Back substitution

The Gaussian elimination method provides a systematic approach for implementation of the described forward reduction and back substitution processes for large systems of algebraic equations. 

Step 6 - the remaining unknowns are found by back substitution using the following formula from the ( - l)st to the first equation in turn... 

To estimate the computational time required in a Gaussian elimination procedure we need to evaluate the number of arithmetic operations during the forward reduction and back substitution processes. Obviously multiplication and division take much longer time than addition and subtraction and hence the total time required for the latter operations, especially in large systems of equations, is relatively small and can be ignored. Let us consider a system of simultaneous algebraic equations, the representative calculation for forward reduction at stage is expressed as... 

Hence the solution is found by back substitution based on... 

A set of n first-degree equations in n unknowns is solved in a similar fashion by multiplication and addition to eliminate n - 1 unknowns and then back substitution. Second-degree equations in 2 unknowns may be solved in the same way when two of the following are given the product of the unknowns, their sum or difference, the sum of their squares. For further solutions, see Numerical Methods. ... 

Back substituting equation 40-lb into equation 40-2, we obtain... 

The previous chapter showed how the reverse Euler method can be used to solve numerically an ordinary first-order linear differential equation. Most problems in geochemical dynamics involve systems of coupled equations describing related properties of the environment in a number of different reservoirs. In this chapter I shall show how such coupled systems may be treated. I consider first a steady-state situation that yields a system of coupled linear algebraic equations. Such a system can readily be solved by a method called Gaussian elimination and back substitution. I shall present a subroutine, GAUSS, that implements this method. 

Program 0GC03 solves the steady state ocean model using Gaussian elimination and back substitution. 

GAUSS Subroutine GAUSS solves a system of simultaneous linear algebraic equations by Gaussian elimination and back substitution. The number of equations (equal to the number of unknowns) is NROW. The coefficients are in array SLEQ(NR0W,NR0W+1), where the last column is the constants. 

Calculate unknowns by back substitution unk(nrow) = sleq(nrow, ncol)... 

The steady-state problem yields a system of simultaneous linear algebraic equations that can be solved by Gaussian elimination and back substitution. I shall turn now to calculating the time evolution of this system, starting from a phosphate distribution that is not in steady state. In this calculation, assume that the phosphate concentration is initially the same in all reservoirs and equal to the value in river water, 10 I 3 mole P/m3. How do the concentrations evolve from this starting value to the steady-state values just calculated ... 

Program DGC04 solves the time-dependent ocean simulation by the reverse Euler method using Gaussian elimination and back substitution, nrow = 5 the number of equations and unknowns... 

The application of the reverse Euler method of solution to a system of coupled differential equations yields a system of coupled algebraic equations that can be solved by the method of Gaussian elimination and back substitution. In this chapter I demonstrated the solution of simultaneous algebraic equations by means of this method and showed how the solution of algebraic equations can be used to solve the related differential equations. In the process, I presented subroutine GAUSS, the computational engine of all of the programs discussed in the chapters that follow. 

In the procedure of back substitution, therefore, by which the final values of the unknowns are calculated, it is necessary to refer only to the previous unknown. Earlier unknowns do not enter into the back substitution, and so this stage of the calculation is a lot shorter also. 

This chapter demonstrated the computational simplification that is possible in systems consisting of a one-dimensional chain of coupled reservoirs, which arise in diffusion and heat conduction problems. In such systems each equation is coupled just to its immediate neighbors, so that much of the work involved in Gaussian elimination and back substitution can be avoided. I presented here two subroutines, GAUSSD and SLOPERD, that deal efficiently with this kind of system. 

A factorization of O of this form can be carried out using standard matrix algorithms such as LU decomposition or Gaussian elimination with back-substitution [2,34,35]. The main advantage is the straightforward evaluation of the stmcture transformation corresponding to Eq. (3), which is just... 

The concentration profile at f = n + 1 is then found through back-substitution from / = /- lto/=2. Rearranging equation (7.32) ... 

Assuming that the original problem was solved using Newton s method, the Jacobian J, evaluated at the solution, is already known and factored into its LU components. The 3F/9a matrix is determined one column at a time by finite-difference perturbation of the parameter, analogous to the procedure described by Eq. 15.36. For each column of dF/da, a column of the sensitivity matrix 3y/3a is determined via a back-substitution using the Jacobian s LU factors. 

By solving for Al3+ and SO4- and back substitution into the mass action expressions, the concentration of each species can be calculated. The proportion of each species as a proportion of the total aluminium is given in Fig. 5.3. The result shows the tendency of SO4 to form complexes in solution with aluminium in the pH range 3-5. 

Use back-substitution to determine the solution whose leading coefficient corresponds to the rightmost undetermined pivot element. Set this coefficient to unity. 

For a tridiagonal linear system, Gaussian elimination simplifies to a simple algebraic factorization followed by back-substitution. The time taken is linearly proportional to the number of equations. 

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