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Re: Cashflow and TimeValue Question

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  • Subject: [mg131382] Re: Cashflow and TimeValue Question
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Tue, 2 Jul 2013 00:48:02 -0400 (EDT)
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The algorithm appears to look forward over the next 12 months to see if
there is an extra day, e.g., in Dec it looks at the following year but in
Jan it looks at the current year.


Table[cf = Cashflow[{
    {{yr, 12, 29}, a},
    {{yr, 12, 30}, b},
    {{yr, 12, 31}, c}}]; {yr, LeapYearQ[cf[[1, 1, 1]]],
  Rationalize[TimeValue[cf, r, cf[[1, 1, 1]]]]},
 {yr, 1999, 2009}]



Table[cf = Cashflow[{
    {{yr, 1, 29}, a},
    {{yr, 1, 30}, b},
    {{yr, 1, 31}, c}}]; {yr, LeapYearQ[cf[[1, 1, 1]]],
  Rationalize[TimeValue[cf, r, cf[[1, 1, 1]]]]},
 {yr, 1999, 2009}]



Bob Hanlon


On Mon, Jul 1, 2013 at 5:47 AM, Dana DeLouis <dana01 at icloud.com> wrote:

> Hello.   When doing Net-Present value types of problems, I seem to
> occasionaly get slightly different answers when compared to a very
> popular spreadsheet program.  The Spreadsheet program uses a standard
> 365 day year, yet I don't see anything in the Mathematica documentation
> on this.
> I've reduced the issue down to these two simple examples.
>
> 1.  3 simple daily cash flows in a row.
>      cf[[1,1,1]] is returning the start date, or time 0.
>       The days per year is a fraction, so I use Rationalize to return
> the values used for easy viewing.
>
> cf=Cashflow[{
> {{2009,12,29},a},
> {{2009,12,30},b},
> {{2009,12,31},c}
> }];
>
> Rationalize[TimeValue[cf,r,cf[[1,1,1]]],0]
>
> a + b/(r+1)^(1/365) + c/(r+1)^(2/365)
>
> It appears the program uses 365 days/year also.  This example would most
> likely match.
> We just note that the year 2009 is not a leap year.
>
> 2.   However, If I change the year to 2007, which is also not a leap
> year, it appears that the program is using a 366 day year.  These types
> of problems are giving slightly different answers then spreadsheets that
> use a constant 365 day year.
>
>
> cf=Cashflow[{
> {{2007,12,29},a},
> {{2007,12,30},b},
> {{2007,12,31},c}
> }];
>
> Rationalize[TimeValue[cf,r,cf[[1,1,1]]],0]
>
> a + b/(r+1)^(1/366) + c/(r+1)^(1/183)
>
> Note:  it's using 366 day year...
> {1,2}/366
> {1/366,1/183}
>
> Does anyone have any insight?  I'm not sure what the standard convention
> is.
> I just note that both years are non leap years, yet different methods of
> calculations.
> It appears to me that the random use of 366 is causing the slightly
> different answers.
>
>
> Thanks in advance.  :>)
> Mac, and Mathematical v 9.
>
>


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