GATE Overflow Test Series | Computer Networks | Test 1 | Question: 19

Question

A $300\;\text{Mbps}$ link is carrying traffic from a number of sources. Each of them generates an on-off traffic stream; when the source is on, the rate of traffic is $12\;\text{Mbps}$, and when the source is off, the traffic rate is zero. The duty cycle, which is the ratio of on-time to off-time, is $1:4$. 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 $X.$ Assuming that all sources are synchronized and the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is $Y.$ The value of $X+ Y = $ ________

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https://gateoverflow.in/3611/gate-it-2006-question-67?show=159119#c159119
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Answer 1

When there is no buffer inorder to ensure no data loss we need
    
    $X = \left \lfloor \dfrac{300}{12}\right \rfloor = 25.$
    
    With buffer, the expected bandwidth utilization $= \dfrac{1}{1+4} \times 12 \times Y.$
    
    So, $\dfrac{1}{5} \times 12 \times Y \leq 300$
    
    $\implies Y\leq 125.$
    
    So, $X+Y = 150.$