Looking for TVM formulas

04042014, 05:23 PM
Post: #21




RE: Looking for TVM formulas
(04042014 02:07 PM)Dave Britten Wrote: Awesome, I used Thomas' suggestion for calculating (1 + i%) ^ n  1 using the more accurate e^x1 formula, 2 * sinh(x / 2) * e ^ (x / 2), and now for the Susan example, my TI95 program is giving me FV = 333,783.3499486. I've used that in Jeff Kearns' program for the HP15C: Code: 001  42,21,15 LBL E Now I get 333,783.3500 instead of 333,783.3508. Cheers Thomas 

04042014, 09:44 PM
Post: #22




RE: Looking for TVM formulas
(04042014 02:07 PM)Dave Britten Wrote: Program size is currently 478 steps, and I figure there will be another 150200 when all that is implemented. Wow! That is quite the program... Just FYI, the total program length (including Solver) for the HP29C or HP11C is 71 steps in length. Granted, it does poorly (if at all) at solving for interest. The program size for calculators with builtin solvers is substantially smaller (and finds solutions for periodic interest rates). What does your program do exactly? I thought you wanted a TVM solver. Regards, Jeff 

04042014, 10:12 PM
(This post was last modified: 04042014 10:13 PM by Dave Britten.)
Post: #23




RE: Looking for TVM formulas
(04042014 09:44 PM)Jeff_Kearns Wrote:(04042014 02:07 PM)Dave Britten Wrote: Program size is currently 478 steps, and I figure there will be another 150200 when all that is implemented. TVM solver, mostly. The TI95 has lots of RAM to work with, for a keystroke programmable calculator (about 8KB), and the default partitioning gives you about 1000 steps/bytes of work RAM to use for coding. Programs can be saved/loaded/executed from the file partition, which defaults to about 5KB (5000 steps or so). There are also instructions to save ranges of registers into the file memory for later retrieval. With default settings, you get 125 storage registers, all addressible directly or indirectly (any register can be used for indirection). I think it's also possible to shrink file space to zero, and rely on an 8/32 KB expansion module for your file storage, but you'd have to be working with some enormous programs to need that. Off the top of my head, these are the routines in the program so far: 1. Main menu  Allows accessing the TVM menu, or selecting Begin/End modes. Amort will be accessible from here when I implement it. 2. TVM menu  Shows the 5 TVM variables, and allows storing a value into each, or requesting solutions. The 95 allows testing for the INV prefix key being used to modify a soft key, so each of the five soft key handlers checks for it, calls one of five "solve" routines if needed, then puts the display value into the selected variable. 3. Five "solve" routines  Three of these are direct solutions (PV, PMT, FV), and the other two prepare... 4. ...The numeric solver  A straight port of the HP 20S solver, with a couple changes specific to this progra. It chokes on overflows trying to solve for n or i% in the Susan example, so I probably ought to replace it. Works okay for 'typical' problems, but isn't particularly fast, as it's lacking any initial guess logic (just starts at 1). 5. ln(1+x)  Uses numeric methods discussed in this thread. 6. e^x  1  Ditto. 7. (1 + i%)^n  Uses the ln(1 + x) routine for greater accuracy. Needed by all of the solution equations. 8. ((1 + i%)^n)  1  Also uses e^x  1, and again needed by the solutions. 9. i% Routine  Returns i% / 100 for calculations. Needs to have some of the repeat calls optimized away. 10. Begin/End Routine  Returns 1 or 0 if Begin or End is selected, respectively. 11. Solver equation  TVM equation with one side solved for zero, for use with numeric solver routine. So, the (somewhat) redundant direct solutions, menu handlers, math utility routines, and included numeric solver explain the size. Also, the TI needs something like three bytes (steps) for goto, gosub, STO, RCL, comparison tests, and other instructions that require an argument, to allow for the full three or four digits used in the addresses. Any functions prefixed with INV will also use another byte for the prefix. 

04052014, 09:58 PM
(This post was last modified: 04052014 10:22 PM by Dieter.)
Post: #24




RE: Looking for TVM formulas
(04042014 10:12 PM)Dave Britten Wrote: 3. Five "solve" routines  Three of these are direct solutions (PV, PMT, FV), and the other two prepare... Hm, why do you think that there is no direct solution for n and an iterative approach is required? Consider the basic TVM formula you already posted: PV * (1 + i)^{n} + PMT * ((1 + i)^{n} – 1)) * ((1 / i) + B) + FV = 0 Then (1 + i)^{n} – 1 = –(PV + FV) / (PMT * (1 / i + B) + PV) And thus, using ln1+x for best accuracy: n = ln1+x(–(PV + FV) / (PMT * (1 / i + B) + PV)) / ln1+x(i) Or a bit more formal: \(\large n = ln (1  \frac{PV + FV}{PV + PMT (\frac{1}{i} + B)}) : ln (1 + i)\) Try the Susan example: (1+i)^{n} – 1 = (–(0 + 333783,35)) / ( 0,01 * (1/3,567351598 E9 + 0) + 0) = 333783,35 / 2803200 = 0,1190722567 ln1+x(0,1190722567) / ln1+x(3,567351598 E9) = 31536000 Dieter 

04052014, 11:38 PM
Post: #25




RE: Looking for TVM formulas
(04052014 09:58 PM)Dieter Wrote: Hm, why do you think that there is no direct solution for n and an iterative approach is required? Oh, I figured there was one, I just hadn't found/derived it yet, so I was feeding it through the numeric solver for now. Thanks for posting that, it should prove useful. 

04062014, 07:46 PM
(This post was last modified: 04062014 07:50 PM by Dieter.)
Post: #26




RE: Looking for TVM formulas
(04022014 05:18 PM)Thomas Klemm Wrote: You could use \(e^x1=2 \cdot sinh(\frac{x}{2}) \cdot e^{\frac{x}{2}}\) instead. Since we're at it: hyperbolics can also be used for ln1+x. \(ln(1+x) = 2 \cdot artanh(\frac{x}{x+2})\) Seems to work well if x is not too large so that the artanhargument does not get too close to 1. Dieter 

04092014, 01:36 PM
(This post was last modified: 04092014 02:15 PM by Dieter.)
Post: #27




RE: Looking for TVM formulas
(04062014 07:46 PM)Dieter Wrote: Since we're at it: hyperbolics can also be used for ln1+x. This one seems even better and does not have problems with very large x: \(ln(1+x) = arsinh(\frac{x^2+2x}{2+2x})\) For better numeric accuracy this can be written as \(y = \frac{x}{2}; z = y + \frac{y}{2y+1}\) \(ln(1+x) = arsinh z\) With \(z \rightarrow \infty, arsinh z \rightarrow ln 2z\) and \(z \rightarrow \frac{x+1}{2}\), so that \(arsinh z \rightarrow ln(1+x)\). Dieter 

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