NON-BLOCKING SWITCHING SYSTEMS 415 



and outputs per swdtch in the fourth stage, then the following equation 

 gives the total number of crosspoints: 



0(5) = (2n - 1) [2N + i2m - 1) (f + £,)] (11) 



The partial derivative of this equation with respect to m when set 

 equal to zero yields: 



n = m^ (12) 



The partial derivative of this equation with respect to n when set 

 equal to zero yields the following equation: 



^j nrn(2n + 2m — 1) , ^^ 



^ = (2m - l)(n - 1) ^^^^ 



Equations (12) and (13) can be solved for n and m in terms of given 

 values of N. For example for N = 240, we obtain n = 6.81 and m = 

 3.56. 



SEARCH FOR THE SMALLEST N FOR A GIVEN U FOR THE THREE- 

 STAGE ARRAY 



For a given value of n, equation (7) furnishes a means for locating 

 that size of three-stage switching array which has N^ or fewer cross- 

 points. This can be done by setting equation (7) equal to N^: 



N' = {2n - 1) (2N + J) (14) 



and solving for N in terms of n. The solution is: 



^ s ?!^(H!^_i) (15) 



{n - 1)2 



Minimum values of N for given values of n are hsted in Table V. 

 This table also hsts the next highest N exactly divisible by n. From 

 this table it appears that when iV = 24, we have the smallest switching 

 array for which it may be possible to have less than N^ crosspoints. 

 However for N = 25, as sho\vn in Table I, equation (2) gives more than 

 N2 crosspoints. The problem is one of finding an array for N = 25 with 

 fewer than N" crosspoints. For this and all cases beyond, the next sec- 

 tion indicates that it is profitable to consider situations where N is not 

 exactly divisible by n. 



