Question

I just need someone to tell me how to solve this.
Chase and Julia are purchasing a townhouse and finance $136,400 with a 20-year 4/1 ARM at 5.15% with a 3/11 cap structure. What will their payments be at the beginning of the fifth year assuming they are charged the maximum interest rate for that year?

Answer 1

must be able to read the notation :)

Answer 2

a strategy is: find the payment needed for a 20 year fixed loan; balance it out after 4 years (thats the 4/1 part, 4 years at the initial rate, adjusted at 1year intervals afterwards).

Answer 3

the remaining balance is termed out for the remaining part of the 20 years: at an extra 3 points added to the interest (added each adjusted year but not to exceed a total of 11 points added overall.)

Answer 4

so, any idea how to find the first set of payments? or do you have a formula for finding the balance after 4 years?

Answer 5

Would you use the balance remaining of ARM formula?

Answer 6

i use my own formulas so i got no idea what your material has :/

i find the payment of the 20 year fixed at the intial rate:
\[Bk^{20*12}\frac{1-k}{1-k^{20*12}}=P\]
for the initial balance, and the compounding factor k=1+r/12

Answer 7

after words, i use that for 4 years of inital payments: to balance it out\[A=Bk^{4*12}-P\frac{1-k^{4*12}}{1-k}\]
this amount is then used to assess the payments at r+3, for the remaining 20-4 years

Answer 8

\[A(k_1)^{16*12}\frac{1-(k_1)}{1-(k_1)^{16*12}}=P_1\]
for the compounding k=1+(r+3)/12 for the adjustment

Answer 9

I'm sorry, i'm trying to understand that formula you use... My brain doesn't seem to want to plug in the numbers. :/

Answer 10

P = 136400k^(240)(1-k)/(1-k^(240)), k=1+.0515/12

so out initial payments are about: 911.52
http://www.wolframalpha.com/input/?i=136400k%5E%28240%29%281-k%29%2F%281-k%5E%28240%29%29%2C+k%3D1%2B.0515%2F12

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balancing that out after 4 years gets us:

A = 136400k^(48)-911.52*(1-k^(48))/(1-k), k=1+.0515/12

A = 119 058
http://www.wolframalpha.com/input/?i=136400k%5E%2848%29-911.52*%281-k%5E%2848%29%29%2F%281-k%29%2C+k%3D1%2B.0515%2F12

soo, readjusting for 16 years left at 8.15 gets us

P = 119058k^(16*12)(1-k)/(1-k^(16*12)), k=1+.0815/12

so im going with something about 1111.70 starting at the beginning of the 5th year.
http://www.wolframalpha.com/input/?i=119058k%5E%2816*12%29%281-k%29%2F%281-k%5E%2816*12%29%29%2C+k%3D1%2B.0815%2F12

youll of course want to verify with whatever formulas your material gives you to play with.

Answer 11

OH! I see it now! Thank you so much! I'll try with my formula and see if the answer is the same.

Answer 12

sometimes my formulas are too precise and can be off from the expected results so thats why i say its best to use the materials formulas ... which i can never remember lol

Answer 13

Yea it's hard to remember all of them. Your formula looks simpler lol Thank you so much for you help! With my formula i actually got the same answer.

Answer 14

same answers are comforting lol ...