engg economy 061911 rev %5bcompatibility mode%5d

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GE 301:

ENGINEERING ECONOMY

Engr. Raymond Marquez

• EEC President 1987 & IIEE CSC Chairman 1987

• BSEE 1988, UST

• MBA 1999, DLSU

• REE, Asean Engr.

• IIEE National President, 2007

• ENPAP National President, 2008

• EE Consultant to ADB, WB/IFC, BPI, KFW/LBP

• Director for Operations/Business Devt – Cofely

Phils.

Background

Methodology

• Lecture 20%

• Applications of Engg Economy 40%

• Quizzes/Prelims/Finals 40%

• References:

Engineering Economy, 3rd ed. (Engr. Hipolito

Sta. Maria)

Engineering Economy, 15th ed.

(Sullivan, Wicks, Koelling)

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Tentative Schedule3EEa 3EEb

Jun 7 Overview Overview

Jun 10 Intro to Engg Economy Intro to Engg Economy

Jun 14 Economic

Environment

Economic

Environment

Jun 17 Time Value of Money Time Value of Money

Jun 21 Time Value of Money Time Value of Money

Jun 24 No classes No classes

Jun 28 No classes No classes

Jul 1 No classes No classes

Intro to Engineering Economy, S1

• Engineering Economy – systematic evaluation

of the economic merits of proposed solutions

to engineering problems.

Intro to Engineering Economy

• Principles of Engineering Economy

1. Develop the alternatives

2. Focus on the difference

3. Use of a consistent viewpoint

4. Use of a common unit of measure

5. Consider all relevant criteria

6. Make risk and uncertainty explicit

7. Revisit your decision

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• Engineering Economy Analysis Procedure

�Problem recognition, definition, and evaluation.

�Development of the feasible alternatives.

�Development of the outcomes and cash flows for each alternative.

�Selection of criteria.

�Analysis and comparison of the alternative.

�Selection of the preferred alternative.

�Performance monitoring and post evaluation of results.


Example 1-1 Defining the Problem and Developing Alternatives

• The management team of a small furniture mfg. company is

under pressure to increase profitability to get a much-needed

loan from the bank to purchase a more modern pattern-

cutting machine. One proposed solution is to sell waste wood

chips and shavings to a local charcoal manufacturer instead of

using them to fuel space heaters for the company’s office and

factory areas.

• A. Define the company’s problem. Next, reformulate the

problem in a variety of creative ways.

• B. Develop at least one potential alternative for your

reformulated problems in A.

Intro to Engineering EconomyExample 1-1 Defining the Problem and Developing

Alternatives

Solution:

• A. The company’s problem appears to be that revenues are

not sufficiently covering costs.

1. The problem is to increase revenues while reducing

costs.

2. The problem is to maintain revenues while reducing

costs.

3. The problem is an accounting system that provides

distorted cost information.

4. The problem is that the new machine is really not

needed.

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Example 1-1 Defining the Problem and Developing Alternatives

Solution:

• B. Based on reformulation 1, an alternative is to sell wood chips and shavings as long as increased revenue exceeds extra expenses that may be required to heat the buildings. Another alternative is to discontinue the manufacture of specialty items and concentrate on standardized, high volume products. Yet another alternative is to pool purchasing, accounting, engineering, and other white-collar support services with other small firms in the area by contracting with a local company involved in proving these services.


Example 1-2 Application of Engineering Economic Analysis Procedure

• A friend of yours bought a small apartment building for P100,000 in a college town. She spent P10,000 of her own money for the building and obtained a mortgage from a local bank for the remaining P90,000. The annual mortgage payment to the bank is P10,500. Your friend also expects that annual maintenance on the building and grounds will be P15,000. There are 4 apartments in the building that can each be rented for P360/month.



• A. Does your friend have a problem? If so, what is it?

• B. What are her alternatives?

• C. Estimate the economic consequences and other required data for the alternatives in B.

• D. Select a criterion for discriminating among alternatives, and use it to advise your friend on which course of action to pursue.

• E. Attempt to analyze and compare the alternatives in view of at least one criterion in addition to cost.

• F. What should your friend do based on the information you and she have generated?

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SolutionA. A quick set of calculations shows that your friend does

indeed have a problem. A lot more money is being spent by your friend each year (P10,500+P15,000 = P25,500) than is being received (4 x P360 x 12 = P17,280). The problem could be that the monthly rent is too low. She’s losing P8,220 per year.

B. Option 1. Raise the rent.

Option 2. Lower maintenance expenses.

Option 3. Sell the apartment building.

Option 4. Abandon the building.

Intro to Engineering EconomyExample 1-2 Application of Engineering Economic Analysis Procedure

C. Option 1. Raise the total monthly rent to P1,440 + P Rent for

the 4 apartments to cover monthly expenses of P2,125. Note

that the minimum increase in rent would be (P2,125 – P1,440)/4

= P171.25 per apartment per month.

Option 2. Lower monthly expenses to P2,125 – P Cost so that

these expenses are covered by the monthly revenue of

P1,440/month. Monthly maintenance expenses would have to

be reduced to (P1,440 – P10,500/12) = P565.

Option 3. Try to sell the apartment building for P X, which

recovers the original P10,000 investment and recovers the

P685/month loss (P8,220/12) on the venture during the time it

was owned.

Option 4. Walk away from the venture and kiss your investment

goodbye. The bank would likely assume possession through

foreclosure and may try to collect fees from your friend.

Intro to Engineering EconomyExample 1-2 Application of Engineering Economic Analysis

Procedure

D. One criterion could be to minimize the expected loss of

money. In this case, you might advise your friend to

pursue Option 1 or 3.

E. Let’s use “credit worthiness” as an additional criterion.

Option 4 is immediately ruled out. Option 3 could also

harm your friend’s credit rating. Thus Option 1 & 2 may

be her only realistic and acceptable alternatives.

F. Your friend should probably do a market analysis of

comparable housing in the area to see if the rent could

be raised. Maybe a fresh coat of paint and new

carpeting would make the apartments more appealing to

prospective renters.

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Economic Environment, S2

• Consumer goods/services – products or

services that are directly used by people to

satisfy their wants.

• Producer goods/services – used to produce

consumer goods and services or other

producer goods.

Economic Environment

• Necessities – products or services that are required to support human life and activities, that will be purchased in somewhat the same quantity even though the price varies considerably.

• Luxuries – products or services that are desired by humans and will be purchased if money is available after the required necessities have been obtained.


• Demand – quantity of a certain commodity that is bought at a certain price at a given place and time.

• Elastic demand – occurs when a decrease in selling price result in a greater than proportionate increase in sales.

• Inelastic demand – occurs when a decrease in the selling price produces a less than proportionate increase in sales.

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Price-Demand Relationship Price-Supply Relationship


• Perfect competition – occurs in a situation where a commodity or service is supplied by a no. of vendors and there is nothing to prevent additional vendors entering the market.

• Monopoly – opposite of perfect competition.

• Oligopoly – exists when there are so few suppliers of a product or service that action by one will almost inevitably result in similar action by the others.


• Law of Supply & Demand – under conditions of perfect competition the price at which a given product will be supplied and purchased is the price that will result in the supply and demand being equal.

• Supply – quantity of a certain commodity that is offered for sale at a certain price at a given place and time.

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Fixed, Variable, & Incremental

Costs• Type of Costs

– Fixed costs – costs which remain constant, whether or not a given change in

operations or policy is adopted.

– Variable costs – costs which vary with output or any change in the activities of

an enterprise.

– Incremental costs – costs that arise as the result of a change in operations or

policy. A very simple example would be a factory making nail where it takes

one employee an hour to make a nail. As a simple

figure, the incremental cost of a nail would be the wages for the employee for

an hour plus the cost of the materials needed to produce a nail. A more

accurate figure could include added costs, such as shipping the additional nail

to a customer, or the electricity used if the factory has to stay open longer.

– Marginal cost – additional cost of producing one more unit of a product.

Marginal cost and average cost can differ greatly. For example, suppose it

costs $1000 to produce 100 units and $1020 to produce 101 units. The

average cost per unit is $10, but the marginal cost of the 101st unit is $20

– Sunk cost – represents money which has been spent or capital which has been

invested and which cannot be recovered due to certain reasons.

Direct, Indirect and Standard Costs

• Direct Cost – costs that can be reasonably measured and allocated to a specific output or work activity, e.g. labor, materials

• Indirect Cost – costs that are difficult to attribute or allocate to a specific output or work activity, e.g. cost of tools, general supplies, electricity (overhead)

• Standard Cost – planned costs per unit of output that are established in advance of actual production or service delivery.

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Opportunity Cost & Life-Cycle Cost

• Opportunity Cost – incurred because of use of limited

resources.

Example: A firm is considering the replacement of an existing

piece of equipment that originally cost $50,000, is presently

shown on the company records with a value of $20,000, but

has a present market value of only $5,000. For purposes of an

engineering economic analysis of whether to replace the

equipment, the present investment in that equipment should

be considered as $5,000, because, by keeping the

equipment, the firm is giving up the opportunity to obtain

$5,000 from its disposal. Thus, the $5,000 immediate selling

price is really the investment cost of not replacing the

equipment and is based on the opportunity cost concept.

Life Cycle Cost (LCC) analysis is a management tool

that can help companies minimize waste and

maximize energy efficiency for many types of systems.

The LCC of any piece of equipment is the total

“lifetime” cost to

purchase, install, operate, maintain, and dispose of

that equipment.

Life Cycle Cost AnalysisLife Cycle Cost Analysis

Life Cycle Cost AnalysisLife Cycle Cost Analysis

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• Initial costs

• Installation and commissioning costs

• Energy costs

• Operation costs

• Repair and maintenance costs

• Down time costs

• Environmental costs

• Decommissioning and disposal costs

Life Cycle Cost Analysis ComponentsLife Cycle Cost Analysis Components

0%

20%

40%

60%

80%

100%

120%

IB 50W CFL 9W

PV Energy

PV Replacement

First Cost

LCCA for IB & CFLLCCA for IB & CFL

LCCA for Hot WaterLCCA for Hot Water

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0%

20%

40%

60%

80%

100%

120%

STACU 1TR Solar Cooling 1TR

PV Energy

PV

Maintenance

First Cost

LCCA for STACU & Solar Assisted CoolingLCCA for STACU & Solar Assisted Cooling

Sample, S2An electrical contractor has a job which should be

completed in 100 days. At present, he has 80 men

on the job and it is estimated that they will finish the

work in 130 days. If of the 80 men, 50 are paid P190

a day, 25 at P220 a day, and 5 at P300 a day and if for

each day beyond the original 100 days, the

contractor has to pay P2000 liquidated damages.

(a) How many more men should the contractor add

so he can complete the work on time?

(b) If the additional men of 5 are paid P220 a day and

the rest at P 190 a day, would the contractor save

money by employing more men and not paying the

fine?

Solution: (a) Let x = no. of men to be added to complete

the job on time(x+80)(100) = (80)(130)

x = 24 men(b) 80 men on the job

Wages: 50xP190x130 = P1,235,00025xP220x130 = P 715,0005xP300x130 = P 195,000

Damages P2000x30 = P 60,000104 men on the jobWages: (50+19)(190)(100) = P1,311,000

(25+5)(190)(100) = P 660,0005xP300x100 = P 150,000

Savings = P2,205,000 – P2,121,000 = P84,000

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Time Value of Money, S3

• A dollar today is worth more than a dollar one

or more years from now (for several reasons).

• Interest and profit are payments for the riskthe investor takes in letting another use his or her capital.

• Any project or venture must provide a sufficient return to be financially attractive to the suppliers of money or property.

Time Value of Money

Simple Interest - when the total interest earned

is linearly proportional to the initial amount of

the loan (principal), the interest rate, and the

number of interest periods for which the

principal is committed.

P = principal amount lent or borrowed

N = number of interest periods (e.g., years)

i = interest rate per interest period

The total amount repaid at the end of N interest

periods is P + I.

Time Value of Money

If P5,000 were loaned for five years at a

simple interest rate of 7% per year, the

interest earned would be

So, the total amount repaid at the end

of five years would be the original

amount (P5,000) plus the interest

(P1,750), or P6,750.

I = P5,000 x 5 x 0.07 = P1,750

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Time Value of MoneyCompound interest reflects both the remaining

principal and any accumulated interest. For P1,000 at

10%…

Period

(1)

Amount owed

at beginning of

period

(2)=(1)x10%

Interest

amount for

period

(3)=(1)+(2)

Amount

owed at end

of period

1 P1,000 P100 P1,100

2 P1,100 P110 P1,210

3 P1,210 P121 P1,331

Compound interest is commonly used in personal and

professional financial transactions.

Time Value of Money

• Notation used in formulas for compound interest calculations.

– i = effective interest rate per interest period

– N = number of compounding (interest) periods

– P = present sum of money; equivalent value of one or more cash flows at a reference point in time; the present

– F = future sum of money; equivalent value of one or more cash flows at a reference point in time; the future

– A = end-of-period cash flows in a uniform series continuing for a certain number of periods, starting at the end of the first period and continuing through the last

Time Value of MoneyA cash flow diagram is an indispensable

tool for clarifying and visualizing a series

of cash flows.

0 1 2 3 4 = N

End of Period

Start ofPeriod 1

End ofPeriod 1

P = ? (outflow or disbursement)

F = ? (inflow or receipt)

i = ?% per period

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Time Value of Money

0 1 2 3 4 = N

End of Year

10,000

2,000

5,310

3,000

Time Value of MoneyCash flow tables are essential to modeling

engineering economy problems in a

spreadsheet

Time Value of Money

Using the standard notation, we find that a

present amount, P, can grow into a future

amount, F, in N time periods at interest rate

i according to the formula below.

In a similar way we can find P given F by

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Time Value of MoneyIt is common to use standard notation

for interest factors.

This is also known as the single payment

compound amount factor. The term on the

right is read “F given P at i% interest per

period for N interest periods.”

is called the single payment present worth

factor.

Time Value of Money

P2,500 at time zero is equivalent to how much after six

years if the interest rate is 8% per year?

P3,000 at the end of year seven is equivalent to how

much today (time zero) if the interest rate is 6% per

year?

F = P2,500 (F/P, 8%, 6) = P2,500 (1.5869) = P3,967

P = P3,000 (P/F, 6%, 7) = P3,000(0.6651) = P1,995

Sample on Compound Interest

A P2000 loan was originally made at 8% simple

interest for 4 years. At the end of this period

the loan was extended for 3 years, without the

interest being paid, but the new interest rate

was made 10% compounded semiannually.

How much should the borrower pay at the

end of 7 years?

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0 1 2 3 4 5 6 7

F7F4

P2000

Simple interest Compound interest

Solution:

F4 = P (1+ni) = P2000 (1+ (4)(0.08)) = P2,640

F7 = F4 (1+i)^n = P2,640 (1+0.05)^6 = P3,537.86

Time Value of MoneyThere are interest factors for a series of

end-of-period cash flows.

How much will you have in 40 years if you

save $3,000 each year and your account

earns 8% interest each year?

Time Value of MoneyFinding the present amount from a

series of end-of-period cash flows.

How much would is needed today to provide

an annual amount of $50,000 each year for 20

years, at 9% interest each year?

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Time Value of MoneyFinding A when given F.

How much would you need to set aside each

year for 25 years, at 10% interest, to have

accumulated $1,000,000 at the end of the 25

years?

Time Value of MoneyFinding A when given P.

If you had $500,000 today in an account

earning 10% each year, how much could you

withdraw each year for 25 years?

Time Value of Money

Acme Steamer purchased a new pump for $75,000. They borrowed the money for the pump from their bank at an interest rate of 0.5% per month and will make a total of 24 equal, monthly payments. How much will Acme’s monthly payments be?

Pause and solve

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Time Value of Money, S4Finding N

Acme borrowed $100,000 from a local bank, which

charges them an interest rate of 7% per year. If Acme

pays the bank $8,000 per year, now many years will it

take to pay off the loan?

So,

Time Value of MoneyFinding i

Jill invested $1,000 each year for five years in a local

company and sold her interest after five years for

$8,000. What annual rate of return did Jill earn?

So,

Sample on Annuity

A businessman needs P50,000 for his

operations. One financial institution is willing

to lend him the money for one year at 12.5%

interest per annum (discounted). Another

lender is charging 14%, with the principal and

interest payable at the end of one year. A

third financier is willing to lend him P50,000

payable in 12 equal monthly installments of

P4,600. Which offer is best for him?

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0 1

P X = P50,000 (1-0.125) = P43,750

First Offer:Rate of interest = (P50,000 – P43,750) /P43,750 = 14.29%/yr (eff rate)

Compare the effective rate of each offer and select the one with the lowest effective rate.

P50,000

Second Offer:14%/yr (eff rate)

0 1

P50,000

P50,000 (1.14)

0 1 12

P50,000

P4,600 P4,600

Third Offer:P/A,i%,12= P/A(1-(1+i)^-12)/i)=10.87Try i=1% & 2%:

i = 1.57%/mo.

Eff rate = (1+0.0157)^12 – 1 = 20.26%/yr

Sample on AnnuityToday, you invest P100,000 into a fund that pays

25% interest compounded annually. Three

years later, you borrow P50,000 from a bank

at 20% annual interest and invest in the fund.

Two years later, you withdraw enough money

from the fund to repay the bank loan and all

interest due on it. Three years from this

withdrawal you start taking P20,000 per year

out of the fund. After five withdrawals, you

withdraw the balance in the fund. How much

was withdrawn?

0 1 2 3 4 5 6 7 8 9 10 11 12

Q

P20,000 each

P100,000

P50,000 (1.20)^2

P50,000 (1.20)^2 (F/P, 25%, 7)

P20,000 (F/A, 25%, 5)

P100,000 (F/P, 25%, 12)

P50,000

P50,000 (F/P, 25%, 9)

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Solution: Let Q = amt. withdrawn after 12 yrs

Using 12 yrs as Focal date: equation of value is

Q + P20,000 (F/A, 25%, 5) + P50,000 (1.20)^2 (F/P, 25%, 7)= P100,000 (F/P, 25%, 12) + P50,000 (F/P, 25%, 9)

Q+P20,000(8.2070)+P50,000(1.20)^2(4.7684) = P100,000(14.5519) + P50,000 (7.4506)

Q = P1,320,255

Sample on AnnuityA certain property is being sold and the owner

received two bids. The first bidder offered to

pay P400,000 each year for 5 years, each

payment is to be made at the beginning of

each year. The second bidder offered to pay

P240,000 first year, P360,000 the second year

and P540,000 each year for the next 3

years, all payments will be made at the

beginning of each year. If money is worth 20%

compounded annually, which bid should the

owner of the property accept?

0 1 2 3 4 5

P1

P400,000 Yr (0 – 4)

Solution:Let P1 = present worth of the 1st bid

P1 = A (1 + P/A, 20%, 4)

= P400,000 (1+2.5887)= P1,435,480

First Bid

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0 1 2 3 4 5

P2

P540,000 Yr (2 – 4)

Solution:Let P2 = present worth of the 2nd bid

P2 = P240,000 + P360,000

(P/F,20%,1)+P540,000(P/A,20%,3)(P/F,20%,1) = P1,487,875 (Choose 2nd bid)

Second Bid

P360,000

P240,000

Time Value of MoneyWe need to be able to handle

cash flows that do not occur until

some time in the future.

• Deferred annuities are uniform series that

do not begin until some time in the future.

• If the annuity is deferred J periods then the

first payment (cash flow) begins at the end

of period J+1.

Time Value of MoneyFinding the value at time 0 of a

deferred annuity is a two-step

process.

1. Use (P/A, i%, N-J) find the value of the deferred annuity at the end of period J(where there are N-J cash flows in the annuity).

2. Use (P/F, i%, J) to find the value of the deferred annuity at time zero.

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Sample on Deferred AnnuityA debt of P40,000, whose interest rate is 15%

compounded semiannually, is to be

discharged by a series of 10 semiannual

payments, the first payment to be made 6

months after consummation of the loan. The

first 6 payments will be P6,000 each, while the

remaining 4 payments will be equal and of

such amount that the final payment will

liquidate the debt. What is the amount of the

last 4 payments?

0 1 2 3 4 5 6 7 8 9 10

P6,000 (P/A, 7.5%, 6)

P40,000

P6,000 Yr (1 – 6) A Yr (7-10)

A (P/A, 7.5%, 4)(P/F, 7.5%, 6) A (P/A, 7.5%, 4)

Solution: Using today as focal date, the equation of value isP40,000 = P6,000 (P/A, 7.5%, 6) + A(P/A,7.5%, 4)(P/F,7.5%,6)

P40,000 = P6,000(4.6938)+A(3.3493)(0.6480)A = P5,454

Sample on AmortizationA debt of P10,000 with interest at the rate of

20% compounded semiannually is to be

amortized by 5 equal payments at the end of

each 6 months, the first payment is to be

made after 3 years. Find the semiannual

payment and construct an amortization

schedule.

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0 1 2 3 4 5 6 7 8 9 10

P10,000

A Yr (6-10)

Solution:

P = A(P/A,10%,5)(P/F,10%,5)

A = P10,000 (0.2638)(1.6105)

A = P4,248.50

Amortization Schedule

Period Outstanding

Principal at

beginning of

period

Interest due at end

of period

Payment Principal

repaid at end

of period

1 10,000.00 1,000.00

2 11,000.00 1,100.00

3 12,100.00 1,210.00

4 13,310.00 1,331.00

5 14,641.00 1,464.10

6 16,105.10 1,610.51 4,248.50 2,637.99

7 13,467.11 1,346.71 4,248.50 2,901.79

8 10,565.32 1,056.53 4,248.50 3,191.97

9 7,373.35 737.34 4,248.50 3,511.16

10 3,862.19 386.22 4,248.50 3,862.28

P11,242.41 P21,242.50 P16,105.19

Time Value of Money

Irene just purchased a new sports car and wants to also set aside cash for future maintenance expenses. The car has a bumper-to-bumper warranty for the first five years. Irene estimates that she will need approximately $2,000 per year in maintenance

expenses for years 6-10, at which time she will sell the vehicle. How much money should Irene deposit into an account today, at 8% per year, so that she will have sufficient funds in that account to cover her projected maintenance expenses?

Pause and solve

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Time Value of Money

Uniform Arithmetic Gradient

Economic analysis problems involve receipts

or disbursements that increase or decrease

by a uniform amount each period. For

example, Repairs & Maintenance expenses on

specific equipment or property may increase

by a relatively constant amount each period.

Time Value of MoneyIt is easy to find the present

value of a uniform gradient

series.Similar to the other types of cash flows, there is a

formula (albeit quite complicated) we can use to find

the present value, and a set of factors developed for

interest tables.

Time Value of MoneyWe can also find A or F

equivalent to a uniform gradient

series.

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Time Value of MoneyEnd of Year Cash Flows ($)

1 2,000

2 3,000

3 4,000

4 5,000

The annual equivalent of this series

of cash flows can be found by

considering an annuity portion of the

cash flows and a gradient portion.

End of Year Annuity ($) Gradient ($)

1 2,000 0

2 2,000 1,000

3 2,000 2,000

4 2,000 3,000

Sample on Uniform Arith. GradientA loan was to be amortized by a group of four

end of year payments forming an ascending

arithmetic progression. The initial payment

was to be P5,000 and the difference between

successive payments was to be P400. But the

loan was renegotiated to provide for the

payment of equal rather than uniformly

varying sums. If the interest rate of the loan

was 15%, what was the annual payment?

0 1 2 3 4

P

P5,000P5,400

P5,800P6,200

0 1 2 3 4

Pa

P5,000 Yr (1 – 4)

= +

0 1 2 3 4

Pg

P400P800

P1,200

=

0 1 2 3 4

P

A’ Yr (1 – 4)

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Solution: Let A’ = annual payment; A=P5,000; G=P400, n=4; i = 15%

P/A,15%,4 = (1-(1.15)^-4)/0.15 = 2.8550

P/G,15%,4 = (1/0.15)[(((1.15)^4-1)/0.15)-4][1/((1.15)^4)] = 3.7865

P = A(P/A,15%,4) + G (P/G,15%,4)= (P5,000)(2.8550) + (P400)(3.7865)= P15,789.60

A’(P/A,15%,4) = P15,789.60A’ = P5,530.51

Time Value of MoneyNominal and effective interest

rates.• More often than not, the time between

successive compounding, or the interest period, is

less than one year (e.g., daily, monthly, quarterly).

• The annual rate is known as a nominal rate.

• A nominal rate of 12%, compounded monthly,

means an interest of 1% (12%/12) would accrue

each month, and the annual rate would be

effectively somewhat greater than 12%.

• The more frequent the compounding the greater

the effective interest.

Time Value of MoneyThe effect of more frequent

compounding can be easily

determined.

Let r be the nominal, annual interest rate and M the

number of compounding periods per year. We can

find, i, the effective interest by using the formula

below.

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Time Value of MoneyFinding effective interest rates.

For an 18% nominal rate, compounded quarterly, the

effective interest is.

For a 7% nominal rate, compounded monthly, the

effective interest is.

Time Value of MoneyContinuous compounding

interest factors.

The other factors can be found from these.

Sample on Interest

Compare the accumulated amounts after 5

years of P1,000 invested at the rate of 10% per

year compounded (a) annually, (b) semi

annually, (c) quarterly, (d) monthly, (e) daily,

and (f) continuously.

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Solution:

Using the formula, F = P (1+i)^n

(a) F = P1000 (1+0.10)^5 = P1,610.51(b) F = P1000 (1+0.10/2)^10 = P1,628.89(c) F = P1000 (1+0.10/4)^20 = P1,638.62(d) F = P1000 (1+0.10/12)^60 = P1,645.31(e) F = P1000 (1+0.10/365)^1825 = P1,648.61(f) F = Pe^rn = P1000 (e)^(0.10x5) = P1,648.72

engg economy 061911 rev %5bcompatibility mode%5d

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