Hello,
Here's how the problem goes "The loss in the group of s = 30 channels is B(30, y) = 0.2. What would be the loss in the selective selection of traffic of the same intensity if in the previous situation you used a) the first 15 common channels b) the last 15 common channels". The part that I don't get is the usage of common channels.
Without "common channels" this problem would sound like this: "Loss in the first half of the group of Su = 30 channels when selectively choosing is b = B (15, y) = 0.2. Determine: Loss in the second half of the group"
From the Erlang Table, for c = 15 and b = 0.2, the offered traffic is y = 15,608 Earl. The loss in the whole system is bsu = B (Su = c + s, y) = B (30; 15,608). According to the Erlang Tables, for Su = 30, y is less of the smallest traffic y1 = 16.684E, at b1 = 0.001.
Look at the graph.
bsu / b1 = y / y1 (graph), ie bsu = 15.6 × 0.001 / 16.7 = 0.00094. The loss in the second half of the beam is bs = B (su, y) / B (c, y) = 0.00094 / 0.2 = 0.0047.
The mentioning of "groups" and "common channels" in the problem confuses me and I don't quite know where to find the answer.
Any help would be appreciated.
Here's how the problem goes "The loss in the group of s = 30 channels is B(30, y) = 0.2. What would be the loss in the selective selection of traffic of the same intensity if in the previous situation you used a) the first 15 common channels b) the last 15 common channels". The part that I don't get is the usage of common channels.
Without "common channels" this problem would sound like this: "Loss in the first half of the group of Su = 30 channels when selectively choosing is b = B (15, y) = 0.2. Determine: Loss in the second half of the group"
From the Erlang Table, for c = 15 and b = 0.2, the offered traffic is y = 15,608 Earl. The loss in the whole system is bsu = B (Su = c + s, y) = B (30; 15,608). According to the Erlang Tables, for Su = 30, y is less of the smallest traffic y1 = 16.684E, at b1 = 0.001.
Look at the graph.
bsu / b1 = y / y1 (graph), ie bsu = 15.6 × 0.001 / 16.7 = 0.00094. The loss in the second half of the beam is bs = B (su, y) / B (c, y) = 0.00094 / 0.2 = 0.0047.
The mentioning of "groups" and "common channels" in the problem confuses me and I don't quite know where to find the answer.
Any help would be appreciated.
Attachments

34.6 KB Views: 1