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ENGINEERING

ECONOMYINTEREST AND MONEY-TIME

RELATIONSHIPSPERPETUITIES & CAPITALIZED

COST

Engr. Armando C. Emata

December 3, 2013

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TOPIC OBJECTIVES:

Board work on Nov 28 and Dec 3assignments

Familiarize with another form of annuitythe Perpetuity

Solve sample problem on Perpetuity Be introduced to a key learning in

succeeding Economy studies whereperpetuity is appliedthe Capitalized

Cost Learn the tree types of Capitalized Cost Solve sample problems on capitalized

CostAssignment for next meeting

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PERPETUITY

Another type of annuity of interest to the engineer, known as aperpetuity, is a uniform series in which the payments continueindefinitely. If there exists a present or principal amount P pesosand this earns interest at the rate of i%per period, then the endof period perpetual payment, A pesos, (like interest earned)which can be made from this principal is

A = P x i

One can then convert this equation to show that the presentworth of a perpetuity of payments ofAcan be found as

P = A / i

The Pin this equation is often spoken of as the capitalized valueor capitalized costofA.

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PERPETUITY

Aperpetuityis an annuity in which the payments continueindefinitely.

P = A --------------- = A ------------------

P = ---- (2-32)

P

A A A

0 1 2 3n -> inf.

Cash f low diagram to f ind P given A

1(1+i)- n

i

- inf.1(1+i)i

A

i

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PERPETUITY

PROBLEM:What amount of money invested today at 15% interest canprovide the following scholarships: PHP30,000 at the end ofeach year for 6 years, PHP40,000 for the next 6 years andPHP50,000 thereafter?

Solution:

0 1 2 6 7 12 138 14

P

P30k P30k P30k

P40k P40k P40k

P50k P50k

P50,000

P30,000(P/A, 15%, 6)

P30,000(P/A, 15%, 6)(P/F, 15%, 6)P40,000(P/A, 15%, 6)

------------------ (P/F, 15%, 12)0.15

P50,000------------------

0.15

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PERPETUITY

Using today as the focal date, the equation of value is

P = P30,000(P/A, 15%, 6) + P40,000(P/A, 15%, 6)(P/F,15%, 6)

+ ------------ (P/F, 15%, 12)

= P30,000(3.7845) + P40,000(3.7845)(0.4323)

+ ----------- (0.1869)

P = PHP241,277

P50,000

0.15

P50,000

0.15

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CAPITALIZED COST

In providing for the perpetual carefor some structure or themaintenance of endowed foundations we often encounter aspecial type of perpetuity. A certain amount Smay be neededevery kperiods to provide for replacement or maintenance.The owner or founder wishes to provide a fund of sufficient

size so that the earnings from it will provide for this periodicdemand.

To be available perpetually, Smust be accumulated in kperiods from the interest Ithat is earned by some amount ofprincipalX, invested at rate i. Thus periodic deposit toward

this accumulation will beXi. And Xcan be computed throughthe relationship

S = I (F/A, i%, k) = Xi (F/A, i%, k)

and

X = ----- x --------------- = ------ x (A/F, i%, k)

S

i

S

i

1

(F/A, i%, k)

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CAPITALIZED COST

Another rationale for computingXin this circumstance is toreason as follows:

What principal amountX, when compounded at i%per periodfor kperiods, will at the end of the kth period equal the Sneeded plusXto be available for accumulating interest toprovide the next Spayments? Algebraically, this can bestated as

X(F/P, i%, k) = S + X

Thus,

X = ---------------------------

S

[(F/P, i%, k)1]

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CAPITALIZED COST

If the first costof the project is added toX, the sum is knownas the capitalized cost. Thus, the capitalized cost of an articleis the amount of sufficient size to purchase the article andalso to provide for its perpetual maintenance.

Specifically, it is the sum of the first cost and the presentworth of all costs of replacement, operation and maintenancefor a long time or forever.

Examples:

Permanent structures like parks, monuments an other landmarks (Lunetain Manila, Central Park in NY, Statue of Liberty in NY, Burnham Park inBaguio)

Buildings (Manila City Hall, Central Bank in Manila, Araneta Coliseum inQ.C.)

Factories and similar facilities (Philippine Match Co, in Manila, KimberlyClark in Laguna)most of these large facilities have already shutdown

and left however.

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CAPITALIZED COST

CASE 1. No replacement, only maintenance and/or operationevery period.

Capitalized cost = First cost + Present worth of perpetual

operation and or maintenance

CASE 1 SAMPLE PROBLEM:Determine the capitalized cost of a structure that requires aninitial investment of PHP1,500,000 and an annualmaintenance of PHP150,000. Interest is 15%

0 1 2

P

P150,000 P150,000

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CAPITALIZED COST

CASE 1 SAMPLE PROBLEM contd:

P = ---- = --------------- = PHP1,000,000

Capitalized cost = First cost + P

= P1,500,000 + P1,000,000

= PHP2,500,000

A

i

P150,000

0.15

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CAPITALIZED COST

CASE 2 contd...

S = Xi (F/A, i%, k)

X = ---- --------------- = ---- ------------------

X = ---------------- (2-33)

S

i

1

F/A, i%, k

S

i (1+i) 1k

i

(1+i) 1k

S

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CAPITALIZED COST

CASE 2 contd...Difference between PandXin a perpetuity

P = ---- X = ----------------

Pis the amount invested now at i%per period whose interestat the end of every period forever isAwhileXis the amountinvested now at i%per period whose interest at the end ofevery kperiod is S. If k=1, then,X=P.

A SA A S S

P X

0 1 2 3 0 k 2k 3k

A

i (1+i) 1k

S

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CAPITALIZED COST

CASE 2 contd... CASE 2 SAMPLE PROBLEM:

A new engine was installed by a textile plant at a cost ofPHP300,000 and projected to have a useful life of 15 years.

At the end of its useful life, it is estimated to have a salvage

value of PHP30,000. Determine its capitalized cost if interestis 18% compounded annually.

Solution:

0 15 30 45

P30,000 P30,000 P30,000

P300,000 P300,000 P300,000 P300,000

Cash f low diagram for the engine

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CAPITALIZED COST

CASE 2 SAMPLE PROBLEM contd...

X = --------------- = -------------------- = PHP24,604

Capitalized cost = First cost + X = P300,000 +P24,604

= PHP324,604

0 15 30 45

P270,000 P270,000 P270,000

XS

(1+i) 1k

(1+0.18) 115

P270,000

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CAPITALIZED COST

CASE 3.Replacement, maintenance and/or operation everyperiod

Capitalized cost = First cost + Present worth of cost of

perpetual operation and/ormaintenance

+ Present worth of cost of perpetual

replacement

CASE 3 SAMPLE PROBLEM:

Determine the capitalized cost of a research laboratory whichrequires PHP5,000,000 for original construction; PHP100,000at the end of every year for the first 6 years and thenPHP120,000 each year thereafter for operating expenses;and PHP500,000 every 5 years for replacement of equipment

with interest at 12% per annum.

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CAPITALIZED COST

CASE 3 SAMPLE PROBLEM contd:Solution:

Operation:

0 1 2 3 4 5 6 7 8 9

P100,000

P100,000

P100,000

P100,000

P100,000

P100,000

Q

P120,000

P120,000

P120,000

P100,000(P/A, 12%, 6)

----------------(P/F, 12%, 6)P120,000

0.12

----------------P120,000

0.12

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CAPITALIZED COST

CASE 3 SAMPLE PROBLEM contd:

Let Q = the present worth of cost of perpetual operation

Q = P100,000(P/A, 12%, 6) + -------------- (P/F,12%, 6)

= P100,000(4.1114) + ------------- (0.5066)

Q = PHP917,740

P120,000

0.12

P120,000

0.12

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CAPITALIZED COST

CASE 3 SAMPLE PROBLEM contd:

Let X= the present worth of cost of perpetual replacement

X = -------------- = ------------------- = P655,910

Capitalized cost = First cost + Q + X

= P5,000,000 + P917,740 + P655,910

= PHP6,753,650

0 5 10 15

P500,000

P500,000

P500,000

X

(1+i) 1kS

(1+0.12)1

P500,000

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AssignmentDec 5, 2013

For next meeting, submit in one sheet of bond paper. Writeyour name, subject/section, date and write the problemstatement. Please write legibly.

Non-compliance will mean non-acceptance of yourassignment.

Calculate the capitalized cost of an infrastructureproject that has an estimated initial cost of $14,000,000and an additional investment cost of $3,000,000 at theend of every ten years. The annual operating cost willbe $200,000 at the end of every year for the first fouryears and $280,000 thereafter. In addition, there isexpected to be a recurring major rework cost of$450,000 every 13 years. Assume interest at 15%.

Note: Correct diagram is worth 10 points.