CHAPTER 9 INVENTORIES: ADDITIONAL VALUATION ISSUES Sommers – ACCT 3311.
MACT 3311 - Chapter 3
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Transcript of MACT 3311 - Chapter 3
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.
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t 𝑃𝑀𝑇𝑡 𝐼𝑡 = 𝑂𝐵𝑡−1 x 𝑖 𝑃𝑅𝑡 = 𝑃𝑀𝑇𝑡 − 𝐼𝑡 𝑂𝐵𝑡 = 𝑂𝐵𝑡−1 − 𝑃𝑅𝑡.
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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|
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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|
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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