Assignmen (1)

ENVE292 – Engineering Economics Fall 2005

ASSIGNMENT #1 SOLUTIONS Due: September 29, 2005 Text: 2.11, 2.13, 2.18, 2.25, 2.30, 3.18, 3.19, 3.20 2.11 Greg will invest $20 000 today in five-year investment certificates that pay 8% nominal

interest, compounded quarterly. How much money will this be in five years? Cash Flow Diagram:

4 compounding periods per year

i = 8% p.a. compounded l

? F =

$20 000

20 compounding periods require a modification of the nominal (quarterly) interest rate to an effective (annual) interest rate.

08243.0

1408.01

11

4

=

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

m

e mri

The effective interest rate for quarterly compounding of a nominal rate of 8% is 8.243% per year. Transfer factor to use is Compound Amount Factor. Here, we use N = 5 because we have converted the interest rate to the equivalent annual interest rate including quarterly compounding.

508243.0

?20000

====

Ni

FP

[ ][ ]65.29718

08243.0120000

,,/5

=+=

= NiPFPF

Therefore Greg will have $29 719 at the end of 5 years from an initial investment of $20 000 (earning $9 719 in interest)

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2.13 You have a bank deposit now worth $5000. How long will it take for your deposit to be worth more than $8000 if:

a) The account pays 5% actual interest every half year (6 months) and is compounded? b) The account pays 5% nominal interest, compounded semi-annually?

Cash Flow Diagram $8000+ i = 5% (nominal/effective)

Semi-annual compounding (unknown number of periods)

$5000 The future value of ~$8000 will be at N (semi-annually) periods. a) To determine the number of compounding periods required, first, we need to understand the interest rate that is given. The account pays 5% actual interest every half year. Therefore every six months, the balance of the account accumulates 5% interest. Assuming it is compounding semi-annually as well, the compounding periods and the interest rates match.

80005000

periods annualsemiannually-semi 05.0

==

−==

FPNie

[ ][ ]

( ) ( )( )( )

periods annual-semi 1063.9

05.1log6.1log

05.1log6.1log05.16.1

05.0150008000

%,5,/

==

=

==

+=

=

N

N

NPFPF

N

N

As there are two periods per year, the number of years to achieve $8000 is 5 (10 compounding periods) years.

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b) The account pays 5% nominal interest and is compounded semi-annually. This is different to that of Part A. The annual effective interest rate is:

050625.0

1205.01

11

2

=

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

m

e mri

80005000

periods annual050625.0

====

FPNie

[ ][ ]

( ) ( )( )

( )

periods annual 10517.9

050625.1log6.1log

050625.1log6.1log050625.16.1

050625.0150008000

%,0625.5,/

==

=

==

+=

=

N

N

NPFPF

N

N

As there is only one period per year, the number of years to achieve $8000 is 10 years (20 compounding periods). 2.18 For a 15% effective annual interest rate, what is the nominal rate if:

a) Interest is compounded monthly? b) Interest is compounded daily? c) Interest is compounded continuously?

The general formula to convert from nominal to effective interest rates is (discrete compounding):

11 −⎟⎠⎞

⎜⎝⎛ +=

m

e mri

a) Monthly compounding of interest results in 12 periods in a year.

3


( )14058.0

115.11212

115.1

12115.1

112

115.0

11

12

12

12

12

=−=

+=

⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

r

r

r

r

mri

m

e

Therefore the equivalent nominal rate is 14.058%. b) Daily compounding of interest results in 365 periods in a year.

( )1398.0

115.1365365

115.1

365115.1

1365

115.0

11

365

365

365

365

=−=

+=

⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

r

r

r

r

mri

m

e

Therefore the equivalent nominal rate is 13.98%. c) The conversion for continuous interest compounding is:

( )

1397.0)15.1ln(

1)exp(15.01exp

==

−=−=

rrrie

Therefore the equivalent nominal rate is 13.97%.

4


2.25 The Bank of Kitchener is offering a new savings account that pays a nominal 7.99%

interest, compounded continuously. Will your money earn more in this account than a daily interest account that pays 8%?

Calculate the annual effective interest rate from the nominal continuous rate of 7.99% over one year.

%32.8083178.0

1)0799.0exp(1)exp(

==

−=−= rie

From here, an annual effective interest rate of 8.32% results. To compare with the nominal interest rate of 8% compounded daily, we need to convert this also to the effective annual interest rate:

%327.808327.0

1365

08.01

11

365

==

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

m

e mri

Based on the results, discrete daily interest rate of 8% will yield higher performance compared to a continuous interest rate of 7.99%. It is important to analyze the effect of comparing a daily interest rate of 8% with a continuous interest rate at 8% also.

%3287.8083287.0

1)08.0exp(1)exp(

==

−=−= rie

We can see a slightly higher annual effective interest rate is the result of continuous compounding. However, the difference is very small. Therefore continuous compounding, in this case, is similar but not equal to that of discrete daily compounding. A slight advantage exists with continuous compounding over discrete daily compounding.

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2.30 You have $50 000 to invest in the stock market and have sought the advice of Adam, an experienced collegue who is willing to advise you for a fee. Adam has told you that he has found a one-year investment for you that provides 15% interest compounded monthly.

a) What is the effective annual interest rate, based on a 15% nominal annual rate and

monthly compounding? b) Adam says he will make the investment for you for a modest fee of 2% of the

investments value one year from now. If you invest $50 000 today, how much will you have at the end of one year? (Before Adam’s fee)

c) What is the effective annual interest rate of this investment including Adam’s fee? Part A To determine the effective annual interest rate we require the use of the conversion formula:

%08.1616075.0

11215.0

1

11

12

==

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

m

e mri

Therefore the effective annual interest rate is estimated at 16.08%. Part B If $50 000 was invested today at the aforementioned scenario, then we would like to know the value after one year (before Adam’s fee). Cash Flow Diagram:

1 year One compounding

period

ie = 16.08% per yearF1

$50 000

[ ][ ]

58037.43 OR 580401608.0150000

1%,08.16,1

1

=+=

= PFPF

6


Part C To determine the effective annual interest rate including Adam’s fee, we can do the following: Firstly, the fee Adam imposes is 2% of F1 ($58 040) = $1160.80 Secondly, we subtract this fee from F1 yielding the net value at the end of year 1.

20.56879

80.1160580401

=

−=NET

F

Now we can represent this in a revised (net) cash flow diagram.

1 year One compounding

period

ie = ?% per year $56879.20

$50 000

[ ][ ]

%76.131376.0

15000020.56879

1,,/1

==

+=

=

e

e

ii

iPFPF

Therefore, with the fee Adam has charged, the effective annual interest rate of this investment is approximately 13.76%.

7


3.18 Yoko has just bought a new computer ($2000), a printer ($350) and a scanner ($210). She wants to take the monthly payment option. There is a monthly interest of 3% on her purchase.

a) If Yoko pays $100 per month, how long does it take to complete her payment? b) If Yoko wants to finish paying in 24 months, how much will her monthly payments be?

Total initial investment of (assume values include taxes):

25602103502000

=++=P

Part A Cash flow diagram A = $100

ip = 3% per month N

$2560 The interest rate is given as 3% per month. From this, we can assume this is the effective period interest rate (ip).

[ ]( )

( )( )

( )

103.103.130208.1

103.0103.0103.02560100

111

%,3,/

−=

⎥⎦

⎤⎢⎣

⎡

−++

=

⎥⎦

⎤⎢⎣

⎡

−++

=

=

N

N

N

N

N

N

iiiPA

NPAPA

To solve, must equate LEFT and RIGHT sides.

If N = 10 months then, 908.3103.1

03.110

10

=−

If N = 20 months then, 241.2103.1

03.120

20

=−

8


If N = 40 months then, 442.1103.1

03.140

40

=−

If N = 50 months then, 295.1103.1

03.150

50

=−

If N = 48 months then, 319.1103.1

03.148

48

=−

Therefore, at 50 months (4+ years), the computer and accessories may be paid off via monthly payments of $100. OR Solving algebraically…

( )( )

( )

( )43.49

03.1log130208.1

30208.1log

130208.130208.103.1

30208.1130208.103.130208.103.130208.103.1

103.130208.103.1

30208.1103.1

03.1

=

⎟⎠⎞

⎜⎝⎛

−=

−=

=−

−=

−=

=−

N

N

N

N

NN

Nn

N

N

Therefore, at 50 months (4 years and 2 months), the computer and accessories may be paid off via monthly payments of $100. Part B Cash Flow Diagram A i = 3% per month

24 months

$2510

9


[ ]

( )( )

( )( )

17.151103.01

03.0103.02560

111

24%,3,/

24

24

=

⎥⎦

⎤⎢⎣

⎡

−++

=

⎥⎦

⎤⎢⎣

⎡

−++

=

=

A

A

iiiPA

PAPA

N

N

If she would like to pay off her computer and accessories in 2 years (24 months) then monthly payments of $151.17 is required. 3.19 Rinku has just finished her first year of university. She wants to go to Europe when

she graduates in three years. By having a part-time job through the school year and a summer job during the summer, she plans to make regular weekly deposits into a savings account at 18% interest compounded monthly.

a) If Rinku deposits $15 per week, how much will she have in three years? How about

$20 per week? b) Find out exactly how much Rinku needs to deposit every week is she wants to save

$5000 in 3 years. It has been assumed the part-time job was not done in Rinku’s first year in University. Therefore only years 2, 3, and 4 will be considered. The textbook solves this question through the assumption of 4 weeks per month and 12 months per year. Unfortunately, this results in only 48 weeks being analyzed, 4 short of the typical 52 week year. Therefore, it is important to transfer all cash flows from monthly to weekly annuities. Part A Regular weekly deposits of $A will be done. A total of 156 weekly installments of $A will be completed over three years. Cash Flow Diagram

i = 18% p.a. compounded monthly F

156 weeks

A

10


First, we calculate an effective annual interest rate with monthly compounding:

%56.19195618.0

11218.01

11

12

==

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

m

e mri

As we are working with weekly installments, we must convert this into a weekly effective interest rate (ip). Notice the exponent n is 52 indicating we would like to achieve an period interest rate for 52 equal periods (i.e. weekly) from the effective annual rate.:

( )( )

%34417.00034417.0

11195618.0

1152

=

=

−+=

−+=

p

p

npe

i

i

ii

Now we use 0.34417% in the following as the weekly interest rate for the calculations. If weekly deposits of $15:

[ ]( )

( )

61.3090

0034417.010034417.0115

11

%,,/

156

=

⎥⎦

⎤⎢⎣

⎡ −+=

⎥⎦

⎤⎢⎣

⎡ −+=

=

iiA

NiAFAFN

If weekly deposits of $20

[ ]( )

( )

82.4120

0034417.010034417.0120

11

%,,/

156

=

⎥⎦

⎤⎢⎣

⎡ −+=

⎥⎦

⎤⎢⎣

⎡ −+=

=

iiA

NiAFAFN

Part B To achieve $5000 at the end of 36 months (or 156 weeks), we must deposit weekly deposits of $B

11


[ ]

( )

( )27.24

10034417.010034417.05000

11

%,,/

156

=

⎥⎦

⎤⎢⎣

⎡

−+=

⎥⎦

⎤⎢⎣

⎡

−+=

=

B

iiF

NiFAFB

N

Therefore, weekly deposits of approximately $24.27 is required to achieve sufficient funds to travel to Europe in three years. 3.20 Seema is looking at an investment in upgrading an inspection line at her plant. The

initial cost would be $140 000 with a salvage value of $37 000 after 5 years. Use the Capital Recovery Formula to determine how much money must be saved each year to justify the investment at a rate of 14%.

Here, we assume the interest rate is compounded annually. The Capital Recovery Formula is:

( )[ ] SiNiPASPA +−= ,,/ To illustrate this equation is the cash flow diagram:

i = 14% p.a. compounded annually

N N

A S

P

In this case, the following values are assigned to the parameters

12


%14537000140000

====

iNSP

( )[ ]

( ) ( )( )

( ) ( )( )

( )

( )[ ] ( )21.35182

14.0370002913.037000140000

14.037000114.1

14.114.037000140000

111

,,/

5

5

=+−=

+⎥⎦

⎤⎢⎣

⎡

−−=

+⎥⎦

⎤⎢⎣

⎡

−++

−=

+−=

Sii

iiSP

SiNiPASPA

N

N

So what does this mean? The company will invest in equipment that will, in a five years, hold a residual (salvage) value. The difference between the purchase price (P = $140000) and salvage value (S = $37000) is the “depreciation” or lost value of the equipment (here it is $103000). The capital recovery formula is simply a way of representing the lost value of the equipment over the analysis period and spreading this loss over (for this case) 5 years. Essentially, the company must save $35182.21 per year to justify the purchase of this machine. If the actual annual savings is greater than $35182.21, then it is a bonus to the company. However, if the annual savings is less than that value, it was not worth purchasing this equipment in the first place.

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