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Engineering Economic AnalysisEngineering Economic Analysis

Chapter 4 More Interest Formulas

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Annual Percentage Rate (APR) or rNominal rate 6% per year is designated ~ i.

Effective interest rate ~ ieff

ieff = (1 + r/m)m – 1 where m is the number of pay periods

APR is 12% compounded monthly

ieff = (1 + 0.12/12)12 – 1 = 12.68% effective yearly rate.

APR is 12% compounded monthly; find effective quarterly rate.ieff = (1 + 0.03/03)3 – 1 = 3.03% effective quarterly rate.

Effective Interest RateEffective Interest Rate

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Effective Interest RateEffective Interest Rate

Annual Percentage Rate (APR) is 12%

If compounded monthly, effective monthly rate is 1%

effective quarterly rate is 3.03%

effective yearly rate is 12.68%

If compounded quarterly,

effective quarterly rate is 3%

effective yearly rate is 12.55%

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Interest RateInterest Rate

A credit card company charges 1.5% interest on the unpaid balance each month.

Nominal annual interest rate is _________________. ans. 12 * 1.5% = 18%

Effective annual interest rate is ________.

ans. (1 + 0.18/12)12 – 1 = 19.56%

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EquivalenceEquivalence

Are equivalent cash flows equivalent at any common point in time? For example, $1000 now at i = 10% for 10 years is equivalent to 1000(1 + 0.10)10 = $2593.74. Are these 2 cash flows equivalent at time 3.37?

Does $1000(1.1)3.37 = $2593.74(1.1)-6.63?

Check: $1378.77 = $1378.77 => Yes

Are these cash flows equivalent at i = 8%? No,

1000(1.08)3.37 = $1296.10 2593.74(1.08)-6.63 = $1557.14

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Monthly PaymentsMonthly Payments

a) What is the monthly payment for a 5-year car loan of $35,000 at 6% compounded monthly?

b) Find the amount of the principal reduction of the 25th payment.

c) After making the 50th monthly payment, you decide to pay off the loan what a check for _________.

d) Find the interest on the 35th payment.

a) A = $35,000(A/P, ½ %, 60) = $676.65.

b) PR25 = 676.65(P/F, ½ %, 60 – 25 + 1) = $565.44.

c) B50 = 676.65(P/A, ½ %, 10) = $6584.08

d) I35 = B34 * i = 676.65(P/A, ½ %, 26) * ½ % = $82.29

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MortgageMortgage

Find the total interest paid on a $300,000 30-year loan at 6% compounded monthly.

A = 300K(A/P, ½%, 360) = $1798.65

Total interest = 360 * $1798.65 - $300K = $347,514

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Arithmetic GradientArithmetic Gradient

0 1 2 3 . . . n -1 n

Gradient begins in Year 2

= A(P/A, i%, n) + G(P/G, i%, n)

G2G

(n-1)G

P

A

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Gradient ExampleGradient Example

What is the present worth at pay period 0 of the following yearly cash flow at 7% compounded annually:

n 0 1 2 3 4 5cf 1000 1300 1600 1900 2200 2500

PW(7%) = 1000 + 1300(P/A, 7%, 5) + 300(P/G, 7%, 5)

= 1000 + 5330.27 + 2294

= $8624.26

(F/P (PGG 1000 300 7 6) 7 1) $8624.26

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Geometric GradientGeometric Gradient

$100 grows geometrically by 10% per year.Compute the growth after n years n Cash Flow1 100 2 100 + 0.10 * 100 = 100(1 + 0.10)1 = 1103 110 + 0.10 * 110 = 100(1 + 0.10)2 = 1214 121 + 0.10 * 121 = 100(1 + 0.10)3 = 133… … … … … … …n 100(1 + 0.10)n – 1

An = A1(1 + g)n-1

Each An must be brought back to year 0 to find the present worth.

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Geometric GradientGeometric Gradient

An = An-1(1+ g) => An = A1(1 + g)n – 1 Then

P = A1(1 + g)n-1(1 + i)-n

P =

=

n - n

1

1-(1+g) (1+i)A i g

i -g

1nA

i=g1+i

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Geometric Gradient ExampleGeometric Gradient Example

A machine’s first year cost is $1,000 and increases 8% per year thereafter for 15 years. Maintenance funds earn 10% per year compounded annually. How much should be deposited in the maintenance fund to cover costs?

P = 1000[1 – (1.08)15(1 + 0.10)-15]/ (0.10 – 0.08)

= $12,030.40

(PGGG 1000 8 10 15) 12030.39

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Problem 4-125Problem 4-125

Find the present worth of a cash flow beginning at $10K and increasing at 8% for 4 years at 6%/yr interest.(PGGG-table 10000 8 6 4)

n Cash-flow 8% PW-factor 6% PWorth

1 10000.00 0.9434 9433.96 2 10800.00 0.8900 9611.96 3 11664.00 0.8396 9793.32 4 12597.12 0.7921 9978.10

$38,817.54

PW = 10K[(1 – (1.08)4)/(1.06)4(0.06 – 0.08)] = $38,817.54

(PGGG 10E3 8 6 4) 38817.54

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Geometric Gradient ExampleGeometric Gradient Example

You want to accumulate $1M 20 years from now by depositing $A1 at year 1 and increasing the deposit by 6% each year for 20 years. Find A1 if the bank pays 8% interest compounded annually.

F = A1(P/A1, g = 6%, i = 8%, n = 20)(F/P, i = 8%, 20)

1M = A1{[1 – 1.0620*1.08-20]/0.02}(1.08)20

A1 = $13,756.85 = (/ 1e6 (FGP (PGGG 1 6 8 20) 8 20)

n - n

1

1-(1+g) (1+i)A i g

i -g

P =

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Effective Interest RatesEffective Interest Rates

$1000 is deposited at 7% compounded monthly. Find the value 5 years from now using monthly, quarterly, semiannually, yearly and bi-yearly effective rates.

Monthly: 1000(F/P, 7/12%, 60) = $1417.63Quarterly: 1000(F/P, 1.76%, 20) = $1417.63Semi-annually 1000(F/P, 3.55%, 10) = $1417.63Annually 1000(F/P, 7.23%, 5) = $1417.63Biennially 1000(F/P, 14.98%, 2.5) = $1417.63.Pentad 1000(F/P, 41.76%, 1) = $1417.63

e.g. Pentad effective rate = [1 + 0.35/60)60 – 1 = 41.7625%

1000(1 + im)nm = 1000(1 + ia)n

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Relationships of Interest FactorsRelationships of Interest Factors

F/P =1/(P/F); A/P = 1/(P/A); F/A = 1(A/F)F/A = 1 + Σ(F/P, i%, n-1)A/P = A/F + i; A/P = P/A – A/F; CRF = (P-S)(A/P. i%, n) + SiP/F * F/A = P/A; P/F = 1 – P/A * iA/F = A/P – i;

7%n F/P P/F A/F A/P F/A P/A A/G P/G 1 1.0700 0.9346 1.0000 1.0700 1.0000 0.9346 0.0000 0.0000 2 1.1449 0.8734 0.4831 0.5531 2.0700 1.8080 0.4831 0.8735 3 1.2250 0.8163 0.3111 0.3811 3.2149 2.6243 0.9549 2.5061 4 1.3108 0.7629 0.2252 0.2952 4.4399 3.3872 1.4155 4.7948 5 1.4026 0.7130 0.1739 0.2439 5.7507 4.1002 1.8650 7.6467

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Continuous CompoundingContinuous Compounding

In the effective interest formula let m = rp and the formula

becomes

ieff = (1 + r/m)m - 1 e

ieff = (1 + 1/p)rp =

= er as p

(F/P, r%, N) = erN for continuous compounding

11

rp

p

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Continuous CompoundingContinuous Compounding

You deposit $100 per month in a savings account with an APR of 6% per year compounded continuously. How much will accumulate in 5 years?

F = 100(F/A, e0.005 -1, 60) = $6979.70

The monthly continuous compounding rate e0.005 -1 = 0.501

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Problem 4-36Problem 4-36

$12K is borrowed at 4% per annum and is to repaid in 5 payments. After the 2nd payment, the borrower was given the option of paying off the loan the following year. How much was then due?

A = 12K(A/P, 4% , 5) = $2695.53

Balance = 2695.53 + 2695.53(P/A, 4%, 2)

= $7779.54

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Problem 4-38Problem 4-38

Sold in 2002 for $150K at 20% down payment and 15-year loan at 8% per year. Buyer makes first payment in 2003. How much will be owed after 2009?

Loan amount = 150K - 0.2 * 150K = $120K

A = 120K(A/P, 8% 15) = $14,019.55

Balance after making 7 payments is

B = 14,019.55 (P/A, 8%, 8)

= $80,565.27

(loan 120E3 8 15)

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Problem 4-50Problem 4-50

A debt of $5K is repaid according to the cash flow below at 8% compound interest. Find X.

n 1 2 3 4 5

cf $500 1000 1500 2000 X

[5K – [500(P/A, 8%, 4) + 500(P/G 8% 4)](F/P, 8% 5) = X => X = $1497.08

(List-pgf '(0 500 1000 1500 2000 1497.08) 8) $5000

(IRR '(-5000 500 1000 1500 2000 1497.06)) 8%

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Capitalized CostCapitalized Cost

You can have 5% interest in perpetuity (forever). You need to generate $10,000 a year for a scholarship fund. How much investment is needed to do so?

P = A/i = 10,000/0.05 = $200,000.

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GradientGradient

Find the equivalent sum at year 7 for the following cash flow at 7% compound interest per year.

n 1 2 3 4 5 6 7

Cf 1000 2000 3000 4050 5000

F7 = [(1000(P/A, 7%, 5) + 1000(P/G, 7%, 5)

+ 50(P/F 7%, 4)](F/P, 7%, 7)

= $18,924.23(F/P (+ (PGG 1000 1000 7 5) (P/F 50 7 3)) 7 7) 18928.52

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Shady DealShady DealYou borrow $1000 to be repaid in 24 monthly installments. The interest rate is a mere 1.5% per month. Further

Amount requested $1000

Credit risk insurance 5

Credit investigation 25Total $1030

Interest: ($1030)(24)(0.015) = $371

Total owed: $1030 + $371 = $1401

Payment: $1401/24 = $58.50

Find the effective annual interest rate charged.

1000 = 58.50(P/A, i%, 24) => im = 2.92% => APR = 34.04%

Iaeff = 41.25%

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Sports ContractsSports Contracts

Headline blares Ace Stacey sings 10-year contract for $50 million paid $5M now and $4M for the first 5 years and $5M for the next 5 years. How much is the contract worth to Ace now if the interest rate is 7%?

PW = 5M + 4M(P/A, 7%, 5) + 5M(P/A, 7%, 5)(P/F, 7%, 5)

= 5M + 16,400,790 + 14,616,921

= $36,017,710

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Capital Recovery Capital Recovery

(A/P, i%, n) – (A/F, i%, n) = i

P

F = Salvage

P(A/P, i%, n) – S(A/F, i%, n)

EUAC = (P - S)(A/P, i%, n) + Si

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Capital Recovery ExampleCapital Recovery Example

A new machine's first cost is $5,000 with a 5-year life and a salvage value of $1000. Compute the annual cost at i = 7%.

1)5000(A/P, 7%, 5) – 1000(A/F, 7%, 5) = 1219.45 – 173.89 = $1045.56

2)(P – S)(A/P, 7%, 5) + Si = 4000(P/A, 7%,5) + 0.07*1000 = 975.56 + 70 = $1045.56

3) (P – F)(A/F, 7%, 5) + Pi = 4000(A/F, 7%, 5) + 5000 * 0.07 = 695.56 + 350 = $1045.56

ReviewReview

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Change in RateChange in Rate

You borrow $20,000 at 7% compounded monthly over 48 months. After making the 24th payment, you negotiate with the bank to pay off the loan in 8 equal quarterly payments.

Determine the quarterly payment at the same interest rate.

Am = 20K(A/P, 7/12 %, 48) = $478.92

B24 = 478.92(P/A, 7/12 %, 24) = $10,696.84

Aq = 10,698.84(A/P, 1.76%, 8) = $1445.17

Exact Rate of ReturnExact Rate of Return

Find the exact rate of return for the following cash flow.

n 0 1 2cf -1200 900 700

(quadratic -12 9 7) 1.225857 => 22.5857%

(list-pgf '(-1200 900 700) 22.585738) 0

(IRR '(-1200 900 700)) 22.585738

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Wright Learning CurveWright Learning Curve

Unit Hours 1 1000.00 2 600.00 3 445.02 4 360.00 5 305.41

a) The time to make the 10th unit is ________.  

b) The learning curve rate in percent is ________.

c) The slope of the learning curve is ________.

d) The time to make the 13th unit is __________.

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MortgageMortgage

You borrow $10,000 at 6% compounded monthly for 24 years. Your monthly payment is closest to

a) $660 b) $550 c) $450 d) $350

Your principal reduction on the 12th payment is closest to

a) $315 b) $415 c) $515  d) not given

Total interest paid after making 12th payment is

a) $370 b) $470 c) $570 d) $670

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Annual WorthAnnual Worth

 You want $100,000 in a fund 10 years from now, the amount to deposit in years 6 through 9 at i = 10% per year is closest to

  a) $19,588 b) $20,614 c) $21,547 d) $22,389 $100K

6 7 8 9 10

 

(AGF (PGF 100E3 10 1) 10 4) 19588.25

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Doubling InvestmentDoubling Investment

If you invest $2,000 at 12% compounded monthly for the same length of time that it takes an investment to double in value at 12% compounded quarterly, you will have

 

a) $3709 b) $4027 c) $4352 d) 4580

 

2 = (1.03)q => q = 23.35 quarters or 70.35 months

2K(1.01)70.35 = $4027.50

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Measuring InvestmentsMeasuring Investments1. Present Worth (PW)

2. Annual Worth (AW)

3. Future Worth

4. Internal Rate of Return

5. External Rate of Return

6. Benefits/Costs ratio

7. Payback Period

8. Capitalized Worth