# Solving Systems of Nonlinear Multivariate Equations

• Aug 30th 2010, 07:06 AM
BlueEnergyServices
Solving Systems of Nonlinear Multivariate Equations
Background: I am working on a project involving blending metallurgical coals, and developing a blending model involves repeatedly solving systems of nonlinear multivariate equations.

What I Need: I need a means (either an excel or other program) that will enable me to repeatedly (and efficiently) solve systems of equations like the following:

(1) (19.4/0.62) = (1.5/a) + (44.1/b) + (27.6/c) + (1.5/d)
(2) (18.0/0.57) = (1.5/a) + (38.1/b) + (29.7/c) + (6.9/d)
(3) (15.7/0.48) = (1.5/a) + (38.0/b) + (38.0/c) + (0.8/d)
(4) (16.8/0.61) = (0/a) + (3.1/b) + (42.9/c) + (32.0/d)

Solving for the values of a, b, c & d (which are constants)

Note: The numbers in the equations have been rounded to one or two decimal places. I am not sure if this will be an issue when trying to solve for the four constants.

• Aug 30th 2010, 07:23 AM
Ackbeet
Here's what I would do: your equations are actually linear in the variables 1/a, 1/b, 1/c, and 1/d. So define new variables equal to those reciprocals. Then use standard methods to solve for the reciprocal variables. Finally, flip them over again to get your a, b, c, and d.

Question: are the coefficients of the matrix going to stay the same? Or do all the coefficients change for your repeat problems?
• Aug 30th 2010, 08:47 AM
BlueEnergyServices
Thank you Adrian! I just tested your suggestion, and it works perfectly. Here's what I did - I set up the matrix calculations using the MMULT and MINVERSE commands in excel, solved for the reciprocal variables, and then flipped them over to get the a,b,c & d values.

To answer your question: At least some (and usually all) of the coefficients (which are % vitrinoid types from the petrographic analysis of a coal sample) will change for repeat problems.

Below are my excel calculations to test your suggestion:

a b c d
2.0 5.0 7.0 2.0 3.32
35.0 7.0 4.0 3.0 13.38
45.0 8.0 36.0 4.0 22.61
4.0 7.0 23.0 11.0 8.65

Inverse Values Recipricals of Values
(0.04) 0.02 0.01 (0.00) a = 0.3 a = 3.3
0.27 0.02 (0.03) (0.05) b = 0.2 b = 4.4
0.01 (0.04) 0.03 (0.00) c = 0.2 c = 5.5
(0.18) 0.06 (0.05) 0.13 d = 0.2 d = 6.6

Test
a = b = c = d =
3.3 4.4 5.5 6.6

2 5 7 2 3.3
35 7 4 3 13.4
45 8 36 4 22.6
4 7 23 11 8.7

As for the four equations in my initial post, I think value of 0 [as in (0/a)] in the 4th equation prevented me from successfully solving for the correct value of a (in theory, it should be between 2 and 3). What I plan to do replace it with another equation that doesn't have 0 as a coefficient. Below are the calculations related to the equations in my initial post:

a b c d
1.5 44.1 27.6 1.5 31.29
1.5 38.1 29.7 6.9 31.58
1.5 35.7 38.0 0.8 32.71
- 3.1 42.9 32.0 27.54

Inverse Values Recipricals of Values
(3.73) 4.06 0.34 (0.71) a = 3.0 a = 0.3
0.14 (0.10) (0.04) 0.02 b = 0.3 b = 3.0
0.02 (0.07) 0.06 0.01 c = 0.4 c = 2.4
(0.03) 0.11 (0.07) 0.01 d = 0.3 d = 3.8

Thanks again!
• Aug 30th 2010, 08:55 AM
Ackbeet