Chapter 3: Loan Repayment

Instructor: Roba Bairakdar, ASA

Semester: Fall 2015

Mathematics of Investment –

MACT 3311

1

Amortization Method of Loan

Repayment

The amortization method is the

distribution of payment of a loan into

multiple cash flow instalments, as

determined by an amortization schedule.

Each repayment instalment consists of both

interest and principal. When a payment is

made, it must be first applied to pay interest

due and then any remaining part of the

payment is applied to pay principal.

2

Exercise

Consider a loan of amount 1,000 with an

interest rate of 10% per year. Suppose that

there is a payment of 200 at the end of 1 year,

a payment of 500 at the end of 2 years, and a

final payment at the end of 3 years to

completely repay the loan.

1 2 3Time t 0

1,000 1,000(1+10%)

= 1,100

900(1+10%)

= 990

490(1+10%)

= 539

-200

=900

-500

=490

-539

=0

Payment

Outstanding

Balance

Exercise

Consider a loan of amount 𝑶𝑩𝟎, which is the

outstanding balance at time 0. Payments of

𝑃𝑀𝑇𝑡 are made at time t until the 𝑛𝑡ℎ payment.

The outstanding balance at time t right after

the payment is made is 𝑶𝑩𝒕.

1 2 nTime t 0

𝑶𝑩𝟎 𝑶𝑩𝟎(1+i)

-𝑷𝑴𝑻𝟏

=𝑶𝑩𝟏

Payment

Outstanding

Balance

𝑶𝑩𝟏(1+i)

-𝑷𝑴𝑻𝟐

=𝑶𝑩𝟐

…

𝑶𝑩𝒏−𝟏(1+i)

-𝑷𝑴𝑻𝒏

=𝑶𝑩𝒏 = 𝟎

n-1

𝑶𝑩𝒏−𝟐(1+i)

-𝑷𝑴𝑻𝒏−𝟏

=𝑶𝑩𝒏−𝟏

…

Amortized Loan

5

An amortized loan of amount L made at time 0

at periodic interest i and to be repaid by n

payments of amounts 𝑃𝑀𝑇1, 𝑃𝑀𝑇2, … 𝑃𝑀𝑇𝑛 at

times 1, 2, … , 𝑛 is based on the equation

𝑳 = 𝑷𝑴𝑻𝟏𝒗 + 𝑷𝑴𝑻𝟐𝒗𝟐 +⋯+𝑷𝑴𝑻𝒏𝒗

𝒏

Exercise

6

A borrower would like to borrow 30,000 at 8%

for 5 years, but would like to pay only 5,000

for the first two years and then catch up with

a higher payment for the final three years.

What is the payment for the final 3 years?

Solution:

30,000 = 5,000𝑎 2| + 𝐾𝑣2𝑎 3|30,000 = 5,000 1.783 + 𝐾 0.857 2.577𝐾 = 9,542.52

Components of a general

Amortized Loan

7

Loan Amount L:

Loan amount L at periodic interest i and to be

repaid by n payments of amounts

𝑃𝑀𝑇1, 𝑃𝑀𝑇2, … 𝑃𝑀𝑇𝑛 at the end of n successive

periods.

Outstanding Balance 𝑶𝑩𝒕:

𝑶𝑩𝒕 is the amount owed on the loan just after

the 𝑡𝑡ℎ payment was made.

𝑶𝑩𝟎 = 𝑳 and 𝑶𝑩𝒏 = 𝟎

Components of a general

Amortized Loan

8

Interest Due 𝑰𝒕:

𝑰𝒕 is the interest on the outstanding balance

since the previous payment was made.

𝑰𝒕 = 𝑶𝑩𝒕−𝟏𝐱 𝒊

Principal Repaid 𝑷𝑹𝒕:

𝑷𝑹𝒕 is the part of the payment 𝑷𝑴𝑻𝒕 that is

applied toward repaying the loan principal

amount.

𝑷𝑹𝒕 = 𝑷𝑴𝑻𝒕 − 𝑰𝒕

Outstanding Balance Calculation

9

Outstanding Balance 𝑶𝑩𝒕:

𝑶𝑩𝒕 = 𝑶𝑩𝒕−𝟏 − 𝑷𝑹𝒕

= 𝑶𝑩𝒕−𝟏 − 𝑷𝑴𝑻𝒕 −𝑶𝑩𝒕−𝟏𝐱 𝒊

= 𝟏 + 𝒊 𝑶𝑩𝒕−𝟏 − 𝑷𝑴𝑻𝒕

Amortization Schedule

10

t Payment Interest Due Principal Repaid Outstanding Balance

0 __ __ __ 1,000

1 200 1,000 x 10%

= 100

200 – 100 = 100 1,000 – 100 = 900

2 500 900 x 10% =

90

500 – 90 = 410 900 – 410 = 490

3 539 490 x 10% =

49

539 – 49 = 490 490 – 490 = 0

Amortization Schedule

11

t Payment Interest Due Principal Repaid Outstanding Balance

0 __ __ __ 𝐿 = 𝑂𝐵0

1 𝑃𝑀𝑇1 𝐼1 = 𝑂𝐵0 x 𝑖 𝑃𝑅1 = 𝑃𝑀𝑇1 − 𝐼1 𝑂𝐵1 = 𝑂𝐵0 − 𝑃𝑅1

2 𝑃𝑀𝑇2 𝐼2 = 𝑂𝐵1 x 𝑖 𝑃𝑅2 = 𝑃𝑀𝑇2 − 𝐼2 𝑂𝐵2 = 𝑂𝐵1 − 𝑃𝑅2.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

t 𝑃𝑀𝑇𝑡 𝐼𝑡 = 𝑂𝐵𝑡−1 x 𝑖 𝑃𝑅𝑡 = 𝑃𝑀𝑇𝑡 − 𝐼𝑡 𝑂𝐵𝑡 = 𝑂𝐵𝑡−1 − 𝑃𝑅𝑡.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

n 𝑃𝑀𝑇𝑛 𝐼𝑛 = 𝑂𝐵𝑛−1 x 𝑖 𝑃𝑅𝑛 = 𝑃𝑀𝑇𝑛 − 𝐼𝑛 𝑂𝐵𝑛 = 𝑂𝐵𝑛−1 − 𝑃𝑅𝑛 = 0

Prospective Method

12

The loan balance after a payment is made is

the present value of the remaining payments

at the time of payment.

1 nTime t 0 …

𝑷𝑴𝑻𝒏

n-1t… t+1

𝑷𝑴𝑻𝒏𝒗𝒏−𝒕

𝑷𝑴𝑻𝒕+𝟏𝑷𝑴𝑻𝒕+𝟏𝒗

𝑶𝑩𝒕 = 𝑷𝑴𝑻𝒕+𝟏𝒗 +⋯+ 𝑷𝑴𝑻𝒏𝒗𝒏−𝒕

Exercise

13

A borrower would like to borrow 30,000 at 8% for 5 years. He would like to repay his loan by making level annual payments at the end of each year. Calculate the outstanding balance after the third payment was made using the prospective method.

Solution:

30,000 = 𝐾𝑎 5|𝐾 = 7,513.69𝑂𝐵3 = 7,513.69 𝑣 + 𝑣2 = 13,398.9

Retrospective Method

The loan balance after a payment is made is

the amount of the loan accumulated to time t,

minus the accumulated value of all payments

to time t, up to and including 𝑃𝑀𝑇𝑡.

1 nTime t 0 …

𝑷𝑴𝑻𝟏

t…

−𝑷𝑴𝑻𝟏(𝟏 + 𝒊)𝒕−𝟏

𝑶𝑩𝟎 𝑶𝑩𝟎(𝟏 + 𝒊)𝒕

𝑶𝑩𝒕 = 𝑶𝑩𝟎(𝟏 + 𝒊)𝒕−⋯− 𝑷𝑴𝑻𝒕−𝟏 𝟏 + 𝒊 − 𝑷𝑴𝑻𝒕

t-1

𝑷𝑴𝑻𝒕−𝟏…

−𝑷𝑴𝑻𝒕−𝟏(𝟏 + 𝒊)

−𝑷𝑴𝑻𝒕

Exercise

15

A borrower would like to borrow 30,000 at 8% for 5 years. He would like to repay his loan by making level annual payments at the end of each year. Calculate the outstanding balance after the third payment was made using the retrospective method.

Solution:

30,000 = 𝐾𝑎 5|𝐾 = 7,513.69𝑂𝐵3= 30,000(1 + 𝑖)3−7,513.69 1 + 𝑖 2 − 7,513.69 1 + 𝑖 2

− 7,513.69 = 13,398.9

Exercise

16

A loan at 10% annually has an initial payment of 100, and 9 further payments. The payment amount increases by 2% each year. Find the loan balance immediately after the fourth payment.

Solutions:

The payments are 100, 100 1.02 ,… , 100(1.02)9.

Immediately after the fourth payments, the remaining payments are 100(1.02)4, … , 100(1.02)9.

𝑂𝐵4 =100(1.02)4

1.1+ ⋯+

100 1.02 9

1.16

=100 1.02 4

1.11 +

1.02

1.1+⋯+

1.02

1.1

5

=100 1.02 4

1.1

1 −1.021.1

6

1 −1.021.1

= 492.93

Exercise

17

A 30-year monthly payment mortgage loan for

250,000 is offered at a rate of 6%. The

borrower would like to have graduated

payments where the first year’s monthly

payment if P, the second year’s monthly

payment is P+100 and all subsequent

payments are P+200. Find the

a. Initial Payment P

b. Balance at the end of one year

Exercise – cont’d

18

a. Using the equation of value

250,000 = 𝑃𝑎360|0.005 + 100𝑣12𝑎348|0.005

+100𝑣24𝑎336|0.005

𝑃 = 1,319.37

b. Using the prospective method

𝑂𝐵12 = (𝑃 + 100)𝑎348|0.005+100𝑣12𝑎336|0.005

= 249,144

Exercise

19

You have a 30,000 loan at 8% annually for 5

years. You agree to pay off the principal in

instalments of 6,000 per year, and to pay

interest on the outstanding balance each year.

What is the interest due in the 4th payment?

Solution:

𝑂𝐵3 = 30,000 − 3 6000 = 12,000𝐼4 = 𝑖 x 𝑂𝐵3 = 0.08 x 12,000 = 960

Exercise

20

A loan at 10% annually has an initial payment

of 100, and 9 further payments. The payment

amount increases by 10 each year. Find the

loan balance immediately after the fourth

payment.

Solution:

𝑂𝐵4 = 140𝑎 6|10% + 10𝑣(𝐼𝑎) 5|10% = 706.57

Level Payment Loan

Prospective / Retrospective Method

21

For a level payment loan of n periods with a

payment PMT, the outstanding balance at time

t can be expressed is

𝑶𝑩𝒕 = 𝑷𝑴𝑻𝒂𝒏−𝒕| = 𝑶𝑩𝟎(𝟏 + 𝒊)𝒕−𝑷𝑴𝑻𝒔 𝒕|

Level Payment Loan Amortization

Schedulet Payment Interest Due Principal

Repaid

Outstanding Balance

0 __ __ __ 𝐿 = 𝑂𝐵0 = 𝑎 𝑛|

1 𝐾 = 1 𝐼1 = 𝑖𝑎 𝑛|

= 1 − 𝑣𝑛𝑃𝑅1 = 𝐾1 − 𝐼1 = 𝑣𝑛 𝑂𝐵1 = 𝑂𝐵0 − 𝑃𝑅1 = 𝑎 𝑛| − 𝑣𝑛

= 𝑎𝑛−1|

2 𝐾 = 1 𝐼2 = 𝑖𝑎𝑛−1|= 1 − 𝑣𝑛−1

𝑃𝑅2 = 𝐾2 − 𝐼2= 𝑣𝑛−1

𝑂𝐵2 = 𝑂𝐵1 − 𝑃𝑅2= 𝑎𝑛−1| − 𝑣𝑛−1 = 𝑎𝑛−2|

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

t 𝐾 = 1 𝐼𝑡 = 𝑖𝑎𝑛−(𝑡−1)|= 1 − 𝑣𝑛−(𝑡−1)

𝑃𝑅𝑡 = 𝐾𝑡 − 𝐼𝑡= 𝑣𝑛−(𝑡−1)

𝑂𝐵𝑡 = 𝑂𝐵𝑡−1 − 𝑃𝑅𝑡= 𝑎𝑛−(𝑡−1)| − 𝑣𝑛−(𝑡−1) = 𝑎𝑛−𝑡|

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

n 𝐾 = 1 𝐼1 = 𝑖𝑎 1|= 1 − 𝑣

𝑃𝑅𝑛 = 𝐾𝑛 − 𝐼𝑛 = 𝑣 𝑂𝐵𝑛 = 𝑂𝐵𝑛−1 − 𝑃𝑅𝑛 == 𝑎 1| − 𝑣 = 0

Level Payment Loan Amortization

Components

23

Interest paid in payment 𝑰𝒕:

𝑰𝒕 = 𝑷𝑴𝑻(𝟏 − 𝒗𝒏− 𝒕−𝟏 )

Total interest paid:

𝑷𝑴𝑻 𝟏 − 𝒗𝒏 + 𝟏 − 𝒗𝒏−𝟏 +⋯+ 𝟏 − 𝒗 = 𝑷𝑴𝑻 𝒏 − 𝒂 𝒏|= 𝑷𝑴𝑻 ∗ 𝒏 − 𝑳

Principal paid 𝑷𝑹𝒕:

𝑷𝑹𝒕 = 𝑷𝑴𝑻𝒗𝒏−(𝒕−𝟏)

𝑷𝑹𝒕+𝒌 = (𝟏 + 𝒊)𝒌𝑷𝑹𝒕

Total principal paid:

𝑶𝑩𝟎 = 𝑳𝑻𝒐𝒕𝒂𝒍 𝒑𝒂𝒚𝒎𝒆𝒏𝒕𝒔 − 𝒕𝒐𝒕𝒂𝒍 𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒑𝒂𝒊𝒅

Exercise

24

A 30-year monthly payment mortgage loan for

250,000 is offered at a nominal rate of 6%

convertible monthly. Find the

a. Monthly Payment

b. The total principal and interest that would

be paid on the loan over 30 years

c. The balance in 5 years

d. The principal and interest paid over the

first 5 years

Exercise – cont’d

25

Solution:

a. 250,000 = 𝑘𝑎360|0.005𝑘 = 1,498

b. Total principal = amount of the loan = 250,000

Total Interest = 360 1,498 − 250,000 = 289,596

c. The balance in 5 years

𝑂𝐵60 = 1,498𝑎300|0.005 = 232,635

c. The principal paid over 5 years =

250,000 − 232,635 = 17,364

The total interest paid in 5 years =

60 1,498 − 17,364 = 72,569

Exercise

26

A borrower would like to borrow 30,000 at 8%

for 5 years. He would like to repay his loan by

making level annual payments at the end of

each year. Calculate the level payment.

Solution:

30,000 = 𝐾𝑎 5|30,000 = 𝐾 3.99271𝐾 = 7,513.69

Exercise

27

Following on the previous example, what is the

principal paid in the 4th payment?

Solution:

𝑃𝑅𝑡 = 𝐾𝑣𝑛−(𝑡−1) = 7,513.69𝑣5−(4−1) = 6,441.78

Exercise

28

For an 8% level payment loan, the amount of

principal in the second payment is 5,522.79.

Find the amount of principal in the 4th

payment.

Solution:

𝑃𝑅𝑡+𝑘 = (1 + 𝑖)𝑘𝑃𝑅𝑡𝑃𝑅4 = (1 + 𝑖)2𝑃𝑅2 = (1 + 0.08)2𝑥5522.79

= 6,441.78

Exercise (SOA)

29

A loan of 1,000 at a nominal rate of 12%

convertible monthly is to be repaid by six

monthly payments with the first payment due

at the end of 1 month.

The first three payments are x each, and the

final three payments are 3x each.

Determine the sum of the principal repaid in

the third payment and the interest in the fifth

payment.

Exercise (SOA) – cont’d

30

1000 = 𝑥𝑎 3|1% + 3𝑥𝑣3𝑎 3|1% By solving for x, 𝑥 = 86.92

Principal repaid in the third payment = 𝑃𝑅3 = 𝑂𝐵2− 𝑂𝐵3

𝑂𝐵2 = 𝑥𝑣 + 3𝑥𝑣𝑎 3|1% = 845.4 (prospectively)

𝑂𝐵2 = 1000(1.01)2−𝑥𝑠 2|1% = 845.4 (retrospectively)

𝑂𝐵3 = 3𝑥𝑎 3|1% = 766.9 (prospectively)

𝑂𝐵3 = 1000(1.01)3−𝑥𝑠3|1% = 766.9 (retrospectively)

Principal repaid in the third payment = 845.4− 766.9 = 78.6

Exercise (SOA) – cont’d

31

Interest paid in the fifth payment = 𝐼5 = 𝑂𝐵4. 𝑖

𝑂𝐵4 = 3𝑥𝑎 2|1% = 513.8 (prospectively)

𝐼5 = 𝑂𝐵4. 𝑖 = 513.8 (1%) = 5.138

𝑃𝑅3 + 𝐼5 = 83.738

Sinking Fund Repayment of a

Loan

32

When you use a sinking fund, you only pay

the lender the interest at his stated rate i on

the loan each period. In addition, you make

level deposits to an account called a sinking

fund that earns interest at a rate j.

The goal is to make a deposit into the sinking

fund that will cause the fund to grow to the

amount of the loan at the end of the loan term

Exercise

33

A 100,000 annual payment loan is made for a term of 10 years at 10% interest. The lender agreed to take only payments of interest until the end of year 10 when the 100,000 must be repaid. The borrower will make level annual year-end payments to a sinking fund earning 8%.

Find the

a. Sinking fund deposit

b. Total annual loan payment

c. Balance in the sinking fund at time 3.

Exercise – cont’d

34

Solution:

a. Sinking Fund annual deposit = K, where

100,000 = 𝐾𝑠10|0.08𝐾 = 6,903

b. Total annual loan payment = 6,903+ 100,000 10% = 16,903

c. Balance at time 3 = 6,903𝑠 3|0.08 = 22,410

Exercise - Sinking Fund

compared to Amortization

35

A borrower would like to borrow 1,000 at 10% for

10 years. What is the level annual payment using

the amortization method?

Assume the lender agreed to take only payments

of interest until the end of year 10 when the 1,000

must be repaid. The borrower will make level

annual year-end payments to a sinking fund.

What is the level annual payment if the sinking

fund earns 8%, 10% or 12%?

Exercise – cont’d

36

Using the amortization method, the level

annual payment is 1,000

𝑎10|10%= 162.75

Using the sinking fund method,

Annual Interest

payment

SF Interest

Rate

SF Deposit Total

annual

payment

1000𝑥10% = 100 8% 1,000

𝑠10|8%= 69.03

169.03

1000𝑥10% = 100 10% 1,000

𝑠10|10%= 62.75

162.75

1000𝑥10% = 100 12% 1,000

𝑠10|12%= 56.98

156.98

Sinking Fund components

37

Total payment each period = 𝑳

𝒔 𝒏|𝐣+ 𝑳. 𝒊 = 𝑺𝑭𝑫 + 𝑳. 𝒊

Balance in the Sinking Fund at time k =

𝑺𝑭𝑩𝒂𝒍𝒌 = 𝑺𝑭𝑫𝒔 𝒌|𝒋

Principal paid in the kth payment=

𝑺𝑭𝑩𝒂𝒍𝒌 − 𝑺𝑭𝑩𝒂𝒍𝒌−𝟏 = 𝑺𝑭𝑫𝒔 𝒌|𝒋 − 𝑺𝑭𝑫𝒔𝒌−𝟏|𝒋= 𝑺𝑭𝑫(𝟏 + 𝒋)𝒌−𝟏

Net interest in the kth payment=

𝑵𝒆𝒕 𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 = 𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒕𝒐 𝒍𝒆𝒏𝒅𝒆𝒓 − 𝑰𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝒊𝒏 𝒔𝒌𝒊𝒏𝒌𝒊𝒏𝒈 𝒇𝒖𝒏𝒅𝑳. 𝒊 − 𝑺𝑭𝑩𝒂𝒍𝒌−𝟏. 𝒋

or𝑵𝒆𝒕 𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 = 𝑻𝒐𝒕𝒂𝒍 𝑷𝒂𝒚𝒎𝒆𝒏𝒕 − 𝑷𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝑷𝒂𝒊𝒅

𝑺𝑭𝑫 + 𝑳. 𝒊 − 𝑺𝑭𝑫 𝟏 + 𝒋 𝒌−𝟏

Exercise

38

For a sinking fund loan, SFD = 6902.95 and

the interest rate for the sinking fund is 8%.

Find the principal paid in the 4th payment?

Solution:

𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑃𝑎𝑖𝑑 = 𝑆𝐹𝐷(1 + 𝑗)𝑘−1

6902.95(1 + 8%)4−1= 8695.73

Exercise

39

For a sinking fund loan, SFD = 5310.76 and

the amount of principal in the third payment

is 5967.17. What is the interest rate?

Solution:

𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑃𝑎𝑖𝑑 = 𝑆𝐹𝐷(1 + 𝑗)𝑘−1

5,916.17 = 5,310.76(1 + 𝑗)3−1

𝑗 = 0.06

Capitalization of interest and

negative amortization

40

It is possible that during repayment of a loan

by the amortization method, the payment is

not large enough to cover the interest due

(𝑃𝑀𝑇𝑡 < 𝐼𝑡).

Capitalization of interest and negative

amortization occur when the payment made is

less than the interest on the loan. In other

words, the outstanding balance increases by

the amount of unpaid interest.

Exercise

41

A borrower would like to borrow 30,000 at 8% for 5 years, but would like to pay only 2,000 for the first two years and then catch up with a higher payment for the final three years. What is the payment for the final 3 years?

Illustrate the payments by using the amortization schedule.

Solution:

𝑂𝐵2 = 30,000(1.08)2−2,000 1.08 − 2,000 = 30,83230,832 = 𝐾𝑎 3|0.08𝐾 = 11,964

Exercise – cont’d

42

t Payment Interest Due Principal

Repaid

Outstanding Balance

0 __ __ __ 30,000

1 2,000 2,400 −400 30,400

2 2,000 2,432 −432 30,832

3 11,964 2,466.56 9,479.29 21,334.71

4 11,964 1,706.78 10,257.07 11,077.64

5 11,964 886.21 11,077.64 0

Exercise

43

Referring to the previous question, find the

principal paid, interest required and the

balance in year 1 if the initial payment were

1,500 instead of 2,000.

Solution:

Interest = 2,400

Principal = -900

Balance = 30,900