12/26/2015rd1 engineering economic analysis chapter 4 more interest formulas
TRANSCRIPT
04/21/2304/21/23 rdrd 11
Engineering Economic AnalysisEngineering Economic Analysis
Chapter 4 More Interest Formulas
04/21/2304/21/23 rdrd 22
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
04/21/2304/21/23 rdrd 33
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%
04/21/2304/21/23 rdrd 44
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%
04/21/2304/21/23 rdrd 55
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
04/21/2304/21/23 rdrd 66
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
04/21/2304/21/23 rdrd 77
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
04/21/2304/21/23 rdrd 88
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
04/21/2304/21/23 rdrd 99
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
04/21/2304/21/23 rdrd 1010
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.
04/21/2304/21/23 rdrd 1111
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
04/21/2304/21/23 rdrd 1212
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
04/21/2304/21/23 rdrd 1313
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
04/21/2304/21/23 rdrd 1414
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 =
04/21/2304/21/23 rdrd 1515
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
04/21/2304/21/23 rdrd 1616
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
04/21/2304/21/23 rdrd 1717
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
04/21/2304/21/23 rdrd 1818
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
04/21/2304/21/23 rdrd 1919
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
04/21/2304/21/23 rdrd 2020
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)
04/21/2304/21/23 rdrd 2121
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%
04/21/2304/21/23 rdrd 2222
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.
04/21/2304/21/23 rdrd 2323
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
04/21/2304/21/23 rdrd 2424
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%
04/21/2304/21/23 rdrd 2525
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
04/21/2304/21/23 rdrd 2626
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
04/21/2304/21/23 rdrd 2727
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
04/21/2304/21/23 rdrd 2828
04/21/2304/21/23 rdrd 2929
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
04/21/2304/21/23 rdrd 3030
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 __________.
04/21/2304/21/23 rdrd 3131
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
04/21/2304/21/23 rdrd 3232
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
04/21/2304/21/23 rdrd 3333
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
04/21/2304/21/23 rdrd 3434
04/21/2304/21/23 rdrd 3535
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