0231-3-nominaleffective-cashflow (1).pdf

Business

1. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
LECTURE NOTES
WEEK 3
Cash Flow Examples
Nominal and Effective Interest Rates
CIVIL ENGINEERING DEPARTMENT
CE 231
ENGINEERING ECONOMY

2. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Highlights
•Money has a “time value” (interest
rates)
•Acash flow diagram(CFD) visually
represents the “timing” of the cash
inflows and/or outflows –we draw CFD
to calculate “economically equivalent”
values.
2

3. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
3

4. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
4
A company wants to start an excavation business, which will continue for 11 years.
They plan to buy an excavator which will cost 620 000 TLwith a useful life of 5 years
and a salvage value of 50 000 TL.
A major repair of 60 000 TLis estimated for the end of the 3
rd
year.
They also expect to get an income of 180 000 TL per year.
At the end of the 5 year period, they are planning to buy a new excavator for 800 000
TLwith a useful life of 6 years and a salvage value of 75 000 TL.
The expected income from this excavator is 200 000 TL per yearduring its useful life.
Maintenance and repair cost for the second excavator is expected to be 80 000 TLin
the 3
rd
yearof its useful life.
Interest rate for the first 5 years is 10%and for the next 6 years interest rate is
expected to be 12%.
Decide whether this is a good investment or not by calculating the present valueof all
expected expenditures and incomes.

5. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
5
A company wants to start an excavation business, which will continue for 11 years.
0 1 2 3 4 5 6 7 8 9 1011

6. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
6
They plan to buy an excavator which will cost 620 000 TLwith a useful life of 5 years
and a salvage value of 50 000 TL.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL

7. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
7
A major repair of 60 000 TL is estimated for the end of the 3rd year.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL

8. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
8
They also expect to get an income of 180 000 TL per year.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL
A1= 180 000 TL/year

9. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
9
At the end of the 5 year period, they are planning to buy a new excavator for 800 000
TLwith a useful life of 6 years and a salvage value of 75 000 TL.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL
A1= 180 000 TL/year
E3 = 800 000 TL
SV2= 75 000 TL

10. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
10
The expected income from this excavator is 200 000 TL per yearduring its useful life.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL
E3 = 800 000 TL
SV2= 75 000 TL
A2= 200 000 TL/year
A1= 180 000 TL/year

11. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
11
Maintenance and repair cost for the second excavator is expected to be 80 000 TLin
the 3
rd
yearof its useful life.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL
E3 = 800 000 TL
SV2= 75 000 TL
E4 = 80 000 TL
A1= 180 000 TL/year
A2= 200 000 TL/year

12. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
12
Interest rate for the first 5 years is 10% and for the next 6 years interest rate is expected to be 12%.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL
A1= 180 000 TL/year
E3 = 800 000 TL
SV2= 75 000 TL
A2= 200 000 TL/year
E4 = 80 000 TL
i= 10%
i= 12%

13. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
13
Decide whether this is a good investment or not by calculating the present value of
all expected expenditures and incomes.
0 1 2 3 4 5 6 7 8 9 1011
E1 = 620 000 TL
SV1= 50 000 TL
E2 = 60 000 TL
A1= 180 000 TL/year
E3 = 800 000 TL
SV2= 75 000 TL
A2= 200 000 TL/year
E4 = 80 000 TL
i= 10%
i= 12%

14. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
14
Incomes:
P1 = A1 x (P/A, 10%, 5) =

15. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
15
Incomes:
P1 = A1 x (P/A, 10%, 5)
P1 = 180 000 x 3.7908
P1 = 682 344 TL

16. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
16
Incomes:
P1 = A1 x (P/A, 10%, 5) = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5)

17. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
17
Incomes:
P1 = A1 x (P/A, 10%, 5) = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5)
P2 = 200 000 x 4.1114 x (P/F, 10%, 5)

18. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
18
Incomes:
P1 = A1 x (P/A, 10%, 5) = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5) =
P2 = 200 000 x 4.1114 x 0.6209 =
P2 = 510 554 TL

19. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
19
Incomes:
P1 = A1 x (P/A, 10%, 5) = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5)
P2 = 510 554 TL
P3= SV1 x (P/F, 10%, 5)
P3= 50 000 x 0.6209
P3 = 31 045 TL

20. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
20
Incomes:
P1 = A1 x (P/A, 10%, 5) = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5)
P2 = 510 554 TL
P3= SV1 x (P/F, 10%, 5) = 31 045 TL
P4= SV2 x (P/F, 12%, 6) x (P/F, 10%, 5) =
P4= 75 000 x 0.5066 x (P/F, 10%, 5) =

21. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
21
Incomes:
P1 = A1 x (P/A, 10%, 5) = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5)
P2 = 510 554 TL
P3= SV1 x (P/F, 10%, 5) = 31 045 TL
P4 = SV2 x (P/F, 12%, 6) x (P/F, 10%, 5) =
P4 = 75 000 x 0.5066 x 0.6209
P4 = 23 591 TL

22. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
22
Incomes:
P1 = A1 x (P/A, 10%, 5) = 180 000 x 3.7908 = 682 344 TL
P2 = A2 x (P/A, 12%, 6) x (P/F, 10%, 5) = 200 000 x 4.1114 x 0.6209 = 510 554 TL
P3= SV1 x (P/F, 10%, 5) = 50 000 x 0.6209 = 31 045 TL
P4= SV2 x (P/F, 12%, 6) x (P/F, 10%, 5) = 75 000 x 0.5066 x 0.6209 = 23 591 TL
Total present worth of incomes: P
inc= 682 344 + 510 554 + 31 045 + 23 591 = 1 247 534 TL

23. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
23
Expenditures:
P1=E1= 620 000 TL
Total present worth of incomes:
P
inc= 1 247 534 TL

24. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
24
Expenditures:
P1=E1= 620 000 TL
P2 = E2 x (P/F, 10%, 3)
P2 = 60 000 x 0.7513
P2 = 45 078 TL
Total present worth of incomes:
P
inc= 1 247 534 TL

25. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
25
Expenditures:
P1=E1= 620 000 TL
P2= E2 x (P/F, 10%, 3) = 45 078 TL
P3= E3 x (P/F, 10%, 5)
P3 = 800 000 x 0.6209
Total present worth of incomes:
P
inc= 1 247 534 TL
P3 = 496 720 TL

26. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
26
Expenditures:
P1=E1= 620 000 TL
P2= E2 x (P/F, 10%, 3) = 45 078 TL
P3= E3 x (P/F, 10%, 5) = 496 720 TL
P4= E4 x (P/F, 12%, 3) x (P/F, 10%, 5) =
Total present worth of incomes:
P
inc= 1 247 534 TL
80 000 x 0.7118x 0.6209 =35 357 TL

27. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
27
Expenditures:
P1=E1= 620 000 TL
P2= E2 x (P/F, 10%, 3) = 60 000 x 0.7513 = 45 078 TL
P3= E3 x (P/F, 10%, 5) = 800 000 x 0.6209 = 496 720 TL
P4= E4 x (P/F, 12%, 3) x (P/F, 10%, 5) = 80 000 x 0.7118 x 0.6209 = 35 357 TL
Total present worth of expenditures: Pexp= 620 000 + 496 720 + 35 357 + 45 078 = 1 197 155 TL
Total present worth of incomes:
P
inc= 1 247 534 TL

28. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cashflow Example 1
28
P
net= NPV = 1 247 534 -1 197 155 = 50 379 TL
As NPV > 0 , this is a good investment.
Total present worth of incomes: Total present worth of expenditures:
P
inc= 1 247 534 TL P
exp= 1 197 155 TL

29. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
•Nominal and Effective Interest Rate
•Effective Interest Rate Formulation
•Cash Flow Calculations
29
NOMINAL AND EFFECTIVE INTEREST
Outline

30. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Nominal and Effective Interest Rates
30
•Nominal and effective interest rates are used when the compounding period
is less than 1 year.
•When an interest rate is expressed over a period of time shorter than a year,
such as 1% per month, the terms nominal and effective interest rates must be
considered.
•Dictionary meaning of nominal: so-called, stated, supposed, assumed
•Nominal interest rate is not a correct or actual rate. Nominal interest rates
should be converted into effective rates in order to accurately reflect time-
value considerations.

31. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Nominal and Effective Interest Rates
31
Why is “nominal interest rate” not correct ?
Because nominal interest rate ignores time value of money and the frequency
with which interest is compounded.
When is an interest rate called as “effective” ?
When the time value of money is taken into consideration in calculating interest
rates from period interest rates, rate is called an effective interest rate.
We have to convert nominal rates to effective rates to solve cash flow
problems !

32. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Nominal and Effective Interest Rates
32
A nominal interest rate can be found as follows:
�=??????�����������������??????�������������??????���
where r is nominal interest rate.
For example : Interest rate = 1% per month (interest compounds 1% every
month)
→�=1%×3=3%→���??????�??????�3%�����??????����
→�=1%×6=6%→���??????�??????�6%������????????????���??????����??????��
→�=1%×12=12%→���??????�??????�12%�����??????�

33. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Nominal and Effective Interest Rates
33
→Effective interest rate of 3% per 6 months is = 3%
0 6 12
is = 3%
→Effective interest rate of 2% per 3 months0 3 126 9
iq = 2% iq = 2% iq = 2% iq = 2%
r
s= 3% + 3% = 6% per year, compounded semi-annually
months
r
q= 2% + 2% + 2% + 2% = 8% per year, compounded quarterly
months

34. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
•Although nominal rates are not exactly correct, they are very close to effective
(real, actual) rates. Nominal interest rate is always smaller than the effective
rate as it ignores compounding of interest. However, nominal rates are
indicative and very simple to calculate !
•Assume that you want to estimate how much money will be accumulated in
your bank account where you deposit your money every 6 months, but the
yearly nominal interest rate is given to you by the bank (such as 12% per year
compounding monthly). Then, you need to calculate effective interest rate for 6
months to calculate future value of your deposits (eg. in 5 years). You have to
consider compounding of interest every month to calculate effective interest
rate per 6 months. Effective rates should be used in cash flow calculations.
Nominal rates should be converted to real/actual/effective rates !
34
Why do we bother with nominal and effective interest rates ?

35. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
35
1.i= 12% per year
i= 1% per month
i=effective 12% per year compounded yearly
i=effective 1% per month compounded monthly
When no compounding period is given, interest rate is an effective rate, assume that
compounding period equals to stated time period.
2.i= 8% per year, compounded monthly
i= 4% per quarter, compounded monthly
i=nominal 8% per year compounded monthly
i=nominal 4% per quarter compounded monthly
When compounding period is given without stating whether the interest rate is nominal or
effective, it is assumed to be nominal. Compounding period is as stated.
3.i= effective 10% per year, compounded monthly
i= effective 16% per quarter
i=effective 10% per year compounded monthly
i=effective 6% per quarter compounded quarterly
Stated as effective, it is an effective interest rate. If compounding period is not given, it is
assumed to be equal to stated time period.
How do we recognize whether the given rate is nominal or effective ?

36. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Examples
36
Interest Statement Nominal or Effective InterestCompounding Period
15% per year, compounded
monthly
15% per year
Effective 15% per year,
compounded monthly
2% per month, compounded
weekly
Effective 6% per quarter
1% per week, compounded daily
Nominal Monthly
Effective Yearly
Effective Monthly
Nominal Weekly
Effective Quarterly
Nominal Daily

37. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Effective Interest Rate Formulation
37
Example 1: You deposit 100 $ in a bank account for 1 year
a)Bank pays 12% interest compounded annually.100$
F=?
1
F=P1+i
n
F=1001+0.12
1
=112$
P =

38. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Effective Interest Rate Formulation
38
Example 1: You deposit 100 $ in a bank account for 1 year
b)Bank pays 12% interest per year, compounded semiannually
(nominal 12%)
i
s = i
effective= 6% semi-annually 2
100$
F=?
1 Interest period
F=1001+0.06
2
=112.36$
F=P1+i
n
P =
(6 months) (12 months)

39. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Effective Interest Rate Formulation
39??????= �+
�
�

�
−� In this formula i and r
have the same periods


where,
i= effective interest rate per period
r = nominal interest rate per period
m = number of compounding periods
??????
�=
annually semi-annually quarterly monthly
In our previous example;
??????
??????=1+
�
�
�
�
−1=1+
0.12
2
2
−1=12.36%
•Bank pays 12% interest compounded semiannually
1+??????
�
2
−1=1+??????
�
4
−1=1+??????
�
12
−1
??????
�=1+
�
�
2
2
−1=1+
�
�
4
4
−1=1+
�
�
12
12
−1
??????
�=
�
�
2
??????
�=
�
�
4
??????
�=
�
�
12

40. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Effective Interest Rate Formulation -Examples
40
•If the nominal interest rate is 6% per year with different compounding periods:
i
s= 3% per half year
??????
??????=1+
�
�
�
�
−1=1+
0.06
2
2
−1=6.09%
❖Compounding period=semiannual
i
q= 1.5% per quarter
??????
??????=1+
�
�
�
�
−1=1+
0.06
4
4
−1=6.136%
❖Compounding period=quarterly

41. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Effective Interest Rate Formulation -Examples
41
•If the nominal interest rate is 6% per year with different compounding periods:
i
m= 0.5% per month
??????
??????=1+
�
�
�
�
−1=1+
0.06
12
12
−1=6.168%
❖Compounding period=monthly
i
q= 0.115% per week
??????
??????=1+
�
??????
�
�
−1=1+
0.06
52
52
−1=6.180%
❖Compounding period=weekly
i
d= 0.016% per day
??????
??????=1+
�
�
�
�
−1=1+
0.06
365
365
−1=6.183%
❖Compounding period=daily

42. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Effective Interest Rate Formulation –Continuous Compounding
42
•If the nominal interest rate is 6% per year with different compounding periods:
❖Compounding period=continuously
Continuouscompounding=??????=�
??????
−1
??????
??????=�
0.06
−1= 6.184 %

43. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Some examples about calculation of effective rates (1)
43
a) Interest rate is 2% per month
Effective rate per semiannual period →i
s ?
���??????�??????��??????��������????????????���??????����??????��=2%×6months=12%=0.12=�
??????
�=1+
�
�
�
=1+
0.12
6
6
−1=12.62%
b) Interest rate is 5% per quarter
Effective rate per semiannual period →i
s ?
�
���????????????���??????�=5%×2quarters=10%=0.10
??????
�=1+
�
�
�
=1+
0.10
2
2
−1=10.25%
Semi-annual period = 6 months
Semi-annual period = 2 quarters

44. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
44
c) Interest rate is 5% per quarter
Effective rate per annual period →i
a?
??????
??????=1+
�
�
�
=1+
0.20
4
4
−1=21.55%
�
??????���??????�=5%×4quarters=20%=0.20
•
�
�
alwaysequalstoeffectiveinterestratepercompoundingperiod.
Some examples about calculation of effective rates (2)
Annual period = 4 quarters

45. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cash flow calculations
45
❖When solving interest problems with nominal interest, there are 2 basic situations:
1.Compounding periods and occurrence of payments coincide
In this case, the effective rate of interest corresponding to that compounding
period (which is, at the same time, the payment period) is calculated and the
interest periods are adopted.
2.Compounding periods and occurrence of payments do not coincide
In this case, the effective annual interest rate is calculated by using the formula
??????=(1+
�
�
)
�
−1; then the effective interest rate corresponding to the payment
period is calculated by using the same formula in reverse, and then we proceed
according to the payment periods.

46. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Cash flow calculations -Effective Interest Rate
46
•Case 1: compounding period = occurrence of payments
Example 1: Deposit 2 000 $ quarterly into a fund that pays a nominal interest of 12%
per year, compounded quarterly. What is the amount accumulated at the end of 5
years?
CP (Compound Period) = quarter
PP(Payment Period) = quarter
i
qrequired!
??????
��??????����=
12
4
=3%������??????��??????�����??????����
�=4×5=20���??????���
??????=??????Τ????????????,3%,20
??????=200026.8704= 53 740 $

47. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
47
•Case 2: compounding period ≠ occurrence of payments (Compounding
period is morefrequent than payment period)
Example 2: Deposit 2 000 $ semiannually into a fund that pays a nominal interest of
12% per year compounded quarterly. What is the amount accumulated at the end of 5
years?
CP = quarter
PP = semiannual
effective ifor semiannual period
should be calculated!CP quarterly PP semiannually
Yearly
ia
r= 12% per year
compounded quarterly is=?
Cash flow calculations -Effective Interest Rate
??????
??????=1+
�
�
�
−1=1+
0.12
4
4
−1=0.1255
0.1255=1+
�
�
2
2
−1
??????
�=1.1255−1=0.061=6.1%
0.1255=1+??????
�
2
−1

48. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Example 2 -continues
48
??????
�=6.1%
??????=??????Τ????????????,6.1%,10
F=2000Τ????????????,6.1%,10
??????
�=1+??????
�
2
−1=1+
0.12
4
2
−1=0.061
AΤ????????????,i,�=??????
(1+??????)
�
−1
??????
F=2000
(1+0.061)
10
−1
0.061
=2000∗13.2429=���??????�.??????$
Solving by using the interest factor formula:

49. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
Example 2 -continues
49
Τ????????????,6%,10=13.181
Τ????????????,7%,10=13.816
6% 7%
13.181
13.816
6.1%
�
x=13.181+
13.816−13.181
7−6
∗0.1
Τ????????????,6.1%,10=13.245
F=2000Τ????????????,6.1%,10= 26 490 $
??????
�=1+??????
�
2
−1=1+
0.12
4
2
−1=0.061
Τ????????????,6.1%,10= x
??????
�=6.1%
??????=??????Τ????????????,6.1%,10
F=2000Τ????????????,6.1%,10
Solving by using the interest factor table and linear interpolation:

50. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
50
•Case 2: compounding period ≠ occurrence of payments (Compounding
period is less frequentthan payment period)
Example 3: Deposit 2000 $ monthly into a fund that pays a nominal interest rate of
12% per year compounded quarterly. What is the amount accumulated at the end of 5
years?
??????
??????=1+
0.12
4
4
−1=0.1255→0.1255=1+??????
�
12
−1→??????
�=0.0099≈1%
??????=??????Τ????????????,1%,60=2000Τ????????????,1%,60=2000∗81.6697=163340$
Cash flow calculations -Effective Interest RateCP quarterly PP monthly
yearly
r= 12% per year
compounded quarterly im=?
ia

51. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
51
Acompanyplanstobuyabulldozeranduseitfor10years.Duringthisperiod,theincomes
andexpendituresareshowninthecashflowgivenbelow.Iftheinterestratesareexpectedto
be10%peryearcompoundedquarterlyforthefirst5years,and12%peryearcompounded
semiannuallyforthelast5years,findthepresentworthofthisinvestment.
0 1 2 3 4 5 6 7 8 9 10 years
E1= 80 000 TL
E2=60 000 TL
A1= 30 000 TL per year
A2= 20 000 TL per 6 months
E4= 50 000 TL
r = 10% compounding quarterly r = 12% compounding semiannually
Cash flow calculations -Example

52. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
52
0 1 2 3 4 5 6 7 8 9 10 years
E1= 80 000 TL
E2=60 000 TL
A1= 30 000 TL per year
A2= 20 000 TL per 6 months
E4= 50 000 TL
r = 10% compounding quarterly r = 12% compounding semiannually
Cash flow calculations –Example (continued)
Annual Effective Interest Rate for
First 5 Years
??????
??????=1+
0.10
4
4
−1=0.1038 ??????
??????=10.38%
Semiannual Effective Interest Rate
for Last 5 Years
??????
�=
0.12
2
=0.06
??????
�=6%
Annual Effective Interest Rate for
Last 5 Years
??????
??????=1+
0.12
2
2
−1=0.1236 ??????
??????=12.36%

53. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
53
0 1 2 3 4 5 6 7 8 9 10 years
E1= 80 000 TL
E2=60 000 TL
A1= 30 000 TL per year
A2= 20 000 TL per 6 months
E4= 50 000 TL
i
a= 10.38%
i
s= 6%
i
a= 12.36%
Incomes:
P1 = A1 x (P/A, 10.38%, 5) = 30 000 x
(�+�.���??????)
�
−�
�.���??????(�+�.���??????)
�
= 30 000 x 3.7543 = 112 628.15 TL
P2 = A2 x (P/A, 6%, 10) x (P/F, 10.38%, 5) = 20 000 x 7.3601 x
�
(�+�.���??????)
�
=147 202 x 0.6103= 89 839.48 TL
Total present worth of incomes: P
inc= 112 628.15 + 89 839.48 = 202 467.63 TL
Cash flow calculations –Example (continued)
(P/A, 6%, 10)
(P/F, 10.38%, 5)
2 semi-annual periods per year x 5 years
= 10 semi-annual periods

54. •
Week 3
Nominal and Effective Interest Rates
METU CIVIL ENGINEERING DEPARTMENT
CE 231 ENGINEERING ECONOMY
54
0 1 2 3 4 5 6 7 8 9 10 years
E1= 80 000 TL
E2=60 000 TL
A1= 30 000 TL per year
A2= 20 000 TL per 6 months
E3= 50 000 TL
i
a= 10.38%
i
s= 6%
i
a= 12.36%
Expenditures:
P1=E1= 80 000 TL
P2= E2 x (P/F, 10.38%, 2) = 60 000 x
�
(�+�.���??????)
�
= 60 000 x 0.8207 = 49 245.94 TL
P3= E3 x (P/F, 12.36%, 3) x (P/F, 10.38%, 5) = 50 000 x
�
(�+�.����)
�
x
�
(�+�.���??????)
�
= 50 000 x 0.7049 x 0.6103 = 21 510.02 TL
Total present worth of expenditures: P
exp= 80 000 + 49 245.94 + 21 510.02 = 150 755.96 TL
Total present worth = P
inc-P
exp= 202 467.63 –150 755.96 = 51 711.67 TL
Cash flow calculations –Example (continued)