How Can You Solve Systems of Four Equations with Four Unknowns?

[A&X]

Homework Statement

Hello,

Although this could be labeled as an EE problem, this is more of an Algebra problem. Anyways, I am having a very hard time solving for the following 4 equations with 4 unknonws.

269.0 = R1*(R2+R3+R4)/(R1+R2+R3+R4)

267.8 = R2*(R1+R3+R4)/(R1+R2+R3+R4)

265.8 = R3*(R1+R2+R4)/(R1+R2+R3+R4)

267.3 = R4*(R1+R2+R3)/(R1+R2+R3+R4)

Homework Equations

Can I set R2 & R3 = 0 to make it easier to simplify it?

The Attempt at a Solution

This is what I was able to narrow it down to:

R4=267.3(R1+R2+R3)/(R1+R2+R3-267.3)
R3=268.8(R1+R2+R4)/(R1+R2+R4-268.8)
R2=267.8(R1+R3+R4)/(R1+R3+R4-267.8)
R1=269.0(R2+R3+R4)/(R2+R3+R4-269.0)

I don't know what to do from here on now. When I tried replacing R1 into R2, R2 into R3, etc. it just kept getting messier and it didn't lead to anything promising.

So I figured I set R2 & R3 = 0 to make it easier, but I don't know if I am allowed to do that. Can someone please lead me to the right direction?

Your help is much appreciated,

Thank you.

 

[jedishrfu]

I guess I'd try a numerical solution (ie write a program) trying all combinations of r1, r2, r3 and r4 within a given range. I can't see any way to reduce it algebraically and I don't think assuming r2=r3=0 will help either.

What if you tried multiplying both sides by (r1+r2+r3+r4) and then tried doing linear algebra on it to cancel righthand side terms like r2*r3 vs r3*r2 (subtract eqn3 from eqn2 ...)? I didn't try it but it probably still gets messy.

 

[wotanub]

I don't think there is a way to do it analytically.

Run it through your favorite computer program, there are many solutions.

 

[A&X]

Thank you for all the response.

Using Matlab to solve, I think this is what the code would be:

[R1,R2,R3,R4]=solve('R1=269.0(R2+R3+R4)/(R2+R3+R4-269.0)','R2=267.8(R1+R3+R4)/(R1+R3+R4-267.8)','R3=268.8(R1+R2+R4)/(R1+R2+R4-268.8)','R4=267.3(R1+R2+R3)/(R1+R2+R3-267.3)','R1','R2','R3','R4')

However, I do not have access to MATLAB at the moment to check. Correct me if I am wrong

 

[jedishrfu]

Okay, MATLAB says:

Undefined function 'solve' for input arguments of type 'char'.

 

[rcgldr]

I created a program to search for values with minimum least square difference, to 4 digits I get:

R1 = 359.6940
R2 = 357.2815
R3 = 353.2965
R4 = 356.2810

but this doesn't actually solve the 4 equations.

 

[gneill]

I created a program to search for values with minimum least square difference, to 4 digits I get:

R1 = 359.6940
R2 = 357.2815
R3 = 353.2965
R4 = 356.2810

but this doesn't actually solve the 4 equations.

It is a numerical solution. There are many cases in practice where sets of equations are intractable algebraically and numerical solutions are required.

 

[Chestermiller]

Homework Statement

Hello,

Although this could be labeled as an EE problem, this is more of an Algebra problem. Anyways, I am having a very hard time solving for the following 4 equations with 4 unknonws.

269.0 = R1*(R2+R3+R4)/(R1+R2+R3+R4)

267.8 = R2*(R1+R3+R4)/(R1+R2+R3+R4)

265.8 = R3*(R1+R2+R4)/(R1+R2+R3+R4)

267.3 = R4*(R1+R2+R3)/(R1+R2+R3+R4)

Homework Equations

Can I set R2 & R3 = 0 to make it easier to simplify it?

The Attempt at a Solution

This is what I was able to narrow it down to:

R4=267.3(R1+R2+R3)/(R1+R2+R3-267.3)
R3=268.8(R1+R2+R4)/(R1+R2+R4-268.8)
R2=267.8(R1+R3+R4)/(R1+R3+R4-267.8)
R1=269.0(R2+R3+R4)/(R2+R3+R4-269.0)

I don't know what to do from here on now. When I tried replacing R1 into R2, R2 into R3, etc. it just kept getting messier and it didn't lead to anything promising.

So I figured I set R2 & R3 = 0 to make it easier, but I don't know if I am allowed to do that. Can someone please lead me to the right direction?

Your help is much appreciated,

Thank you.

I have a different way of approaching this problem that may work. Rewrite the first equation as:
[tex]269.0 = R1*(R2+R3+R4)/(R1+R2+R3+R4)=R1*(R1+R2+R3+R4-R1)/(R1+R2+R3+R4)[/tex][tex]269.0=R1-\frac{(R1)^2}{(R1+R2+R3+R4)}[/tex]
Next:
[tex]\frac{269.0}{(R1+R2+R3+R4)}=\frac{R1}{(R1+R2+R3+R4)}-\frac{(R1)^2}{(R1+R2+R3+R4)^2}[/tex]
Equivalently:
[tex]\frac{269.0}{y}=x_1-(x_1)^2[/tex]
where y=(R1+R2+R3+R4) and [itex]x_1=\frac{R1}{(R1+R2+R3+R4)}=\frac{R1}{y}[/itex]
From the other equations, we have:
[tex]\frac{267.8}{y}=x_2-(x_2)^2[/tex][tex]\frac{265.8}{y}=x_3-(x_3)^2[/tex][tex]\frac{267.3}{y}=x_4-(x_4)^2[/tex]
Clearly, [itex]x_1+x_2+x_3+x_4=1[/itex]
If we solve the first equation for x₁ in terms of y, we obtain:
[tex]x_1=\frac{1-\sqrt{1-\frac{(4)(269.0)}{y}}}{2}[/tex]
(I'm guessing that the minus sign is the correct one to use.)
We obtain similar equations for x₂, x₃, and x₄. If we add the four solutions for the x's together, we eliminate the x's using [itex]x_1+x_2+x_3+x_4=1[/itex], and obtain a single equation in one unknown for y.

 

[rcgldr]

... obtain a single equation in one unknown for y.

an equation with 4 roots, which could be another issue.

Here is the source code to the program I used:

/* 269.0 = R1*(R2+R3+R4)/(R1+R2+R3+R4) */
/* 267.8 = R2*(R1+R3+R4)/(R1+R2+R3+R4) */
/* 265.8 = R3*(R1+R2+R4)/(R1+R2+R3+R4) */
/* 267.3 = R4*(R1+R2+R3)/(R1+R2+R3+R4) */

#include <stdio.h>
#include <math.h>

static double A[4] =            /* equation values, actual */
    {269.0, 267.8, 265.8, 267.3};
static double B[4], C[4];       /* equation values, best, current */
static double R[4], S[4];       /* variables, best, current */
static double SMIN[4], SMAX[4]; /* range for current variables */
static double BLSQ, CLSQ;       /* least squares, best, current */
static double DS;               /* delta step */

int main(void)
{
    BLSQ = 100000000.;          /* init best least squares to large value */
    DS   = 1000.;               /* init delta step */
    R[0] = 1000.;               /* init variables */
    R[1] = 1000.;
    R[2] = 1000.;
    R[3] = 1000.;
    while(DS > .0000005){
        SMIN[0] = R[0]-DS;      /* set ranges */
        SMIN[1] = R[1]-DS;
        SMIN[2] = R[2]-DS;
        SMIN[3] = R[3]-DS;
        SMAX[0] = R[0]+DS;
        SMAX[1] = R[1]+DS;
        SMAX[2] = R[2]+DS;
        SMAX[3] = R[3]+DS;
        DS /= 10.;
        for(S[0] = SMIN[0]; S[0] <= SMAX[0]; S[0] += DS){    
            for(S[1] = SMIN[1]; S[1] <= SMAX[1]; S[1] += DS){    
                for(S[2] = SMIN[2]; S[2] <= SMAX[2]; S[2] += DS){    
                    for(S[3] = SMIN[3]; S[3] <= SMAX[3]; S[3] += DS){    
                        C[0] = S[0]*(S[1]+S[2]+S[3])/(S[0]+S[1]+S[2]+S[3]);
                        C[1] = S[1]*(S[0]+S[2]+S[3])/(S[0]+S[1]+S[2]+S[3]);
                        C[2] = S[2]*(S[0]+S[1]+S[3])/(S[0]+S[1]+S[2]+S[3]);
                        C[3] = S[3]*(S[0]+S[1]+S[2])/(S[0]+S[1]+S[2]+S[3]);
                        CLSQ =  (A[0]-C[0])*(A[0]-C[0]) + 
                                (A[1]-C[1])*(A[1]-C[1]) +
                                (A[2]-C[2])*(A[2]-C[2]) +
                                (A[3]-C[3])*(A[3]-C[3]);
                        if(CLSQ < BLSQ){    /* if new best */
                            BLSQ = CLSQ;    /*   update params */
                            B[0] = C[0];
                            B[1] = C[1];
                            B[2] = C[2];
                            B[3] = C[3];
                            R[0] = S[0];
                            R[1] = S[1];
                            R[2] = S[2];
                            R[3] = S[3];
                        }
                    }
                }
            }
        }
        printf("A = %12.7lf %12.7lf %12.7lf %12.7lf\n",A[0],A[1],A[2],A[3]);
        printf("B = %12.7lf %12.7lf %12.7lf %12.7lf\n",B[0],B[1],B[2],B[3]);
        printf("R = %12.7lf %12.7lf %12.7lf %12.7lf\n",R[0],R[1],R[2],R[3]);
        printf("\n");
    }
    return(0);
}

 

[The Electrician]

Mathematica is able to find exact solutions, but they are VERY complicated. Here is an example of just one solution for R1:

Numerically, here is a solution good to quite a few decimal places, with the four expressions evaluated with the solution, showing its accuracy:

 

[Chestermiller]

an equation with 4 roots, which could be another issue.

This physical problem is going to have only one physically realistic answer. I can obtain a very accurate numerical solution to that answer by hand with very little effort. From my previous post, the algebraic equation I get for y is as follows:
[tex]\sqrt{\frac{1}{4}-\frac{(269.0)}{y}}+\sqrt{\frac{1}{4}-\frac{(267.8)}{y}}+\sqrt{\frac{1}{4}-\frac{(265.8)}{y}}+\sqrt{\frac{1}{4}-\frac{(267.3)}{y}}=1[/tex]
To solve this numerically, all I need is a good initial guess for y:
[tex]y^{(0)}=\frac{4s}{3}[/tex]
where s = (269.0+267.8+265.8+267.3)
and then I write:
[tex]y=\frac{4s}{3}+δ[/tex]
where δ is the parameter I will be solving for.
I also write:
[tex]269.0=\frac{s}{4}+(269.0-\frac{s}{4})[/tex]
where [itex](269.0-\frac{s}{4})[/itex] is small compared to s/4
I write similar equations for the other three values. I then substitute into the equation for y, and linearize with respect to δ. I then solve the simple linear equation for δ. I guarantee this will give an accurate answer for y.

 

[epenguin]

.

So I figured I set R2 & R3 = 0 to make it easier, but I don't know if I am allowed to do that. Can someone please lead me to the right direction?

.

Because the LHS all nearly equal, if you are doing it by successive approximations R1=R2=R3=R4 will give you an R s that are all close to a solution.

They will probably be the engineering one, not obvious whether it is the only one

 

[A&X]

Thank you all for all the inputs. This problem is for a strain gage with values in a closed bridge with 4 measurements (for those who were curious what this problem was about).

Here's how I was able to solve it using Matlab:

>>equ1='269.0-R1*(R2+R3+R4)/(R1+R2+R3+R4)';
>>equ2='267.8-R2*(R1+R3+R4)/(R1+R2+R3+R4)';
>>equ3='265.8- R3*(R1+R2+R4)/(R1+R2+R3+R4)';
>>equ4='267.3 - R4*(R1+R2+R3)/(R1+R2+R3+R4)';
>>sol=solve(equ1,equ2,equ3,equ4);
>>R1=sol.R1
>>R2=sol.R2
>>R3=sol.R3
>>R4=sol.R4

R1 =

359.69396536017785860423709040818
1.3507021132304585888310244089084
-0.15281241991439544414763284998392
1.8490642922477590181409957542694


R2 =

357.28145453217212584831447055658
-1.3544948044533407508506455121324
0.15238453051809368598400290995311
1.8449211357365443870556345198613


R3 =

353.29653057117303395043663397118
-1.3494402487364598096284603749859
-0.1519010380142547092615828782759
-1.8507933465933516936061609699429


R4 =

356.28105339639228188360351205007
1.3464165716255193771005390620164
0.15224216764211807620525039756688
-1.8559902351679833508578661774425

And since R's cannot be negative, the only possible solution would be:
R1= 359.69396536017785860423709040818
R2=357.28145453217212584831447055658
R3=353.29653057117303395043663397118
R4=356.28105339639228188360351205007

 

[Chestermiller]

For whatever it's worth, I completed the linearized analysis of this problem that I outlined in my two previous posts in this thread. The final analytic results came out very simple, and the calculations can be readily done on a hand calculator:
[tex]R_j=2L_j-\frac{(L_1+L_2+L_3+L_4)}{6}[/tex]
where, in this example,

L₁=269.0
L₂=267.8
L₃=265.8
L₄=267.3

The results agree with the Matlab values to one unit in the fifth significant figure.

Chet

 

[A&X]

I checked your formula with different values for L's and compared with the MATLAB code and it is actually pretty accurate.

In fact, with this formula I can make a spreadsheet on excel which is much more user friendly. Thank you Chestermiller!

 

[A&X]

By the way Chestermiller, I tried following your work and I was able to follow along for the most part. However, one step you did that i cannot figure it out how you did it is the third step.

How do you go from: 269.0=R1∗(R1+R2+R3+R4−R1)/(R1+R2+R3+R4)
To: 269.0= R1− (R1)^2/(R1+R2+R3+R4)

 

[Chestermiller]

By the way Chestermiller, I tried following your work and I was able to follow along for the most part. However, one step you did that i cannot figure it out how you did it is the third step.

How do you go from: 269.0=R1∗(R1+R2+R3+R4−R1)/(R1+R2+R3+R4)
To: 269.0= R1− (R1)^2/(R1+R2+R3+R4)

You are going to kick yourself when you see this.
[tex]269.0=R1\frac{(R1+R2+R3+R4−R1)}{(R1+R2+R3+R4)}=R1\left(\frac{(R1+R2+R3+R4)}{(R1+R2+R3+R4)}-\frac{R1}{(R1+R2+R3+R4)}\right)[/tex]
Chet

 

[A&X]

Wow...how emberassing of me :shy:

 

[rcgldr]

[tex]R_j=2L_j-\frac{(L_1+L_2+L_3+L_4)}{6}[/tex]

I tried this and there are some differences, but it could be the the formula is correct, and that the L values have a relationship that doesn't quite match the stated values due to just 4 digits of precision. The first set of values are what I get from my program and match the MATLAB results to 10 digits. The second set of values are what I get from the linear formula:

R =  359.6939654  357.2814545  353.2965306  356.2810534
L =  269.0000000  267.8000000  265.8000000  267.3000000

R =  359.6833333  357.2833333  353.2833333  356.2833333
L =  268.9934789  267.7997038  265.7921330  267.2999141

If you substitute the L's for the R's in the original equation, say just the first one, does it work out?

L1 = R1*(R2+R3+R4) / (R1+R2+R3+R4) = (2*L1 - (L1+L2+L3+L4)/6)(...) / (...) ?

 

[Chestermiller]

I tried this and there are some differences, but it could be the the formula is correct, and that the L values have a relationship that doesn't quite match the stated values due to just 4 digits of precision. The first set of values are what I get from my program and match the MATLAB results to 10 digits. The second set of values are what I get from the linear formula:

R =  359.6939654  357.2814545  353.2965306  356.2810534
L =  269.0000000  267.8000000  265.8000000  267.3000000

R =  359.6833333  357.2833333  353.2833333  356.2833333
L =  268.9934789  267.7997038  265.7921330  267.2999141

If you substitute the L's for the R's in the original equation, say just the first one, does it work out?

L1 = R1*(R2+R3+R4) / (R1+R2+R3+R4) = (2*L1 - (L1+L2+L3+L4)/6)(...) / (...) ?

The equation I gave is an approximate solution to the equations, obtained by linearizing them. I'm sure you will agree that getting a solution that is accurate to 4 significant figures isn't too bad, especially since the input data is only accurate to 4 significant figures and since the approximate solution is so simple to implement (not requiring a computer).

Chet