time value for money

1. Chapter 4:
The Time Value of Money

2. The objective of Chapter 4 is to
explain time value of money
calculations and to illustrate
economic equivalence.

3. Money has a time value.
 Capital refers to wealth in the form of money or
property that can be used to produce more wealth.
 Engineering economy studies involve the
commitment of capital for extended periods of
time.
 A dollar today is worth more than a dollar one or
more years from now (for several reasons).

4. Simple interest is used infrequently.
When the total interest earned or charged is
linearly proportional to the initial amount of the
loan (principal), the interest rate, and the number
of interest periods, the interest and interest rate
are said to be simple.

5. Computation of simple interest
The total interest, I, earned or paid may be computed
using the formula below.
P = principal amount lent or borrowed
N = number of interest periods (e.g., years)
i = interest rate per interest period
The total amount repaid at the end of N interest
periods is P + I.

6. If $5,000 were loaned for five years at a
simple interest rate of 7% per year, the
interest earned would be
So, the total amount repaid at the end
of five years would be the original
amount ($5,000) plus the interest
($1,750), or $6,750.

7. Ice Breaking
In 2005, total debit (credit cards, auto loans, home
mortgages, etc.) amounted to more than 100% of total
disposable income for the average U.S household. If
your total disposable income is $50,000, how much
sum can you expect to pay at the end of one year
period when the average interest rate on your debit is
12% per year?
Simple Interest Calculation,
I = P*N* I
I= $50,000*1*(0.12)
Therefore, I = $ 6,000

8. Compound interest reflects both the remaining principal
and any accumulated interest. For $1,000 at 10%…
Period
(1)
Amount owed
at beginning of
period
(2)=(1)x10%
Interest
amount for
period
(3)=(1)+(2)
Amount
owed at end
of period
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331
Compound interest is commonly used in personal and
professional financial transactions.

9. Economic equivalence allows us to compare
alternatives on a common basis.
 Each alternative can be reduced to an equivalent
basis dependent on
 interest rate,
 amount of money involved, and
 timing of monetary receipts or expenses.
 Using these elements we can “move” cash flows so
that we can compare them at particular points in
time.

10. We need some tools to find economic
equivalence.
 Notation used in formulas for compound
interest calculations.
 i = effective interest rate per interest period
 N = number of compounding (interest) periods
 P = present sum of money; equivalent value of one or
more cash flows at a reference point in time; the present
 F = future sum of money; equivalent value of one or more
cash flows at a reference point in time; the future
 A = end-of-period cash flows in a uniform series
continuing for a certain number of periods, starting at the
end of the first period and continuing through the last

11. A cash flow diagram is an indispensable tool
for clarifying and visualizing a series of cash
flows.

12. Cash flow tables are essential to modeling
engineering economy problems in a
spreadsheet

13. Active Learning-1
Before evaluating the economic merits of a proposed
investment, the XYZ Corporation insists that its engineers
develop a cash-flow diagram of the proposal. An
investment of $10,000 can be made that will produce
uniform annual revenue of $ 5,310 for five years and then
have a market (recovery) value of $ 2000 at the end of
the year five. Annual expenses will be $3000 at the end
of each year for operating and maintaining the project.
Draw a cash-flow diagram for the five year life of the
project.

14. We can apply compound interest formulas to
“move” cash flows along the cash flow diagram.
Using the standard notation, we find that a
present amount, P, can grow into a future
amount, F, in N time periods at interest rate
i according to the formula below.
In a similar way we can find P given F by

15. It is common to use standard notation for
interest factors.
This is also known as the single payment
compound amount factor. The term on the
right read as “F given P at i% interest per
period for N interest periods.”
is called the single payment present worth
factor.

16. Active Learning- 2
$2,500 at time zero is equivalent to how much after six years if
the interest rate is 8% per year?
$3,000 at the end of year seven is equivalent to how much today
(time zero) if the interest rate is 6% per year?

17. Betty will need $12,000 in five years to pay for a major
overhaul on her tractor engine. She has found an
investment that will provide a 5% return on her
invested funds. How much does Betty need to invest
today so she will have her overhaul funds in five
years?
Active Learning- 3

18. Uniform Series Annuity and Future
Equivalent Values
How much will you have in 40 years if you
save $3,000 each year and your account
earns 8% interest each year?

19. Active Learning - 4
A recent government study reported that a college
degree is worth an extra $23,000 per year in income (A)
compared to what a high school graduate makes. If the
interest rate (i) is 6 % per year and you work for 40
years (N) , What is the future compound amount (F) of
this extra income ? Please draw Cash Flow Diagram
and Solve.

20. Uniform Series Annuity and Present
Equivalent Values
How much would be needed today to provide
an annual amount of $50,000 each year for 20
years, at 9% interest each year?

21. Active Learning – 5
If a certain machine undergoes a major overhaul now,
its output can be increased by 20%- which translates
into additional cash flow of $20,000 at the end of each
year for five years. If i= 15% per year, how much can
we afford to invest to overhaul this machine? Draw a
Cash Flow Diagram and Solve?

22. Finding A when given F.
How much would you need to set aside each
year for 25 years, at 10% interest, to have
accumulated $1,000,000 at the end of the 25
years?

23. Finding A when given P.
If you had $500,000 today in an account/fund
earning 10% each year, how much could you
withdraw each year for 25 years?

24. Active Learning- 6
You borrow $15,000 from your credit union to purchase
a used car. The interest rate on your loan is 0.25% per
month and you will make a total of 36 monthly
payments. What is your monthly payment? Make a
Cash Flow Diagram from the perspective of a bank and
Solve.

25. Acme Steamer purchased a new pump for $75,000.
They borrowed the money for the pump from their bank
at an interest rate of 0.5% per month and will make a
total of 24 equal, monthly payments. How much will
Acme’s monthly payments be? Make a Cash Flow
Diagram from the perspective of Acme Steamer and
Solve.
Active Learning- 7

26. It can be challenging to solve for N or i.
 We may know P, A, and i and want to find N.
 We may know P, A, and N and want to find i.
 These problems present special challenges that are
best handled on a spreadsheet.

27. Finding N
Acme borrowed $100,000 from a local bank, which
charges them an interest rate of 7% per year. If Acme
pays the bank $8,000 per year, now many years will it
take to pay off the loan?
So,
This can be solved by using the interest tables and interpolation,
but we generally resort to a computer solution.
(P/A, 7%, 30) = 12.40 (P/A, 7%, 31) = 12.53

28. Finding i
The car you want to buy will cost $60,000 in eight years. You
are going to put aside $6,000 each year for eight years. At what
interest rate must you invest your money now to achieve your
goal ?
So,
Again, this can be solved using the interest tables and
interpolation, but we generally resort to a computer solution.
(F/A, i%, 8) = 10
(F/A, 6%, 8) = 9.8975
(F/A, 7%, 8) = 10.2598

29. Finding i

30. Spreadsheet/Excel functions to find N and i.
The Excel function used to solve for N is
NPER(rate, pmt, pv), which will compute the
number of payments of magnitude pmt required to
pay off a present amount (pv) at a fixed interest
rate (rate).
One Excel function used to solve for i is
RATE(nper, pmt, pv, fv), which returns a fixed
interest rate for an annuity of pmt that lasts for nper
periods to either its present value (pv) or future value
(fv).

32. Deferred Annuities (Uniform Series)
 Deferred annuities are uniform series that do not
begin until some time in the future.
 If the annuity is deferred J periods then the first
payment (cash flow) begins at the end of period
J+1.

33. Finding the value at time 0 of a deferred
annuity is a two-step process.
1. Use (P/A, i%, N-J) find the value of the deferred
annuity at the end of period J (where there are N-
J cash flows in the annuity).
2. Use (P/F, i%, J) to find the value of the deferred
annuity at time zero.

34. Irene just purchased a new sports car and wants to also set
aside cash for future maintenance expenses. The car has a
bumper-to-bumper warranty for the first five years. Irene
estimates that she will need approximately $2,000 per year in
maintenance expenses for years 6-10, at which time she will
sell the vehicle. How much money should Irene deposit into
an account today, at 8% per year, so that she will have
sufficient funds in that account to cover her projected
maintenance expenses?
Active Learning-8
P5 = A(P/A, 8%, 5) = $2,000*(3.9927)= $7,985.40
P0 = F5 (P/F, 8%, 5)= $7,985.40*(0.6806)= $5,434.86

35. Active Learning 9
A father on the day his son is born, wishes to determine
what lump sum amount would have to be paid into an
account bearing interest of 12% per year to provide
withdrawals of $2000 on each of the son’s 18th, 19th and
20th and 21st birthday?
P17 = A(P/A, 12%, 4) = %2,000*(3.0373)= $6,074.60
P0 = F17 (P/F, 12%, 17)= $6,074.60*(0.1456) = $884.46

36. Active Learning 10
When you take your first job, you decide to start saving right
away for your retirement. You put $5000 per year into the
company’s 410(k) plan, which averages 8% per year. Five
years later, you move to another job and start a new 401(k)
plan. You never get around to merging the funds in the two
plans. If the first plan continued to earn interest at the rate of
8% per year for 35 years after you stopped making
contributions, how much is the account worth.
F5 = $5,000(F/A, 8%,5)= $5,000*(5.8666)= $29,333
F40 = P5 (F/P, 8%, 35) = $ 29,333*(14.7853)= $433,697
P0= $5,000(P/A,8%,5)= $5,000*(3.9927)= $19,963.5
F40 = P0 (F/P, 8%,40) = $ 19,963.50*(21.7245)= $433,697
Alternatively,

37. Active Learning 11
The cash flows extends over 8 years. The amounts are $100
for the first year, $200 for the second year, $500 for the third
year and $400 for each year from the fourth year through the
eighth. These cash flows represent something like the
expenditures for a certain piece of equipment or payments
into a fund. The annual interest rate is 20%. Find;
a. The present equivalent expenditure, P0
b. The future equivalent expenditure, F8
c. The annual equivalent expenditure, A
(P0= $ 1203.82, F8=$5176.19 & A=$313.73)

38. Active Learning 12
Two receipts of $1,000 each are desired at the EOYs 10 and
11. To make these receipts possible, four equal yearly annuity
amounts will be deposited in a bank. At EOYs 2,3,4, and 5.
The bank’s interest rate I = 12% per year.
a. Draw a cash-flow diagram for this situation.
b. Determine the value of A that establishes equivalence in
your cash flow diagram.
c. Determine the lump-sum value at the end of year 11 of the
completed cash flow diagram based on your answers to
Part(a) and Part(b).
(A=$224.75 & F= 0)

39. Deferred Annuities (Uniform Series)
F0 = A(F/A, i%, N-J) (F/P, i%, J)
P0 = A(P/A, i%, N-J) (P/F, i%, J)
ൗ
𝑷
𝑨 =
(𝟏 + 𝒊)𝑵
− 𝟏
𝒊(𝟏 + 𝒊)𝑵 ൗ
𝑭
𝑨 =
(𝟏 + 𝒊)𝑵
− 𝟏
(𝟏 + 𝒊)𝑵
ൗ
𝑭
𝑷 = (𝟏 + 𝒊)𝑵
ൗ
𝑷
𝑭 =
𝟏
(𝟏 + 𝒊)𝑵

40. Sometimes cash flows change by a constant
amount each period: Arithmetic Gradient
We can model these situations as a uniform
gradient of cash flows. The table below shows
such a gradient.
End of Period Cash Flows
1 0
2 G
3 2G
: :
N (N-1)G

41. It is easy to find the present value of a
uniform gradient series.
Similar to the other types of cash flows, there is a
formula (albeit quite complicated) we can use to find
the present value, and a set of factors developed for
interest tables.
Present Value = G* (P/G, i%, N)

42. We can also find A or F equivalent to a
uniform gradient series.
Annuity = G (A/G, i%, N)
Future Value= G (F/G, i%, N)

43. Active Learning 13
Suppose that the cash flows are expected to be $1000 for the
second year, $2000 for the third year and $3000 for the fourth
year and that if the interest is 15% per year, it is desire to find
a. Present equivalent value
b. Uniform annual equivalent value at the end of the four
years.
P0 = G(P/G, 15%, 4)= $1000(3.79) = $3,790.00
A = G(A/G, 15%, 4)= $1000(1.3263) = $1,326.30

44. End of Year Cash Flows ($)
1 5,000
2 6,000
3 7,000
4 8,000
Find the Annual Equivalent
of this arithmetic gradient
series of cash flows . i =
15%
End of Year Annuity ($) Gradient ($)
1 5,000 0
2 5,000 1,000
3 5,000 2,000
4 5,000 3,000
A = $5,000 + $1,000 (A/G, 15%,4) = $6326.30
P=A(P/A,15%,4)+G(P/G,15%,4) = $18,065

45. End of Year Cash Flows ($)
1 8,000
2 7,000
3 6,000
4 5,000
Active Learning 15
The cash flows are timed in
reverse gradient. Calculate the
present equivalent at i= 15%
per year using arithmetic
gradient interest factors.
End of Year Annuity ($) Gradient ($)
1 8,000 0
2 8,000 -1,000
3 8,000 -2,000
4 8,000 -3,000
A = $8,000 - $1,000 (A/G, 15% ,4) = $ 6,673.70
P= $8,000(P/A, 15% ,4) -1000(P/G, 15%, 4) = Solve

46. Sometimes cash flows change by a
constant rate, , each period- Geometric
Gradient
End of Year Cash Flows ($)
1 1,000
2 1,200
3 1,440
4 1,728
This table presents a
geometric gradient series. It
begins at the end of year 1
and has a rate of growth, ,
of 20%.

48. Present value of a Geometric Series
Where is the initial cash flow in the series.
i

49. Acme Miracle projects good things for their new weight
loss pill, LoseIt. Revenues this year are expected to
be $1.1 million, and Acme believes they will increase
15% per year for the next 5 years. What are the
present value and equivalent annual amount for the
anticipated revenues? Acme uses an interest rate of
20%.
Active Learning - 16
P = $1.1million 1- (P/F, 20%, 5) (F/P, 15%, 5) ] / (0.20 – 0.15)

50. Active Learning 17
Consider the Geometric sequence of cash flows and
determine P, A and F equivalent values. The rate of
decrease is 20% per year after the first year which
continues for 4th year, and the rate of interest is 25%
per year
P = $1000 1- (P/F, 25%,4) (F/P, 20%, 4) ] / (0.25 – (-0.20))
Therefore, P = $1,849.38
A = $1,849.38 (A/P, 25%, 4) = $783.03
F = $1,849.38 (F/P, 25%, 4) = $4,515.08

51. When interest rates vary with time
different procedures are necessary.
 Interest rates often change with time (e.g., a
variable rate mortgage).
 We often must resort to moving cash flows one
period at a time, reflecting the interest rate for that
single period.

52. Activity Learning 18
Mr. X borrowed $4000 four years ago when the interest rate
was 4.06% per year. $5,000 was borrowed three years ago at
3.42%. Two years ago she borrowed $6,000 at 5.23% and last
year $7,000 was borrowed at 6.03% per year. Now Mr. X
would like to consolidate her debt into a single 20 year loan
with a 5% fixed annual interest rate. If Mr. X makes annual
payments (starting in first year) to repay her total loan, what
is the amount of each payment?

53. The present equivalent of a cash flow occurring at
the end of period N can be computed with the
equation below, where ik is the interest rate for the
kth period.
If F4 = $2,500 and i1=8%, i2=10%, and i3=11%, then

54. Nominal and effective interest rates.
 More often than not, the time between successive
compounding, or the interest period, is less than one
year (e.g., daily, monthly, quarterly).
 The annual rate is known as a nominal rate.
 A nominal rate of 12%, compounded monthly, means an
interest of 1% (12%/12) would accrue each month, and
the annual rate would be effectively somewhat greater
than 12%.
 The more frequent the compounding the greater the
effective interest.

55. The effect of more frequent
compounding can be easily determined.
Let r be the nominal, annual interest rate and M the
number of compounding periods per year. We can
find, i, the effective interest by using the formula
below.

56. Effective Interest Rate: Some Examples
Compounding Frequency Number of Compounding
Periods per year, M
Annually 1
Semi Annually 2
Quarterly 4
Bi Monthly 6
Monthly 12
Daily 365
For a 6% nominal interest rate, compounded monthly,
the effective interest is?
I = 6.18%

57. Active Learning 19
Suppose that $100 amount is invested for 10 years at
a nominal interest rate of 18% compounded
quarterly, How much is the worth at the end of the
10th year?
For a 7% nominal rate, compounded monthly, the
effective interest is.
F = $100(F/P, 19.25%, 10) = Find F

58. Interest can be compounded continuously.
 Interest is typically compounded at the end of
discrete periods.
 In most companies cash is always flowing, and
should be immediately put to use.
 We can allow compounding to occur continuously
throughout the period.
 The effect of this compared to discrete
compounding is small in most cases.

59. Continuous Compounding
 Continuous compounding is the mathematical limit
that compound interest can reach. It is an extreme
case of compounding since most interest is
compounded on a monthly, quarterly
or semiannual basis. Hypothetically, with continuous
compounding, interest is calculated and added to the
account's balance every infinitesimally small instant.
While this is not possible in practice, the concept of
continuously compounded interest is important in
finance.

60. We can use the effective interest formula
to derive the interest factors: Continuous
Compounding
As the number of compounding periods gets
larger (M gets larger), we find that

61. Continuous compounding interest factors.
The other factors can be found from these. Please refer book
( F/P, r%, N ) = e r N

63. Active Learning 20
You have $1000 to invest for two years. Your bank offers 5%
interest, compounded continuously for funds in a money
market account. Assuming no additional deposits or
withdrawals, how much money will be in that account at the
end of the two years?
F = $10,000 (F/P, 5%, 2) = $10,000*e (0.05) (2) = $11,052
If 5% were not compounded continuously, then the value of F
would have been
F = $10,000 (F/P, 5%, 2) = $10,000*(1.1025) = $11,025

64. THANK YOU