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MS-291: Engineering Economy(3 Credit Hours)

ARITHMETIC GRADIENT

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Arithmetic Gradient Factors(P/G, A/G)

• Cash flows that increase or decrease by a constant amountare considered arithmetic gradient cash flows.

• The amount of increase (or decrease) is called the gradient

CFn = base amount + (n-1)GCash Flow Formula

G = $25Base = $100

$100$125

$150

$175

0 1 2 3 4

$500

G = -$500Base = $2000

$2000$1500

$1000

0 1 2 3 4

Gradientseriescould beboth: cashinflow (asgivenhere) orOutflows

• When we have a “Gradient” Series wecannot apply Single Amount PresentWorth/Future Worth factors or UniformSeries factors

• We have to use a different methodology toaddress problems related to gradient cashflows.

Arithmetic Gradient Factors(P/G, A/G)

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Solving Arithmetic Gradientrelated problems

Present value of the Arithmetic Gradient series can becalculated as follows:

1. Find the gradient and base2. Cash flow diagram maybe helpful if you draw it3. Break the gradient series into a Uniform series and a

Gradient Series as shown on next slide4. The formula for calculating present value of the

Arithmetic Gradient series is as follows;

5. Calculate PA and PG and use the above formula toget the present value of the Arithmetic Gradient

PT = PA + PG

Arithmetic Gradient Factors(P/G, A/G)

• The base amount is “A” and the “Gradient is “G” in thefollowing graph

CFn = base amount + (n-1)GCash Flow Formula

A A A A AAA+G

A+2GA+3G

A+(n-1) G

0 1 2 3 4 n 0 1 2 3 4 n

0 G2G

3G(n-1)G

0 1 2 3 4 n

= +

PT = PA + PG

Important!!!PG series startwith year 2

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Arithmetic Gradient Factors(P/G, A/G)

• PA = A(P/A, i, n) or Uniform Series Present worthFactor

• PG = G(P/G, i, n) or Arithmetic Gradient PresentWorth Factor … you can use table for it too.

• Alternatively, PG can also be calculated byfollowing formula

PT = PA + PG

n

n

Gii

ini

i

GP

)1(

1)1(2

nn

n

Gi

n

ii

i

i

GP

)1()1(

1)1( Or

Arithmetic Gradient Factors(P/G, A/G)

$100$125

$150$175

G = $25

Base = $100

0 1 2 3 4 0 1 2 3 4

$100

0 1 2 3 4

$25$50

$75

$P = $100(P/A,i,4) + $25(P/G,i,4)

PT = PA + PG

Where PA = Present worth uniform series (P/A, i,n) and PG = present worth of the gradient series (P/G,i, n)

nn

n

G i

n

ii

i

i

GP

)1()1(

1)1(

Equivalent cash flows:

=> +

Note: the gradient series(PG) by convention startsin year 2.

Note: Annuity series (PA)starts from year 1.

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Example (Problem 2.25)Profits from recycling paper, cardboard, aluminium,and glass at a liberal arts college have increased at aconstant rate of $1100 in each of the last 3 years.If this year’s profit (end of year 1) is expected to be$6000 and the profit trend continues through year 5,(a) what will the profit be at the end of year 5 and(b) what is the present worth of the profit at an interest

rate of 8% per year?

G = $1100, Base = $6000

Example (Problem 2.25)(a) what will the profit be at the end of year 5 &(b) what is the present worth of the profit at an interest rate of 8% per

year?

G = $1100 Base = $6000

$6000$7100

$8200

$9300

0 1 2 3 4 5

Find the cash flows as follows:CF = Base + G(n-1)CF1 = 6000 + 1100(1-1)= 6000CF2 = 6000 + 1100(2-1)= 7100CF3 = 6000 + 1100(3-1)= 8200CF4 = 6000 + 1100(4-1)= 9300CF5 = 6000 + 1100(5-1)= 10400

$10400

=>0 1 2 3 4 5

$6000

+0 1 2 3 4 5

$1100$2200

$3300

$4400

P = A(P/A, i, n) G(P/G, i, n)+P = 6000(P/A, 8%, 5) + 1100(P/G, 8%, 5)

P = 6000(3.9927) + 1100(7.3724)

P = 32066

PT = PA + PG

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Practice Question: 5 minutesNeighboring parishes in Louisiana have agreed to poolroad tax resources already designated for bridgerefurbishment. At a recent meeting, the engineersestimated that a total of $500,000 will be deposited atthe end of next year into an account for the repair ofold and safety-questionable bridges throughout the area.Further, they estimate that the deposits will increaseby $100,000 per year for only 9 year thereafter, thencease. Determine the equivalent: present worth, if publicfunds earn at a rate of 5% per year.

5% Uniform Series Factors Athematic Gradientn Sinking

Fund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

GradientPresent Worth(P/G)

GradientUniformSeries (A/G)

9 0.09069 11.0266 0.14069 7.1078 26.1268 3.675810 0.07950 12.5779. 0.12950 7.7217 31.6520 4.0991

Solution

• Base = 500,000• Gradient = 100,000• Taking units in 1000• Base = 500• Gradient =100• i= 5%• n=1+9 = 10

0 1 2 3 4 5 6 7 8 9 10

$500$600

$700$800

$900$1000 $1100

$1200$1300

$1400

PT = 500(P/A,5%,10) + 100(P/G,5%,10)= 500(7.7217) + 100(31.6520)=$7026.05 or ….. ($7,026,050)

PT = PA + PG

P = ?

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…. What about A/G or F/G?

Arithmetic Gradient UniformSeries Factor (A/G)

• Similar procedure as done for Arithmetic Gradient Presentworth Factor

• Following formula:

AT = AA + AG

= 1 − (1 + ) −1AG can be get fromfactor tables or throughgiven formula in box

AA = A (Annual Worth) Given as base value of G series and AG = G(A/G, i, n)

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Using Factor Tables

Example: A/G factor

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Example: A/G Factor

Neighboring parishes in Louisiana have agreed to poolroad tax resources already designated for bridgerefurbishment. At a recent meeting, the engineersestimated that a total of $500,000 will be deposited atthe end of next year into an account for the repair ofold and safety-questionable bridges throughout the area.Further, they estimate that the deposits will increaseby $100,000 per year for only 9 year thereafter, thencease.Determine the equivalent: Annual series amount, ifpublic funds earn at a rate of 5% per year.

Solution

• Base = 500,000• Gradient = 100,000• Taking units in 1000• Base = 500• Gradient =100• i= 5%• n=1+9 = 10

0 1 2 3 4 5 6 7 8 9 10

$500$600

$700$800

$900$1000 $1100

$1200$1300

$1400

AT = 500 + 100(A/G,5%,10)= 500 + 100(4.0991)=$909.91 or ….. ($909,910)

AT = AA + AG

0 1 2 3 4 5 6 7 8 9 10

A= $909,910

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Arithmetic Gradient FutureWorth Factor (F/G)

• Another factor in “Gradient family” is “Future” worthof an Arithmetic Gradient series (F/G)

• It can be obtained by multiply (P/G) and (F/P) factors

× =

Note: To get future value of Arithmetic Gradient …. We do not need todivide the gradient into two separate cash flows like Present worth ofArithmetic gradient series…. This F/G will be the future value of entiregradient series.

/ = 1 (1 + ) −1 −No factor table values is available so only formula can be use for calculating “F/G” factor

Final Words about ArithmeticGradient

Present Worth(PW) or Annual Worth(AW) of Arithmetic Gradient (P/Gor A/G)…. Base and Gradient considered separately for both P/G and A/G…. Get Two series… a PA/AA series and one PG/AG series….use the factor tables to get values for PA/AA & PG/AG

…… Add both to get PT/AT.

Future worth of Arithmetic Gradient (F/G)… Base and Gradient are not considered separately…. No factor values are available so have to relay on formula….formula directly calculate the future worth of Arithmetic Gradient/ = 1 (1 + ) −1 −

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Geometric Gradient Factors(Pg /A)

• A Geometric gradient is when the periodic payment isincreasing (decreasing) by a constant percentage:

• the rate at which the cash flow is increasing is “g”• The initial amount of Geometric Gradient is A1• Pg is the present worth of entire Gradient Series including A1

• It is important to note that “Initial amount is not consideredseparately while working with “Geometric Gradient”

Note: If g is negative, change signs in front of both g values

= 1 − 1 +1 +−for g ≠ i: for g = i:= 1 +

Geometric Gradient Factors(Pg /A)

• A Geometric gradient is when the periodic payment isincreasing (decreasing) by a constant percentage:

A1 = $100, g = 10% or 0.1A2 = $100(1+g)A3 = $100(1+g)2

An = $100(1+g)n-10 1 2 3 4 ……n

where: A1 = cash flow in period 1 and g = rate of increase

Note: If g is negative, change signs in front of both g values

= 1 − 1 +1 +−for g ≠ i:

for g = i:= 1 +

A1

A1 (1+g) A1 (1+g)2

A1 (1+g)n-1

It maybe noted that A1is not consideredseparately in geometricgradients

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Future Worth (F/G) and Annuity(A/G) from Geometric Gradient

Series

• We just learned how to get “Present Value” of aGeometric Gradient

• We can first derive the Present Worth of the GeometricGradient and then can use F/P factor forcalculating future value of a geometric gradient

• Similarly, A/P factor can be applied to P/G factor tocalculate the Annual worth/Annuity series fromGeometric Gradient

Class Practice: 4 Minutes

Determine the present worth of a geometricgradient series with a cash flow of $50,000 inyear 1 and increases of 6% each yearthrough year 8. The interest rate is 10% peryear.

= 1 − 1 +1 +−for g ≠ i:

for g = i:= 1 +

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Class Practice: 4 Minutes

Determine the present worth of a geometricgradient series with a cash flow of $50,000 inyear 1 and increases of 6% each yearthrough year 8. The interest rate is 10% peryear.

gii

g

APg

n

11

1

06.010.010.016.01

150000

8

04.0

257.0000,50 )425.6(000,50

250,321$

04.0

743.01000,50

Summary of allFactors!!!

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Single payment Factors

F/P FactorF= P(F/P, i%, n)

P/F FactorP= F(P/F, i%, n)

P is given

F = ?

n and i is given

F = given

n and i is given

P =?

Uniform Series Factors

P/A FactorP = A(P/A, i%, n)

A/P FactorA = P(A/P, i%, n)

F/A FactorF = A(F/A, i%, n)

A/F FactorA = F(A/F, i%, n)

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Athematic Gradient

PA = A(P/A, i%, n)

PG = G(P/G, i%, n)

FG = ?

F = PT(F/P, i%, n)

or

/ = 1 (1 + ) −1 −

Athematic Gradient

PA = A(P/A, i%, n)

PG = G(P/G, i%, n)

A = PT(A/P, i%, n)or

AT = AA + AG

= 1 − (1 + ) −1AA = A (Annual Worth) &AG = G(A/G, i, n)

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Geometric Gradient

Pg = ?

= 1 − 1 +1 +−for g ≠ i:

for g = i:= 1 +

Fg = ?

F = Pg(F/P, i%, n)Similarly …

A = Pg(A/P, i%, n)

Engineering Economy

Chapter 3Combining Factorsand Spreadsheet

Functions

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This Chapter Objectives

1. Shifted uniform series

2. Shifted series and single cash flows

3. Shifted gradients

Example

0 62 3 4 51 7 8 9 10 11 12 13 Year

A = $50P = ?

• Use the P/F factor to find the present worth of each disbursement at year 0 and addthem.

• Use the F/P factor to find the future worth of each disbursement in year 13, addthem, and then find the present worth of the total, using P/F= F( P/F, i ,13).

• Use the F/A factor to find the future amount F/A =A( F/A, i ,10), and then computethe present worth, using P/F=F(P/F, i ,13).

• Use the P/A factor to compute the “present worth” P3 =A( P/A , i ,10) (which will belocated in year 3, not year 0), and then find the present worth in year 0 by using the(P/F , i ,3) factor.

How can we get “Present worth of this series” ?P3 = ? F

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Shifted Uniform Series• Typically the last method is used for calculating the present worth

of a uniform series that does not begin at the end of period 1.

• Note that a P value is always located 1 year or period prior to thebeginning of the first series amount. Why? Because the P/A factorwas derived with P in time period 0 and A beginning at the end ofperiod 1.

• The most common mistake made in working problems of this typeis improper placement of P .

Remember:• When using P/A or A/P factor, PA is always one year ahead of first A• When using F/A or A/F factor, FA is in same year as last A• The number of periods n in the P/A or F/A factor is equal to the number of uniform

series values

0 62 3 4 51 7 8 9 10 11 12 13 Year

A = $50P3 = ?

0 62 3 4 51 7 8 9 10 11 12 13 Year

A = $50 F = ?

PA is always one year aheadof first A

FA is in same year as last A

The number of periods n in the P/A orF/A factor is equal to the number ofuniform series values

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Steps for applying factors toShifted Cash Flows

1. Draw a diagram of the positive and negative cash flows.2. Locate the present worth or future worth of each series on

the cash flow diagram.3. Determine n for each series by renumbering the cash flow

diagram.4. Draw another cash flow diagram representing the desired

equivalent cash flow. (Optional)5. Set up and solve the equations.

ExampleThe offshore design group at Bechtel just purchasedupgraded CAD software for $5000 now and annualpayments of $500 per year for 6 years starting 3 yearsfrom now for annual upgrades. What is the presentworth in year 0 of the payments if the interest rate is8% per year?

1. Draw a diagram of the positive and negative cash flows.Solution

0 62 3 4 51 7 8

P0 = $5000

A = $500PT = ?

2. Locate the presentworth or future worth ofeach series on the cashflow diagram.

P’A = ?PA = ?

i= 8% per year

3. Determine n for eachseries by renumbering thecash flow diagram.

0 62 3 4 51 nYear

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P' A = $500( P /A ,8%,6)

PA = P' A ( P /F ,8%, 2)

P T = P 0 + P A=5000 + 500( P /A ,8%,6)( P / F ,8%,2)=5000 +500(4.6229)(0.8573)$6981.60

5. Set upand solvetheequations.

PA = $500( P /A ,8%,6) ( P /F ,8%, 2)

Class Practice 5 Minutes Time

A = $10,000

0 1 2 3 4 5 6

i = 10%

Calculate the present worth of the cash flow shown below at i = 10%

Actual year

10% Single Payments Uniform Series Factors

n Compound Amount(F/P)

PresentWorth(P/F)

SinkingFund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

1 1.1000 0.9091 1.00000 1.0000 1.10000 0.90915 1.6105 0.6209 0.16380 6.1051 0.26380 3.7908

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Class Practice 5 Minutes Time

A = $10,000PT = ?

0 1 2 3 4 5 6

i = 10%

Calculate the present worth of the cash flow shown below at i = 10%

PT = A(P/A,10%, 5) (P/F,10%,1)

$34462

0 1 2 3 4 5

P’A = ?

Actual yearSeries year

(2) Use P/F factor with n = 1 to move P’A back for PT in year 0

= 10,000(3.7908)(0.9091)

(1) Use P/A factor with n = 5 (for 5 arrows) to get P’A in year 1Solution

A(P/A,10%, 5)----

(P/F,10%, 1)----

Shifted Series and RandomSingle Amounts

• For cash flows that include uniform series and randomly placedsingle amounts:

Uniform series procedures are applied to the series amounts

Single amount formulas are applied to the one-time cash flows

• The resulting values are then combined per the problem statement

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Example

Find the present worth in year 0 for the cash flows shown using an interestrate of 10% per year.

0 1 2 3 4 5 6 7 8 9 10

A = $5000

i = 10%

• Find the cash flows both positive and negatives

$2000

0 1 2 3 4 5 6 7 8 9 10

PT = ?A = $5000

i = 10%

$2000

0 1 2 3 4 5 6 7 8

Solution:

Actual year

Series year

• Locate the present worth/ future worth• Determine the “n” by re-numbering the cash flows series• Uniform series procedures are applied to the series amounts. Single amount

formulas are applied to the one-time cash flows• The resulting values are then combined per the problem statement

Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26,675

Move PA back to year 0 using P/F: P0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22,044Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933Now, add P0 and P2000 to get PT: PT = 22,044 + 933 = $22,977

Example:

PA = ?PT = ?

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Class Practice: 8 MinutesAn engineering company lease the mineral rights to a miningcompany on its land. The engineering company makes aproposal to the mining company that it pay $20,000 per year for20 years beginning 1 year from now, plus $10,000 six years fromnow and $15,000 sixteen years from now. If the mining companywants to pay off its lease immediately, how much should it paynow if the investment is to make 16% per year?

16% Single Payments Uniform Series Factors

n CompoundAmount (F/P)

Present Worth(P/F)

CapitalRecovery (A/P)

Present Worth(P/A)

6 2.4364 0.4104 0.27139 3.68477 2.8262 0.3538 0.24761 4.038616 10.7480 0.0930 0.17641 5.668517 12.4677 0.0802 0.17395 5.748720 19.4608 0.0514 0.16867 5.9228

A =$20,000

P = ?

0 1 2 3 4 5 6 7 16 17 18 19 20

$10,000 $15,000

P = 20,000(P/A ,16%,20)+

= $124,075

10,000( P /F ,16%,6) +15,000(P/F,16%,16)

P = $20,000(5.9288)+ $ 10,000( 0.4104) + $ 15,000(0.0930)

Solution

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Important Points for P and Aof Shifted Gradient Series

• Must use multiple factors to find P in actual year 0, for shiftedgradient series

• The present worth (P) of an arithmetic gradient will always belocated two periods before the gradient starts.

• To find the equivalent A series of a shifted gradient throughall the n periods, first find the present worth of the gradient atactual time 0, then apply the (A/P, i, n) factor.

• F from gradient series can also be find by first calculating Pand then using F/P factor

SolutionFor the cash flows shown, find the future value in year 7 at i = 10% per year

F=?

0 1 2 3 4 5 6 7

700 650

500 450550600

G = $-50Solution:

Actual years0 1 2 3 4 5 6 Gradient years

PG is located in gradient year 0 (actual year 1); base amount of $700 is in gradient years 1-6

F = P’G(F/P,10%,6) = 2565(1.7716) = $4544

i = 10%

P’GPG = ?

P’G = A(P/A,10%,6) – G(P/G,10%,6)P’G = 700(P/A,10%,6) – 50(P/G,10%,6) = 700(4.3553) – 50(9.6842) = $2565

PG= P’G(F/P,10%,1) = 2565(0.9091) = $2331.84

F = P’G(F/P,10%,7) = 2331.84(1.9487) = $4544

Set up theequationsonly…

Method 1

Method 2

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Using Single Amount factors (Correctbut not Standard methods)

Method 3

Method 4

So for ….1. Introduction What is Economics? Economics for Engineers ? What is Engineering Economy ? Performing Engineering Economy

Study ? Some Basic Concepts Utility & Various cost concept, Time value of

money (TVM), Interest rate and Rate of Returns, Cash Flow, EconomicEquivalence, Minimum Attractive Rate of Return, Cost of Capital andMARR, Simple and compound interest rates

2. Various Type of Factors

Factors Single payment Factors P/F, F/P

Uniform Series Factors P/A, A/P, F/A, A/F

Gradient Series Factors Arithmetic Gradient and Geometric Gradient

3. Dealing with Shifted Series Shifted uniform series Shifted series and single cash flows Shifted gradients

These were three“FoundationalPillars” we needfor using “variousengineeringeconomy criteria”for decisionmaking

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Lets go for the “Finalfoundational pillar” beforestudying formal engineeringeconomy EVAUALTING criteriaof decision making

MS291: Engineering Economy

Chapter 4Nominal and Effective

Interest Rates

Course Instructor: Dr. Muhammad Sabir

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Content of the Chapter

• Interest Rate: important terminologies• Nominal and Effective Rate of Interest• Effective Annual Interest Rate• Converting Nominal rate into Effective Rate• Calculating Effective Interest rates• Equivalence Relations: PP and CP• Continuous Compounding• Varying Interest Rates

Lets start with aSimple Example

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15% per year

Compounded daily

Paid $1000 from credit cardHow muchyou going topay after 1year ?

1000+150 = $1150 ?

But ..is Due amountafter a year is really$1150 ? Lets docheck!!!

Rate is 15% per year but compounding is daily … so the rate at per day is 0.15/365 =0.000411 per day or 0.0411% per day

Days

1

Interest earned

Amount x r =0.411

Amount ($)

1000

Total due ($)

1000.411

2 0.4111691000.411 1000. 82269

3 0.4111691000. 82269 1001.233507

- -------------- ------

365 0.477311161.338553 1161.815863

1161.815863 …..But this is around

16.81% rate … ratherthan 15% stated

Lets Continue onnext slide

1161.815863 …. But this isaround 16.81% rate … rather

than 15% stated

Nominal Interest Rate (15%) Effective Interest Rate (16.81%)• denoted by (r)• does not include any consideration of

the compounding ofinterest(frequency)

• It is given as: r = interest rate perperiod x number of compoundingperiods

• Denoted by (i)• take accounts of the effect of the

compounding period• commonly express on an annual

basis (however any time maybeused)

Interest rate is samefor each period

But interest “due” isincreasing in everyperiod

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Previous Learning

• Our learning so for is based “one” interest rate that’scompounded annually

• Interest rates on loans, mortgages, bonds & stocks arecommonly based upon interest rates compounded morefrequently than annually

• When amount is compounded more than once annually,distinction need to be made between nominal andeffective rate of interests

Interest Rate:important terminologies

Interest period (t) – period of time over which interest is expressed.For example, 1% per month.

New time-based definitions to understand and remember

Compounding period (CP) – The time unit over which interest is charged or earned.For example,10% per year, here CP is a year.

Compounding frequency (m) – Number of times compounding occurs within theinterest period t.

For example, at i = 10% per year, compounded monthly, interest would becompounded 12 times during the one year interest period.

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Examples of interest rateStatements

= 1 year= 1 month

= 12

Annual interest rate of 8% compounded monthly …interest period (t)

compounding period (CP)compounding frequency (m)

Annual interest rate of 6% compounded weekly …= 1 year= 1 Week

= 52

interest period (t)compounding period (CP)

compounding frequency (m)

IMPORTANT: CompoundingPeriod and Interest Rate

• Some times, Compounding period is not mentioned inInterest statement

• For example, an interest rate of “1.5% per month”………..It means that interest is compounded eachmonth; i.e., Compounding Period is 1 month.

• REMEMBER: If the Compounding Period is notmentioned it is understood to be the same as the timeperiod mentioned with the interest rate.

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Calculating EffectiveInterest Rate

• Effective interest rate per compoundingperiod can be calculated as follows:

=

=

Example:

Three different bank loan rates for electricgeneration equipment are listed below.Determine the effective rate on the basis of thecompounding period for each rate

(a) 9% per year, compounded quarterly(b) 9% per year, compounded monthly(c) 4.5% per 6 months, compounded weekly

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Example: CalculatingEffective Interest rates per CP

a. 9% per year, compounded quarterly.b. 9% per year, compounded monthly.c. 4.5% per 6 months, compounded weekly.

Class Practice 1:

For nominal interest rate of 18% per yearcalculate the effective interest rate

i. If compounding period is yearlyii. If compounding period is semi-annuallyiii. If compounding period is quarterlyiv. If compounding period is monthlyv. If compounding period is weeklyvi. If compounding period is daily

18%

9%

4.5%

1.5%

0.346%

0.0493 %

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Effective Annual InterestRates

• Nominal rates are converted into Effective Annual Interest Rates (EAIR)via the equation:

whereia = effective annual interest ratei = effective rate for one compounding period (r/m)m = number times interest is compounded per year

• When we talk about “Annual” we consider year as theinterest period t , and the compounding period CP can beany time unit less than 1 year

= (1 + ) −1 = (1 + ) −1Effective Interest Rate for any

time period

r = 18% per year,compounded CP-ly

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ExampleFor a nominal interest rate of 12% per year, determine the nominal and effectiverates per year for(a) quarterly, and(b) monthly compounding

Solution:

(a) Nominal r per year = 12% per year

Effective i per year = (1 + 0.03)4 – 1 = 12.55% per year(b) Nominal r per month = 12/12 = 1.0% per month

Effective I per year = (1 + 0.01)12 – 1 = 12.68% per year

Nominal r per quarter = 12/4 = 3.0% per quarter

ia = (1 + i)m – 1

where ia = effective annual interest ratei = effective rate for one compounding

period (r/m)m = number times interest is compounded

per year

15% per yearCompounded daily

Effective I per year = (1 + 0.15/365)365 – 1 = 16.81% per year

Economic Equivalence:From Chapter 1

• Different sums of money at different times may be equalin economic value at a given rate

01

$100 now

$110Rate of return = 10% per year

$100 now is economically equivalent to $110 one year fromnow, if the $100 is invested at a rate of 10% per year

Year

Economic Equivalence: Combination of interest rate (rate of return) andtime value of money to determine different amounts of money at differentpoints in time that are economically equivalent …..Compounding/Discounting (F/P, P/F, F/A, P/G etc.)

1

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Equivalence Relations: PaymentPeriod(PP) & Compounding

Period(CP)

• The payment period (PP) is the length of timebetween cash flows (inflows or outflows)

• Assume the monthly payments as shown below

Months0 1 2 3 4 5 6 7 8 9 10 11 12

│PP │1 month

E.g., r = nominal 8% per year, compounded semi-annuallyCP

6 monthsCP

6 months

PP = CP, PP >CP, or PP<CP

Equivalence Relations: PaymentPeriod(PP) and Compounding Period

• It is common that the lengths of the payment period and thecompounding period (CP) do not coincide

• To do correct calculation of Economic Equivalence…Interest rate must coincide with compounding period

• It is important to determine if PP = CP, PP >CP, or PP<CPLength of Time Involves Single

Amount(P and F Only)

Involves GradientSeries (A, G, or g)

PP = CPP/F , F/P P/A, P/G, P/g

F/A etc.PP > CP

PP < CP P/F, F/P P/A, P/G, F/A etc.

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Thank You