Chapter 3
Interest and Equivalence
EGN 3615
ENGINEERING ECONOMICS
WITH SOCIAL AND GLOBAL
IMPLICATIONS
1
Learning Objectives
 Understand the concept of “time value of money”
 Distinguish between simple and compound interest
 Understand the concept of “equivalence” of cash flows
 Solve problems using Single Payment Compound
Interest Formulas
Slide 2
Computing cash flow
 To purchase a new $30,000 machine, two alternatives:
 Pay the full price now minus a 3% discount or
 Pay $5000 now; $8000 at the end of year 1; and $6000 at
the end of each of the next 4 years (A total of $37,000!)
 Why do the second?
0
1
2
3
4
5
0
$5,000
$29,100
1
2
3
4
5
$8,000 $6,000 $6,000 $6,000 $6,000
Slide 3
Computing Cash flow (cont…)
 To repay a loan of $1,000 at 8% interest in 2 years
 Repay half of $1000 plus interest at the end of year 1
 Repay the remaining $500 plus interest at the end of yr 2
$1000
Yr Interest Balance Repayment
0
1000
Cash
Flow
1000
1
2
-580
-540
80
40
500
0
500
500
0
1
$580
2
$540
Slide 4
Time value of money
 Money has purchasing power
 Money has earning power
 Money is a valuable asset
 People are willing to pay some charges (interest) to have
money available for their use
Money loses value in time!
Or money has a time value!
The time value lost over time is made up by “interest”!
Slide 5
Interest Calculations
 Simple interest
Total interest earned =
P in
(Eq. 3-1)
where P = Principal (loan)
i = Simple annual interest rate
n = Number of years
F  P  P in
(Eq. 3-2)
where F = Amount due at the end of n years
Interest in computed on just the principal in each year!
Slide 6
Simple interest (example)
 Loan of $10,000 for 5 yrs at simple interest rate of 6%
 Total interest earned = $10000(6%)(5) = $3000
 Amount due at end of loan = $10000 + $3000 = $13000
 Although it is “simple”, it is rarely used.
Slide 7
Compound interest
 Interest is computed on the unpaid balance (not just
the principal), which includes the principal and any
unpaid interest from the preceding period
 Common practice for interest calculation, unless
specifically stated otherwise
Slide 8
Compound interest (example)
Loan of $10K for 5 yrs at compound interest rate of 8%
Year
Balance
Beginning of
year
Interest
Balance at end of year
1
$10,000.00
$10000 x 0.08 = $800.00
$10,000 + 800 =10,800.00
2
$10,800.00
$10,800 x 0.08 = $864.00
$10,800 + 864 =11,664.00
$11,664.00
$11,664 x 0.08 = $933.12
$11,664 + 933.12
=12,597.12
$12,597.12
$12,597.12
x 0.08 = $1007.77
$12,597.12 +
$1007.77=13,604.89
$13,604.89
$13,604.89
x 0.08 = $1088.39
$13,604.89 + $
1088.39=14,693.28
3
4
5
Slide 9
Repaying a debt
 Repay of a loan of $5000 in 5 yrs at interest rate of 8%
 Plan #1: Pay $1000 principal plus interest due
Yr
Balance at
the
Beginning
of Year
Interest
Balance at
the end of
yr before
payment
Interest
Payment
Principal
Payment
Total
Payment
1
$5,000.00
$400.00
$5,400.00
$400.00
$1,000.00
$1,400.00
2
$4,000.00
$320.00
$4,320.00
$320.00
$1,000.00
$1,320.00
3
$3,000.00
$240.00
$3,240.00
$240.00
$1,000.00
$1,240.00
4
$2,000.00
$160.00
$2,160.00
$160.00
$1,000.00
$1,160.00
5
$1,000.00
$80.00
$1,080.00
$80.00
$1,000.00
$1,080.00
$1,200.00
$5,000.00
$6,200.00
Subtotal
Slide 10
Repaying a debt
 Repay of a loan of $5000 in 5 yrs at interest rate of 8%
 Plan #2: Pay interest due at end of each year and
principal at end of 5 years
Yr
Balance at
Beginning
of Year
Interest
Balance at
the end of
Year
Interest
Payment
Principal
Payment
Total
Payment
1
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
2
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
3
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
4
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
5
$5,000.00
$400.00
$5,400.00
$400.00
$5,000.00
$5,400.00
$2,000.00
$5,000.00
$7,000.00
Subtotal
Slide 11
Repaying a debt
 Repay of a loan of $5000 in 5 yrs at interest rate of 8%
 Plan #3: Pay in 5 equal end-of-year payments
Yr
Balance at
the
Beginning
of year
Interest
Balance at
the end of
yr before
payment
1
$5,000.00
$400.00
$5,400.00
$400.00
$852.28
$1,252.28
2
$4,147.72
$331.82
$4,479.54
$331.82
$920.46
$1,252.28
3
$3,227.25
$258.18
$3,485.43
$258.18
$994.10
$1,252.28
4
$2,233.15
$178.65
$2,411.80
$178.65
$1,073.63
$1,252.28
5
$1,159.52
$92.76
$1,252.28
$92.76
$1,159.52
$1,252.28
$1,261.41
$5,000.00
$6,261.41
Subtotal
Interest
Payment
Principal
Payment
Total
Payment
Slide 12
Repaying a debt
 Repay of a loan of $5000 in 5 yrs at interest rate of 8%
 Plan #4: Pay principal and interest in one payment at
end of 5 years
Yr
Balance at
Beginning
of year
Interest
Balance at
the end of
year
1
$5,000.00
$400.00
$5,400.00
$0.00
$0.00
$0.00
2
$5,400.00
$432.00
$5,832.00
$0.00
$0.00
$0.00
3
$5,832.00
$466.56
$6,298.56
$0.00
$0.00
$0.00
4
$6,298.56
$503.88
$6,802.44
$0.00
$0.00
$0.00
5
$6,802.44
$544.20
$7,346.64
$2,346.64
$5,000.00
$7,346.64
$2,346.64
$5,000.00
$7,346.64
Subtotal
Interest
Payment
Principal
Payment
Total
Payment
Slide 13
Equivalence
Total payment summary:
Plan #1
Plan #2
Plan #3
Plan #4
Pay $1K principal plus interest due
$6,200.00
Pay interest at yr 1-4, & balance at yr 5
$7,000.00
Make 5 equal payments at end of yr 1-5
$6,261.41
Make one payment = balance at yr 5
$7,346.64
Slide 14
Equivalence
 The 4 repayment plans are equivalent to one
another and to $5000 now at 8% interest
 Using the concept of equivalence, one can
convert different types of cash flows at
different points of time to an equivalent
value at a common reference point
 Equivalence is dependent on interest rate
Slide 15
Single payment.
Compound interest formulas
 Notation:
 i = interest rate per compounding period
 n = number of compounding periods
 P = a present sum of money
 F = a future sum of money
Single Payment Compound Amount Formula
F  P(1  i)
n
F  P(F/P, i, n)
(Eq. 3-3)
(Eq. 3-4)
The above notation is read as Find F, given P, at i, over n
Slide 16
Single payment.
Compound interest formulas
Solving for P in Eq. 3-3 yields
Single Payment Present Worth Formula
P  F(1  i)
-n
P  F(P/F, i, n)
(Eq. 3-5)
(Eq. 3-6)
The above notation is read as Find P, given F, at i, over n
Slide 17
Simple payment.
Compound interest formulas (ex.)
 Wish to have $800 at the end of 4 years, how
much should be deposited in an account
that pays 5% annually?
F = P(1+i)n
F=800
P = F(1+i)-n = 800(1+0.05)-4
Eq 3-5
Eq. 3-5
= $658.16
0
1
2
i=5%
P=?
3
4
Or
P = F×(P/F, i, n)
Eq. 3-6
= 800(0.8227) = $658.16
Slide 18
Example 3-5
$500 were deposited in a saving account, paying 6%,
compounded quarterly, for 3 years — i, n, F
F=?
i = 6%/4 = 1.5% - quarterly interest rate
n = 3 x 4 = 12 quarters
0
1
2
11
i=1.5%
$500
12
F = P(1+i)n = P(F/P, i, n)
= 500(1+0.015)12 = 500(F/P,1.5%,12)
= 500(1.196) = $598.00
“…paying 6%...” means paying annually.
Slide 19
How much can you borrow if you can repay it in the two
payments shown?
Assume a 12% interest rate
Solve for P, using Appendix C (or Eq. 3-6)
Year
0
1
2
3
4
5
Cash Flow
+P
0
0
-400
0
-600
 P = 400(P/F,12%,3)
+ 600(P/F,12%,5)
 = 400(0.7118) + 600(0.5674)
 = 625.16
P=?
0
2
1
i=12%
3
400
4
5
600
Slide 20
With the same two payments shown, if the interest is
increased to 15%, Will the value of P be larger or
smaller?
Solve for P
Year
0
1
2
3
4
5
 P = 400(P/F,15%,3)
+ 600(P/F,15%,5)
Cash Flow
+P
0
0
-400
0
-600
 = 400(0.6575) + 600(0.4972)
 = 561.32 at 15%
 P = 625.16 at 12%
P=?
0
2
1
i=15%
3
400
4
→ P is smaller at 15% than at 12%
→ Given future payments, P
5
decreases, as i increases!
600
Slide 21
Problem 3-2
Borrow $2000
Repay at end of 3 yrs, with 10% simple interest/yr
How much to pay?
Solution
P = $2000, i = 10%, n = 3 yrs
F = P(1+in)
= 2000[1+0.1(3)]
= $2,600
Slide 22
Problem 3-11
Borrow $5M
Repay at end of 5 yrs, with 10% compound interest/yr
How much to pay?
Solution
P = $5,000,000, i = 10%, n = 5 yrs
F = P(1+i)n = P(F/P, i, n)
= 5M(1+0.1)5 = 5M(F/P, 0.1, 5)
= 5M(1.61051)
= $8,052,550
Slide 23
Problem 3-18
Deposit $2K in an account, & earn 6% interest
How much is in the account
(a) after 5 years?
(b) after 10 years?
(c) after 20 years?
(d) after 50 years?
(e) after 100 years?
Slide 24
Problem 3-18
Solution
P = $2k, and i = 6%
(a) n = 5
F = P(1+i)n = P(F/P, i, n)
= 2000(1+0.06)5 = 2000(F/P, 0.06, 5)
= 2000(1.338226)
= $2,676.45
Slide 25
Problem 3-18
Solution
(b) n = 10
F = P(1+i)n = P(F/P, i, n)
= 2000(1+0.06)10 = 2000(F/P, 0.06, 10)
= $3,581.70
(c)
n = 20
F = 2000(1+0.06)20 = 2000(F/P, 0.06, 20)
= $4,414.27
Slide 26
Problem 3-18
Solution
(d) n = 50
F = 2000(1+0.06)50 = 2000(F/P, 0.06, 50)
= $36,840.31
(e)
n = 100
F = 2000(1+0.06)100 = 2000(F/P, 0.06, 100)
= $678,604.17
Slide 27
Problem 3-19
Inheritance amount of $20k, interest rate is 7%
How much is it worth, if it is received
(a) after 5 years?
(b) after 10 years?
(c) after 20 years?
(d) after 50 years?
Slide 28
Problem 3-19
Solution
F = $20k, and i = 7%
(a) n = 5
P = F(1+i)-n = F(P/F, i, n)
= 20000(1+0.07)-5 = 20000(P/F, 0.07, 5)
= 20000(0.712986)
= $14,259.72
Slide 29
Problem 3-19
Solution
(b) n = 10
P = F(1+i)-n = P(F/P, i, n)
= 20000(1+0.07)-10 = 20000(P/F, 0.07, 10)
= $10,166.99
(c)
n = 20
P = 20000(1+0.07)-20 = 20000(P/F, 0.07, 20)
= $5,168.38
Slide 30
Problem 3-19
Solution
(d) n = 50
P = 20000(1+0.07)-50 = 20000(P/F, 0.07, 50)
= $678.96
(*)
n = 100
P = 20000(1+0.07)-100 = 20000(P/F, 0.07, 100)
= $23.05
Slide 31
Single Payment Formulas
Single payment compound amount formula
F = P(1+i)n - given P, i, & n, find F.
(3-3)
Single payment present worth formula
P = F(1+i)-n - given F, i, & n, find P.
(3-5)
Single payment interest formula
i = (F/P)1/n – 1 - given P, F, & n, find i.
Single payment time formula
n = log(F/P)/log(1+i) - given P, F, & i, find n.
Slide 32
Nominal & Effective Interest
 Nominal interest rate (per year), r
Annual interest rate, without compounding,
also called APR (annual percentage rate)
 Effective interest rate (per year), ia
Annual interest rate with compounding
ia = (1 + r/m)m – 1
(3-7)
m = # of compounding periods in a year
m = 4, if compounding quarterly
m = 12, if compounding monthly
Slide 33
Nominal and Effective Interest Rate
 Question: What is the effective interest rate
for 12% interest compounded monthly?
m
12
r

 0.12 
ia  1    1  ia  1 
 1
12 
 m

 0.12683  12.683%
r :12%
m :12(monthly compoundin g)
34
Nominal & Effective Interest
 Example 3-9
A Bank pays 1.5% interest quarterly
What are r and ia?
 Solution
Nominal annual interest rate, r = 4(1.5%) = 6%
Effective interest rate, ia = (1 + 0.015)4 – 1 = 6.14%
Slide 35
Nominal and Effective Interest Rate
 Question: If you deposit $1000 in a bank that pays a 12%
interest compounded quarterly what will be the amount you
can withdraw in five equal yearly sums? In other words, how
much should you withdraw each year?
Solution
r = 12% and m = 4
m
4
r

 0.12 
ia  1    1  ia  1 
 1
4 
 m

 0.12551  12.551%
36
QUESTION CONTINUES
0
W
W
W
W
W
1
2
3
4
5
Given P, ia, n, find W.
ia : %12.551 per year
$1000
n : 5 years
 i (1  i ) n 
 0.12551(1  0.12551) 5 
W  P
  1000 

n
5
 (1  i )  1
 (1  0.12551)  1 
 1000 (0.2812)  281.2
37
Nominal & Effective Interest
 Example 3-10
Loan shark: “If I give you $50 on Monday, you own
me $60 on the following Monday.”
(a) What are r and ia?
(b) Starting with $50, how much is it worth after 1 yr?
Slide 38
Nominal & Effective Interest
 Solution
(a)
Weekly interest rate = 60/50 – 1 = 0.2
Nominal annual interest rate, r = 52(0.2) = 10.4
(b) Effective annual interest rate,
ia = (1 + 0.2)52 – 1 = 13,103.63 = 1,310,363%
Year end worth F = Pia = 50(13103.63) = $655,181.5
Slide 39
Continuous Compounding
For nominal annual interest rate r, compounded
continuously, effective annual interest rate is
i a = er – 1
(3-14)
Compound amount F = P(er)n = P[F/P, r, n]
(3-15)
P = F(er)-n = F[P/F, r, n]
(3-16)
Present Worth
Slide 40
Continuous Compounding
Again, F = P(er)n has quantities P, F, r, and n.
Given any three, you can find the other one!
1) Given P, r & n, F = Pern.
2) Given F, r & n, P = Fe-rn.
3) Given P, F, & r, n = (1/r)ln(F/P).
4) Given P, F, & r, r = (1/n)ln(F/P).
(3-15)
(3-16)
Slide 41
Continuous Compounding
Example 3-11
If a bank pays 5% nominal interest, compounded
continuously, how much would $2000 be worth
after 2 years?
We can solve it by two methods!
Slide 42
Example 3-11
Solution
Method 1: Use Eq. (3-3) or (3-4)
(single payment compound interest formula)
r = 5%, P = $2000, n = 2
ia = er – 1 = e0.05 – 1 = 0.051271
F = P(F/P, ia, n) = 2000(1+0.051271)2 = $2,210.34
Method 2: Use Eq. (3-15)
F = P[F/P, r, n] = 2000(e)0.05(2) = $2,210.34
Slide 43
Continuous Compounding
Example 3-13
How long will it take for money to double at 10%
nominal interest, compounded continuously?
Solution
This is the case that given P, F, & r, find n.
In particular, F = 2P, & r = 10%,
n = (1/r)ln(F/P) = (1/0.1)ln2
= (10)(0.693) = 6.93 years
Slide 44
Nominal and Effective Interest Rate
 Table below shows effective interest rates for
various compounding periods with r = 12%
Number of
Compounding
Periods per year
Length of
Periods
Effective Interest
Rate
1
Annual
0.12
2
Semiannual
0.1236
4
Quarterly
0.12551
12
Monthly
0.12683
52
Weekly
0.12734
365
Daily
0.12747
Continuous
 (t )
0.12749
45
End of Chapter 3
Slide 46