Parametric Systems of Equations

# Solving Linear Parametric RHSSystem of Equations with Applications

This site is a part of the JavaScript E-labs learning objects for decision making. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this page.

Professor Hossein Arsham

In many applied decision making problems such as sensitivity analysis of linear optimization, one needs to solve a linear parametric right-hand-side (RHS) system of equations. Often varieties of parametric RHS systems of equations can be reduced to the following standard form:

AX = b + Dr,

where A is an n by n non-singular matrix, the column matrix b the numerical value of the RHS, and diagonal matrix D contains the coefficients of the parameters r = [r1, r2, …rn]T, respectively.

The following JavaScript solve parametric systems of up to 7 equations. Starting at the upper left table, enter matrix A and then the RHS column, followed by an n by n diagonal matrix D, then click on Calculate button. The JavaScript returns the constant term in the solution as the elements of the (n+1) column, followed by coefficient of r as a square matrix.

Application 1: Suppose we wish to solve the following RHD parametric systems of 2 equations with 2 unknowns:

2 X1 + X2 = 40 + r1
X1 + 2 X2 = 50 + r2

By entering the matrix [A, b, D]:

2    1    40    1    0
1    2    50    0    1

and after clicking on the Calculate button, we get the following parametric solution:

X1 = 10 + 2/3r1 - 1/3 r2
X2 = 20 - 1/3r1 + 2/3r2

Application 2: Under what conditions on values of r1 and r2 the solution to the following parametric system of equations remains non-negative?

2 X1 + X2 = 40 + .4r1
X1 + 2 X2 = 50 + .5r2

Following the above data entering instruction, the entering table is [A, b, D]:

2    1    40    0.4    0
1    2    50    0    0.5

The parametric solution is:

X1 = 10 + 4/15r1 - 1/6r2
X2 = 20 - 2/15r1 + 1/3r2

Therefore, in order to have non-negative solution, the feasible region for r1 and r2 is the set:

S = {r1, r2 | 10 + 4/15r1 -1/6r2 ³ 0, 20 -2/15r1 + 1/3r2 ³ 0}.

Application 3: Note that one may use this JavaScript to solve the ordinary, i.e., non-parametric system of equations by setting the matrix D to zero matrix, and further more not entry matrix D in the data matrix.

A straightforward JavaScript for solving systems of equations up to 10 equations and be found at a Solving System of Linear Equations .

Application 4: To find the inverse of a square matrix A enter the matrix A and then an identity matrix of the same size, then click on Calculate button. The following numerical example illustrates the process:

Suppose we wish to find the inverse (A-1 ) of the following matrix (if it exists) A:

 2 1 A = 1 -1

Entring the matrix A and then the 2 by 2 identity matrix, the output is: This gives X1 = 1/3 , X2 = -2/3. Therefore, A-1 is:

 1/3 1/3 A-1 = 1/3 -2/3

For an alternative approach, use Solving System of Linear Equations which computes the inverse of up-to 10 by 10 matrix.

Warning: In all applications and cases, after clicking on the Calculate button, the output must contain an identity matrix appearing on the left-hand-side of the table. Otherwise, the entering matrix might have been a singular matrix.

For Technical Details, Back to:
Construction of the Sensitivity Region for LP Models

Professor Hossein Arsham

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