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按楼主所说的计算公式 L9605和L9705算出来的Gb口下行峰值占用率不一样，求解释！谢谢！
NSVCL9705:NSVC接收NS PDU的总字节数(字节)GBGb口下行峰值占用率
NSVC索引=1, NSVC标识=1094, NSE标识=10925346999950.584865551
NSVC索引=0, NSVC标识=1093, NSE标识=10926377783270.697614692
NSVC索引=1, NSVC标识=1092, NSE标识=10915809848140.635492812
NSVC索引=0, NSVC标识=1091, NSE标识=10916360734890.695749907
NSVC索引=3, NSVC标识=1100, NSE标识=10927111049710.777820843
NSVC索引=2, NSVC标识=1099, NSE标识=10926666988040.729248489
NSVC索引=5, NSVC标识=1096, NSE标识=10915731005140.626868807
NSVC索引=6, NSVC标识=1098, NSE标识=10916496968240.710651383
NSVC索引=3, NSVC标识=1095, NSE标识=10916147909840.672470677
NSVC索引=2, NSVC标识=1097, NSE标识=10916448972640.705401528
BCL9606:配置的BC链路带宽(无)L9605:接收NS PDU峰值字节数(字节)GBGb口下行峰值占用率
承载信道标识=1, 槽位号=14, 框号=0192011841190.481819255
承载信道标识=0, 槽位号=14, 框号=0192011894280.483979492
承载信道标识=1, 槽位号=14, 框号=0198412088210.476003733
承载信道标识=0, 槽位号=14, 框号=0198412305570.484562831
承载信道标识=3, 槽位号=14, 框号=0192011880460.483417155
承载信道标识=2, 槽位号=14, 框号=0192011881370.483454183
承载信道标识=5, 槽位号=14, 框号=0198412270550.48318383
承载信道标识=4, 槽位号=14, 框号=0198412277700.483465379
承载信道标识=3, 槽位号=14, 框号=0198412265160.482971585
承载信道标识=2, 槽位号=14, 框号=0198412277150.483443722
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