My Math Forum Annuity, calculate N

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 March 3rd, 2015, 02:17 PM #1 Newbie   Joined: Oct 2014 From: Canada Posts: 4 Thanks: 0 Math Focus: none Annuity, calculate N Leased equipment valued at 23000, the terms of the lease require 1800/month. If the first payment is due in nine months after the lease was signed and interest is 11% compounded monthly, what is the term? The answer is 13.61927598. PY = 12 and according to the book is one year and 3 months. I understand the 1 year but why 3 months? I don’t see that. Is there a way to calculate these kind of months into years conversions, I mean is there a tool online available somewhere? Last edited by skipjack; March 3rd, 2015 at 11:17 PM.
March 3rd, 2015, 08:33 PM   #2
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DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.
Quote:
 Originally Posted by Surfboard Leased equipment valued at 23000, the terms of the lease require 1800/month. If the first payment is due in nine months after the lease was signed and interest is 11% compounded monthly, what is the term? The answer is 13.61927598. PY = 12 and according to the book is one year and 3 months. I understand the 1 year but why 3 months? I don’t see that. Is there a way to calculate these kind of months into years conversions, I mean is there a tool online available somewhere?
I also get one year and 3 months, but your quoted answer seems way off.
If my interpretation of this problem is correct, then we should have
$$23,000\left( {1 + {\textstyle{{0.11} \over {12}}}} \right)^8 = 1,800 \times {{1 - \left( {1 + {\textstyle{{0.11} \over {12}}}} \right){}^{ - n}} \over {{\textstyle{{0.11} \over {12}}}}} \Leftrightarrow n = {\rm{14}}{\rm{.75902091304825552506632274824692250 5497485916298569}}$$

Last edited by skipjack; March 3rd, 2015 at 11:18 PM.

 March 5th, 2015, 08:49 AM #3 Newbie   Joined: Oct 2014 From: Canada Posts: 4 Thanks: 0 Math Focus: none Dang, I need a serious answer I have an exam next week and ran into a similar one: Ali deposits 450 dollar at the beginning of every 3 months (put calculator in BGN?). He wants to build up his account so that he can withdraw $1000 every 3 months starting 3 months after the last deposit (put calculator to END?) If he wants to make withdrawals for 15 years and interest is 10% compounded quarterly, find N. I came to 40.481038030 = 10.12025951, or 10 years and... months? However the correct answer should be: 10 years & 3 months. 1. Do I change my calculation for the second part to END or keep it at BGN? I think END is correct. 2. The conversion of N to years & months, what's up with that? Last edited by Surfboard; March 5th, 2015 at 09:03 AM.  March 5th, 2015, 12:55 PM #4 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs We can use the formula:$\displaystyle P=\frac{Ai}{1-(1+i)^{-n}}$where P = monthly payment, A = amount borrowed, i = monthly interest rate, and n = the number of payments. Now, if you wish to derive the given formula, then consider the recursion:$\displaystyle D_{n}=(1+i)D_{n-1}-P\tag{1}\displaystyle D_{n+1}=(1+i)D_{n}-P\tag{2}$Subtracting (1) from (2) yields the homogeneous recursion:$\displaystyle D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}$whose associated auxiliary equation is:$\displaystyle r^2-(2+i)r+(1+i)=0\displaystyle \left(r-(1+i)\right)\left(r-1\right)=0$Thus, the closed-form for our recursion is:$\displaystyle D_n=k_1(1+i)^n+k_2$Using initial values, we may determine the coefficients$\displaystyle k_i$:$\displaystyle D_0=k_1+k_2=A\displaystyle D_1=k_1(1+i)+k_2=(1+i)A-P$Solving this system, we find:$\displaystyle k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}$and so we have:$\displaystyle D_n=\left(\frac{Ai-P}{i}\right)(1+i)^n+\left(\frac{P}{i}\right)=\frac {(Ai-P)(1+i)^n+P}{i}$Now, equating this to zero, we can solve for P:$\displaystyle \frac{(Ai-P)(1+i)^n+P}{i}=0\displaystyle (Ai-P)(1+i)^n+P=0\displaystyle (P-Ai)(1+i)^n=P\displaystyle P\left((1+i)^n-1\right)=Ai(1+i)^n\displaystyle P=\frac{Ai(1+i)^n}{(1+i)^n-1}\displaystyle P=\frac{Ai}{1-(1+i)^{-n}}$Now, let's solve for$A$and$n$:$\displaystyle A=\frac{P}{i}\left(1-(1+i)^{-n}\right)\tag{3}\displaystyle n=\frac{\ln\left(\dfrac{P}{P-Ai}\right)}{\ln(1+i)}\tag{4}$We can now use (3) to determine how much he needs to accumulate to be able to make the specified withdrawals, and then we can use (4) to determine how many of the given deposits he must make...can you proceed? March 5th, 2015, 10:49 PM #5 Senior Member Joined: May 2008 Posts: 299 Thanks: 81 DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views. Quote:  Originally Posted by Surfboard Dang, I need a serious answer ....... You didn't read much into my absinthe powered post carefully. No worries. I still love you. Had you bothered to take note of my solution to your 1st problem, you might have noticed that the reason why it should be one year and 3 months according to the book is that my theoretical answer of n = 14.76 denotes months. Thus, 12 months (or one year) + 2.76 months (or 3 months). Quote:  Originally Posted by Surfboard The answer is 13.61927598. I see by your reference to the phrases "put calculator in BGN" and "put calculator to END" in your 2nd post that you're quite fond of your financial calculator. Quote:  Originally Posted by Surfboard Dang, I need a serious answer I have an exam next week and ran into a similar one: Ali deposits 450 dollar at the beginning of every 3 months (put calculator in BGN?). He wants to build up his account so that he can withdraw$1000 every 3 months starting 3 months after the last deposit (put calculator to END?) If he wants to make withdrawals for 15 years and interest is 10% compounded quarterly, find N. I came to 40.481038030 = 10.12025951, or 10 years and... months? However the correct answer should be: 10 years & 3 months. 1. Do I change my calculation for the second part to END or keep it at BGN? I think END is correct. 2. The conversion of N to years & months, what's up with that?
Another annuity problem whose solution must need be a theoretical one.
Quote:
 Originally Posted by Surfboard I came to 40.481038030 = 10.12025951, or 10 years and... months? However the correct answer should be: 10 years & 3 months.
I beg to differ. In theory, it should be 39.85195112
In practice, the end of the 40th quarterly period (or 10 years times 4), should do the trick.
Accordingly, the most efficient way to solve an annuity problem is to make a time diagram, determine the type of annuity, and then apply the proper formula(s).
Bearing this in mind, you really need to lose your dependence on your financial calculator and rely more on basic algebra. You also need to familiarize yourself with those basic annuity formulas.
Allow me to enlighten you with my absinthe powered views.
In order to withdraw 1000 dollars every 3 months starting 3 months after the last deposit for 15 years, or 60 quarterly periods (15*4) and interest is 10% compounded quarterly, you simply apply the amortization formula as quoted by Sir MarkFL. Thus,
$$A = 1000{{1 - \left( {1 + {\textstyle{{0.10} \over 4}}} \right)^{ - (15*4)} } \over {{\textstyle{{0.10} \over 4}}}}$$
This then is the amount of money that Ali must have in the bank 3 months after the last deposit in order for him to withdraw 1000 dollars every 3 months for 15 years.
You then equate this amount with the formula for the Future value of an annuity due (in direct contrast to Sir MarkFL's suggestion) given here.
Thus,
$$450{{\left( {1 + {\textstyle{{0.10} \over 4}}} \right)^n - 1} \over {{\textstyle{{0.10} \over 4}}}}\left( {1 + {\textstyle{{0.10} \over 4}}} \right) = 1000{{1 - \left( {1 + {\textstyle{{0.10} \over 4}}} \right)^{ - (15*4)} } \over {{\textstyle{{0.10} \over 4}}}}$$
If you're familiar with exponential equations, solving for n in this equation of value should be easy. If not then I'm afraid that you may need a refresher.
I'm sure Sir Denis will come along shortly to enlighten us why this theoretical solution is just the 1st step for a more practical solution using integer values of n. He might even delight us with his bank account format proof.

Last edited by jonah; March 5th, 2015 at 11:15 PM. Reason: Added disclaimer

March 6th, 2015, 02:11 AM   #6
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Joined: May 2008

Posts: 299
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Quote:
Originally Posted by jonah
DISCLAIMER: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.
Quote:
 Originally Posted by Surfboard Dang, I need a serious answer .......
You didn't read much into my absinthe powered post carefully.
No worries. I still love you.
Had you bothered to take note of my solution to your 1st problem, you might have noticed that the reason why it should be one year and 3 months according to the book is that my theoretical answer of n = 14.76 denotes months. Thus, 12 months (or one year) + 2.76 months (or 3 months).
Quote:
 Originally Posted by Surfboard The answer is 13.61927598.
I see by your reference to the phrases "put calculator in BGN" and "put calculator to END" in your 2nd post that you're quite fond of your financial calculator.
Quote:
 Originally Posted by Surfboard Dang, I need a serious answer I have an exam next week and ran into a similar one: Ali deposits 450 dollar at the beginning of every 3 months (put calculator in BGN?). He wants to build up his account so that he can withdraw $1000 every 3 months starting 3 months after the last deposit (put calculator to END?) If he wants to make withdrawals for 15 years and interest is 10% compounded quarterly, find N. I came to 40.481038030 = 10.12025951, or 10 years and... months? However the correct answer should be: 10 years & 3 months. 1. Do I change my calculation for the second part to END or keep it at BGN? I think END is correct. 2. The conversion of N to years & months, what's up with that? Another annuity problem whose solution must need be a theoretical one. Quote:  Originally Posted by Surfboard I came to 40.481038030 = 10.12025951, or 10 years and... months? However the correct answer should be: 10 years & 3 months. I beg to differ. In theory, it should be 39.85195112 In practice, the end of the 40th quarterly period (or 10 years times 4), should do the trick. Had me some beer to moderate my absinthe visions. Turned out I was off the mark. EDIT: In theory, it should be 40.48104037. In practice, the end of the 41st quarterly period (or 10.25 years times 4), should do the trick. Accordingly, the most efficient way to solve an annuity problem is to make a time diagram, determine the type of annuity, and then apply the proper formula(s). Bearing this in mind, you really need to lose your dependence on your financial calculator and rely more on basic algebra. You also need to familiarize yourself with those basic annuity formulas. Allow me to enlighten you with my absinthe powered views. In order to withdraw 1000 dollars every 3 months starting 3 months after the last deposit for 15 years, or 60 quarterly periods (15*4) and interest is 10% compounded quarterly, you simply apply the amortization formula as quoted by Sir MarkFL. Thus, $$A = 1000{{1 - \left( {1 + {\textstyle{{0.10} \over 4}}} \right)^{ - (15*4)} } \over {{\textstyle{{0.10} \over 4}}}}$$ This then is the amount of money that Ali must have in the bank 3 months after the last deposit in order for him to withdraw 1000 dollars every 3 months for 15 years. EDIT: Accordingly, the most efficient way to solve an annuity problem is to make a time diagram, determine the type of annuity, and then apply the proper formula(s). Bearing this in mind, you really need to lose your dependence on your financial calculator and rely more on basic algebra. You also need to familiarize yourself with those basic annuity formulas. Allow me to enlighten you with my absinthe powered views. In order to withdraw 1000 dollars every 3 months starting 3 months after the last deposit for 15 years, or 60 quarterly periods (15*4) and interest is 10% compounded quarterly, you simply apply the present value of an annuity due formula. Thus, $$A = 1000{{1 - \left( {1 + {\textstyle{{0.10} \over 4}}} \right)^{ - (15*4)} } \over {{\textstyle{{0.10} \over 4}}}}\left( {1 + {\textstyle{{0.10} \over 4}}} \right)$$ This then is the amount of money that Ali must have in the bank 3 months after the last deposit in order for him to withdraw 1000 dollars every 3 months for 15 years. You then equate this amount with the formula for the Future value of an annuity due (in direct contrast to Sir MarkFL's suggestion) given here. Thus, $$450{{\left( {1 + {\textstyle{{0.10} \over 4}}} \right)^n - 1} \over {{\textstyle{{0.10} \over 4}}}}\left( {1 + {\textstyle{{0.10} \over 4}}} \right) = 1000{{1 - \left( {1 + {\textstyle{{0.10} \over 4}}} \right)^{ - (15*4)} } \over {{\textstyle{{0.10} \over 4}}}}$$ EDIT: You then equate this amount with the formula for the Future value of an annuity due (in direct contrast to Sir MarkFL's suggestion) given here. Thus, $$450{{\left( {1 + {\textstyle{{0.10} \over 4}}} \right)^n - 1} \over {{\textstyle{{0.10} \over 4}}}}\left( {1 + {\textstyle{{0.10} \over 4}}} \right) = 1000{{1 - \left( {1 + {\textstyle{{0.10} \over 4}}} \right)^{ - (15*4)} } \over {{\textstyle{{0.10} \over 4}}}}\left( {1 + {\textstyle{{0.10} \over 4}}} \right)$$ If you're familiar with exponential equations, solving for n in this equation of value should be easy. If not then I'm afraid that you may need a refresher. I'm sure Sir Denis will come along shortly to enlighten us why this theoretical solution is just the 1st step for a more practical solution using integer values of n. He might even delight us with his bank account format proof. EDIT: From this equation of value, as noted in my 1st edit, we get n = 40.48104037 Not much difference from your answer I guess. Cheers. March 6th, 2015, 01:49 PM #7 Newbie Joined: Oct 2014 From: Canada Posts: 4 Thanks: 0 Math Focus: none Quote:  Originally Posted by jonah You didn't read much into my absinthe powered post carefully. No worries. I still love you. I'm sorry it was a bit overwhelming. Quote:  Originally Posted by jonah Had you bothered to take note of my solution to your 1st problem, you might have noticed that the reason why it should be one year and 3 months according to the book is that my theoretical answer of n = 14.76 denotes months. Thus, 12 months (or one year) + 2.76 months (or 3 months). I'm gonna have to take a look at that, I still get to 39.85something Quote:  Originally Posted by jonah I see by your reference to the phrases "put calculator in BGN" and "put calculator to END" in your 2nd post that you're quite fond of your financial calculator. Yeah I know, I certainly don't have the knowledge you guys have. At least: not yet. I get the formulas and how it works but it's still fairly new to me. I can ride a bike I just need to know what sort of bike I'm on before I need to apply brakes March 7th, 2015, 09:34 AM #8 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,107 Thanks: 1001 Sir Jonah bravely shows that a part pay't equivalent to .76 of a month is required, and Sir Jonah is correct again, and deserves a stiff belt of Vodka. Quote:  Originally Posted by surfboard leased equipment valued at 23000, the terms of the lease require 1800/month. If the first payment is due in nine months after the lease was signed and interest is 11% compounded monthly, what is the term? Code: month payment interest balance 00 23,000.00 01 .00 210.83 23,210.83 ... 08 .00 224.74 24,741.78 09(01) -1800.00 226.80 23,168.58 ... 22(14) -1800.00 28.66 1,355.31 22.76(14.76) 9.44 1,364.75 ***1 23(15) 12.42 1,367.73 ***2 ***1: This is amount owing at month # 23.76 Assuming this is a 30day month (make it April), then you theoretically have to come in exactly at 7:12pm on Apr. 22nd to remit this amount. ***2: However, in real life, the payment is called for on Apr. 30th, interest increases accordingly, for a final payment of 1,367.73. The legal verbiage something like: ...in amount of$1,800.00 for 14 months...
...then a final payment of 1,367.73 at end of 15th month...

March 7th, 2015, 09:45 AM   #9
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Quote:
 Originally Posted by Surfboard Do I change my calculation for the second part to END or keep it at BGN? I think END is correct.
Get used to NOT using your calculator to get solutions:
use it to "see" if your calculations are correct.

On an exam, you will probably NOT be allowed a calculator.

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