solve for the (1,1) variable in matrix c) the (1,5) and (1,9) variables

must be torn and converged. To do this involves evaluating all of the

functions, since the order of the rows in the incidence matrix is the

order of solving the equations. To solve for the same variable in the

second matrix involves evaluating only three functions until conver-

gence. Since the equations in both incidence matrices are the same,

the extra function evaluations in the first method would be wasted. The

same number of iterations would be required to converge the first

variable for both methods, but the number of function evaluations would

not be the same.

Theorem 3-3: Let il be a function index whose IDM does not fully

precedence order and i2 be a function index whose IDM does fully

precedence order. If the two indices are adjacent in index ordering,

the number of function evaluations necessary for convergence of the set

of functions they describe for the ordering i2il will be less than for

the ordering ili2.

Proof: Convergence is necessary for blocks of variables with i2

constant, iI ranging over its allowable values. Call the set of

variables in a block with a particular value of i2 an is block. Let

the number of iterations required to converge each i2 block be k, .
 '2
 For the ordering i2il the number of function evaluations to converge

 the entire set of variables would be
 M2 M2
 NI = (k i2. ) 1= M1 k. 3-5
 i2=1 i= =1

 Mi and M2 are the number of rows (and columns) in the IDM's for i1 and

 i2 respectively. Equation 3-5 states that the total number of function

 evaluations is the sum of the number of iterations for each i2 block