Chapter 6 Amortization And Sinking Funds

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					Chapter 6 – Amortization Schedules And Sinking Funds
6.1 (page 166) Introduction
 2 general methods of repaying loans
 amortization
   installment payments at periodic intervals (usually level)
   e.g. biweekly, monthly, quarterly, semi-annually, annually
   e.g. mortgages, car loans, furniture loans
 sinking funds
   lump sum payment at the end of the term of the loan
   periodic interest payments PLUS
   payments into a sinking fund that is to accumulate to the amount of
     the loan to be repaid
   payments may be into a trust for security reasons
   payments into the sinking fund may be a requirement of the issue of
     the original loan amount
   sometimes the sinking fund balance is used to retire part of the loan
     on the open market (usually because the after tax interest being
     earned on the sinking fund is substantially less than the cost of the
     interest payments on the loan – may be interest penalties to pay in
     order to retire in this fashion)
6.2 (page 167) Finding The Outstanding Balance
Amortization
 installment payments are in the form of an annuity
 synonymous terms
   outstanding loan balance
   outstanding principal
   unpaid balance
   remaining loan indebtedness
 prospective method
   look into the future



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      what is the present value of the remaining payments when
       calculated at the loan rate
   retrospective method
     original loan accumulated to the current time less payments
       accumulated to current time (both at the loan rate)
   retrospective = prospective
   payments are first applied to pay interest on the previous outstanding
    loan balance
   remainder used to reduce the outstanding loan balance

Lender In Flow:               P            P             P         P

Time:          0              1            2            n-1        n
Loan:          L

Equation Of Value: @ i, L = PV(Out Flow) = PV(In Flow)
                         = P ∙ an|i
                              therefore, P = L / an|i
Loan Balance Immediately After A Payment:
Bk = Outstanding balance just after kth payment
= P ∙ ank|i (“prospective” method – present value of remaining payments)
= L ∙ (1+i)k – P ∙ sk|i (“retrospective” method – accumulated value at time
k of L less accumulated value of past k payments)
Note: Bk+t = Bk ∙ (1+i)t for 0 < t < 1

BtP  balance from prospective viewpoint = P ∙ ank|i
BtR  balance from retrospective viewpoint = L ∙ (1+i)k – P ∙ sk|i
We want to show they are equivalent
P ∙ ank|i = L ∙ (1+i)k – P ∙ sk|i which can be rewritten as
P ∙ ank|i = P ∙ an|i ∙ (1+i)k – P ∙ sk|i
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The payments cancel out and you are left with
         1  vn                (1  i ) k  1
ank|i =            ∙ (1+i)k –
            i                         i
         (1  i ) k  v nk  (1  i ) k  1 1  v nk
       =                                               ank|i
                          i                      i
6.3 (page 169) Amortization Schedules
 each payment is part interest and part principal
 split varies with each payment
 interest portion is larger in early payments and smaller later
 Pk = principal portion of kth payment
 Ik = interest portion of kth payment
 by definition, P = Pk + Ik
 Ik = Bk-1 ∙ i (interest on previous Outstanding Loan Balance)
 Pk = P ∙ v
             n+1-k
                   (P1, P2, ..., Pn is a geometric sequence which is linear if i
   = 0)
Example Of Amortization Of 3 Year Loan Of $10,000 @ 7%
Time       Payment           Interest        Principal    Outstanding
                             Portion         Portion      Balance
0                                                         10,000.00
1          3,810.52          700.00          3,110.52       6,889.48
2          3,810.52          482.26          3,328.26       3,561.22
3          3,810.51          249.29          3,561.22           0.00
Using The Calculator:
i) Finding The Payment
3 N 10,000 PV 7 %I CPT PMT ==>> 3810.5167
ii) Finding O/S Balance @ End Of Period
Enter 1, 2, or 3 to find balance and press BAL
Note: does the calculation without rounding to pennies
2 BAL ==>> 3561.2305
iii) Finding Interest Portion Of Payment At End Of Period
3 I/P ==>> 249.2861
iv) Finding Principal Portion – Use The X  Y Key
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Table 6.1
    Loan of a n|i
    By definition, payment of 1 at end of each period
    Repaid over n periods at interest rate i
    Period     Payment Interest Paid               Principal Repaid                            Outstanding Loan
      0                                                                                                      an|i
      1                 1             i  an|i  1  v n        vn                         an|i  vn  an 1|i
      2                 1           i  an 1|i  1  v n 1   v n 1              an 1|i  vn 1  an  2|i

       ∙                ∙                      ∙                 ∙                               ∙
       ∙                ∙                      ∙                 ∙                               ∙
       ∙                ∙                      ∙                 ∙                               ∙
       t                1             i  at |i  1  vt        vt               an t 1|i  v n t 1  an t |i
      ∙                 ∙                      ∙                 ∙                               ∙
      ∙                 ∙                      ∙                 ∙                               ∙
      ∙                 ∙                      ∙                 ∙                               ∙
     n-1                1             i  a2|i  1  v2         v2                           a2|i  v 2  a1|i
      n                 1              i  a1|i  1  v         v                               a1|i  v  0
    Total               n                  n  an|i            an|i

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Another view of the amortization
   an|i is the loan value at time 0
   one period later, you have accumulated it with one period’s interest
    and taken out a payment of 1
                                      (1  i )  v n 1      1  v n 1
 B1  an|i            (1  i )  1                    1 
                                              i                  i

An additional view of the amortization
At time zero, you have a series of payments of 1 that have a value of
an|i :

               v       v2     v3           ...        vn-3 vn-2 vn-1 vn

     0 1       2     3         ...      n-3 n-2 n-1 n
     
One year later, their value has grown with one years interest and the
payment at time 1 has been paid and so is no longer outstanding so the
outstanding loan balance is now the sum of:

               1       v      v2                 . . . vn-4 vn-3 vn-2 vn-1

       0       1       2      3            ...        n-3 n-2 n-1 n
               
which equals an 1|i .


Assumptions To Date
 constant i
 payment period and interest conversion period are the same
 payments are level

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Example
A loan of $100,000 is to be repaid by 40 annual payments with the first
payment one year from now. Assuming interest is at 6%,

a) What is the annual payment needed to amortize the loan?

P = $100,000 / a40|.06 = $100,000 / 15.0463 = $6,646.15 or
100000 PV 0 FV 6 %I 40 N CPT PMT = 6646.1536

b) What is the Outstanding Loan Balance at t = 25?

Assuming you already have the above input into your calculator, input
25 2nd BAL = $64,549.08        otherwise,
B25 = $6,646.15 ∙ a15|.06 = $6,646.15 ∙ 9.7122 = $64,548.74
Note: using the interest tables loses accuracy in about the 5th digit

c) What is the interest portion of the 15th payment?

Once again assuming you have the inputs in a) still in the calculator,
input 15 2nd INT = $5,185.26       otherwise,
i∙B14 =.06∙$6,646.15∙ a14|.06 =.06∙$6,646.15∙13.0032 = $5,185.27


6.4 (page 175) Sinking Funds
 sometimes lender insists that sinking fund be built up to reduce the
  amount lost on default
 we will deal with regular contributions to the sinking fund
 the borrower is paying regular interest on the outstanding balance of
  the loan plus making regular contributions to a sinking fund to retire
  the loan at the end of the term
 net loan outstanding is the loan less the sinking fund balance



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   if interest earned on the sinking fund equals interest being paid on the
    loan, the two methods are equivalent
   you will remember our formula from chapter 3
                           1                        1
                                            i 
                          s n         i
                                                   a n   i

                  amount to       interest    amount needed to
                 accumulate + on the = amortize the loan
                  to the loan       loan
   even though the interest paid on the loan is level, the net interest is
    declining as the sinking fund earns interest
   net interest is same under the two methods IF sinking fund is earning
    the same rate as is being paid on the loan

        Sinking Fund For A $10,000 5 Year Loan At 10% Interest
      Year Interest   Sinking    Interest Sinking    Net Loan
            Paid On Fund         Earned Fund
            Loan      Deposit    On
                                 Sinking
                                 Fund
          0                                            10,000.00
          1 1,000.00 1,637.97                1,637.97 8,362.03
          2 1,000.00 1,637.97 163.80         3,439.75 6,560.25
          3 1,000.00 1,637.97 343.97         5,421.70 4,578.30
          4 1,000.00 1,637.97 542.17         7,601.84 2,398.16
          5 1,000.00 1,637.97 760.18 10,000.00               0.00
Check that equivalent to amortization: $10,000 / a5|.10 = $2,637.97

   IF i on loan = i on sinking fund:
     total payment is same
     annual net interest is same
     annual net repayment is same

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   In practice, rate earned on sinking fund (j) usually < i
   total payment then becomes
                                     1                      1
                                               i 
                                s    n     j
                                                       a    n i& j

                                  which can be reexpressed as
                        1                                           1
                                       (i  j ) 
                       an        j
                                                                a   n i& j




                                                  or
                                                       an   j
                           a n i& j 
                                               1  (i  j )  a n       j

                                  which equals a n i when i = j
      Sinking Fund For A $10,000 5 Year Loan At 10% Interest But
                            Sinking Fund Earning 8%
     Year Interest         Sinking     Interest Sinking      Net Loan
             Paid On Fund              Earned Fund
             Loan          Deposit     On
                                       Sinking
                                       Fund
           0                                                   10,000.00
           1 1,000.00 1,704.56                      1,704.56 8,295.44
           2 1,000.00 1,704.56 136.37               3,545.49 6,454.51
           3 1,000.00 1,704.56 283.64               5,533.70 4,466.30
           4 1,000.00 1,704.56 442.70               7,680.96 2,319.04
           5 1,000.00 1,704.56 614.48 10,000.00                      0.00
   This is (i-j) ∙ loan higher than the payment needed for a 5 year
    amortization at 8% ($10,000 / 3.9927 = $2,504.56)

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   That is, the schedule above is the same as if the loan and sinking fund
    had been both at 8% but the additional amount of (.10 - .08) times the
    loan ($200 in this case) is being paid in loan interest
                                Total Payment
                           $100,000 40 Year Loan
                                  Loan Rate
        Sinking      4%         6%       8%         10%        12%
          Fund
          Rate
              4% 5,052.35 7,052.35 9,052.35 11,052.35 13,052.35
              6% 4,646.15 6,646.15 8,646.15 10,646.15 12,646.15
              8% 4,386.02 6,386.02 8,386.02 10,386.02 12,386.02
             10% 4,225.94 6,225.94 8,225.94 10,225.94 12,225.94
             12% 4,130.36 6,130.36 8,130.36 10,130.36 12,130.36
6.5 (page 181) Differing Payment Periods And Interest Periods
Payments Less Frequent Than Interest Convertible
 payment every kth period
 interest compounds k times in that period
 n is the number of measurement periods over which the loan is being
  amortized
 (m / k) ∙ n payments to amortize the loan (or build sinking fund)

            i (m) k
 use
       (1       )  1 = j as your interest rate
              m

Payments More Frequent Than Interest Convertible
 k payments before interest compounds
 m is the number of times interest compounds in the normal
  measurement period
 n is the number of measurement periods over which the loan is being
  amortized
 k ∙ m ∙ n payments to amortize the loan (or build sinking fund)

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            i ( m ) 1/ k
   use (1  m )  1 = j as your interest rate and




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Example
Create the sinking fund table for the following: a 3 year loan of $10,000,
with interest payable semiannually at the nominal interest rate of 8.00%
is to be retired by a sinking fund funded by quarterly deposits earning an
effective semi-annual interest rate of 3.00%

   Interest of $400 (.08/2 ∙ $10,000) is paid on the loan
   The quarterly rate on the sinking fund is 1.4889% (1.03.5-1)
   $10,000 / s12|.014889 = $767.28

      t               Interest         Sinking   Interest   Amount     Net Loan
                      Paid             Fund      Earned     In Sinking
                                       Deposit   on         Fund
                                                 Sinking
                                                 Fund
    0.00                                                     10,000.00
    0.25                    767.28                  767.28 9,232.72
    0.50       400.00       767.28      11.42 1,545.98 8,454.02
    0.75                    767.28      23.02 2,336.27 7,663.73
    1.00       400.00       767.28      34.79 3,138.33 6,861.67
    1.25                    767.28      46.73 3,952.33 6,047.67
    1.50       400.00       767.28      58.85 4,778.45 5,221.55
    1.75                    767.28      71.15 5,616.88 4,383.12
    2.00       400.00       767.28      83.63 6,467.78 3,532.22
    2.25                    767.28      96.30 7,331.36 2,668.64
    2.50       400.00       767.28    109.16 8,207.79 1,792.21
    2.75                    767.28    122.21 9,097.27            902.73
    3.00       400.00       767.28    135.45 10,000.00              0.00
Note: sinking fund interest is shown as it accrues – in actual fact, it is
only being credited semi-annually.


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