Starting in 5 years you want to be able to withdraw $6220 every six months for 7.5 years. You want to deposit a single amount immediately and then let it grow at a rate of 4.88% compounded quarterly. How much do you have deposit? 2. Try$62,116.62

3. Originally Posted by aponi23
Starting in 5 years you want to be able to withdraw $6220 every six months for 7.5 years. You want to deposit a single amount immediately and then let it grow at a rate of 4.88% compounded quarterly. How much do you have deposit? I am puzzled by this question. Are you asking "How can I look at this problem and immediately know the answer without doing any calculations"? I can't help you there! I certainly wouldn't be able to do that. But if you know what these words mean, then you should be able to do it by just cranking the numbers. Suppose you start with "A" dollars to begin with. In 3 months you earn .0122A in interest and so have 1.022A total. In another 3 months you again earn that interest but now the "Base amount" was 1.022A rather than A so you have (1.0122)^2A. But now you deduct$6220 so have (1.022)^2A- 6220. Now repeat the process using "(1.022)^2A- 6220" instead of A. At the end of another six months (one year) you will have (1.022^2((1.022)^2A- 6220)- 6220= (1.022)^4A- (1.022)^2(6220)- 6220.

Now we can do the same for the whole second year, replacing the "A" in that last formula with (1.022)^A- (1.022)^2(6220)- 6220: (1.022)^4(1.022)^4A- (1.022)^2(6220)- 6220)- 6220= (1.022)^8A- (1.022)^6(6220)- (1.022)^4(6220)- 6220.

And now we can do the next two years, using that formula for two year, replacing "A" in it with (1.022)^8A-(1.022)^6(6220)- (1.022)^4(6220)- 6220: At then end of 4 years you will have (1.022)^8((1.022)^8A- (1.022)^6(6220)- (1.022)^4(6220)- 6220)= (1.022)^16A- (1.022)^14(6.220)- (1.022)^12(6220)- (1.022)^8(6220)- (1.022)^4(6220)- 6220.

Finally, we have one year left to do and we can use our "end of one year" formula,
(1.022)^4A- (1.022)^2(6220)- 6220 with A equal to that last formula: at the end of the fifth year you will have (1.022)^20A- (1.022)^16(6220)- (1.022)^12(6.220)- (1.022)^8(6220)- (1.022)^4(6220)- 6220.

Now, one thing you did NOT say in the statement of the problem- what do you want to happen at the end of the five years? Do you want to have withdrawn all the money? If so, set that last formula equal to 0 and solve for A.

4. Jonah's answer is correct, assuming 1st withdrawal is 5 years after the initial deposit.

i = (1 + .0488/4)^2 - 1 = .02454884...

u = 6220[1 - 1/(1 + i)^14] / i = 72945.22893751... (value after 5 years, including 1st withdrawal

p = (u + 6220) / (1 + i)^10 = 62116.61949767... (amount of opening deposit)

5. Originally Posted by HallsofIvy
I am puzzled by this question. Are you asking "How can I look at this problem and immediately know the answer without doing any calculations"? I can't help you there! I certainly wouldn't be able to do that. But if you know what these words mean, then you should be able to do it by just cranking the numbers.

Suppose you start with "A" dollars to begin with. In 3 months you earn .0122A in interest and so have 1.022A total. In another 3 months you again earn that interest but now the "Base amount" was 1.022A rather than A so you have (1.0122)^2A. But now you deduct $6220 so have (1.022)^2A- 6220. Now repeat the process using "(1.022)^2A- 6220" instead of A. At the end of another six months (one year) you will have (1.022^2((1.022)^2A- 6220)- 6220= (1.022)^4A- (1.022)^2(6220)- 6220. Now we can do the same for the whole second year, replacing the "A" in that last formula with (1.022)^A- (1.022)^2(6220)- 6220: (1.022)^4(1.022)^4A- (1.022)^2(6220)- 6220)- 6220= (1.022)^8A- (1.022)^6(6220)- (1.022)^4(6220)- 6220. And now we can do the next two years, using that formula for two year, replacing "A" in it with (1.022)^8A-(1.022)^6(6220)- (1.022)^4(6220)- 6220: At then end of 4 years you will have (1.022)^8((1.022)^8A- (1.022)^6(6220)- (1.022)^4(6220)- 6220)= (1.022)^16A- (1.022)^14(6.220)- (1.022)^12(6220)- (1.022)^8(6220)- (1.022)^4(6220)- 6220. Finally, we have one year left to do and we can use our "end of one year" formula, (1.022)^4A- (1.022)^2(6220)- 6220 with A equal to that last formula: at the end of the fifth year you will have (1.022)^20A- (1.022)^16(6220)- (1.022)^12(6.220)- (1.022)^8(6220)- (1.022)^4(6220)- 6220. Now, one thing you did NOT say in the statement of the problem- what do you want to happen at the end of the five years? Do you want to have withdrawn all the money? If so, set that last formula equal to 0 and solve for A. From what I could gather, Sir Hallsoflvy, it seems to me that you're trying to write an equation of value with the end of 5 years as the comparison date. With A as the initial deposit, j=.0488/4=.0122 ,b=1+j and R=6220, the final form of your equation of value should be something like: Ab^20-Rb^18-Rb^16-Rb^14-Rb^12-Rb^10-Rb^8-Rb^6-Rb^4-Rb^2-R=0 This then would amount to A=$54564.97.
Equivalently, we could just as well write 6220*(1-(1.02454884)^(-10))/.02454884
This is of course valid only if the first withdrawal was to be made in 6 months after the initial deposit.
The problem statement however is almost clear in stating that the first withdrawal will be in 5 years after the initial deposit and that such withdrawals will go on for 7.5 years. This is the usual interpretation of deferred annuities.
Your post also contained a number of inconsistencies, omissions and typos. Heavily intoxicated as I am at the moment, I can't help but feel that you are a kindred spirit and that you were wearing beer goggles at the time when you posted or you were without your your reading glasses at the time. Consider the following:
Originally Posted by HallsofIvy
In 3 months you earn .0122A in interest and so have 1.022A total
Typo/inconsistency: 1.022A should be 1.0122A
Originally Posted by HallsofIvy
In another 3 months you again earn that interest but now the "Base amount" was 1.022A rather than A so you have (1.0122)^2A.
Same typo/inconsistency. The incorrect factor 1.022 was then used all throughout.

"Now we can do the same for the whole second year, replacing the "A" in that last formula with (1.022)^A- (1.022)^2(6220)- 6220: (1.022)^4(1.022)^4A- (1.022)^2(6220)- 6220)- 6220= (1.022)^8A- (1.022)^6(6220)- (1.022)^4(6220)- 6220."

Incorrect factor and the withdrawal whose exponent is 2 is omitted.

"And now we can do the next two years, using that formula for two year, replacing "A" in it with (1.022)^8A-(1.022)^6(6220)- (1.022)^4(6220)- 6220: At then end of 4 years you will have (1.022)^8((1.022)^8A- (1.022)^6(6220)- (1.022)^4(6220)- 6220)= (1.022)^16A- (1.022)^14(6.220)- (1.022)^12(6220)- (1.022)^8(6220)- (1.022)^4(6220)- 6220."

Incorrect factor, incorrect skipping of the 3rd year, withdrawals whose exponents are 10, 6, & 2 are omitted/missing. Withdrawal whose exponent is 14 is only $6.220 Finally, we have one year left to do and we can use our "end of one year" formula, (1.022)^4A- (1.022)^2(6220)- 6220 with A equal to that last formula: at the end of the fifth year you will have (1.022)^20A- (1.022)^16(6220)- (1.022)^12(6.220)- (1.022)^8(6220)- (1.022)^4(6220)- 6220. Incorrect factor, withdrawals whose exponents are 18, 14, 10, 6, 2 are omitted / missing. Withdrawal whose exponent is 12 is only$6.220

Copying and pasting that final formula (after setting it equal to 0) into Wolfram Alpha results in a valuation of \$18913.40

But then again, this drunken analysis could also be incorrect and I could have made worst typos.
Cheers.