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UNIT 11  :  MATHEMATICS OF INVESTMENT

LESSON 6: MORTGAGES HOMEWORK QUESTIONS

Homework Questions:  (Solutions below)

1  a)  Greg and Beth are house hunting.  They have been pre-approved for a maximum mortgage of \$180 000.  If they use the full amount (\$180 000), and will repay it in monthly instalments over 35 years, find the monthly payment if the interest rate is 7.2%/a, compounded semi-annually.

b)  Determine the total interest paid over the 35 year period.

2.  In the question above, determine the outstanding balance at the end of 10 years.

3.  The Sturrocks purchased a cottage on Little Straggle Lake for \$140 000.  They paid 25% down, financing the rest with a mortgage over 20 years with interest at 7.8%/a, compounded semi-annually.

a)  Determine the monthly payment.

b)  Determine the total amount of interest paid.

4.  In question 3 above, suppose the initial term of the mortgage was 5 years at which time the terms are renegotiated.

a)  Find the amount still owing at this time.

b)  Determine the new monthly payment to maintain the mortgage if the rate changes to 8.4%/a, compounded semi-annually.

5  a)  Nora has just purchased a new home. For \$220 000.  If she will pay \$100 000 down and mortgage the rest it in weekly instalments over 20 years, find the weekly payment if the interest rate is 7.2%/a, compounded semi-annually.

b)  If she had chosen monthly payments at the same rate, what would the payment be?

1    a)  Greg and Beth are house hunting.  They have been pre-approved for a maximum mortgage of \$180 000.  If they use the full amount (\$180 000), and will repay it in monthly instalments over 35 years, find the monthly payment if the interest rate is 7.2%/a, compounded semi-annually.

b)      Determine the total interest paid over the 35 year period.

Solution:

Here the payment interval (monthly ) is different than the interest period (semi-annual).  This is a general annuity.

We must match the interest period to the payment interval.

Ie. We must find the monthly rate that is equivalent to 7.2%/a, compounded semi-annually.

Step 1:  Using the formula  A = P(1 + i)n, find the value of \$1 invested at 7.2%/a, compounded semi-annually after 1 year.

Step 2:  Let the equivalent monthly rate be i %.  (Note the equivalent yearly rate would be 12i %.)

Now find the value of \$1 invested at i % per month after 1 year.

A = 1(1 + i)12                           ** n = 12, the number of times interest is compounded per year.

Step 3:  These two amounts must be equal.  Hence

The money in question is borrowed now – at point 0 on the time line.  Hence this is a PV general annuity question

Interest Period   0          1         2          3                                                                                                                                        418   419  420

Payment                          R         R        R                                                                                                                                          R      R      R

R(1.005911931)-1

R(1.005911931)-2

.

.

R(1.005911931)-418

R(1.005911931)-419

R(1.005911931)-420

This forms the following geometric series:

R(1.005911931)-420 + R(1.005911931)-419 + . . . + R(1.005911931)-2 + R(1.005911931)-1

b)  Determine the total interest paid over the 25 year period.

Total amount repaid = 1013.85 x 300 = \$304 095.00

Mortgage amount                               = \$150 000

Interest paid = \$304 095 - !50 000    =\$154 095

Hence The total interest paid over 25 years is \$154 095.

2.  In the question above, determine the outstanding balance at the end of 10 years.

Solution:

Step 1: Using A = P(1 + i)n, find the accumulated amount of \$150 000 after 10 years at 0.5911931% per month

This is the amount owing after 10 years if nothing was repaid.

Step 2: Find the amount of the annuity of monthly payments of \$1161.87 after 10 years at 0.5911931% per month.

Step 3:  Subtract above 2 amounts to determine the amount still owing after 10 years.

\$304 289.085 - \$202 149.2461 = \$102 139.84

Hence \$102 139.84 is still owing after 10 years.

The chart below shows the amortization schedule through months 117 to 122.  Note after 120 months or 10 years the amount \$102 139.84 is still owing.

Month         Payment         Interest      Principal      Balance

 117 1161.87 616.85 545.02 103,794.32 118 1161.87 613.62 548.25 103,246.07 119 1161.87 610.38 551.49 102,694.59 120 1161.87 607.12 554.75 102,139.84 121 1161.87 603.84 558.03 101,581.81 122 1161.87 600.54 561.33 101,020.49

Note:

Interest in month 118 calculated as follows:   I = \$103 794.32 x 0.005911931 = \$613.62

Principal                                                       P = \$1161.87 - \$613.62 = \$548.25

Balance (still owing)                                      B = \$103 794.32 - \$548.25 = \$103 246.07

3.  The Sturrocks purchased a cottage on Little Straggle Lake for \$140 000.  They paid 25% down, financing the rest with a mortgage over 20 years with interest at 7.8%/a, compounded semi-annually.

a)  Determine the monthly payment.

Solution:

Here the payment interval( monthly ) is different than the interest period ( semi-annual).  This is a general annuity.

We must match the interest period to the payment interval.

Ie. We must find the monthly rate that is equivalent to 7.8%/a, compounded semi-annually.

Step 1:  Using the formula  A = P(1 + i)n, find the value of \$1 invested at 7.8%/a, compounded semi-annually after 1 year.

Step 2:  Let the equivalent monthly rate be i %.  (Note the equivalent yearly rate would be 12i %.)

Now find the value of \$1 invested at i % per month after 1 year.

A = 1(1 + i)12                           ** n = 12, the number of times interest is compounded per year.

Step 3:  These two amounts must be equal.  Hence

Down Payment = 0.25 x \$140 000 = \$35 000

Mortgage amount = \$140 000 - \$35 000 = \$105 000.

The money in question is borrowed now – at point 0 on the time line.  Hence this is a PV general annuity question

Interest Period   0          1         2          3                                                                                                                                        238   239  240

Payment                          R         R        R                                                                                                                                          R      R      R

R(1.006395825)-1

R(1.006395825)-2

.

.

R(1.006395825)-238

R(1.006395825)-239

R(1.006395825)-240

This forms the following geometric series:

R(1.006395825)-240 + R(1.006395825)-239 + . . . + R(1.006395825)-2 + R(1.006395825)-1

b)  Determine the total interest paid over the 25 year period.

Total amount repaid = 857.14 x 240 = \$205713.60

Mortgage amount                               = \$105 000

Interest paid = \$205 713.60 - !05 000    = \$100 713.60

Hence the total interest paid over 20 years is \$100 713.60.

4.  In question 3 above, suppose the initial term of the mortgage was 5 years at which time the terms are renegotiated.

a)      Find the amount still owing at this time.

Solution:

Step 1: Using A = P(1 + i)n, find the accumulated amount of \$135 000 after 5 years at 0.6395825% per month

This is the amount owing after 5 years if nothing was repaid.

Step 2: Find the amount of the annuity of monthly payments of \$857.14 after 5 years at 0.6395825% per month.

Step 3:  Subtract above 2 amounts to determine the amount still owing after 10 years.

\$153 928.45 - \$62 449.26 = \$91 479.19

Hence \$91 479.19 is still owing after 5 years.

The chart below shows the amortization schedule through months 57 to 62.  Note after 60 months or 5 years the amount \$91 479.18 is still owing.

(difference of \$0.01 due to rounding in above solution).

Month         Payment         Interest      Principal      Balance

 57 857.14 591.94 265.2 92,285.02 58 857.14 590.24 266.9 92,018.12 59 857.14 588.53 268.61 91,749.51 60 857.14 586.81 270.33 91,479.18 61 857.14 585.08 272.06 91,207.13 62 857.14 583.34 273.8 90,933.33

Note:

Interest in month 60 calculated as follows:   I = \$91 749.51 x 0.006395825 = \$586.81

Principal                                                       P = \$857.14 - \$586.81 = \$270.33

Balance (still owing)                                      B = \$91 749.51 - \$270.33 = \$91 479.18

b)  Determine the new monthly payment to maintain the mortgage if the rate changes to 8.4%/a, compounded semi-annually.

Solution:

We must find the monthly rate that is equivalent to 8.4%/a, compounded semi-annually.

Step 1:  Using the formula  A = P(1 + i)n, find the value of \$1 invested at 8.4%/a, compounded semi-annually after 1 year.

Step 2:  Let the equivalent monthly rate be i %.  (Note the equivalent yearly rate would be 12i %.)

Now find the value of \$1 invested at i % per month after 1 year.

A = 1(1 + i)12                           ** n = 12, the number of times interest is compounded per year.

Step 3:  These two amounts must be equal.  Hence

The money in question is borrowed now – at point 0 on the time line.  Hence this is a PV general annuity with 15 years still to go.

Interest Period   0          1         2          3                                                                                                                                        178   179  180

Payment                          R         R        R                                                                                                                                          R      R      R

R(1.00671944)-1

R(1.00671944)-2

.

.

R(1.00671944)-178

R(1.00671944)-179

R(1.00671944)-180

This forms the following geometric series:

R(1.005911931)-180 + R(1.005911931)-1799 + . . . + R(1.005911931)-2 + R(1.005911931)-1

5  a)  Nora has just purchased a new home. For \$220 000.  If she will pay \$100 000 down and mortgage the rest it in weekly instalments over 20 years, find the weekly payment if the interest rate is 7.2%/a, compounded semi-annually.

Solution:

We must find the weekly  rate that is equivalent to 7.2%/a, compounded semi-annually.

Step 1:  Using the formula  A = P(1 + i)n, find the value of \$1 invested at 7.2%/a, compounded semi-annually after 1 year.

Step 2:  Let the equivalent weekly rate be i %.  (Note the equivalent yearly rate would be 52i %.)

Now find the value of \$1 invested at i % per week after 1 year.

A = 1(1 + i)52                           ** n = 52, the number of times interest is compounded per year.

Step 3:  These two amounts must be equal.  Hence

The money in question is borrowed now – at point 0 on the time line.  Hence this is a PV general annuity with a term of 20 years.

Interest Period   0          1         2          3                                                                                                                                       1038  1039  1040      weeks

Payment                          R         R        R                                                                                                                                          R      R      R

R(1.0013612)-1

R(1.0013612)-2

.

.

R(1.0013612)-1038

R(1.0013612)-1039

R(1.0013612)-1040

This forms the following geometric series:

R(1.005911931)-1040 + R(1.005911931)-1039 + . . . + R(1.005911931)-2 + R(1.005911931)-1

b)  If she had chosen monthly payments at the same rate, what would the payment be?

Solution:

We must find the monthly  rate that is equivalent to 7.2%/a, compounded semi-annually.

Step 1:  Using the formula  A = P(1 + i)n, find the value of \$1 invested at 7.2%/a, compounded semi-annually after 1 year.

Step 2:  Let the equivalent monthly rate be i %.  (Note the equivalent yearly rate would be 12i %.)

Now find the value of \$1 invested at i % per month after 1 year.

A = 1(1 + i)12                           ** n = 12, the number of times interest is compounded per year.

Step 3:  These two amounts must be equal.  Hence

The money in question is borrowed now – at point 0 on the time line.  Hence this is a PV general annuity with a term of 20 years.

Interest Period   0          1         2          3                                                                                                                                       238   239   240      months

Payment                          R         R        R                                                                                                                                          R      R      R

R(1.005911531)-1

R(1.005911531)-2

.

.

R(1.005911531)-238

R(1.005911531)-239

R(1.005911531)-240

This forms the following geometric series:

R(1.005911931)-240 + R(1.005911931)-239 + . . . + R(1.005911931)-2 + R(1.005911931)-1