Here are the key steps to solve this problem: * Future sum (F) = Rs. 1,00,000 * Interest rate (i) = 15% * Number of periods (n) = 10 years * Interest is compounded annually * To find the present worth (P): * P = F / (1 + i)^n * P = Rs. 1,00,000 / (1 + 0.15)^10 * P = Rs. 1,00,000 / 2.472 * P = Rs. Rs. 40,320 Therefore, the single payment that should be made now to receive Rs. 1,00,000 after 10 years at

1. UNIT3. Time Value of Money
3.1. Introduction, Simple and Compound Interest,
Concept of Equivalence, Cash-Flow Diagrams,
Calculation of Present, Future and Annual Equivalent -
Single Cash Flows, Uniform Series (Annuity, Deferred
Annuity, Uniform and Geometric Gradient Series),
Interest Rates that Vary with Time.
3.2. Nominal and Effective Interest Rate, Effective
Interest Rate per Payment Period, Continuous
Compounding.

2. Time Value of Money
• Introduction:
• Time value of money means that a sum of money is
worth more now than the same sum of money in the
future.
• This is because money can grow only through
investing. An investment delayed is an opportunity
lost.
• The formula for computing the time value of money
considers the amount of money, its future value, the
amount it can earn, and the time frame.
• For savings accounts, the number of compounding
periods is an important determinant as well.

3. Time Value of Money
• The most fundamental TVM formula takes into account the following
variables:
• FV = Future value of money
• PV = Present value of money
• i = interest rate
• n = number of compounding periods per year
• t = number of years
• Based on these variables, the formula for TVM is:
• FV = PV x [ 1 + (i / n) ] (n x t)
• Time Value of Money Examples
• Assume a sum of $10,000 is invested for one year at 10% interest
compounded annually. The future value of that money is:
• FV = $10,000 x [1 + (10% / 1)] ^ (1 x 1) = $11,000

4. Time Value of Money
• The formula can also be rearranged to find the
value of the future sum in present day dollars.
For example, the present day dollar amount
compounded annually at 7% interest that
would be worth $5,000 one year from today
is:
• PV = $5,000 / [1 + (7% / 1)] ^ (1 x 1) = $4,673

5. Simple Interest
• Simple interest is calculated on the principal,
or original, amount of a loan.
• Simple Interest is calculated using the
following formula:
• S.I.= P × R × T

6. Simple Interest
• Question 1: A sum of Rs 4000 is borrowed and
the rate is 7%. What is the simple interest for 2
years?
• Solution:
• Simple Interest = Principle × Rate × Time =
PTR/100
• ⇒ Simple Interest = 4000 × (7 ⁄ 100) × 2
• ⇒ Simple Interest = 560
• ∴ The simple Interest for 2 years is Rs. 560

7. Compound Interest
• Compound interest is calculated on the
principal amount and the accumulated interest
of previous periods, and thus can be regarded
as “interest on interest.”
• Compound Interest is calculated using the
following formula: C.I.= P × (1+r)t
- P

8. Compound Interest
• Question 1: A sum of Rs 4000 is borrowed and the
rate is 7%. What is the compound interest for 2
years?
• Compound Interest = Principal × (1 + Rate)Time
−
Principal
• So, Compound Interest = 4000 × (1 + 7 ⁄ 100)2
− 4000
• ⇒ Compound Interest = (4000 × 1.14) − 4000
• ⇒ Compound Interest = 560
• ∴ The compound interest for 2 years is Rs. 560

9. Question 3: Find the compound interest
on Rs. 13000 at 10% for 2 years,
compounded annually
• Solution:
Given,
Principal = P = Rs. 13000
Rate of interest = r = 10%
Time = t = 2 years
Amount on CI = P(1 + r/100)2
= 13000(1 + 10/100)2
= 13000 (1 + 0.1)2
= 13000(1.1)2
= 13000 × 1.21
= 15730
CI = Amount on CI – Principal
= Rs. 15730 – Rs. 13000
= Rs. 2730
Therefore, the compound interest = Rs. 2730

10. Concept of Equivalence
• Economic equivalence is a fundamental concept
upon which engineering economy computations
are based.
• Economic equivalence is a combination of
interest rate and time value of money to
determine the different amounts of money at
different points in time that are equal in
economic value.

11. Concept of Equivalence
• Two things are said to be equivalent when they
have the same effect. Economic equivalence
refers to the fact that a cash flow - whether
single payment or series of payments - can be
converted to an equivalent cash flow at any
point in time.

12. GENERAL PRINCIPLES
• PRINCIPLE 1 :- Equivalence calculations made to compare
alternatives require common time basis. When selecting
a point in time at which to compare the value of
alternative cash flows we commonly use either the
present time & calculate present worth (PW) of the cash
flow, or the present time & calculate future worth (FW) of
the cash flow. The choice of time depends on the
circumstances surrounding a particular decision, or it may
be chosen for convenience.
• PRINCIPLE 2 : - Equivalence depends on interest rate. The
equivalence between cash flows is a function of the
magnitude and timing of individual cash flows and the
interest rate or rates that operate on those flows. This
principle is easy to grasp in relation to principle 1.

13. GENERAL PRINCIPLES
• PRINCIPLE 3 :- Equivalence calculations may require
converting multiple cash flow to a single cash flow.
Convert the given cash flows of alternatives consisting
various type of cash flows to a particular type of cash
flow. Different alternatives consist of various types of
cash flow according to nature of work. For comparison,
convert them into one particular type of cash flow.
• PRINCIPLE 4 :- Equivalence is maintained regardless of
point of view. Cash flow diagram are drawn with
different point of view. However, as long as we use the
same interest rate in equivalence calculations,
equivalence can be maintained regardless of point of
view.

14. CASH FLOW
• A present sum P invested now for N interest periods at
interest rate i% per period.
• Its future worth F would be F = P * (1 + i)N ( N is in power)
• The factor (1 + i%) N is termed as Single Payment
Compound Amount Factor.
• We may also express that factor in functional notation as
• (F/P, i, N), which is read as "Find F, when given P, i, and N."
• It is expressed as
• F = P * (1 + i%)N (N is in power)
• = P * (F/P, i%, N) meaning "Find F, when given P, i, and N."

15. Example
• EXAMPLE 3.1 A person deposits a sum of Rs. 20,000 at the interest
rate of 18% compounded annually for 10 years.
• Find the maturity value after 10 years.
• Solution P = Rs. 20,000
• i = 18%
• N = 10 Years
• compounded annually n = 10 years
• F = P(1 + i%) n (n is in power)
• = P(F/P, i%, n) "Find F, when given P, i, and N."
• = 20,000 (F/P, 18%, 10)
• = 20,000 (1+ 18%) 10 (10 is in power)
• = 20,000 (5.234)
• = Rs. 1,04,680
• The maturity value of Rs. 20,000 invested now at 18%
compounded yearly is equal to Rs. 1,04,680 after 10 years.

16. EXAMPLE
• If you invest Rs.10,000 now for 10 years at
10% per annum, how much would it be worth
at the end of 10 years?
• F = P * (1 + i%)n (n is in power)
• = 10,000 * (1 + 0.1)10
• = Rs.25,937
• Alternately, F = P * (F/P, i%, N)
• = 10,000 * (P/F, 10, 10)
• = 10,000 * (2.5937) = Rs.25,937

17. Solve
• EXAMPLE 3.2 A person wishes to have a future sum of Rs. 1,00,000
for his son’s education after 10 years from now. What is the
single-payment that he should deposit now so that he gets the
desired amount after 10 years?
• The bank gives 15% interest rate compounded annually.
• Solution F = Rs. 1,00,000 i = 15%, compounded annually n = 10
years
• P = F/(1 + i) n
• = F(P/F, i, n) "Find P when given F, i, and n."
• = 1,00,000 (P/F, 15%, 10)
• = 1,00,000 (0.2472)
• = Rs. 24,720
• The person has to invest Rs. 24,720 now so that he will get a sum
of Rs. 1,00,000 after 10 years at 15% interest rate compounded
annually.

18. REVENUE-DOMINATED CASH FLOW
DIAGRAM
• P represents an initial investment
• Rj the net revenue at the end of the jth year.
• The interest rate is i, compounded annually.
• S is the salvage value at the end of the nth year.
• To find the present worth of the above cash flow diagram for a
given interest rate, the formula is
• PW(i) = – P + R1[1/(1 + i)^1 ] + R2[1/(1 + i) ^2 ] + ... + Rj[1/(1 + i)j ]
+ Rn[1/(1 + i) n ] + S[1/(1 + i) n ]
• In this formula, expenditure is assigned a negative sign and
revenues are assigned a positive sign. If we have some more
alternatives which are to be compared with this alternative, then
the corresponding present worth amounts are to be computed
and compared. Finally, the alternative with the maximum present
worth amount should be selected as the best alternative.

19. Example Tech. 1
• Initial outlay,
• P = Rs. 12,00,000
• Annual revenue, A = Rs. 4,00,000
• Interest rate, i = 20%,
• compounded annually Life of this technology, n = 10 years
• The present worth expression for this technology is
PW(20%)1
• Find Present worth (PW)
• If PW = – P + R1[1/(1 + i)^n ]
• = –12,00,000 + 4,00,000 (P/A, 20%, 10)
• = –12,00,000 + 4,00,000 (4.1925)
• = –12,00,000 + 16,77,000
• = Rs. 4,77,000

20. COST-DOMINATED CASH FLOW
DIAGRAM
• P represents an initial investment,
• Cj the net cost of operation and maintenance at the end of the jth
year, and
• S is the salvage value at the end of the nth year.
• To compute the present worth amount of the above cash flow
diagram for a given interest rate i,
• we have the formula PW(i) = P + C1[1/(1 + i)^1 ] + C2[1/(1 + i) 2 ] + ...
+ Cj[1/(1 + i) j ] + Cn[1/(1 + i) n ] – S[1/(1 + i) n ]
• In the above formula, the expenditure is assigned a positive sign and
the revenue a negative sign. If we have some more alternatives
which are to be compared with this alternative, then the
corresponding present worth amounts are to be computed and
compared. Finally, the alternative with the minimum present worth
amount should be selected as the best alternative.

21. Bid 1
• Initial cost, P = Rs. 4,50,000 Annual operation and
maintenance cost, A = Rs. 27,000 Life = 15 years Interest
rate, i = 15%, compounded annually.
• Solution:
• The present worth of the above cash flow diagram is
computed as follows:
• PW(15%)
• PW(i) = P + C1[1/(1 + i)^n
• = 4,50,000 + 27,000(P/A, 15%, 15)
• = 4,50,000 + 27,000 ( 5.8474 )
• = 4,50,000 + 1,57,879.80
• = Rs. 6,07,879.80

22. Single-Payment Present Worth Amount
• The objective is to find the present worth
amount (P) of a single future sum (F) which
will be received after n periods at an interest
rate of i compounded at the end of every
interest period.

23. Single-Payment Present Worth Amount
• Questions: A person wishes to have a future
sum of Rs. 1,00,000 for his son’s education
after 10 years from now. What is the
single-payment that he should deposit now so
that he gets the desired amount after 10
years? The bank gives 15% interest rate
compounded annually.

24. Single-Payment Present Worth Amount
• Solution:
F = Rs. 1,00,000
i = 15%,
Compounded annually n = 10 years
P = F/(1 + i) n
• = F(P/F, i, n)
• = 1,00,000 (P/F, 15%, 10)
• = 1,00,000 0.2472
• = Rs. 24,720 The person has to invest Rs. 24,720 now
so that he will get a sum of Rs. 1,00,000 after 10 years
at 15% interest rate compounded annually.

25. Equal-Payment Series Compound
Amount
• In this type of investment mode, the objective is to find the future
worth of n equal payments which are made at the end of every
interest period till the end of the nth interest period at an interest
rate of i compounded at the end of each interest period.
• A = equal amount to be deposited at the end of each interest
period
• n = No. of interest periods
• i = rate of interest
• F = single future amount at the end of the nth period
•
• The formula to obtain the present worth is
• F = [A (1 + i)n
– 1] / i = A (F/A, i, n)
• Where
• (F/A, i, n) is termed as equal-payment series compound amount
factor.

26. Equal-Payment Series Compound
Amount
• A person who is now 40 years old is planning for his
retired life. He plans to invest an equal sum of Rs.
7,500 at the end of every year for the next 20 years
starting from the end of the next year. The bank
gives 15% interest rate, compounded annually. Find
the maturity value of his account when he is 60
years old.
Solve:
A = Rs. 7,500
n = 20 years
i = 15%
To find
F ?

27. Equal-Payment Series Compound
Amount
Formula used
F = A (F/A, i, n)
Solution
F = A (F/A, i, n)
= 7,500 (F/A, 15%, 20)
= 7,500 * 102.444
F = Rs. 7,68,330
Result
The maturity value is Rs. 7,68,330 in his account at
15% interest rate after 20 years.

28. Questions
• A person who is now 35 years old is planning
for his retired life. He plans to invest an equal
sum of Rs. 10,000 at the end of every year for
the next 25 years starting from the end of the
next year. The bank gives 20% interest rate,
compounded annually. Find the maturity value
of his account when he is 60 years old.

29. Solve
• Solution:
• A = Rs. 10,000
• n = 25 years
• i = 20%
• F = ?
• Ans: Rs. 47,19,810
• The future sum of the annual equal payments
after 25 years is equal to Rs. 47,19,810.

30. Equal-Payment Series Sinking Fund
A sinking fund is a fund established by an economic entity by
setting aside revenue over a period of time to fund a future
capital expense, or repayment of a long-term debt.
In this type of investment mode, the objective is to find the
equivalent amount (A) that should be deposited at the end
of every interest period for n interest periods to realize a
future sum (F) at the end of the nth interest period at an
interest rate of i.

31. Equal-Payment Series Sinking Fund
The formula to get F is
A = (F * i / [(1 + i)n
– 1] =F (A/F, i, n)
Where
(A/F, i, n) is called as equal-payment series sinking fund factor.
Here,
A = equal amount to be deposited at the end of each interest
period
n = No. of interest periods
i = rate of interest
F = single future amount at the end of the nth period

32. Equal-Payment Series Sinking Fund
• A company has to replace a present facility after 12
years at an outlay of Rs. 4,00,000. It plans to deposit
an equal amount at the end of every year for the next
12 years at an interest rate of 16% compounded
annually. Find the equivalent amount that must be
deposited at the end of every year for the next 12
years.
• Given data
• n = 12 years
• F = Rs. 4,00,000
• i = 16%

33. Equal-Payment Series Sinking Fund
• To find
• A ?
• Formula used
• A = F (A/F, i, n)
• Solution
• A = F (A/F, i, n)
• = 4,00,000 (A/F, 16%, 12)
• = 4,00,000 * 0.0324
• A = Rs. 12,960
• Result
• The company should deposit an equal amount Rs. 12,960 at
the end of every year for the next 12 years at an interest
rate of 16% to get an outlay Rs. 4,00,000.

34. Equal-Payment Series Sinking Fund
• A company has to replace a present facility after
15 years at an outlay of Rs. 5,00,000. It plans to
deposit an equal amount at the end of every year
for the next 15 years at an interest rate of 18%
compounded annually. Find the equivalent
amount that must be deposited at the end of
every year for the next 15 years.
• Solution: F = Rs. 5,00,000 n = 15 years i = 18% A =
? ANS:
• Rs. 8,200

35. Equal-Payment Series Present Worth
Amount
• The objective of this mode of investment is to find the present worth of an
equal payment made at the end of every interest period for n interest
periods at an interest rate of i compounded at the end of every interest
period.
• The formula to compute P is
• P = A (1 + i)^n − 1
• = A(P/A, i, n)
• Here,
• P = present worth
• A = annual equivalent payment
• i = interest rate
• n = No. of interest periods
• Where
• (P/A, i, n) is called equal- payment series present worth factor.
•

36. Equal-Payment Series Present Worth
Amount
• A company wants to set up a reserve which will help it
is have an amount equivalent amount of Rs 18,00,000
for the next 20 years towards its employees welfares
measures. The reserve is assumed to grow at the rate
of 18% annually. Find the single payment that must be
made as the reserve amount.
Given data
A = Rs. 18,00,000
n = 20 years
i = 18%
To find P

37. Equal-Payment Series Present Worth
Amount
• Formula used
• P = A (P/A, i, n)
• Solution
• P = A (P/A, i, n)
• = 18,00,000 (P/A, 18%, 20)
• = 18,00,000 * 5.3837
• = Rs. 96,90,660
• Result
• A company should set up a reserve Rs. 96,90,660
which will help it is have an amount equivalent
amount of Rs 18,00,000 for the next 20 years.

38. Solve
• A company wants to set up a reserve which
will help the company to have an annual
equivalent amount of Rs. 10,00,000 for the
next 20 years towards its employees welfare
measures. The reserve is assumed to grow at
the rate of 15% annually. Find the
single-payment that must be made now as the
reserve amount.

39. P = A (1 + i)^n − 1 = A(P/A, i, n)
• A = Rs. 10,00,000
• i = 15%
• n = 20 years
• P = ?
• Ans; Rs. 62,59,300

40. Equal-Payment Series Capital Recovery
Amount
• The objective of this mode of investment is to
find the annual equivalent amount (A) which is
to be recovered at the end of every interest
period for n interest periods for a loan (P)
which is sanctioned now at an interest rate of i
compounded at the end of every interest
period

41. Equal-Payment Series Capital Recovery
Amount
The formula to compute P is as follows:
A = P* i(1 + i)n
/ (1 + i)n
– 1 = P (A/P, i, n)
Where,
(A/P, i, n) is called equal-payment series capital
recovery factor.
P = present worth (loan amount)
A = annual equivalent payment (recovery amount)
i = interest rate
n = No. of interest periods

42. Equal-Payment Series Capital Recovery
Amount
• A company takes a loan of Rs 35,00,000 to modernize
its boiler sections. The loan is to be repaid in 25 equal
installments at 10 % interest rate compounded
annually. Find the equal installment amount that
should be paid for the next 25 years.
Given data
P = Rs. 35,00,000
i = 10%
n = 25 years
To find A

43. Equal-Payment Series Capital Recovery
Amount
• Formula used
• A = P (A/P, i, n)
• Solution
• A = P (A/P, i, n)
• A = P* i(1 + i)n
/ (1 + i)n
– 1
• = 35,00,000 (A/P, 10%, 25)
• = 35,00,000 * 0.1102
• A = Rs. 3,85,700
• Result
• The company should pay an equal amount Rs. 3,85,700 at
the end of every year for the next 20 years at an interest
rate of 10% to repay the loan Rs. 35,00,000.

44. Equal-Payment Series Capital Recovery
Amount
• A bank gives a loan to a company to purchase an
equipment worth Rs. 10,00,000 at an interest rate
of 18% compounded annually. This amount should
be repaid in 15 yearly equal installments. Find the
installment amount that the company has to pay
to the bank.
• Solution P = Rs. 10,00,000 i = 18% n = 15 years A =
?
• Ans: Rs. 1,96,400
• A = P* i(1 + i)n
/ (1 + i)n
– 1

45. Calculations functions and formulas

46. Uniform Gradient Series Annual
Equivalent Amount
• The objective of this mode of investment is to
find the annual equivalent amount of a series
with an amount A1 at the end of the first year
and with an equal increment (G) at the end of
each of the following n – 1 years with an
interest rate i compounded annually.
• A = A1 + G * i(1 + i)n
-in -1/ i(1 + i)n
– 1
• A = A1 + G (A/G,i,n)
• (A/G, i, n) is called uniform gradient series
factor

47. Uniform Gradient Series Annual
Equivalent Amount
• A person is planning for his retired life. He has
10 more years of service. He would like to
deposit 20% of his salary, which is Rs. 4,000, at
the end of the first year, and thereafter he
wishes to deposit the amount with an annual
increase of Rs. 500 for the next 9 years with an
interest rate of 15%. Find the total amount at
the end of the 10th year of the above series.

48. Uniform Gradient Series Annual
Equivalent Amount
• Solution Here, A1 = Rs. 4,000 G = Rs. 500 i = 15% n
= 10 years A = ? & F = ?
• A1 + G(A/G, i, n)
• A = A1 + G * i(1 + i)n
-in -1/ i(1 + i)n
– 1
• Rs. 5,691.60
• F = A(F/A, i, n)
• = Rs. 1,15,562.25
• At the end of the 10th year, the compound
amount of all his payments will be Rs.
1,15,562.25.

49. Uniform Gradient Series Annual
Equivalent Amount
• A person is planning for his retired life. He has
10 more years of service. He would like to
deposit Rs. 8,500 at the end of the first year
and thereafter he wishes to deposit the
amount with an annual decrease of Rs. 500 for
the next 9 years with an interest rate of 15%.
Find the total amount at the end of the 10th
year of the above series.
• Solution Here, A1 = Rs. 8,500 G = –Rs. 500 i =
15% n = 10 years A = ? & F = ?

50. Geometric Gradient present worth
• A geometric gradient series is a cash flow
series that either increases or decreases by a
constant percentage each period. The
uniform change is called the rate of change.
• g = constant rate of change, in decimal form,
by which cash flow values increase or
decrease from one period to the next.

51. Geometric Gradient
Increasing Geometric Gradient
• P= A [ 1- (1+g%/1-i%)^n]/i-g
• When g is not equal to I
• Where, P= present of all class flow between
periods 1 and n
A= Cash flow in period one
g = rate of change per period
i = interest rate per period
n= time period

52. Geometric Gradient
• Technical contractor is trying to calculate the
present worth of personnel salaries over the next
five years. He has four employees whose
combined salaries through the end of this year are
$150,000. If he expects to give each employee a
raise of 5% each year, the present worth of his
employees' salaries at an interest rate of 12% per
year is nearest to: $ 591008.
Cash flow at the end of year 1 is $150,000,
increasing by g=5% per year. What is the present
worth (p) ? P= A1(P/A1,g,I,n)

53. Geometric Gradient
• Decreasing Geometric Gradient
• P= A * N/1+i
• When g is equal to i

54. Annuity
• An annuity is a fixed amount of money that you will
get each year for the rest of your life.
• An annuity is a contract between you and an
insurance company that requires the insurer to
make payments to you, either immediately or in the
future.
• An annuity is a series of payments made at equal
intervals. Examples of annuities are regular deposits
to a savings account, monthly home mortgage
payments, monthly insurance payments and
pension payments. Annuities can be classified by
the frequency of payment dates.

55. Annuity
• Jane won a lottery worth $20,000,000 and has opted for an annuity
payment at the end of each year for the next 10 years as a payout
option. Determine the amount that Jane will be paid as annuity
payment if the constant rate of interest in the market is 5%.
• Solution:
• Given:
• PVA (ordinary) = $20,000,000 (since the annuity to be paid at the
end of each year)
• r = 5%
• n = 10 years
• Using the Annuity Formula,
• Annuity = r * PVA Ordinary / [1 – (1 + r)n
]
• Annuity = 5% × 20000000 / [1 - (1 + 0.05)10
• Annuity = $2,590091.499
• Therefore, Jane will pay an annuity amount of $2,590091.499

56. Annuity
• Dan was getting $100 for 5 years every year at an interest rate
of 5%. Find the future value of this annuity at the end of 5 years?
Calculate it by using the annuity formula.
• Solution
• The future value
• Given: r = 0.05, 5 years = 5 yearly payments, so n = 5, and P =
$100
• FV = P×((1+r)n
−1) / r
• FV = $100 × ((1+0.05)5
−1) / 0.05
• FV = 100 × 55.256
• FV = $552.56
• Therefore, the future value of annuity after the end of 5 years is

57. Annuity
• If the present value of the annuity is $20,000. Assuming a monthly
interest rate of 0.5%, find the value of each payment after every month
for 10 years. Calculate it by using the annuity formula.
• Solution:
• Given:
• r = 0.5% = 0.005
• n = 10 years x 12 months = 120, and PV = $20,000
• Using formula for present value
• PV = P×(1−(1+r)n
) / r
• Or, P = PV × ( r / (1−(1+r)n
))
• P = $20,000 × (0.005 / (1−(1.005)120
))
• P = $20,000 × (0.005/ (1−0.54963))
• P = $20,000 × 0.011...
• P = $222
• Therefore, the value of each payment is $220.

58. Deferred Annuity
• A deferred annuity is an insurance contract that
generates income for retirement.
In exchange for one-time or recurring deposits held for
at least a year, an annuity company provides
incremental repayments of your investment plus some
amount of returns.
• Deferred Annuity = P Due
* [1 – (1 + r)n
] / [(1 + r)t-1
* r]
• P Due
= Annuity payment due.
• r = Effective rate of interest.
• n = No. of periods.
• t = Deferred periods.

59. Deferred Annuity
• Let us take the example of John, who got a deal to lend $60,000
today, and in return, he will receive twenty-five annual payments
of $6,000 each. The annuity will start five years from now, and
the effective rate of interest will be 6%. Determine whether the
deal is a feasible one for John if the payment is an ordinary
annuity and annuity due.
• Given, P Ordinary
= $60,000
• r = 6%
• n = 25 years
• t = 5 years
• Deferred Annuity = $6,000 * [1 – (1 + 6%)-25
] / [(1 + 6%)5
* 6%]
• Deferred Annuity = $57,314.80 ~ $57,315

60. Future worth
• Future value is the money amount that will
accrue over time when that sum is invested.

61. Future worth
• A = F * i / (1 +i )^n − 1 = F(A/F, i, n)
• Initial investment,
• P = Rs. 50,00,000
• Annual equivalent revenue, A = Rs. 20,00,000
Interest rate, i = 18%,
• compounded annually Life of alternative A = 4
years
• The future worth amount of alternative A is
computed as FWA(18%) = –50,00,000(F/P, 18%, 4)
+ 20,00,000(F/A, 18%, 4) = –50,00,000(1.939) +
20,00,000(5.215) = Rs. 7,35,000

62. Future worth
Initial investment, P = Rs. 45,00,000 Annual
equivalent revenue, A = Rs. 18,00,000 Interest rate,
i = 18%, compounded annually Life of alternative B
= 4 years
The future worth amount of alternative B is computed
as
FWB(18%) =
–45,00,000(F/P, 18%, 4) + 18,00,000 (F/A, 18%, 4)
= –45,00,000(1.939) + 18,00,000(5.215) = Rs. 6,61,500
The future worth of alternative A is greater than
that of alternative B. Thus, alternative A should be
selected

63. Annual Equivalent
• The annual equivalent rate is the actual
interest rate on investment, loan, or savings
account will yield after accounting for
compounding.

64. Annual equivalent
• Down payment, P = Rs. 5,00,000 Yearly equal
installment, A = Rs. 2,00,000 n = 15 years i = 20%,
compounded annually
• AE1(20%) = 5,00,000(A/P, 20%, 15) + 2,00,000 =
5,00,000(0.2139) + 2,00,000
• = 3,06,950
• Down payment, P = Rs. 4,00,000 Yearly equal
installment, A = Rs. 3,00,000 n = 15 years i = 20%,
compounded annually AE2(20%) = 4,00,000(A/P,
20%, 15) + 3,00,000 = 4,00,000(0.2139) + 3,00,000
• = Rs. 3,85,560.

65. Unit -3
• 3.2. Nominal and Effective Interest Rate,
Effective Interest Rate per Payment Period,
Continuous Compounding.

66. Nominal Interest Rate
• The nominal interest rate is the stated interest
rate of a bond or loan, which signifies the actual
monetary price borrowers pay lenders to use
their money. If the nominal rate on a loan is 5%,
borrowers can expect to pay $5 of interest for
every $100 loaned to them. This is often referred
to as the coupon rate because it was traditionally
stamped on the coupons redeemed
by bondholders.

67. Effective Interest Rate,
• Investors and borrowers should also be aware of the
effective interest rate, which takes the concept of
compounding into account.
• For example, if a bond pays 6% annually and
compounds semiannually, an investor who
places $1,000 in this bond will receive $30 of interest
payments after the first 6 months ($1,000 x .03), and
$30.90 of interest after the next six months ($1,030 x
.03). In total, this investor receives $60.90 for the
year. In this scenario, while the nominal rate is 6%,
the effective rate is 6.09%.

68. Effective Interest Rate per Payment
Period,
• The effective interest rate is calculated
through a simple formula: r = (1 + i/n)^n - 1. In
this formula, r represents the effective interest
rate, i represents the stated interest rate, and
n represents the number of compounding
periods per year.

69. Continuous Compounding
• Formula; Pt = P0 e^rt
P(t) = value at time
Po = original principal sum
r = nominal annual interest rate
t =length of time the interest is applied
E = is a mathematical constant where e ≈ 2.7183.

70. Continuous Compounding
• Example 1: Tina invested $3000 in a bank that
pays an annual interest rate of 7% compounded
continuously. What is the amount she can get
after 5 years from the bank? Round your answer
to the nearest integer.
• Solution:
• The amount after 5 years.
• The initial amount is P = $3000.
• The interest rate is, r = 7% = 7/100 = 0.07.
• Time is, t = 5 years.

71. Continuous Compounding
• Substitute these values in the continuous
compounding formula,
• A = Pert
• A = 3000 × e0.07(5)
≈ 4257
• The answer is calculated using the calculator
and is rounded to the nearest integer.

72. Questions
• 1. Explain the time value of money
• 2. A person deposits a sum of Rs. 1,00,000 in a bank for his
son’s education who will be admitted to a professional course
after 6 years. The bank pays 15% interest rate, compounded
annually. Find the future amount of the deposited money at the
time of admitting his son in the professional course.
• 3. A person needs a sum of Rs. 2,00,000 for his daughter’s
marriage which will take place 15 years from now. Find the
amount of money that he should deposit now in a bank if the
bank gives 18% interest, compounded annually.
• 4. A person who is just 30 years old is planning for his retired
life. He plans to invest an equal sum of Rs. 10,000 at the end of
every year for the next 30 years starting from the end of next
year. The bank gives 15% interest rate, compounded annually.
Find the maturity value of his account when he is 60 years old

73. Questions
• 5. A company is planning to expand its business after 5 years from now. The
expected money required for the expansion programme is Rs. 5,00,00,000.
The company can invest Rs. 50,00,000 at the end of every year for the next
five years. If the assured rate of return of investment is 18% for the
company, check whether the accumulated sum in the account would be
sufficient to meet the fund for the expansion programme. If not, find the
difference in amounts for which the company should make some other
arrangement after 5 years.
6. A financial institution introduces a plan to pay a sum of Rs. 15,00,000 after
10 years at the rate of 18%, compounded annually. Find the annual
equivalent amount that a person should invest at the end of every year for
the next 10 years to receive Rs. 15,00,000 after 10 years from the institution.
• 7. A company is planning to expand its business after 5 years from now. The
money required for the expansion programme is Rs. 4,00,00,000. What
annual equivalent amount should the company deposit at the end of every
year at an interest rate of 15% compounded annually to get Rs. 4,00,00,000
after 5 years from now?

74. Questions
• 8. A company wants to set-up a reserve which will help it
to have an annual equivalent amount of Rs. 15,00,000 for
the next 20 years towards its employees welfare
measures. The reserve is assumed to grow at the rate of
15% annually. Find the single-payment that must be made
as the reserve amount now.
• 9. An automobile company recently advertised its car for a
down payment of Rs. 1,50,000. Alternatively, the car can
be taken home by customers without making any
payment, but they have to pay an equal yearly amount of
Rs. 25,000 for 15 years at an interest rate of 18%,
compounded annually. Suggest the best alternative to the
customers.

75. Questions
10. A company takes a loan of Rs. 20,00,000 to modernize its boiler
section. The loan is to be repaid in 20 equal installments at 12%
interest rate, compounded annually. Find the equal installment
amount that should be paid for the next 20 years.
11. A bank gives loan to a company to purchase an equipment which is
worth of Rs. 5,00,000, at an interest rate of 18% compounded
annually. This amount should be repaid in 25 yearly equal
installments. Find the installment amount that the company has to
pay to the bank. 13. A working woman is planning for her retired life.
She has 20 more years of service. She would like to deposit 10% of
her salary which is Rs. 5,000 at the end of the first year and
thereafter she wishes to deposit the same amount (Rs. 5,000) with
an annual increase of Rs. 1,000 for the next 14 years with an interest
rate of 18%. Find the total amount at the end of the 15th year of the
above series.

76. Questions
• A person is planning for his retired life. He has
10 more years of service. He would like to
deposit 20% of his salary, which is Rs. 10,000,
at the end of the first year and thereafter he
wishes to deposit the same amount (Rs.
10,000) with an annual increase of Rs. 2,000
for the next 9 years with an interest rate of
20%. Find the total amount at the end of the
10th year of the above series.

77. Questions
• A person is planning for his retired life. He has 10
more years of service. He would like to deposit Rs.
30,000 at the end of the first year and thereafter
he wishes to deposit the same amount (Rs.
30,000) with an annual decrease of Rs. 2,000 for
the next 9 years with an interest rate of 18%. Find
the total amount at the end of the 10th year of
the above series.
• A person invests a sum of Rs. 50,000 in a bank at
a nominal interest rate of 18% for 15 years. The
compounding is monthly. Find the maturity
amount of the deposit after 15 years.

78. Present worth questions
• The cost of erecting an oil well is Rs. 1,50,00,000. The
annual equivalent yield from the oil well is Rs. 30,00,000.
The salvage value after its useful life of 10 years is Rs.
2,00,000. Assuming an interest rate of 18%, compounded
annually, find out whether the erection of the oil well is
financially feasible, based on the present worth method.
• The details of the feasibility report of a project are as shown
below. Check the feasibility of the project based on present
worth method, using i = 20%. Initial outlay = Rs. 50,00,000
Life of the project = 20 years. Annual equivalent revenue =
Rs. 15,00,000 Modernizing cost at the end of the 10th year
= Rs. 20,00,000 Salvage value at the end of project life = Rs.
5,00,000.

79. Present worth questions
• An automobile company recently advertised its car for a down
payment of Rs. 1,50,000. Alternatively, the car can be taken home
by customers without making any payment, but they have to pay
an equal yearly amount of Rs. 25,000 for 15 years at an interest
rate of 18%, compounded annually. You are asked to advise the
best alternative for the customers based on the present worth
method of comparison.
• A working woman is planning for her retired life. She has 20 more
years of service. She would like to have an annual equivalent
amount of Rs. 3,00,000, starting from the end of the first year of
her retirement. Find the single amount that should be deposited
now so that she receives the above mentioned annual equivalent
amount at the end of every year for 20 years after her retirement.
Assume i = 15%, compounded annually.

80. Future worth questions
• A motorcycle is sold for Rs. 50,000. The motorcycle
dealer is willing to sell it on the following terms: (a)
Make no down payment but pay Rs. 1,500 at the
end of each of the first four months and Rs. 3,000
at the end of each month after that for 18
continuous months. (b) Make no down payment
but pay a total amount of Rs. 90,000 at the end of
the 22nd month; till that time the buyer should
mortgage property worth of Rs. 50,000, at present.
Based on these terms and a 12% annual interest
rate compounded monthly, find the best alternative
for the buyer based on the future worth method of
comparison.

81. Future worth questions
• Alpha Finance Company is coming with an option of accepting Rs. 10,000
now and paying a sum of Rs. 1,60,000 after 20 years. Beta Finance Company
is coming with a similar option of accepting Rs. 10,000 now and paying a sum
of Rs. 3,00,000 after 25 years. Compare and select the best alternative based
on the future worth method of comparison with 15% interest rate,
compounded annually.
• An insurance company gives an endowment policy for a person aged 30
years. The yearly premium for an insured sum of Rs. 1,00,000 is Rs. 4,000.
The policy will mature after 25 years. Also, the person is entitled for a bonus
of Rs. 75 per thousand per year at the end of the policy. If a person survives
till the end of the 25th year: (a) What will be the total sum that he will get
from the insurance company at that time? (b) Instead of paying the
premiums for the insurance policy, if the person invests an equal sum of Rs.
4,000 at the end of each year for the next 25 years in some other scheme
which is having similar tax benefit, find the future worth of the investment at
15% interest rate, compounded annually. (c) Rate the above alternatives
assuming that the person is sure of living for the next 25 years.

82. Future worth questions
• A small-scale industry is in the process of buying a
milling machine. The purchase value of the milling
machine is Rs. 60,000. It has identified two banks for
loan to purchase the milling machine. The banks can
give only 80% of the purchase value of the milling
machine as loan. In Urban Bank, the loan is to be repaid
in 60 equal monthly installments of Rs. 2,500 each. In
State Bank, the loan is to be repaid in 40 equal monthly
installments of Rs. 4,500 each. Suggest the most
economical loan scheme for the company, based on the
annual equivalent method of comparison. Assume a
nominal rate of 24%, compounded monthly.

83. Future worth questions
• A company receives two options for purchasing a
copier machine for its office.
• Option 1 Make a down payment of Rs. 30,000 and
take delivery of the copier machine. The remaining
money is to be paid in 24 equal monthly
installments of Rs. 4,500 each.
• Option 2 Make a full payment of Rs. 1,00,000 and
take delivery of the copier machine. Suggest the
best option for the company to buy the copier
machine based on the annual equivalent method of
comparison by assuming 15% interest rate,
compounded annually.