(34C) Accurate TVM for HP-34C - Printable Version +- HP Forums ( https://www.hpmuseum.org/forum)+-- Forum: HP Software Libraries ( /forum-10.html)+--- Forum: General Software Library ( /forum-13.html)+--- Thread: (34C) Accurate TVM for HP-34C ( /thread-386.html) |

(34C) Accurate TVM for HP-34C - Jeff_Kearns - 01-10-2014 04:10 AM
This is an adaptation of the Pioneer's (42S/35S/33S/32Sii/32S) Accurate TVM routine for the HP-34C using Karl Schneider's technique for invoking SOLVE with the routine written as a MISO (multiple-input, single-output) function, using indirect addressing. This post has been edited subsequent to feedback on the forum, reducing the size from 66 lines to 40. HP-34C TVM Routine 001 h LBL A 002 STO f (i) 003 RCL 2 004 EEX 005 2 006 ÷ 007 ENTER 008 ENTER 009 1 010 + 011 LN 012 X<>Y 013 LSTx 014 1 015 X≠Y 016 - 017 ÷ 018 * 019 RCL 1 020 * 021 e^x 022 RCL 3 023 X<>Y 024 * 025 LSTx 026 1 027 - 028 RCL 4 029 * 030 EEX 031 2 032 RCL 2 033 ÷ 034 RCL 6 035 + 036 * 037 + 038 RCL 5 039 + 040 RTN Usage instructions: 1. Store 4 of the following 5 variables as follows, using appropriate cash flow conventions: - N STO 1 --- Number of compounding periods
- I STO 2 --- Interest rate (periodic) expressed as a %
- B STO 3 --- Initial Balance or Present Value
- P STO 4 --- Periodic Payment
- F STO 5 --- Future Value
- B/E STO 6 --- Begin/End Mode. The default is 0 for regular annuity or End Mode.
2. Store the register number containing the floating variable to the indirect storage register (i). 3. f SOLVE A Example from the HP-15C Advanced Functions Handbook- "Many Pennies: A corporation retains Susan as a scientific and engineering consultant at a fee of one penny per second for her thoughts, paid every second of every day for a year. Rather than distract her with the sounds of pennies dropping, the corporation proposes to deposit them for her into a bank account in which interest accrues at the rate of 11.25 percent per annum compounded every second. At year's end these pennies will accumulate to a sum total = (payment) X ((1+i/n)^n-1)/(i/n) where payment = $0.01 = one penny per second, i = 0.1125 = 11.25 percent per annum interest rate, n = 60 X 60 X 24 X 365 = number of seconds in a year. Using her HP-15C, Susan reckons that the total will be $376,877.67. But at year's end the bank account is found to hold $333,783.35 . Is Susan entitled to the $43,094.32 difference?" - 31,536,000 STO 1
- (11.25/31,536,000) STO 2
- 0 STO 3
- -0.01 STO 4
- 5 STO f I
- f SOLVE A
The HP-34C gives the correct result: $333,783.35. Many thanks to Katie Wasserman, Thomas Klemm and Dieter for adapting the above routine, suggesting the workaround for the lack of Recall Arithmetic in the HP-34C, and testing the routine. Jeff Kearns |