Future value of lump sum - use years or months?












0












$begingroup$


I'm trying to calculate the future value of a $1000 lump sum payment at the end of 12 years at 9% annual interest. I used this formula: FVn = PV * (1+r)^n.



I got this result: $2812.66 = 1000 * (1+.09)^12.



But the example I'm working from says the FV is $2,932.84. After some tinkering I realized that the example used the monthly rate and 144 periods. Why is the result different and which is correct?






finance







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$endgroup$












  • $begingroup$
    It depends on the exercise which result is correct. If the payment is monthly you have to convert the yearly interest rate into a monthly interest rate, otherwise you use the yearly interest rate.
    $endgroup$
    – callculus
    Dec 17 '18 at 0:59








  • 2




    $begingroup$
    The result is different because with monthly interest you compound partial increments more frequently.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:06
















0












$begingroup$


I'm trying to calculate the future value of a $1000 lump sum payment at the end of 12 years at 9% annual interest. I used this formula: FVn = PV * (1+r)^n.



I got this result: $2812.66 = 1000 * (1+.09)^12.



But the example I'm working from says the FV is $2,932.84. After some tinkering I realized that the example used the monthly rate and 144 periods. Why is the result different and which is correct?






finance







share|cite|improve this question











$endgroup$












  • $begingroup$
    It depends on the exercise which result is correct. If the payment is monthly you have to convert the yearly interest rate into a monthly interest rate, otherwise you use the yearly interest rate.
    $endgroup$
    – callculus
    Dec 17 '18 at 0:59








  • 2




    $begingroup$
    The result is different because with monthly interest you compound partial increments more frequently.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:06














0












0








0





$begingroup$


I'm trying to calculate the future value of a $1000 lump sum payment at the end of 12 years at 9% annual interest. I used this formula: FVn = PV * (1+r)^n.



I got this result: $2812.66 = 1000 * (1+.09)^12.



But the example I'm working from says the FV is $2,932.84. After some tinkering I realized that the example used the monthly rate and 144 periods. Why is the result different and which is correct?






finance







share|cite|improve this question











$endgroup$




I'm trying to calculate the future value of a $1000 lump sum payment at the end of 12 years at 9% annual interest. I used this formula: FVn = PV * (1+r)^n.



I got this result: $2812.66 = 1000 * (1+.09)^12.



But the example I'm working from says the FV is $2,932.84. After some tinkering I realized that the example used the monthly rate and 144 periods. Why is the result different and which is correct?






finance




finance






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 0:57







Userq49874

















asked Dec 17 '18 at 0:51









Userq49874Userq49874

4015




4015












  • $begingroup$
    It depends on the exercise which result is correct. If the payment is monthly you have to convert the yearly interest rate into a monthly interest rate, otherwise you use the yearly interest rate.
    $endgroup$
    – callculus
    Dec 17 '18 at 0:59








  • 2




    $begingroup$
    The result is different because with monthly interest you compound partial increments more frequently.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:06


















  • $begingroup$
    It depends on the exercise which result is correct. If the payment is monthly you have to convert the yearly interest rate into a monthly interest rate, otherwise you use the yearly interest rate.
    $endgroup$
    – callculus
    Dec 17 '18 at 0:59








  • 2




    $begingroup$
    The result is different because with monthly interest you compound partial increments more frequently.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:06
















$begingroup$
It depends on the exercise which result is correct. If the payment is monthly you have to convert the yearly interest rate into a monthly interest rate, otherwise you use the yearly interest rate.
$endgroup$
– callculus
Dec 17 '18 at 0:59






$begingroup$
It depends on the exercise which result is correct. If the payment is monthly you have to convert the yearly interest rate into a monthly interest rate, otherwise you use the yearly interest rate.
$endgroup$
– callculus
Dec 17 '18 at 0:59






2




2




$begingroup$
The result is different because with monthly interest you compound partial increments more frequently.
$endgroup$
– Ethan Bolker
Dec 17 '18 at 1:06




$begingroup$
The result is different because with monthly interest you compound partial increments more frequently.
$endgroup$
– Ethan Bolker
Dec 17 '18 at 1:06










2 Answers
2






active

oldest

votes


















0












$begingroup$


Why is the result different?




You can use the binomial theorem to see that a monthly compounded lump sum is larger than a yearly compounded one. We can compare one year.



$$left(1+frac{i}{12}right)^{12}=sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k} $$



The first two summands ($k=0,1$) are



$sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k}=1+12cdot left(frac{i}{12}right)^1cdot 1^{11}+dots=1+i+ldots$



It can be seen that the sum of the first two summands already equals $1+i$. But if we regard the other $11$ summands then a monthly compounded lump sum is always greater than a yearly compounded, if $i>0$.



$$left(1+frac{i}{12}right)^{12}>1+i$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
    $endgroup$
    – Userq49874
    Dec 17 '18 at 1:40






  • 1




    $begingroup$
    @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
    $endgroup$
    – callculus
    Dec 17 '18 at 1:42





















0












$begingroup$

Why is the result different? Let's consider just the first year.



If you're paid interest annually, then your $$1000$ dollars at $9%$ will get you $$90$ at the end of the first year.



If you're paid monthly, you'll get just one twelfth of that, $$7.50$, at the end of January, so you now have $$1007.50$ in your account. For the rest of the year, you'll be collecting interest on that amount. So at the end of February, you'll get slightly more than another $$7.50$ added to your account; if I've done the arithmetic right, this second monthly interest payment will be between $$7.55$ and $$7.56$ (or maybe it'll be rounded up or down to exactly one of these amounts). So now you've got $$1015.05$ or so in your account. Your interest payment at the end of March will be based on that. And so forth, for the rest of the year.



So you get $$7.50$ in interest at the end of January and somewhat more than $$7.50$ at the end of each subsequent month. If you didn't get the "somewhat more" then your interest over the first year would be $$7.50times12=$90$, just as if you were paid annually. But you do get the "somewhat more", so at the end of the first year you've made more than $$90$.



That's why annual payments and monthly payments give different results for the first year. The same idea applies to the subsequent years, but "even more so", because with monthly payments you've got more money at the start of the second year, and you'll be getting interest on that in all the subsequent years.






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    2 Answers
    2






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    2 Answers
    2






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    active

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    active

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    0












    $begingroup$


    Why is the result different?




    You can use the binomial theorem to see that a monthly compounded lump sum is larger than a yearly compounded one. We can compare one year.



    $$left(1+frac{i}{12}right)^{12}=sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k} $$



    The first two summands ($k=0,1$) are



    $sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k}=1+12cdot left(frac{i}{12}right)^1cdot 1^{11}+dots=1+i+ldots$



    It can be seen that the sum of the first two summands already equals $1+i$. But if we regard the other $11$ summands then a monthly compounded lump sum is always greater than a yearly compounded, if $i>0$.



    $$left(1+frac{i}{12}right)^{12}>1+i$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
      $endgroup$
      – Userq49874
      Dec 17 '18 at 1:40






    • 1




      $begingroup$
      @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
      $endgroup$
      – callculus
      Dec 17 '18 at 1:42


















    0












    $begingroup$


    Why is the result different?




    You can use the binomial theorem to see that a monthly compounded lump sum is larger than a yearly compounded one. We can compare one year.



    $$left(1+frac{i}{12}right)^{12}=sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k} $$



    The first two summands ($k=0,1$) are



    $sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k}=1+12cdot left(frac{i}{12}right)^1cdot 1^{11}+dots=1+i+ldots$



    It can be seen that the sum of the first two summands already equals $1+i$. But if we regard the other $11$ summands then a monthly compounded lump sum is always greater than a yearly compounded, if $i>0$.



    $$left(1+frac{i}{12}right)^{12}>1+i$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
      $endgroup$
      – Userq49874
      Dec 17 '18 at 1:40






    • 1




      $begingroup$
      @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
      $endgroup$
      – callculus
      Dec 17 '18 at 1:42
















    0












    0








    0





    $begingroup$


    Why is the result different?




    You can use the binomial theorem to see that a monthly compounded lump sum is larger than a yearly compounded one. We can compare one year.



    $$left(1+frac{i}{12}right)^{12}=sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k} $$



    The first two summands ($k=0,1$) are



    $sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k}=1+12cdot left(frac{i}{12}right)^1cdot 1^{11}+dots=1+i+ldots$



    It can be seen that the sum of the first two summands already equals $1+i$. But if we regard the other $11$ summands then a monthly compounded lump sum is always greater than a yearly compounded, if $i>0$.



    $$left(1+frac{i}{12}right)^{12}>1+i$$






    share|cite|improve this answer









    $endgroup$




    Why is the result different?




    You can use the binomial theorem to see that a monthly compounded lump sum is larger than a yearly compounded one. We can compare one year.



    $$left(1+frac{i}{12}right)^{12}=sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k} $$



    The first two summands ($k=0,1$) are



    $sum_{k=0}^{12} binom{12}{k}cdot left(frac{i}{12}right)^kcdot 1^{12-k}=1+12cdot left(frac{i}{12}right)^1cdot 1^{11}+dots=1+i+ldots$



    It can be seen that the sum of the first two summands already equals $1+i$. But if we regard the other $11$ summands then a monthly compounded lump sum is always greater than a yearly compounded, if $i>0$.



    $$left(1+frac{i}{12}right)^{12}>1+i$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 17 '18 at 1:27









    callculuscallculus

    18.1k31427




    18.1k31427












    • $begingroup$
      The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
      $endgroup$
      – Userq49874
      Dec 17 '18 at 1:40






    • 1




      $begingroup$
      @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
      $endgroup$
      – callculus
      Dec 17 '18 at 1:42




















    • $begingroup$
      The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
      $endgroup$
      – Userq49874
      Dec 17 '18 at 1:40






    • 1




      $begingroup$
      @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
      $endgroup$
      – callculus
      Dec 17 '18 at 1:42


















    $begingroup$
    The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
    $endgroup$
    – Userq49874
    Dec 17 '18 at 1:40




    $begingroup$
    The only reason I haven't marked this as the accepted answer is because it's way over my head so I can't say whether I agree. But that's my problem - not callculus'
    $endgroup$
    – Userq49874
    Dec 17 '18 at 1:40




    1




    1




    $begingroup$
    @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
    $endgroup$
    – callculus
    Dec 17 '18 at 1:42






    $begingroup$
    @Userq49874 You can ask if you have a question. No problem. It isn´t so difficult as it might looks like at the first view.
    $endgroup$
    – callculus
    Dec 17 '18 at 1:42













    0












    $begingroup$

    Why is the result different? Let's consider just the first year.



    If you're paid interest annually, then your $$1000$ dollars at $9%$ will get you $$90$ at the end of the first year.



    If you're paid monthly, you'll get just one twelfth of that, $$7.50$, at the end of January, so you now have $$1007.50$ in your account. For the rest of the year, you'll be collecting interest on that amount. So at the end of February, you'll get slightly more than another $$7.50$ added to your account; if I've done the arithmetic right, this second monthly interest payment will be between $$7.55$ and $$7.56$ (or maybe it'll be rounded up or down to exactly one of these amounts). So now you've got $$1015.05$ or so in your account. Your interest payment at the end of March will be based on that. And so forth, for the rest of the year.



    So you get $$7.50$ in interest at the end of January and somewhat more than $$7.50$ at the end of each subsequent month. If you didn't get the "somewhat more" then your interest over the first year would be $$7.50times12=$90$, just as if you were paid annually. But you do get the "somewhat more", so at the end of the first year you've made more than $$90$.



    That's why annual payments and monthly payments give different results for the first year. The same idea applies to the subsequent years, but "even more so", because with monthly payments you've got more money at the start of the second year, and you'll be getting interest on that in all the subsequent years.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Why is the result different? Let's consider just the first year.



      If you're paid interest annually, then your $$1000$ dollars at $9%$ will get you $$90$ at the end of the first year.



      If you're paid monthly, you'll get just one twelfth of that, $$7.50$, at the end of January, so you now have $$1007.50$ in your account. For the rest of the year, you'll be collecting interest on that amount. So at the end of February, you'll get slightly more than another $$7.50$ added to your account; if I've done the arithmetic right, this second monthly interest payment will be between $$7.55$ and $$7.56$ (or maybe it'll be rounded up or down to exactly one of these amounts). So now you've got $$1015.05$ or so in your account. Your interest payment at the end of March will be based on that. And so forth, for the rest of the year.



      So you get $$7.50$ in interest at the end of January and somewhat more than $$7.50$ at the end of each subsequent month. If you didn't get the "somewhat more" then your interest over the first year would be $$7.50times12=$90$, just as if you were paid annually. But you do get the "somewhat more", so at the end of the first year you've made more than $$90$.



      That's why annual payments and monthly payments give different results for the first year. The same idea applies to the subsequent years, but "even more so", because with monthly payments you've got more money at the start of the second year, and you'll be getting interest on that in all the subsequent years.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Why is the result different? Let's consider just the first year.



        If you're paid interest annually, then your $$1000$ dollars at $9%$ will get you $$90$ at the end of the first year.



        If you're paid monthly, you'll get just one twelfth of that, $$7.50$, at the end of January, so you now have $$1007.50$ in your account. For the rest of the year, you'll be collecting interest on that amount. So at the end of February, you'll get slightly more than another $$7.50$ added to your account; if I've done the arithmetic right, this second monthly interest payment will be between $$7.55$ and $$7.56$ (or maybe it'll be rounded up or down to exactly one of these amounts). So now you've got $$1015.05$ or so in your account. Your interest payment at the end of March will be based on that. And so forth, for the rest of the year.



        So you get $$7.50$ in interest at the end of January and somewhat more than $$7.50$ at the end of each subsequent month. If you didn't get the "somewhat more" then your interest over the first year would be $$7.50times12=$90$, just as if you were paid annually. But you do get the "somewhat more", so at the end of the first year you've made more than $$90$.



        That's why annual payments and monthly payments give different results for the first year. The same idea applies to the subsequent years, but "even more so", because with monthly payments you've got more money at the start of the second year, and you'll be getting interest on that in all the subsequent years.






        share|cite|improve this answer









        $endgroup$



        Why is the result different? Let's consider just the first year.



        If you're paid interest annually, then your $$1000$ dollars at $9%$ will get you $$90$ at the end of the first year.



        If you're paid monthly, you'll get just one twelfth of that, $$7.50$, at the end of January, so you now have $$1007.50$ in your account. For the rest of the year, you'll be collecting interest on that amount. So at the end of February, you'll get slightly more than another $$7.50$ added to your account; if I've done the arithmetic right, this second monthly interest payment will be between $$7.55$ and $$7.56$ (or maybe it'll be rounded up or down to exactly one of these amounts). So now you've got $$1015.05$ or so in your account. Your interest payment at the end of March will be based on that. And so forth, for the rest of the year.



        So you get $$7.50$ in interest at the end of January and somewhat more than $$7.50$ at the end of each subsequent month. If you didn't get the "somewhat more" then your interest over the first year would be $$7.50times12=$90$, just as if you were paid annually. But you do get the "somewhat more", so at the end of the first year you've made more than $$90$.



        That's why annual payments and monthly payments give different results for the first year. The same idea applies to the subsequent years, but "even more so", because with monthly payments you've got more money at the start of the second year, and you'll be getting interest on that in all the subsequent years.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 3:18









        Andreas BlassAndreas Blass

        49.7k451108




        49.7k451108






























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