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Annuity with irregular payments

Posted 11 years ago

For some reason I can't figure out how to solve this problem.

How much money do I need to deposit today, earning 4% interest, to make the following semi-annual irregular payments over the next 5 years - $500, $673, $245, $345, $245, $674, $425, $145, $37, $423?

Here's what I've come up with but it doesn't like the variable payment

payments={500, 673, 245, 345, 245, 674, 425, 145, 37, 423};
Solve[TimeValue[Annuity[{payements, {-onePay}}, 5, 1/2], EffectiveInterest[.04, 1/2], 5] == 0, onePay]

Thanks!

Tim

POSTED BY: Tim Heger
9 Replies
Posted 11 years ago
POSTED BY: Brett A Collins
Posted 11 years ago
POSTED BY: Brett A Collins

Given a nominal interest rate 0.04 the effective interest rate e is given by ((1+e)^2)-1 = 0.04 where there are 2 periods in one year.

True, e is not 1.04/2, of course. Thanks.

POSTED BY: Udo Krause
Posted 11 years ago
POSTED BY: Brett A Collins

Why this effectiverate should be used, Brett?

 In[13]:= Clear[pmt]
    pmt = {500, 673, 245, 345, 245, 674, 425, 145, 37, 423};
    Solve[Fold[Sqrt[1.04] #1 - #2 &, x, pmt] == 0, x]
 Out[15]= {{x -> 3381.8}}

If somebody calls something effective, some modeling happened: With Daniel's model one has to deposit 3375.65, with my model 3378.77 and with your model one needs 3381.8 to get the payments done.


Ludwig Schläfli sollte wie sein Vater Geschäftsmann werden. Aber er machte die denkbar schlechtesten Geschäfte, da er nicht begreifen konnte, dass man einen Gegenstand teurer verkaufte, als dass man ihn einkaufte.


Ludwig Schläfli should become a businessman like his father. But he made the worst possible deals because of his inability to understand that people buy cheap and sell dear.


POSTED BY: Udo Krause
Posted 11 years ago

@Udo - The typo was a mistake on my part but even with it corrected it doesn't work.

POSTED BY: Tim Heger
POSTED BY: Udo Krause

Doesn't

In[13]:= Solve[Fold[1.02 #1 - #2 &, x, {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}] == 0, x]
Out[13]= {{x -> 3378.77}}

help? (The interest is per year, the payment starts 6 month after the deposit and in the end no money remains.)

And this shows the account history just after a payment resp. the initial deposit

ListPlot[Transpose[
  {Range[0, 5, 0.5], FoldList[1.02 #1 - #2 &, 3378.77, {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}]}
 ], PlotLabel -> "Tim pays the bill"]

Tim pays the bill

and this is what has been payed every half year:

In[32]:= (1.02 Most[#] - Rest[#])&[FoldList[1.02 #1 - #2 &, 3378.77, {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}]]
Out[32]= {500., 673., 245., 345., 245., 674., 425., 145., 37., 423.}

probe it

In[33]:= % - {500, 673, 245, 345, 245, 674, 425, 145, 37, 423}
Out[33]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}
POSTED BY: Udo Krause

I'm not certain but I suspect you want Cashflow rather than Annuity for this purpose. TimeValue should then suffice for the rest; I don't think Solve is needed.

In particular it might be along the lines below.

payments = {500, 673, 245, 345, 245, 674, 425, 145, 37, 423};
TimeValue[Cashflow[payments, 1/2], 
 EffectiveInterest[.04, 0], -1/2]

(* Out[64]= 3375.65382702 *)

Some parameters might need to be adjusted in case the time period alignments are not as intended.

POSTED BY: Daniel Lichtblau
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