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  #21  
Old 05-20-2019, 11:44 AM
jubair07 jubair07 is offline
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Another problem: You want to accumulate $1,000,000. You intend to do this by making deposits of $5,000 into an investment account at the end of each month, until
your account balance equals $1,000,000. The account earns 8% convertible
quarterly. Determine the number of monthly deposits you will need to make
to achieve your goal.

i(4)= 8%
j= i(12)/12
1+j = 1.00662271 or j = .662271%

FV=1,000,000
PMT=-5,000
I/Y=.662271%
PV=0

N= 127.79. But answer on the book says 127. But you can't reach 1,000,000 until 128th deposit is made. Where did I go wrong?
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  #22  
Old 05-20-2019, 12:22 PM
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Maybe the wording is a little ambiguous, but their answer is reasonable, probably best.

This is how they are thinking about it. At the end of every month, you plan to deposit 5000 into the account unless you already have 1 million.

At the end of 127 months, you do not have 1 million, as your calculations shows. At the end of month 128, you’ve earned some interest. Your balance is over 1 million, so you don´t make the 128th deposit.

Alternatively and equivalently, you assume that you are checking your balance on line every day. At some point during month 28, you see 1 million, so you just end the process, before making the 128th payment.

The only way to justify 128 is to say that you will make the 128th payment if you have not reached 1 million the moment after making the 127th.
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Old 05-20-2019, 05:45 PM
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Quote:
Originally Posted by Gandalf View Post
Maybe the wording is a little ambiguous, but their answer is reasonable, probably best.

This is how they are thinking about it. At the end of every month, you plan to deposit 5000 into the account unless you already have 1 million.

At the end of 127 months, you do not have 1 million, as your calculations shows. At the end of month 128, you’ve earned some interest. Your balance is over 1 million, so you don´t make the 128th deposit.

Alternatively and equivalently, you assume that you are checking your balance on line every day. At some point during month 28, you see 1 million, so you just end the process, before making the 128th payment.

The only way to justify 128 is to say that you will make the 128th payment if you have not reached 1 million the moment after making the 127th.
Thanks mate! I thought the same too even though it is hard to make sense of the questions sometimes.
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Old 05-21-2019, 07:23 AM
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There is $40,000 in a fund which is accumulating at 4% per annum convertible
continuously. If money is withdrawn continuously at the rate of $2,400 per
annum, how long will the fund last?

A: e^δ = 1 + i = 1 + 0.04 = 1.04
δ = ln(1.04)

40,000 e^nδ = 2400 (e^nδ-1)/δ
Since, e^nδ = 1.04^n

giving n = 27.0364
Answer on the book is 27.0367

Just inquiring if my method is correct and its just rounding error by any chance on the book?
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Old 05-21-2019, 08:12 AM
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Yes, rounding
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Old 05-21-2019, 02:26 PM
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Hi, any idea how to approach this problem?

I integrated a|n from 0 to n and got : n^2/2 - 4n

Is integration of a|n = s|n? Then how do I approach this problem?
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Old 05-21-2019, 02:42 PM
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Do they tell you what the answer is supposed to be?

You are misinterpreting the given information. N is an unknown constant. The first expression is not a formula true for all n (a 1-year annuity could not have PV 1-4=-3). So that expression, with the info that delta = 10%, and the standard formula for the PV of an annuity, lets you solve for n. Then integrate the standard formula for an annuity from 0 to that known value of n.

Maybe (probably) that second integral turns out to be the formula for a continuously increasing annuity.
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Old 05-21-2019, 02:45 PM
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Quote:
Originally Posted by Gandalf View Post
Do they tell you what the answer is supposed to be?

You are misinterpreting the given information. N is an unknown constant. The first expression is not a formula true for all n (a 1-year annuity could not have PV 1-4=-3). So that expression, with the info that delta = 10%, and the standard formula for the PV of an annuity, lets you solve for n. Then integrate the standard formula for an annuity from 0 to that known value of n.

Maybe (probably) that second integral turns out to be the formula for a continuously increasing annuity.
Answer is given as 40.

If I set up an= (1-e-^01.n)/(δ) = n-4. That equation is almost unsolvable.
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Old 05-21-2019, 02:52 PM
jubair07 jubair07 is offline
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Find an expression for t; 0 < t < 1; such that 1 paid at time t is equivalent
to 1 paid continuously between 0 and 1.

I assumed sn = sn

So, (e^δ-1)/δ = (1+i)^t

But then I don't the answer on the book.

Answer: 1 − 1/δ*ln(i/δ)
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Old 05-21-2019, 04:02 PM
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Quote:
Originally Posted by jubair07 View Post
Answer is given as 40.

If I set up an= (1-e-^01.n)/(δ) = n-4. That equation is almost unsolvable.
Not so unsolvable. You young-uns. there’s always trial and error. Suppose you were solving the equivalent equation (1-e-^01.n)/(δ) - n= -4
With the TI-30X multiview, it would be pretty easy to put in the left side, and try values of n until you got close to -4. You would conclude n has to be close to 10.5. Then if you did the second integral with an upper limit of 10.5, you would be close enough to 40 to choose it in a multiple choice test.

OTOH, there´s a better way to get it exactly with calculations you could do without a calculator. Realize that the third item (the integral from 0 to n you’re supposed to calculate) is the PV of an n-year continuously decreasing annuity. (In my first post about the question, I hypothesized increasing, but further thought says decreasing.) If you know that formula and plug the first two givens into it, you’ll see it works out very simply.

BTW, what’s the source of the question? I feel sure it’s been discussed on the AO before. I don´t see anything in your post that would be good to search for, but maybe the source would help in a search.
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