Remove redundant equation to solve the needed variables?

Questions
1

Hi,

I have revised the equations and fixed the errors. I need to solve the needed variables. I think there are redundant equations since I am looking for seven variables and there are nine equations. I don’t guess which ones to be removed? The ones that are not to be removed for sure are:

PPA0 + PPA1 + PPA2 + PPA3 == 1 and PPB0 + PPB1 + PPB2 == 1

Please find the file attached. Appreciate your kind help.

Seven_varibles_nine_equations_case3.nb (46 KB)

2

Your equations seem to be inconsistent:

Eliminate[{D11 PPA0 == (1 - d1) P11 PPA1,
  (D22 + (1 - d1) P11) PPA1 == D11 PPA0 + (1 - d2) P22 PPA2,
  (2 (1 - d2) P22 + D33) PPA2 == 
   D22 PPA1 + D22 PPB1 + (1 - d3) P33 PPA3,
  (1 - d3) P33 PPA3 == D33 PPA2,
  PPA0 + PPA1 + PPA2 + PPA3 == 1,
  D11 PPB0 == (1 - d1) P11 PPB1,
  (2 D22 + (1 - d1) P11) PPB1 == 
   D11 PPB0 + (1 - d2) P22 PPA2 + (1 - d2) P22 PPB2,
  D22 PPB1 == (1 - d2) P22 PPB2,
  PPB0 + PPB1 + PPB2 == 1},
 {PPA0, PPA1, PPA2, PPA3, PPB0, PPB1, PPB2}]

D11^2 (-1 + d2) D22^2 D33 P22 == 0
3

When I run Eliminate it returns: D11 (-1 + d2) D22 D33 P22 == 0 ??
Actually d2 cannot be zero. Is there other approximation technique to estimate the variables ?

4

If I used only PPA0 + PPA1 + PPA2 + PPA3 ==1 I got the following:

{{PPA0 -> 0.367931, PPA1 -> 0.272443, PPA2 -> 0.203368, 
  PPA3 -> 0.156258, PPB0 -> 0.367931, PPB1 -> 0.272443, 
  PPB2 -> 0.203368}}

If I used only PPB0 + PPB1 + PPB2 == 1 I got the follwing:

{{PPA0 -> 0.436071, PPA1 -> 0.322898, PPA2 -> 0.241031, 
  PPA3 -> 0.185196, PPB0 -> 0.436071, PPB1 -> 0.322898, 
  PPB2 -> 0.241031}}

But I need both conditions !

5

The equations are inconsistent so redundancy is more or less meaningless in this setting. Do you want a least-squares solution? Saying that two equations need to be removed gives no useful guidance as to which two should go.

6

To add more, I got those equations too:

PPA0 -> -(((-P11 P22 + d1 P11 P22 + d2 P11 P22 - d1 d2 P11 P22) PPA2)/(D11 D22))
PPA1 -> -(((-P22 + d2 P22) PPA2)/D22)
PPA3 -> -((D33 PPA2)/((-1 + d3) P33))
PPB0 -> ((-1 + d2) (-P11 + d1 P11) P22 PPA2)/(D11 D22)
PPB1 -> -(((-1 + d2) P22 PPA2)/D22)
PPB2 -> PPA2
7

Yes if it gives me a solution it will help me.

8

When I used PPA0 + PPA1 + PPA2 + PPA3 + PPB0 + PPB1 + PPB2 == 1 I got the following:

{{PPA0 -> 0.199557, PPA1 -> 0.147766, PPA2 -> 0.110302, 
  PPA3 -> 0.0847503, PPB0 -> 0.199557, PPB1 -> 0.147766, 
  PPB2 -> 0.110302}}