546 



BELL SYSTEM TECHNICAL JOURNAL 



multiple is 



A = (1.13)(.529)(105) = 62.77. 



To obtain the desired load per subgroup we need only divide by 4 

 . . 62.77 



giving a = 



15.69. In terms of one-hundred-second-calls this 



average per subgroup then equals 36a — 565 calls per hour. 



As a second example we may set the question: Given 58 trunks to 

 grade on an access oi x -^ y = 20 with either three or six subgroups, 

 which arrangement is to be preferred? We shall have to assume, 

 since we have no additional information, that the decision rests merely 

 upon the relative efficiencies of the two schemes. From the given 

 data, gx -{- y = 58, x -\- y = 20, and g =^ 3 or g = 6. We read 



opposite 



gx + y _ 5S 



= ^r- = 2.90 for g = 3 and g = 6 on both the 



x -\- y 20 



P — .01 and the P = .001 charts (since the probability was not 

 specified) and construct the following table (Table II): 



TABLE II 



It is clear then, that as far as efficiency of arrangement goes, the 

 six-subgroup plan is in the order of 20 per cent superior to the three- 

 subgroup plan for ordinary grades of service. There is one difficulty 

 here, however, and that is that in the symmetrical multiple of six 

 subgroups the calculated number of individual trunks in each sub- 

 group and the number of commons are not integers. We may get 

 around this trouble by either unbalancing the grade slightly or chang- 

 ing the total number of trunks. For instance, we could use four 

 subgroups of seven trunks, each pair of subgroups feeding into a 

 single trunk common to them, before reaching the 12 through-commons 

 into which the remaining two subgroups of eight trunks would work 

 directly. The total of 58 trunks would then be disposed of at an 

 efficiency gain probably not differing markedly from that estimated 

 in our table above. 



Secondly, we could have reduced the total trunks to 55, making 

 thereby a symmetrical arrangement of x — 7 and >' = 13. Had we 



