; fixed rate mortgages
Documents
User Generated
Resources
Learning Center

# fixed rate mortgages

VIEWS: 7 PAGES: 82

• pg 1
```									   Fixed Rate Mortgage
Mechanics
 Recall that to the investor, the fixed rate
mortgage is a type of annuity.
 The investor pays the borrower an up-front
amount in return for a promised stream of
future cash flows.
 At time zero (i.e. origination) the present
value of the annuity must equal the cash the
investor pays the borrower.
Fixed Rate Mortgage
Mechanics
 Casho = PV0(Future Cash Flows)
 If the cash were worth more than the PV of
the future cash flows, the bank would not be
willing to make the loan they would be
paying more for the annuity than it was
worth.
Fixed Rate Mortgage
Mechanics
 Casho = PV0(Future Cash Flows)
 If the cash were worth less than the PV of the
future cash flows, the borrower would not be
willing to accept the loan because they would
be taking on a liability that was worth more
than the asset they would receive (the cash),
reducing their wealth.
Fixed Rate Mortgage
Mechanics
 Casho = PV0(Future Cash Flows)
 Thus, at time 0, the only way the two parties
will come to an agreement is if the exchange
is equal: the lender must give the investor an
amount in cash that is equal to the present
value of the remaining future cash flows.
 After time 0, of course, this relationship does
not hold.
Fixed Rate Mortgage
Mechanics
 The mortgage contract specifies how to
calculate the various cash flows associated
with the mortgage. This will include:
 The “Principal” amount of the loan
determines the monthly payments. This is
normally set to the amount of cash the
investor gives the borrower at time 0.
(unless the loan includes points).
Fixed Rate Mortgage
Mechanics
 The other terms specified in the mortgage
contract include:
   r - the contract rate of the mortgage,
   n - the number of monthly payments,
   Pmt – the monthly payment on the mortgage.
Fixed Rate Mortgage
Mechanics - Balance
 At time 0 we know that the value of the
mortgage is equal to the cash received. For
now, assume that the principal is set to that
same amount.
 Thus, the value of the mortgage must have
this relationship:            
        1           
 1               n 

Prin  Pmt * 
1 c
12        

c
12
Fixed Rate Mortgage
Mechanics - Balance
 Thus, we know if the contract rate were 8%,
with 20 years (240 payments) term and
monthly payments of \$850, the principal
amount must be 101,621.15

              
       1      
1


 1  .08


240 

12 

101,621.15 850 *
.08
12
Fixed Rate Mortgage
Mechanics - Balance
 Note that this formula actually works for any point
during the life of the mortgage – that is, if you tell
me the remaining term, the contract rate, and the
monthly payment, this formula tells you the
currently outstanding principal.
            
       1    
1


 1 c


n 

12 

Prin  Pmt *
c
12
Fixed Rate Mortgage
Mechanics - Payments
 While knowing how to determine the
principal amount is important, it is perhaps
more interesting (from a potential
homeowners standpoint) to know how to
calculate the payment that will be required
given a known balance.
 This just requires simple algebraic
manipulation of the balance formula.
Fixed Rate Mortgage
Mechanics – Payments
Pmt  Prin *
c / 12
          
    1     
1



 1 c
n 

12 
  

 So, for a \$100,000 loan at 10% for 30 years,
the payment is \$877.57.
877.57  100,000*
.10 / 12
              
       1      
1



 1  .10
360 

12 
   
Fixed Rate Mortgages
Mechanics - Payments
Pmt  Prin *
c / 12
          
    1     
1



 1 c
n 

12 
  
 This formula also works at any point in time.
That is, if you know the balance, remaining
term, and contract rate, you can plug those
numbers into the above formula and
determine the monthly payment.
Fixed Rate Mortgage
Mechanics - Amortization
 The mortgage contract will state the order in
which payments are attributed to the account.
The usual way this occurs is:
 Overdue interest and penalties are paid first,
 Current interest is paid second,

 Overdue principal is paid third,

 Current principal is paid fourth,

 Any remaining cash pre-pays principal.
Fixed Rate Mortgage
Mechanics - Amortization
 Thus, normally (i.e. when scheduled
payments are made on time), the investor
takes the interest out of the payment first,
and then takes the principal.
 The interest amount is found by multiplying
the balance at the beginning of the month by
the monthly interest rate:
   Interest due = Beginning Balance * c/12.
Fixed Rate Mortgage
Mechanics - Amortization
 The principal due can then be found by
subtracting the interest due from the
payment:
   Principal Due = Pmt – Interest Due
 From this information we can create an
amortization chart.
Fixed Rate Mortgage
Mechanics - Amortization
 For a 30 year, 9% mortgage, original balance
of \$200,000.
Principal         \$200,000.00 Payments                    360 Contract rate                 9.00%

Beginning             Interest                 Ending
Month           Balance      Payment Due         Principal Due Balance
1     \$200,000.00 1609.245 \$1,500.00       \$109.25 \$199,890.75
2     \$199,890.75 1609.245 \$1,499.18       \$110.06 \$199,780.69
3     \$199,780.69 1609.245 \$1,498.36       \$110.89 \$199,669.80
4     \$199,669.80 1609.245 \$1,497.52       \$111.72 \$199,558.08
5     \$199,558.08 1609.245 \$1,496.69       \$112.56 \$199,445.52
6     \$199,445.52 1609.245 \$1,495.84       \$113.40 \$199,332.11
7     \$199,332.11 1609.245 \$1,494.99       \$114.25 \$199,217.86
8     \$199,217.86 1609.245 \$1,494.13       \$115.11 \$199,102.75
9     \$199,102.75 1609.245 \$1,493.27       \$115.97 \$198,986.77

Note that the above is an Excel spreadsheet – you should be able to “click” on it and actually use it.
Fixed Rate Mortgage
Mechanics - Amortization
 Notice the relationship between principal
payment, interest payment and total payment.
1800
1600
1400
1200                               Payment
1000
Interest Due
800
600                               Principal Due
400
200
0
0   100   200   300   400
Fixed Rate Mortgage
Mechanics - Price
 At origination the contract rate of the
mortgage will equal the market interest rate
for the type of loan and creditworthiness of
the borrower.
 It is the equality of the market and contract
rates which forces the balance and value of
the mortgage to be the same at time 0.
Fixed Rate Mortgage
Mechanics - Price
 Over time, since the contract rate is fixed, the
contract and mortgage rates will diverge.
Thus, the value and balance of the mortgage
will diverge over time.
 This means we have to concern ourselves
with determining the value (price) of the
mortgage at times other than time t.
Fixed Rate Mortgage
Mechanics - Price
 To do this we simply take the present value
of the remaining payments using the current
market rate:
             
        1    
1



1 r   
n 

12 

Value  Pmt *
r
12
Fixed Rate Mortgage
Mechanics - Price
 Of course this is the same basic formula as
the one we used to calculate the balance,
with the difference that we use the market
rate, r, instead of the contract rate, c.

                                             
        1                               1    
1                                 1

Value  Pmt *




1  r  n 

12 



 1 c


n 

12 

Balance  Pmt *
r                                 c
12                                12
Fixed Rate Mortgage
Mechanics - Example
 At this point it might be useful to look at an
extended example.
   Consider that a borrower originally took out a
\$200,000 loan for 30 years at 9%. Five years have
passed and the market rate is now 7%.
 What is the monthly payment on the loan?
 What is the balance of the loan?
 What is the value of the loan?
Fixed Rate Mortgage
Mechanics - Example
 Example (continued)
   The monthly payment is \$1,609.25:

\$1,609.25 200,000*
.09 / 12
                  
          1       
1



1  .09
12

360 



   After 5 years the balance is: \$191,760:
              
       1      
1



 1  .09
300 

12 
      
191,760.27 1609.25*
.09
12
Fixed Rate Mortgage
Mechanics - Example
 Example (continued)
   Note that I can determine the payment from
ONLY the current balance, contract rate, and
remaining term:
\$1,609.25 191,760*
.09 / 12
                 
         1       
1




1  .09
12

300 


Fixed Rate Mortgage
Mechanics - Example
 Example (continued)
   The value of the mortgage, at the 7% contract rate is:
\$227,687.12,
              
       1      
1


 1  .07


300 

12 

227,687.12 1609.25*
.07
12

   Contrast this with the balance, which is still
              
       1      
1


 1  .09


300 

12 

191,760.27 1609.25*
.09
12
Fixed Rate Mortgage
Mechanics - Effective Yield
 Frequently, we will know the price of a
mortgage, and its contractual details, but we
will not know the market discount rate.
 Fortunately, we can use the present value of
an annuity formula to solve for the discount
rate.
Fixed Rate Mortgage
Mechanics - Effective Yield
 We simply have to solve for effective yield
(y) in the equation below. This can be done
through a search algorithm or by use of a
financial calculator.

              
              
1       1    
  y      n

 1  12  
       
Known Price  Pmt * 
y
12
Fixed Rate Mortgage
Mechanics - Effective Yield
 In the previous example, let us say the a bank
could purchase the mortgage for \$180,000.
What would be the effective yield if a bank
purchased it at that price? It would be 9.79%
                
        1       
1


 1  .0979


300 

12 

\$180,000 1609.25*
.0979
12
Fixed Rate Mortgage
Mechanics - Effective Yield
 Note that in the absence of prepayment
penalties or points, the effective yield on a
mortgage (to the borrower) is always the
contract rate of the loan.
Fixed Rate Mortgage
Mechanics - Prepayment
 This extended example raises an interesting
point. The borrower is scheduled to make
payments that are worth, at the current
market rate of 7%, \$227,687.12. The
mortgage contract, however, grants them the
right to pay off that loan at any time by
repaying the balance, which is the
\$191,760.27.
Fixed Rate Mortgage
Mechanics - Prepayment
 Thus, by taking out a new loan for
\$191,760.27, at the current market rate (7%)
and used the proceeds to pay off the original
loan, they would increase their wealth by
\$35,926.85.
 In essence they would be replacing one
liability worth \$227,687.12 with one worth
\$191,760.27.
Fixed Rate Mortgage
Mechanics - Prepayment
 Discounting the remaining payments at the
market rate and comparing that to the
balance allows us to quantify the benefits to
prepaying the loan.
 Frequently it is costly to refinance a loan.
Optimally, one will not refinance if the gain
to refinancing is less than the refinancing
costs, i.e. (Value – Balance> Cost of Refi).
Fixed Rate Mortgage
Mechanics - Prepayment
 When we talk about the “value” of this loan
to the lender, we have to realize that they
factor in the borrower’s right to “call” the
loan.
   In the previous example the “value” of the loan
is not really \$227,687.12 because the lender
knows the borrower is going to prepay it. They
realize the value is probably no more than
\$191,760.27.
Fixed Rate Mortgage
Mechanics - Prepayment
 If we denote the value of the promised
payments as “A”, and the value of the call
option as “C”, and any transaction costs of
refinancing as “T”, then the true value of the
mortgage will be:
   V = A – (C-T).
   Since in the previous example we had no
transaction costs, i.e. T=0, then
   \$191,760.27 = 227,687.12 – 35,926.85
Fixed Rate Mortgage
Mechanics - Prepayment
 It is useful to examine what happens to the
value of the mortgage if rates changed
instantaneously.
   To do this let’s use the same data from our
previous example but assume it will cost the
borrower \$2500 to refinance.
   We assume the borrower will only prepay when
it is financially beneficial to do so, i.e. when:
 A – Balance – T > 0
Fixed Rate Mortgage
Mechanics - Prepayment
   Graphically, the value of A, i.e. the PV of the
remaining payments, (V if you ignore the value
of C), looks like this (to the bank!)
\$450,000.00

\$400,000.00

\$350,000.00

\$300,000.00

\$250,000.00

\$200,000.00

\$150,000.00

\$100,000.00

\$50,000.00

\$0.00
0   0.02   0.04   0.06   0.08   0.1   0.12   0.14   0.16
Fixed Rate Mortgage
Mechanics - Prepayment
   Graphically, the value of C, i.e. value of the
borrower exercising their call option, is given
(again, to the bank!):
\$50,000.00

\$0.00
0   0.02   0.04   0.06   0.08   0.1   0.12   0.14   0.16

(\$50,000.00)

(\$100,000.00)

(\$150,000.00)

(\$200,000.00)

(\$250,000.00)
Fixed Rate Mortgage
Mechanics - Prepayment
 Combining these two shows the value of the mortgage to the
bank (V). Note the spike in value just below the contract rate.
\$300,000.00

\$250,000.00

\$200,000.00

\$150,000.00

\$100,000.00

\$50,000.00

\$0.00
0.06   0.07   0.08   0.09   0.1   0.11   0.12

(\$50,000.00)

(\$100,000.00)
Fixed Rate Mortgage
Mechanics - Prepayment
 It may be easier to see this by looking only at graph of V.

\$200,000.00

\$190,000.00

\$180,000.00

\$170,000.00

\$160,000.00

\$150,000.00

\$140,000.00

\$130,000.00

\$120,000.00

\$110,000.00

\$100,000.00
0   0.02   0.04   0.06   0.08   0.1   0.12   0.14   0.16
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 One idea to remember is that banks
understand, and explicitly build into
mortgage rates, the risk of prepayments.
 Some borrowers, primarily commercial
borrowers but increasingly residential
borrowers, are willing to contractually agree
not to prepay in order to secure a lower
contract rate.
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 A common way for the borrower to signal to
the lender their willingness to forgo the
prepayment option is by accepting a
prepayment penalty.
 A prepayment penalty is simply an additional
fee that the borrower agrees to pay, in
addition to the outstanding balance, should
they prepay the loan.
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 Frequently these prepayment penalties end
after some specified period of time (5, 10 or
15 years for example).
 Some common prepayment penalties include
   A flat fee,
   A percentage of the outstanding balance,
   The sum of the previous six months interest.
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 The real effect of the prepayment penalty is
to raise the borrower’s effective interest rate
should they prepay.
 Consider the following example.
   A borrower takes out a loan for with a contract rate of
10%, a term of 30 years, and an initial balance of
\$100,000. There is a prepayment penalty of 2% of the
outstanding balance if they prepay the loan.
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 If the borrower prepays after 5 years, what is
the effective interest rate on the loan?
   To determine this we must first determine the
cash flows.
   The original payment is simply \$877.57/month.

877.57  100,000*
.10 / 12
              
       1      
1



 1  .10
360 

12 
 
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 After 5 years the balance will be 96,574.14.
                   
           1       
1


 1  .10


300 

12 

96,574.14  877.57 *
.10 / 12

 Thus, to pay off the loan the borrower will
have to pay a lump sum of 98,505.63
98,505.63 = 96,574.14 * 1.02
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 Thus, to the borrower, the cash flows are:
 Positive cash flow of \$100,000 at time 0
 60 negative cash flows of \$877.57

 One final negative cash flow at month 60 of 98,505.63

 Their effective yield is the yield that makes
this equation true, which is 10.30%.
           1      
1             60 
100,000  877.57 * 
1  y / 12   98,505.63
      y / 12       (1  y / 12)60
                  
                  
Fixed Rate Mortgage
Mechanics – Prepay Penalties
 One has to be careful in this analysis,
however. To the borrower making the
decision to prepay at time 60, the previous 59
must be ignored. Where this enters the
borrower’s decision making is at origination.
A borrower that expects to prepay, would opt
not to take out a mortgage with a prepayment
penalty.
Fixed Rate Mortgage
Mechanics – Prepay Penalties
   Consider at origination if the borrower suspected
that they would likely prepay within 5 years. If
they were offered two loans, one at a contract
rate of 10% and a 2% prepayment penalty or one
at 10.25% and no prepayment penalty, then they
should select the 10.25% loan.
   The reason borrower charge prepayment
penalties is to induce borrower to reveal their
patterns.
Fixed Rate Mortgage
Mechanics - Points
 One unusual feature of the mortgage market
related to prepayment penalties is the
practice of charging borrowers “points”.
 Technically a point is a fee that the borrower
pays the bank at origination. For each point
charged, the borrower pays one percent of
the initial loan balance to the bank.
Fixed Rate Mortgage
Mechanics - Points
 The effect of this, of course, is to reduce the
actual cash received by the borrower.
   Thus if a borrower took out a \$100,000 loan, but
was charged 2 points, they would receive
\$100,000 from the bank and then write a check
to the bank for \$2000, making their net proceeds
\$98,000.
Fixed Rate Mortgage
Mechanics - Points
 Of course since the borrower only received,
net, \$98,000, at origination, the value of the
mortgage at origination can only be \$98,000.
 The payments, however, will be based on the
nominal principal amount of \$100,000. The
only way for the PV of the future payments
to be worth \$98,000, therefore, is to reduce
the contract rate.
Fixed Rate Mortgage
Mechanics - Points
 This is, of course, exactly what happens – the
more points you pay, the lower your contract
rate.
 Note, however, that the effective interest rate
is not constant, it is a function of when the
loan is paid off.
Fixed Rate Mortgage
Mechanics - Points
 To calculate the effective interest rate on a
mortgage with points you must go through
multiple steps:
   First, use the contract parameters (i.e. contract
principal, term, and contract rate) to determine
the cash flows.
   Second, find the effective yield which equates
the present value of the future cash flows to the
amount of cash (net) received at the closing.
Fixed Rate Mortgage
Mechanics - Points
 Again, this may be best illustrated with an
example.
   A borrower takes out a loan with a 10% contract
rate, a balance of \$150,000, and 30 year term.
The bank charges two points.
   If the borrower never prepays, what is the
effective interest rate on the loan?
Fixed Rate Mortgage
Mechanics - Points
 First, determine the actual cash flows
   The monthly payment is

\$1,316.36 150,000*
.10 / 12
                 
         1       
1




1  .10
12

360 



 Since the borrower does not prepay the loan,
the next step is to determine the yield based
on the cash actually received at the closing.
Fixed Rate Mortgage
Mechanics - Points
 Since the bank charges 2 points on a
147,000.
   147,000 = 150,000 * (1-.02)
 The final step is to determine the yield which
equates the cash received at time 0 with the
present value of the monthly payments.
Fixed Rate Mortgage
Mechanics - Points
 That is, determine:
1
1                   360
1  y 

        12 

\$147,000  1,316.36*
 y / 12

 The answer is y = 10.24%
Fixed Rate Mortgage
Mechanics - Points
 What would be the yield if the borrower
prepaid after 10 years?
   Obviously the time 0 cash and the monthly
payments are the same. The only additional item
we need to know is the balance of the loan after
10 years, which is given by discounting the
remaining payments at the contract rate.
1
1
\$136,407.0 2  1,316.36 *

1  .10
12

240

.10 / 12 
Fixed Rate Mortgage
Mechanics - Points
 Now we again determine the yield which sets
present value of the future cash flows equal
to the cash received at time 0:
1
1               120
1  y 

        12 
         136,407.02
\$147,000  1,316.36*                          
 y / 12           (1  y/12)120

 The answer is y=10.33159 %
Fixed Rate Mortgage
Mechanics - Points
 The chart below illustrates the effective yield given
the date at which the borrower prepays the loan
35
30
Effective Yield

25
20
15
10
5
0
0   60   120   180    240   300   360
Prepayment Month
Fixed Rate Mortgage
Mechanics - Points
 Finally, consider if at origination the
borrower had a choice between two loans.
   Loan A is the mortgage with points we just
examined.
   Loan B is for \$147,000, at 10.3% and no points.
 Which loan should the borrower take?
   The answer to that depends upon the borrowers
expectations regarding their tenure in the
mortgage.
Fixed Rate Mortgage
Mechanics - Points
   Clearly if the borrower expects to never prepay
the mortgage, they should take loan A, because
the effective rate on the loan will be 10.24%,
well below the 10.3% of loan B.
   If, however, the borrower expects to prepay after
10 years (or before), they should take loan B,
since with a 10 year prepayment horizon loan A
has an effective interest rate of 10.33%.
Rate Mortgage
Mechanics – Incremental Cost
 The final issue we will examine is the
incremental cost of financing.
 Frequently borrowers of equal
creditworthiness will observe different
interest rates for different sized loans. That
is, an 80% loan to value (LTV) mortgage
will have a lower contract rate than a 90%
LTV loan.
Rate Mortgage
Mechanics – Incremental Cost
 The question is, what is the effective interest
rate on that differential.
   For example in February of 2000, 80% LTV 30
year mortgages had a contract rate of
approximately 8.25% while 95% LTV mortgages
had a contract rate of approximately 8.75%.
   If you were a borrower with a \$200,000 house
you could borrow \$160,000 at 8.25% or
\$190,000 at 8.75%.
Rate Mortgage
Mechanics – Incremental Cost
   So to borrow the incremental \$30,000 your
overall interest rate goes up by .5%.
   One way of looking at this is that you borrowing
the first \$160,000 at 8.25%, and the remaining
\$30,000 at some effective rate. The question is,
what is that effective rate?
   To solve this, let’s consider the cash flows.
Rate Mortgage
Mechanics – Incremental Cost
   The payments on the 80% and 95% loans are
(respectively)
Payment80% loan  160,000*
.0825 / 12        1,202.02
1
1

1  .0825
12
360


Payment95% loan  190,000*
.0875 / 12         1,494.73
1
1

1  .0875
12
360


Thus there is a \$292.70/month differential
between the two
Rate Mortgage
Mechanics – Incremental Cost
   One way to view this is that you are paying
292.70/month to borrow \$30,000 for 30 years. This
implies the following statement must be true:

1
1                   360
1  y 

        12 

30,000  292.70 *
 y / 12

Solving for y we find that the incremental cost of
financing is: 11.308%.
Rate Mortgage
Mechanics – Incremental Cost
   What this means is that if you can borrow
\$30,000 for less than 11.308%, you should take
the 80% LTV loan and then borrow the
remaining funds from that other source. If you
cannot borrow \$30,000 for less than 11.308%,
you should take the 95% loan.
Rate Mortgage
Mechanics – Second Mortgages
   It is not at all uncommon for a borrower to take
out two mortgages. The first will typically be for
80 or 90% LTV, with the second mortgage being
for 20%, 10% or 5%.
   Occasionally, it will be the case that it is cheaper,
in terms of the effective cost of financing, to take
out an 80% LTV first loan, and then a 10%
second loan, than it would be to take out a single
90% LTV loan.
Rate Mortgage
Mechanics – Second Mortgages
   To calculate the effective cost of financing when
there are two mortgages is not particularly
difficult. You first determine each mortgage’s
monthly payments individually, then you
combine their initial balances and monthly
payments and solve for the effective interest rate.
   An example may make this easy to see.
Rate Mortgage
Mechanics – Second Mortgages
   Example: Bob wishes to buy
a house for \$100,000. He
will put 10% down, take out      Pmt1  80,000 *
.055 / 12          454.23
         1          
an 80% LTV first loan at                           1              360 
 1  .055 / 12 
5.5%, and a 10% second
loan at 7%. Each mortgage is                             .07 / 12
for 30 years. What will be       Pmt2  10,000 *                            66.53
         1         
1             360 
his total cost of financing if                      1  .07 / 12 
he keeps each mortgage for
the full 30 year term?
   First, let’s calculate each
mortgage payment:
Rate Mortgage
Mechanics – Second Mortgages
   Now we can combine the two mortgages, and find the
interest rate that sets the initial balances equal to the
present value of the total monthly payments.

          1      
1
 1  r / 12360 
90,000  (520.76) *                   
    (r / 12)     

                 

Solving for r, yields :
r  5.67%
Rate Mortgage
Mechanics – Second Mortgages
   Notice that the effective interest rate is not simply the
average, or even weighted average of the two
contract rates!!!
   You must use the procedure on the previous two slides
to determine the effective interest rate. If you try to
take an arithmetic average of the two contract rates you
   This is because the mortgage payment equations are
non-linear due to the exponents in the formulas.
Rate Mortgage
Mechanics – Second Mortgages
   Now, what happens if the two mortgages are for
unequal terms?
   You simply have to deal with two sets of cash flows.
This means you will have to use your cash flow keys
Money keys, but once you become familiar with that
procedure its not too difficult.
that the second mortgage was only for 10 years.
Rate Mortgage
Mechanics – Second Mortgages
   Of course this does not
change the first                                 .055 / 12
Pmt1  80,000 *                             454.23
payment, but it does                        
1
1          
                360 
change the second                            1  .055 / 12 

payment:                                         .07 / 12
Pmt2  10,000 *                            116.11
         1         
1             120 
 1  .07 / 12 
Rate Mortgage
Mechanics – Second Mortgages
   Once again, we find the interest rate that sets the
present value of the payments equal to the combined
balances. Notice that we now have to deal with two
streams of cash flows:
                     1      
           1           240   
  454.23 *  1  r / 12   
          1                      (r / 12)     
1
 1  r / 12120                              
                            
90,000  (570.34) *                                                     
                  
120
(r / 12)                       r                
                             1                   
                               12 
                                  
                                  

                                  

Solving for r, yields :
r  5.57%
Rate Mortgage
Mechanics – Second Mortgages
   Note that the second annuity (the 454.23/month one)
starts in 121 months, so using the PVA formula tells us
its value at month 120, so we have to discount that
value back to time 0.
                     1      
           1           240   
  454.23 *  1  r / 12   
          1                      (r / 12)     
1
 1  r / 12120                              
                            
90,000  (570.34) *                                                     
                  
120
(r / 12)                       r                
                             1                   
                               12 
                                  
                                  

                                  

Solving for r, yields :
r  5.57%
Rate Mortgage
Mechanics – Second Mortgages
   Its actually pretty easy to do this on your calculator.
Simply use your cash flow keys:

CF0 = -90,000
CF1= 570.34
N1=120
CF2= 454.23
N2=240
   And the solve for IRR. Note that you IRR will be in
monthly terms, don’t forget to multiply by 12. You will
lose points if you forget to multiply by 12!
Rate Mortgage
Mechanics – Second Mortgages
   What if Bob prepaid both mortgages after 5 years?
Let’s go back to the assumption that Bob had a 30 year
second mortgage. Remember that our two mortgage
payments, then, are: Pmt1=454.23, Pmt2 = 66.53.
   We need the balance of each mortgage after 60 months:
         1          
1              300 

Balance  454.23 *  1  .055 / 12   73,968.48
1
.055 / 12

          1        
1             300 

Balance2  66.53 *  1  .07 / 12   9,413.16
.07 / 12
Rate Mortgage
Mechanics – Second Mortgages
   So now we can combine all of the cash flows and
determine the effective interest rate:
        1        
1            60 

90 ,000  520 .76 *  1  r / 12    83,381 .64
r / 12          1  r / 12 60
Solving for r yields :
r  5.67%

   Notice that you can use your time value of money keys for
this: n=60; PV=90,000; PMT=-520.76; FV=-83,381.64
and solve for r.
Rate Mortgage
Mechanics – Second Mortgages
   Finally, what if Bob had only paid off the second mortgage
after 5 years, but had held the first mortgage for the full 30
years?
   Again, all we really have to do is lay out the cash flows on a
month by month basis:
                     1     
          1
 1  r / 12 300  
                              
454.23 *                   
         1                                             r / 12     
1
 1  r / 1259 
90,000  520.76 *                  
520.76  9413.16  
         
                  

    r / 12           1  r / 1260
1  r / 12 60


                

solving for r yields :
5.569%
Rate Mortgage
Mechanics – Second Mortgages
   Again, its easiest to do this using your calculator’s cash
flow keys.
   The only real trick is to realize that you get 59 payments of
520.76, then one payment of (520.76+9,413.16) in month
60 when the second mortgage is paid off, followed by 300
payments of 454.23.To enter this do the following:
CF0=-90,000
CF1=520.76       N1=59
CF2=9,933.92 N2=1
CF3=454.23       N3=300