Solution manual engineering economic analysis 9th edition ch04

The first three payments were $648.90 each.Balance due after 3rd payment equals the Present Worth of the originally planned last three payments of $648.90... This indicates that there w

Chapter 4: More Interest Formulas

$150

0 1 2 3 4

$50

90

Performing linear interpolation:

$150

x

$30

0

$400

$50

0 $400

$300

From the 10% interest table, n = 16

$300

$300

$30

0 $20

0 $100

$20

0 $100

$200

$300

E E P

$400

4B 3B

2B BP

So the equation is:

…………

(P/A, 6.5%, 10) = 7.189

Why do the values differ? Since the compound interest factor is non-linear, linear interpolation will not produce an exact solution

To have sufficient money to pay the four $4,000 disbursements,

x = $4,000 (P/A, 5%, 4) = $4,000 (3.546)

= $14,184This $14,184 must be accumulated by the two series of deposits

The four $600 deposits will accumulate by x (17th birthday):

F = $600 (F/A, 5%, 4) (F/P, 5%, 10)

= $600 (4.310) (1.629)

= $4,212.59Thus, the annual deposits between 8 and 17 must accumulate a future sum:

= $14,184 - $4,212.59

= $9,971.41The series of ten deposits must be:

From the 1.5% interest table, n is between 17 and 18 Therefore, it takes 18 months to

repay the loan

The first three payments were $648.90 each.

Balance due after 3rd payment equals the Present Worth of the originally planned last three payments of $648.90

A’ = ?

($150 - $15) = $10 (P/A, 1.5%, n)

(P/A, 1.5%, n) = $135/$10 = 13.5

From the 1.5% interest table we see that n is between 15 and 16 This indicates that there

will be 15 payments of $10 plus a last payment of a sum less than $10

Compute how much of the purchase price will be paid by the fifteen $10 payments:

The final payment is the present worth of the three unpaid payments

Final Payment = $2,695.20 + $2,695.20 (P/A, 4%, 2)

Column 1 shows the number of interest periods.

Column 2 shows the equal annual amount as computed in part (a) above

The amount $14,019.55 is the total payment which includes the principal and interest portions for each of the 15 years To compute the interest portion for year one, we must first multiply the interest rate in decimal by the remaining balance:

PRINCIPALPORTION

REMAININGBALANCE

at the end of the previous year (y) results in the remaining balance after the first

payment is made in year 1 (y1), of $115,580.45 This completes the year 1 row The other row quantities are computed in the same fashion The interest portion for row two, year 2 is:

It is necessary to have $29,483 at the end of 1997 in order to provide $8,000 at the end of

1998, 1999, 2000, and 2001 It is now necessary to determine what yearly deposits should have been over the period 1981–1997 to build a fund of $29,483

Amortization schedule for a $4,500 loan at 6%

Paid monthly for 24 months

P = $4,500 i = 6%/12 mo = 1/2% per month

Pmt # Amt Owed Int Owed Total Owed Principal Monthly

BOP (this pmt.) (EOP) (This pmt) Pmt

B13 = B12 - E12 (amount owed BOP- principal in this payment)

Column C = amount owed BOP * 0.005

Column D = Column B + Column C (principal + interest)

Column E = Column F - Column C (payment - interest owed)

Column F = Uniform Monthly Payment (from formula for A/P)

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Amortization schedule for a $4,500 loan at 6%

Paid monthly for 24 months

P = $4,500 i = 6%/12 mo = 1/2% per month

Pmt # Amt Owed Int Owed Total Owed Principal Monthly

BOP (this pmt.) (EOP) (This pmt) Pmt

23 0.00 0.00 0.00 0.00 0.00

B12 = $4,500.00 (principal amount)

B13 = B12 - E12 (amount owed BOP- principal in this payment)

Column C = amount owed BOP * 0.005

Column D = Column B + Column C (principal + interest)

Column E = Column F - Column C (payment - interest owed)

Column F = Uniform Monthly Payment (from formula for A/P)

Payment 22 is the final payment Payment amount = $187.59

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Interest Rate per Month = 0.07/12 = 0.00583/month

Interest Rate per Day = 0.07/365 = 0.000192/day

= $79.934.37Interest for 33 days = Pin = $79,934 (33) (0.000192) = $506.46Principal in 2nd payment = $532.03 – 506.46 = $25.57

4-43

(a) F16 = $10,000 (1 + 0.055/4)16

= $12,442.11F10 = $12,442.11 (1 + 0.065/4)24

= $18,319.24(b) $18,319.24 = (1 + i)10 ($10,000)

(1 + i)10= $18,319.24/$10,000= 1.8319

10 ln (1 + i) = ln (1.8319)

Number of yearly investments = (59 – 20 + 1) = 40

The diagram indicates that the problem is not in the form of the uniform series compound amount factor Thus, find F that is equivalent to $1,000,000 one year hence:

F = $1,000,000 (P/F, 15%, 1) = $1,000,000 (0.8696)

20th

Birthday

59th Birthday

-$50

$450

$300

$150

$1,000 (P/A, 8%, 8) - $150 (P/G, 8%, 8)

$10,000

$200

$300

P

i = 8% per year $5,000

$500

$1,500

P

F

$1,50

0

i = 1% / month

Nominal Interest Rate = 12 months/ year (1% / month)

= 12% / yearEffective Interest Rate = (1 + i)m – 1 = (1.01)12 – 1

Nominal Interest Rate = 12 (1.5%) = 18%

Effective Interest Rate = (1 + 0.015)12= 0.1956 = 19.56%

(a) Effective Interest Rate = (1 + i)m – 1 = (1 + 0.025)4 – 1 = 0.1038

= 10.38%

(b) Since the effective interest rate is 10.38%, we can look backwards to

compute an equivalent i for 1/252 of a year

(1 + i)252 – 1 = 0.1038

(1 + i)252 = 1.1038

(1 + i) = 1.10381/252 = 1.000392

Equivalent i = 0.0392% per 1/252 of a year

(c) Subscriber’s Cost per Copy:

A = P (A/P, i%, n) = P [(i (1 + i)n)/((1 + i)n – 1)]

From compound interest tables, i = 1.5%

Nominal Interest Rate = 1.5% (12) = 18%

Nominal Interest Rate = 13.3% (2) = 26.6%

Effective Interest Rate = (1 + 0.133)2 – 1 = 0.284 = 28.4%

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Effective Interest Rate = (1 + 0.0175)12 – 1 = 0.2314 = 23.14%

4-70

Nominal Interest Rate = 1% (12) = 12%

Effective Interest Rate = (1 + 0.01)12 – 1 = 0.1268 = 12.7%

4-71

Effective Interest Rate = (1 + i)m – 1

0.0931 = (1 + i)4 – 1

1.0931 = (1 + i)41.09310.25 = (1 + i) 1.0225 = (1 + i)

Compute F equivalent to the five $10,000 withdrawals:

P n = 19 years P’ n = 30 yearsA = $1,000

The diagram illustrates a problem that can be solved directly

$10

0 $80

$50

$50

$60

$140

Effective interest rate = (1 + i)m – 1 = (1.02)4 – 1 = 0.0824 = 8.24%

Since F = $1,000,000 we can find the equivalent P for i = 8.24% and n = 40

1998 We can use either interest rate, the quarterly or the semiannual Let’s use the

quarterly with n = 27.

P = F (1 + i)-n

= $37,852.04 (1.035)-27

= $14,952

This particular example illustrates the concept of these problems being similar to putting a puzzle together There was no simple formula, or even a complicated formula, to arrive at the solution While the actual calculations were not difficult, there were several steps required to arrive at the correct solution.

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i = interest rate/interest period = 0.13/52 = 0.0025 = 0.25%

Paco’s Account: 63 deposits of $38,000 each, equivalent weekly deposit

Tisha’s Account: 18 deposits of $18,000 each

Equivalent weekly deposit:

Present Worth P1/1/2006= $671.40 (P/A, 0.25%, 18x26)

Deposits

Fdeposits = $2,100 (F/A, 1%, 80)

= $255,509Withdrawals:

Equivalent quarterly interest i quarterly = (1.01)3 – 1

= 0.0303 = 3.03%Fwithdrawals = $5,000 (F/A, 3.03%, 26)

= $5,000 [((1.0303)26 – 1)/0.0303]

= $193,561Amount remaining in the account on January 1, 2005:

1/1/2005

F = A (F/A, i%, n) = $300 (F/A, 1.5%, 12) = $300 (13.041)

= 19.3%

4-88

Monthly Payment = $10,000 (A/P, 0.75%, 12) = $10,000 (0.0875)

= $875.00Total Interest Per Year = $875.00 x 12 - $10,000 = $500.00

Rule of 78s

With early repayment:

Interest Charge = ((12 + 11 + 10) / 78) ($500) = $211.54

Additional Sum (in addition to the 3rd $875.00 payment)

Additional Sum = $10,000 + $211.54 interest – 3 ($875.00) = $7,586.54

Exact Method

Additional Sum equals present worth of the nine future payments that would have been made:

Additional Sum = $875.00 (P/A, 0.75%, 9) = $875.00 (8.672) = $7,588.00

4-89

P = $25,000 n = 60 months i = 18% per year

= 1.5% per month(a) A = $25,000 (A/P, 1.5%, 60)

= $635(b) P = $25,000 (0.98) = $24,500

= $357,887(c) A = $400,000 [(e(0.06/12)(360))(e(0.06/12) – 1)/(e(0.06/12)(360) – 1)]

The 4 3/8% interest (a) has the highest effective interest rate

F = $10,000 (1 + 0.12/365)365x4

= $16,159.47(3) 12.01% compounded monthly

F = $10,000 (1 + 0.1201/12)12x4

= $16,128.65(4) 12.02% compounded quarterly

F = $10,000 (1 + 0.1202/4)4x4

= $16,059.53(5) 12.03% compounded yearly

F = $10,000 (1 + 0.1203)4

= $15,752.06Decision: Choose Alternative (2)

= $2,000 (F/P, 3%, 5)

= $2,000 (1.159)

= $2318The uniform payment:

= $2,318 (A/P, 3%, 4)

= $2,318 (0.2690)

= $623.54 every 6 months(c) Total interest paid:

Nominal Interest Rate = (1.75%) 12 = 21%

Effective Interest Rate = ern – 1 = e(0.21x1) – 1 = 0.2337 = 23.37%

Nominal Interest Rate = 5.26% (2) = 10.52%

Effective Interest Rate = (1 + 0526)2 – 1 = 0.10797 = 10.80%

4-108

West Bank

F = P (1 + i)n = $10,000 (1 + (0.065/365))365 = $10,671.53

(b) Deposits of A = $250 x 106 occur four times a month

So it pays $625,000 a month to move quickly!

A = $1,200 r = 0.14/12

= 0.01167

n = 7 x 12 = 84 compounding periods

First Bank- Continous Compounding

Effective interest rate i a = er – 1 = e0.045 – 1 = 0.04603

= 4.603%

Second Bank- Monthly Compounding

Effective interest rate i a = (1 + r/m)m – 1 = (1 + 0.046/12)12 – 1

No, Barry should have selected the Second Bank

4-115

A = P (A/P, i%, 24)

(A/P, i%, 24) = A/P = 499/10,000 = 0.499

From the compound interest tables we see that the interest rate per month is exactly 1.5%

= 3.93% per quarter year

Effective rate of interest = (1 + i)m – 1 = (1.0393)4 – 1

From compound interest tables, i = 1.25% per month

For an $800 down payment, unpaid balance is $2775

(b) In 1929, the Consumer Price Index was 17 compared to about 126 in

1990 So $165,000 in 1929 dollars is roughly equivalent to $165,000 (126/17) = $1,223,000 in 1990 dollars The real rate of return is closer to 6.9%

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FW= FW

$1000 (F/A, i%, 10) (F/P, i%, 4) = $28,000

By trial and error:

Try i = 12% $1,000 (17.549) (1.574) = $27,622 i too low

($365.357 for exact calculations)

Month 1% Interest $365.36 Principal Balance Due

0.83% $1,053.09 Balance Due 0.83% $1,053.09 Balance Due

Month Interest Principal Month Interest Principal

Year Cash Flow PW 9% Cumulative PW

(b) See Excel output below:

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See Excel output below: