# GATE | GATE IT 2006 | Question 67

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

Explanation: 10 Mbps * n <= 100 Mbps {n= number of stations}
n = 10
if n is very large:
a station transmits with probability 1/(1+2 = 1/3, with a capacity of 10Mbps
so, for n stations:
(1/3)*10+(1/3)*10+(1/3)*10+ ……… + n times <= 100 Mbps
(1/3)*10[1+1+1+1+…….+ n times] <= 100
(1/3)*10*n<=100
n=30

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