Mortgage Problem: series of ballooning house payments

Shorterman

New member
Joined
Apr 22, 2008
Messages
1
A couple take out a mortgage of $200,000. An annual interest rate of 7.5% compounded monthly is offered for a 30 year loan (360 payments). Two different payment schemes are offered. One consists of a series of ballooning house payments which will increase by exactly the same amount each month until the loan is paid off. The first month's payment is only $1250, which is the interest due during the first month. Determine how much the house payment is to increase each month. List the sequence of payments. list the sequence of balances over the life of the loan.


I couldn't figure out how to come up with equations to solve this problem. If someone could help me I would be very greatful, Thank You.
 
A couple take out a mortgage of $200,000. An annual interest rate of 7.5% compounded monthly is offered for a 30 year loan (360 payments). Two different payment schemes are offered. One consists of a series of ballooning house payments which will increase by exactly the same amount each month until the loan is paid off. The first month's payment is only $1250, which is the interest due during the first month. Determine how much the house payment is to increase each month. List the sequence of payments. list the sequence of balances over the life of the loan.

Not seeing anything in your description that indicates that the total paid over the 30 years is not equal by either method of payment, I will assume tat it is true.

The monthly loan payment derives from R = Pi/[(1+i)^n] where R = the monthly payment, P = the loan amount = 200,000, n = the number of interest bearing periods = 360 and i = the monthly decimal interst = I/100(12) = .0065.

Therefore, R = 200,000(.00625)[(1.00625)^360] = $1398.43 per month.
The total paid over the 30 years is 360(1398.43) = $503,434.44.

Under the alternative method of paying, $1250 is paid in the first month and a constant amount is added to the monthly payment each month for 360 months such that the final total paid is $503,434.44.

The $503.434.44 is therefore the sum of an arithmetic progression with firat term a = 1250, last term(payment) = L, the number of terms(payments) = 360.
The sum of such a progression is S = n(a + L)/2 or
503.434.44 = 360(1250 + L)/2 making L = $1546.86.

Therefore, the constant amount added to each monthly payment is (1546.86 - 1250)/360 = $0.8246.

Check: L = a + (n-1)d where a = 1250, n = 360 and d = .8246 or
L = 1250 + (359).8246 = 1546.86.
S = 360(1250 + 1546.86)/2 = $503,434.44.

If you meant something else by your ballooning payment statement, please repost with a clarification.
 
A monthly payment increase of 1.27 (last payment = 1705.93) will do it.

i = .075/12 ; x = increase; need to solve this for x:

(1250+0*x)(1+i)^359 + (1250+1*x)(1+i)^358 + (1250+2*x)^357 + .... + (1250+358*x)(1+i)^1 + (1250+359*x)(1+i)^0 =
200000(1+i)^360

Right, TK?

Don't think I've ever seen a sillier problem. I can see calculating say an annual amount that will reduce the term from
30 years to 25 years, but, but a buck 27...gimme a break!
 
Denis said:
(1250+0*x)(1+i)^359 + (1250+1*x)(1+i)^358 + (1250+2*x)^357 + .... + (1250+358*x)(1+i)^1 + (1250+359*x)(1+i)^0 =
200000(1+i)^360
Well, sure, but can our questioner add that up?

360 is a rather large number of increases. $1.27 does seem reasonable, although quite silly.

I just created an amortization schedule for a large medical bill. I get 10 months to pay it off. I created a simple schedule with exponentially increasing payments. I'm sure my creditor thinks I'm nuts. "Why did he pay THAT amount?" "Why did he increase it by THAT much?" Sometimes I'm bored. Maybe it is this sort of thinking that created the problem. :D
 
Denis said:
(1250+0*x)(1+i)^359 + (1250+1*x)(1+i)^358 + (1250+2*x)^357 + .... + (1250+358*x)(1+i)^1 + (1250+359*x)(1+i)^0 =
200000(1+i)^360
I wish I thought of this first. There's a slight typo though. The 3rd term from the left was missing the (1+i). It should have been
(1250+0*x)(1+i)^359 + (1250+1*x)(1+i)^358 + (1250+2*x)(1+i)^357 + .... + (1250+358*x)(1+i)^1 + (1250+359*x)(1+i)^0 =
200000(1+i)^360
tkhunny said:
Well, sure, but can our questioner add that up?
Sure he can.
It just needs a little tweaking.
Accordingly,

(1250+0*x)(1+i)^359 + (1250+1*x)(1+i)^358 + (1250+2*x)(1+i)^357 + .... + (1250+358*x)(1+i)^1 + (1250+359*x)(1+i)^0 =
200000(1+i)^360
?
[(1250)(1+i)^359 + (0*x)(1+i)^359] + [(1250)(1+i)^358 + (1*x)(1+i)^358] + [(1250)(1+i)^357 + (2*x)(1+i) ^357] + … + [(1250)(1+i)^1 + (358*x)(1+i)^1] + [(1250)(1+i)^0 + (359*x)(1+i)^0] =
200000(1+i)^360
?
[(1250)(1+i)^359 + (1250)(1+i)^358 + (1250)(1+i)^357 + … + (1250)(1+i)^1 + (1250)(1+i)^0] + [(0*x)(1+i)^359 + (1*x)(1+i)^358 + (2*x)(1+i) ^357 + … + (358*x)(1+i)^1 + (359*x)(1+i)^0] =
200000(1+i)^360
?
[(1250)(1+i)^359 + (1250)(1+i)^358 + (1250)(1+i)^357 + … + (1250)(1+i)^1 + (1250)] +
[(1*x)(1+i)^358 + (2*x)(1+i) ^357 + … + (358*x)(1+i)^1 + (359*x)] =
200000(1+i)^360

\(\displaystyle \Leftrightarrow\)
\(\displaystyle 1,250 \cdot\)\(\displaystyle s_{\left. {\overline {{360}}}\! \right| } _i\)\(\displaystyle + x \cdot\)\(\displaystyle \frac{{(1 + i) \cdot s_{\left. {\overline {{359}}}\! \right| } _i - 359}}{i}\)\(\displaystyle = 200,000(1 + i)^{360}\)

\(\displaystyle \Leftrightarrow\)

\(\displaystyle 1,250 \cdot \frac{{(1 + i)^{360} - 1}}{i}\)\(\displaystyle + x \cdot \frac{{(1 + i) \cdot \frac{{(1 + i)^{359} - 1}}{i} - 359}}{i}\)\(\displaystyle = 200,000(1 + i)^{360}\)

\(\displaystyle \Leftrightarrow\)

\(\displaystyle x \approx 1.265892746\)

A monthly payment increase of 1.27 (last payment = 1705.93 = 1,250 + 1.27*359) in theory should do it. Rounding all computations to the nearest cent, however, will result in a final outstanding liability (principal + interest) of approximately 1,057.02 on an amortization schedule.

tkhunny said:
I just created an amortization schedule for a large medical bill. I get 10 months to pay it off. I created a simple schedule with exponentially increasing payments. I'm sure my creditor thinks I'm nuts.

Is it something like the growing annuity mentioned in Wikipedia entry for the Time Value of money?
 
jonah said:
A monthly payment increase of 1.27 (last payment = 1705.93 = 1,250 + 1.27*359) in theory should do it. Rounding all computations to the nearest cent, however, will result in a final outstanding liability (principal + interest) of approximately 1,057.02 on an amortization schedule.
How did you arrive at that?
I get 1050.93 owing after 359th payment, hence -648.43 after 360th.

Since your 1.265892746 is "exact" (zero owing after 360th payment),
then 1.27 has to result in an overpayment...right?
 
Denis said:
How did you arrive at that?
I get 1050.93 owing after 359th payment, hence -648.43 after 360th.

Since your 1.265892746 is "exact" (zero owing after 360th payment),
then 1.27 has to result in an overpayment...right?

I made a mistake. I checked with my draft and schedule, and saw that the final outstanding liability was 1,057.63 and not 1,057.02 as posted. I can’t understand how it got mistyped. I guess Murphy’s law does happen all the time. I got 1,051.06 as the amount owing after the 359th payment. 1,051.06*(.075)/12 = 6.569125 or 6.57. Thus, the couple in Shorterman’s problem will need to pay only $1,057.63 (1,051.06 + 6.57) as their last payment. 1.27 did indeed resulted in an overpayment to the tune of -648.30, at least in my amortization schedule. With an amortization schedule that made no provision for rounding (where “rounding” was just visual), 1.265892746 resulted to zero owing after the 360th payment. I tried to attach the excel amortization schedule I constructed for Shorterman’s problem in my first post but I was informed that the forum does not allow an xls document. I’ll just leave a message in your pm box later as I’m hard pressed for time already.
 
Top