The Museum of HP Calculators

HP-15C: Time Value of Money

Copyright (C) 2002 Glen Kilpatrick

Distributed under GNU General Public License

This program is supplied without representation or warranty of any kind. The author and The Museum of HP Calculators therefore assume no responsibility and shall have no liability, consequential or otherwise, of any kind arising from the use of this program material or any part thereof.


Hewlett Packard's implementation of the Time Value of Money relies upon the following principles:

This program computes the Time Value of Money on an HP-15C, and contains logic for a payment at the beginning or end of the period. It has a "user interface" similar to that of the HP-12C in that a single push of one of the TVM keys will store the X-register to the TVM variable, and two consecutive pushes will first store a guess, then compute a new value based upon that guess and other previously stored variables.



Alas, I haven't had an HP-15C for many years. I could supply examples from many sources, but not be certain that any would show you exactly what you'd see IRL (in real life :). However, something official might suffice....

An example derived from the HP-10B's Owner's Manual (Edition 6, Part Number 00010-90037, November 1994), "5: Time Value of Money Calculations", page 53 (modified as to format, and to show key sequences on the HP-15C):

Example: A Car Loan. You are financing a new car with a three year loan at 10.5% annual nominal interest, compounded monthly. The price of the car is $7,250. Your down payment is $1,500.

Part 1. What are your monthly payments at 10.5% interest? (Assume your payments start one month after the purchase or at the end of the first period.)

f FIX 2               only to match example
f PRGM R/S            repeat until RAD is set then unset, to set
                      END mode (payment at then end of the period)
3 ENTER 12 × A        stores 3 years
10.5 ENTER 12 ÷ B     stores 10.5% per year nominal interest
7250 ENTER 1500 - C   stores loan amount (present value)
0 E                   stores future value (no balloon payment)
D D                   computes monthly payment of -186.89

Part 2. At a price of $7,250.00, what interest rate is necesary to lower your payment by $10.00, to -176.89?

RCL D                 recalls monthly payment to the stack
10 + D                decreases it by $10 to -176.89, then stores
RCL B B B             computes annual nominal interest of 6.75%
                      (the RCL B may not be necessary, but I'm not
                      sure that an initial interest rate of less
                      than zero would be a reasonable guess, and I
                      can't test it myself :)

Part 3. If interest is 10.5%, what is the maximum you can spend on the car to lower your payment to $175,00?

10.5 B                (re)stores 10.5% annual nominal interest
175 +/- D             stores preferred monthly payment
RCL C C C             computes maximum loan of 5384.21 (likewise)
1500 +                adds down payment for maximum purchase price,

Program Listing

I(PMT<>0) is SOLVEd by iteration (LBL 5), with initial estimate I0/100=|PMT/(|PV|+|FV|)|+|(|PV|+|FV|)/(N3×PMT)|



Read and enter program from left to right, line by line:

000-          [comment]             MAT0       1          ENTER
   DIMA       DIMB       DIMC       DIMD       DIME       STOgA
   FIX2       MAT1       [F7 toggle/display]   LBL0       F?7
   SF9        SF7        F?9        CF7        CF9        DEG
   F?7        RAD        R/S        GTO0       [N]        LBLA
   STOA       R/S        GTO0       LBLA       GSB3       ENTER
   RCL-E      X⇔Y                   RCL+C      ÷          LN
   EEX        2          CHS        RCL×B      1          +
   LN         ÷          GTOA       [%I]       LBLB       STOB
   R/S        GTO0       LBLB       RCLD       TEST0      GTO4
   RCLE       RCL÷C      CHS        RCLA       1/X        yx
   1          -          EEX        2          ×          GTOB
   [PV]       LBLC       STOC       R/S        GTO0       LBLC
   GSB3       GSB1       ×          RCL+E      1          LSTX
   +          ÷          CHS        GTOC       [PMT]      LBLD
   STOD       R/S        GTO0       LBLD       GSB2       GSB1
   ×          LSTX       1          +          RCL×C      RCL+E
   CHS        X⇔Y        ÷          GTOD       [FV]       LBLE
   STOE       R/S        GTO0       LBLE       GSB3       RCL+C
   GSB1       ×          RCL+C      CHS        GTOE       [A]
   LBL1       EEX        2          CHS        RCL×B      1
   +          RCLA       yx         1          -          RTN
   [B]        LBL2       0          F?7        1          RCL×B
   EEX        2          +          RCL÷B      RTN        [C]
   LBL3       GSB2       RCL×D      RTN        [I0/100]   LBL4
   RCLC       ABS        RCLE       ABS        +          RCLD
   X⇔Y        ÷          LSTX       RCLA       3          yx
   RCL×D      ÷          ABS        X⇔Y        ABS        +
   ENTER      SF9        SOLVE5     CF9        EEX        2
   ×          GTOB       [I,PMT<>0] LBL5       EEX        2
   ×          STOB       GSB3       RCL+C      GSB1       ×
   RCL+C      RCL+E
[load_check:__172-45,40,15≤>gMEM=...29-0]  [gak(C)22AUG86]

Resources Used

Labels A→E&0→5; Flags 7&9 (user should ~CF9 before all program requests); Memories 0&1 (=1 at all times); matrices A→E; trig. indicator (display only, user may reset at any time).

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