15c challenge: accuracy of a complex equation system

[J-F Garnier]

I recently proposed a simpler solution of a problem example from the 15c Advanced Function Handbook (AFH),
solving a complex 4x4 system that represents an electric circuit driven by an AC source:
   
(See the AHF page 128 for details)

The problem can be expressed numerically by the complex matrix equation:

| (100,-350/3) (0,800/3)    (0,0)       (0,0)       |   | I1 |   | 10 |
| (0,800/3)    (1E6,-350/3) (-1E6,0)    (0,0)       |   | I2 |   | 0  |
| (0,0)        (-1E6,0)     (1E6,442/3) (0,-150)    | x | I3 | = | 0  |
| (0,0)        (0,0)        (0,-150)    (1E5,442/3) |   | I4 |   | 0  |

My solution is using a direct system solving (matrix "division") and is running only on an extended memory 15C such as the CE.
I was happy to write:

(07-29-2023 09:15 AM)J-F Garnier Wrote:  The results are in line with the AFH results obtained with the alternate method, with a minor difference in the output phase.

However, later on I wasn't satisfied by this minor difference that was after all not so minor since in the 5th place of the result.
So I run again both solutions and recorded the full results (rectangular forms):

AFH's solution:
I1 =  1.995031379 e-4  ,  4.096402192 e-3
I2 = -1.448873903 e-3  , -3.563298144 e-2
I3 = -1.454123461 e-3  , -3.563275918 e-2
I4 =  5.344584749 e-5  , -2.259869709 e-6

My solution:
I1 =  1.995210750 e-4  ,  4.096401801 e-3
I2 = -1.448866059 e-3  , -3.563300697 e-2
I3 = -1.454115617 e-3  , -3.563278473 e-2
I4 =  5.344584748 e-5  , -2.259916891 e-6

The results are indeed a bit different. To find out which is better, I compared with Free42:

Solution on Free42 (rounded to 10 digits):
I1 =  1.995795134 e-4  ,  4.096399075 e-3
I2 = -1.448833616 e-3  , -3.563298308 e-2
I3 = -1.454083170 e-3  , -3.563276083 e-2
I4 =  5.344581171 e-5  , -2.259868251 e-6

I was expecting my results to be more accurate but to my surprise, both 15C results are inaccurate !


So the challenge is:

Improve the solution on the 15c to get 9-10 correct digits for each term,
that is with an error of a few ULP on each term.
Solutions for the 41C (for instance with the Advantage ROM) are welcome too.

I will post my simple 15c solution and comments in a few days.

Note: the Free42 reference solution is obtained by using exactly the same system matrix with 10 significant digits, by doing SCI 9 RND on the system matrix before solving it.

J-F