of investment alternatives can commonly evaluated by

comparing their equivalence-uniform-annual-worth (over the life of the investment)

2. Unlike other method, for evaluating alternatives with different lives, EUAW does not require comparison over the least common multiple periods.


0 1 2 3

O&M

SV

R R R R R RR R R

SVSV

I I I

I = Rp. 240 millionsO&M = Rp. 25 millions/monthR = Rp. 190 millions/4 monthsSV = Rp 75 millions

O&M

0 1

R R RI

SV

A

B

EUAWA = EUAWB


Example Consider a project with $3000 annual

operating cost and a $5000 investment required each 5 years. i = 10%

For one cycle EAC = 3000 + 5000(A/P,10%,5) = $4319/yr


$5,000

0 1 2 3 4 5

A 1-5 = $3,000


Multiple cycle..same result!


For two cycles EAC = 3000 + 5000 (1+(P|F, .10, 5))(A|

P, .10, 10) = 3000 + 1319 = $4319/yr

0 1 2 3 4 5

6 7 8 9 10

$5,000

$5,000

A 1-10 = $3,000


(A) Salvage Sinking-Fund Method


1. Initial investment and salvage value are to be converted into an equivalent uniform annual value

2. Any other cash flows are to be converted first to present value (P) or future value (F), then converted into equivalent uniform annual value

3. Combine all equivalent uniform annual values

EUAW = – P(A/P, i, 36) +SV(A/F, i, 36) – A1 – O1(P/F, i, 9)(A/P, i, 36) + I1(F/P, i, 14)(A/F, i, 36)

0 1 2 3

A1

SVI1P O1


(B) Salvage Present Worth Method


1. Calculate the present worth of salvage value using (P/F) factor, and subtract that value from the initial cost, P

2. Convert that value (2) into equivalent uniform worth



0 1 2 3

A1

SVI1P O1

EUAW = – [P-SV(P/F, i, 36)](A/P, i, 36) – A1 – O1(P/F, i, 9)(A/P, i, 36) + I1(F/P, i, 14)(A/F, i, 36)


(C) Capital-Recovery-Plus-Interest Method


1. Subtract the salvage value from initial cost, then convert that value into equivalent uniform worth

2. Multiply the salvage value by the interest rate

3. Add value obtained from (1) and (2)



0 1 2 3

A1

SVI1P O1

EUAW = – (P-SV)(A/P, i, 36) - SV(i) – A1 – O1(P/F, i, 9)(A/P, i, 36) + I1(F/P, i, 14)(A/F, i, 36)

SI-4251 Ekonomi Teknik5-9

Example:

Cash Flow Diagram is:

1 2 3 4 5

P=-23,000

S = +$1500

-$650-$700

-$750-$800

-$850

A = +$1200/yr



Example:

The Capital Recovery component is:

1 2 3 4 5

P=-23,000

S = +$1500

CR(10%) = -23,000(A/P,10%,5) +1500(A/F,10%,5) = -$5822



Example:

(Revenue – Operating Costs) are:

1 2 3 4 5

A = +$1200/yr

$650$700

$750$800

$850



Example:

Cost/Revenue component is seen to equal:

=+550 –50(A/G,10%,5)= 550 – 90.50= $459.50

1.8101



Example

Total Annual worth (CR + Cost/Rev) CR(10%) = -$5822 Revenue/Cost Annual amount: $459.50 AW(10%) = -$5822+$459.50 AW(10%) = $5,362.50

This amount would be required to recover the investment and operating costs at the 10% rate on a per year basis



AW of a Perpetual Investment

EAC of a perpetual investment

If an investment has no finite cycle it is called a perpetual investment. If “P” is the present worth of the cost of that investment, then EAC is P times i, the interest P would have earned each year.

EAC=A = P(i)Remember: P = A/iFrom the previous chapter



Example: Perpetual Investment

EXAMPLE

Two alternatives are considered for covering a football field. The first is to plant natural grass and the second is to install AstroTurf. Interest rate is 10%/year.Assume the field is to last a “long time”.



Example: Continued

Alternative A:

Natural Grass - Replanting will be required each 10 years at a cost of $10,000. Annual cost for maintenance is $5,000. Equipment must be purchased for $50,000 which will be replaced after 5 years with a salvage value of $5,000



Example: Natural Grass

Alternative A:

0 1 2 3 4 5 6 7 8 9 10

$10,000

A = $5,000

F5 = $5,000

Since cost is predominate, let (+) = cost and (-) = salvage values

P = $50,000+ $10,000

F5 = $5,000

F5=$50,000



Example: Natural Grass: Analysis

(+) $60,000(A/P,10%,10) (+) $5,000 (already an annual cost) (+) $50,000(P/F,10%,5)(A/P,10%,10) (-) $5,000(P/F,10%,5)(A/P,10%,10) (+) $10,000(A/F,10%,10) (-) $5,000(A/F,10%,10) = $ 19,046/year



Example: Artificial Carpet (Surface)

A = P(i) for a perpetual life project Annual Cost of Installation: =$150,000 (.10) = $15,000/ year Annual Maintenance = $5,000/year

Total: $15,000 + $5,000 = $20,000/YrChoose A, cost less per year!



Comparing Alternatives by EUAW

5-20

Exercise:Two types of production systems are being considered based on interest rate of 9% and the following characteristics:

system A system B

Initial cost Rp 860.000.000,- Rp 1.360. 000.000,-

Annual O & M expenses Rp 89.000.000,- Rp 73.000.000.-

Annual receipts Rp 121.500.000,- Rp 133.400.000,-

Salvage value Rp 200.000.000,- Rp 320.000.000,-

Life 6 years 10 years(A/P, 9%, 6) = 0.21632

(A/F, 9%, 6) = 0.13632

(P/F, 9%, 6) = 0.60320

(A/P, 9%, 10) = 0.14903

(A/F, 9%, 10) = 0.06903

(P/F, 9%, 10) = 0.46320



Homework #51. 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%

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:

5-21

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 YFirst cost Rp. 3.800.000.000,- Rp. 1.675.000.000,-Annual operating cost Rp. 289.700.000,- Rp. 315.000.000,-Salvage value Rp. 461.000.000,- Rp. 205.000.000,-Life 5 years 3 years


Module 5 Annual Worth Analysis

Documents

Transcript of Module 5 Annual Worth Analysis