A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1 : 2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,

A) 10 and 30

B) 12 and 25

C) 5 and 33

D) 15 and 22

I have solved the part (i) of the problem to find the minimum no. of stations & got 10 as there is no bucket,so for no data loss I equated the incoming traffic rate by S1 stations = Maximum capacity of the channel.In 2nd part I'm having trouble what will happen when a buffer is placed.Please help.Make me correct if 1st part is not correct.


A is correct. As 1st part is already described i will explain derivation of S2 only. Let there are N stations. When buffer is added then even when source is off, data will persist so considering duty cycle here we say that the actual data transmitted per unit time during the whole duty cycle is (1/3)10 [for 1 unit of on-time data is transmitted and for another 2 unit of off-time it stays in buffer so in 3 unit time 10Mbps data is tranferred]. This is the case will all other N-1 stations also. So adding all should be = Max capacity of channel. N(1/3)*10 = 100 N=30


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