[求助]基站最大允许同时通话用户数问题

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<p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">根据上述</span><span lang="EN-US">4</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">×</span><span lang="EN-US">3</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">频率规划方法，对于中国联通</span><span lang="EN-US">6MHz</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">的频率（</span><span lang="EN-US">96</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">～</span><span lang="EN-US">124</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">），采用</span><span lang="EN-US">4</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">×</span><span lang="EN-US">3</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">频率复用模式，</span><span lang="EN-US">BCCH</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">为</span><span lang="EN-US">111</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">～</span><span lang="EN-US">124</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">，共</span><span lang="EN-US">14</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">个频点，其余全部分配给</span><span lang="EN-US">TCH</span><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">，不考虑微蜂窝预留频点。则频率规划方案如下表：</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">频率组号&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;A1&nbsp;&nbsp; B1&nbsp;&nbsp;&nbsp; C1&nbsp;&nbsp;&nbsp;D1&nbsp;&nbsp; &nbsp;A2&nbsp;&nbsp;&nbsp; B2&nbsp;&nbsp;&nbsp; C2&nbsp;&nbsp;&nbsp; D2&nbsp;&nbsp;&nbsp; A3&nbsp;&nbsp;&nbsp; B3&nbsp;&nbsp;&nbsp; C3&nbsp;&nbsp;&nbsp; D3&nbsp;</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">各频率组频点号&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;124&nbsp; 123&nbsp; 122&nbsp;&nbsp;121&nbsp;&nbsp; 120&nbsp;&nbsp; 119&nbsp;&nbsp; 118&nbsp;&nbsp; 117&nbsp;&nbsp; 116&nbsp;&nbsp; 115&nbsp;&nbsp;&nbsp;114&nbsp;&nbsp; 113</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;110&nbsp; 109&nbsp; 108&nbsp; 107&nbsp;&nbsp; 106&nbsp;&nbsp; 105&nbsp;&nbsp; 104&nbsp;&nbsp; 103&nbsp;&nbsp; 102&nbsp;&nbsp;&nbsp;101&nbsp;&nbsp; 100&nbsp;&nbsp;&nbsp;&nbsp;99</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 98&nbsp;&nbsp; 97&nbsp;&nbsp; 96</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">最大复用模式为S3/2/2&nbsp; 那么请各问高手给我解答下&nbsp; 理论和实际上最大允许同时通话用户数是多少？？？？</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;"></span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">S3/2/2指的是一个基站三个小区&nbsp; 分别为3个 2个 和2个载频&nbsp; 一个载频率8个TS&nbsp; 是不是说上面的S3/2/2站型理论上最大允许（2+1+1）* 8 = 32 既全速率32个&nbsp; 半速率 64个对吗？？？？？</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;"></span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">那么一个TCH载频率上的8个TS是不是同时可以分给全速率8个用户通话啊 ？？？</span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;"></span></p><p class="MsoNormal" style="MARGIN: 4pt 0cm 4pt 31.5pt; tab-stops: 26.25pt 409.5pt; mso-para-margin-left: 3.0gd;"><span style="FONT-FAMILY: 宋体; mso-ascii-font-family: Arial; mso-hansi-font-family: Arial;">急切&nbsp; 望各位老手新手 只要知道的给我一个详细点的解答&nbsp; 先感谢了</span></p>