brucegoose
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- Jan 14, 2013
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$95000 yielding 6% annually withdrawing $1000 monthly when will funds be depleted?
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bruce
- Thread starter brucegoose
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Does your book give a formula for the present value of an annuity?$95000 yielding 6% annually withdrawing $1000 monthly when will funds be depleted?
What is the relationship between your initial balance and an annuity?
what are the values to put into the formula for the annuity?
brucegoose
New member
- Joined
- Jan 14, 2013
- Messages
- 2
there is no annuity
Does your book give a formula for the present value of an annuity?
What is the relationship between your initial balance and an annuity?
what are the values to put into the formula for the annuity?[/QUOTE
there is no annuity
OK Please tell me what course you are taking and what exactly the problem asks.Does your book give a formula for the present value of an annuity?
What is the relationship between your initial balance and an annuity?
what are the values to put into the formula for the annuity?[/QUOTE
Can you use excel or some other spreadsheet software?
I'll be gone for an hour or two.
Hello, brucegoose!
Okay, it's not an annuity. .What do you call it?
. . Reverse Amortization? .Sinking Fund?
Formula: .\(\displaystyle A \;=\;P\,\dfrac{i(1+1)^n}{(1+i)^n-1} \)
. . where: .\(\displaystyle \begin{Bmatrix}A &=& \text{periodic withdrawl} \\ P &=& \text{principal invested} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}\)
We have: .\(\displaystyle A = \$1000,\;P = \$95,\!000,\;i = \frac{6%}{12} = 0.005 \)
Substitute: .\(\displaystyle 1000 \;=\;95,\!000\,\dfrac{(0.005)(1.005)^n}{(1.005)^n - 1} \quad\Rightarrow\quad 1 \;=\;\dfrac{(95)(0.005)(1.005)^n}{(1.005)^n-1}\)
. . \(\displaystyle (1.005)^n - 1 \;=\;0.475(1.005)^n \quad\Rightarrow\quad (1.005)^n - 0.475(1.005)^n \;=\;1\)
. . \(\displaystyle (1 - 0.475)(1.005)^n \;=\;1 \quad\Rightarrow\quad 0.525(1.005)^n \;=\;1 \quad\Rightarrow\quad (1.005)^n \:=\:\dfrac{1}{0.525} \)
Take logs: .\(\displaystyle \ln(1.005)^n \:=\:\ln\left(\frac{1}{0.525}\right) \quad\Rightarrow\quad n\ln(1.005) \:=\:-\ln(0.525) \)
\(\displaystyle \text{Therefore: }\:n \:=\:-\frac{\ln(0.525)}{\ln(1.005)} \:=\:129.193313\)
\(\displaystyle \text{The funds will be depleted in about }129\text{ months }= 10\!\frac{3}{4}\,\text{years.}\)
$95,000 yielding 6% annually withdrawing $1000 monthly.
When will funds be depleted?
Okay, it's not an annuity. .What do you call it?
. . Reverse Amortization? .Sinking Fund?
Formula: .\(\displaystyle A \;=\;P\,\dfrac{i(1+1)^n}{(1+i)^n-1} \)
. . where: .\(\displaystyle \begin{Bmatrix}A &=& \text{periodic withdrawl} \\ P &=& \text{principal invested} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}\)
We have: .\(\displaystyle A = \$1000,\;P = \$95,\!000,\;i = \frac{6%}{12} = 0.005 \)
Substitute: .\(\displaystyle 1000 \;=\;95,\!000\,\dfrac{(0.005)(1.005)^n}{(1.005)^n - 1} \quad\Rightarrow\quad 1 \;=\;\dfrac{(95)(0.005)(1.005)^n}{(1.005)^n-1}\)
. . \(\displaystyle (1.005)^n - 1 \;=\;0.475(1.005)^n \quad\Rightarrow\quad (1.005)^n - 0.475(1.005)^n \;=\;1\)
. . \(\displaystyle (1 - 0.475)(1.005)^n \;=\;1 \quad\Rightarrow\quad 0.525(1.005)^n \;=\;1 \quad\Rightarrow\quad (1.005)^n \:=\:\dfrac{1}{0.525} \)
Take logs: .\(\displaystyle \ln(1.005)^n \:=\:\ln\left(\frac{1}{0.525}\right) \quad\Rightarrow\quad n\ln(1.005) \:=\:-\ln(0.525) \)
\(\displaystyle \text{Therefore: }\:n \:=\:-\frac{\ln(0.525)}{\ln(1.005)} \:=\:129.193313\)
\(\displaystyle \text{The funds will be depleted in about }129\text{ months }= 10\!\frac{3}{4}\,\text{years.}\)
In the US, "yield" frequently means the rate of simple interest that is equivalent to the contractual rate compounded over a year assuming no withdrawals. The idea is to give consumers a number that is roughly comparable regardless of compounding frequency, 360/365 day year issues, etc.You stated "yielding 6% annually"; this usually means $1 = $1.06
after 1 year (the 6% does not compound monthly).
So the "i" has to be adjusted accordingly:
(1 + i)^12 = 1.06
i = 1.06^(1/12) - 1 = ~.00486755...
Follow Soroban's step: you should end up with ~127.822
Furthermore, the problem as stated is ambiguous on how to calculate interest on a declining balance. It would help if the original poster gave the exact language of the problem.