The second situation is an annuity. See if you can work with the information given here...
Hello Everyone, This is my first post here.
I have a hypothetical problem regarding the sale of a house. There are two scenarios involved and I would like to calculate the best outcome for the seller. OK, so here's the situation. I have a house for sale for $400,000. Scenario 1 is where a buyer approaches me and says he'll give me 10% more than the asking price but wants to pay it off over 10 years at zero interest. Payments are to made monthly. So what we have is 440,000 divided by 120. This gives a monthly payment of $3667.00 for ten years.
Scenario two is where I sell the house for the asking price of $400,000, put the funds into an interest bearing account that pays 5%. This interest is calculated daily and is paid monthly. Now after the first month in this account I will be drawing out $3667.00 and every month after that until the funds are depleted.
The question is, will the funds from scenario two run out first, and what would be the outcome after ten years of this arrangement? Can anyone give me spreadsheet details on this? I find this most interesting.
Thanks again for the help.
Payment formula: P = Ai / [1 - 1/(1 + i)^n]
P = Payment (3667)
A = Amount (400000)
n = number of payments (?)
i = monthly interest (.05/12)
You need to rewrite in terms of n:
n = LOG[P / (P - Ai)] / LOG(1 + i)
n = LOG[3667 / (3667 - 400000(.05/12)) / LOG(1 + .05/12) = 145.757...
So not quite 146 months.
Note: calculating interest daily and adding monthly means same results as monthly cpd.
(well, within peanuts!)
I know this is a very basic question, and I'm somewhat embarrassed to ask, but can you explain the steps involved in transposing that equation to make n the subject? I'd really like to understand the process. I'm OK with simple transposing. Well I guess you guys would consider this simple anyway!! Thanks again.
Well, do you understand this simpler example:
if a^p = b then p = LOG(b) / LOG(a) ?
4^p = 64
p = LOG(64) / LOG(4) = 4.158883.... / 1.386294.... = 3
So try to rewrite equation in that format; (something)^n = (somethingelse)
You can "see" the (1 + i)^n in the payment formula, right? Isolate it!