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6/22/2011
1
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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Intro to Engineering Economy
• 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.
Intro to Engineering Economy
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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Intro to Engineering Economy
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.
Intro to Engineering Economy
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.
Intro to Engineering Economy
Example 1-2 Application of Engineering Economic Analysis Procedure
• 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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Intro to Engineering Economy
Example 1-2 Application of Engineering Economic Analysis Procedure
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.
Economic Environment
• 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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Economic Environment
Price-Demand Relationship Price-Supply Relationship
Economic Environment
• 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.
Economic Environment
• 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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Economic Environment
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 $2,000 per year in maintenance
expenses for years 6-10, at which time she will sell the vehicle. How much money should Irene deposit into an account today, at 8% per year, so that she will have sufficient funds in that account to cover her projected maintenance expenses?
Pause and solve
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Time Value of Money
Uniform Arithmetic Gradient
Economic analysis problems involve receipts
or disbursements that increase or decrease
by a uniform amount each period. For
example, Repairs & Maintenance expenses on
specific equipment or property may increase
by a relatively constant amount each period.
Time Value of MoneyIt is easy to find the present
value of a uniform gradient
series.Similar to the other types of cash flows, there is a
formula (albeit quite complicated) we can use to find
the present value, and a set of factors developed for
interest tables.
Time Value of MoneyWe can also find A or F
equivalent to a uniform gradient
series.
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25
Time Value of MoneyEnd of Year Cash Flows ($)
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
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