GATE Overflow Test Series | Computer Networks | Test 1 | Question: 17

Question

A link has transmission speed of $5\times 10^5$ bits/sec. It uses data packets of size $2000\;\text{bytes}$ each. Assume that the acknowledgment has negligible transmission delay and that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly $35\%.$ The value of the one way propagation delay (in milliseconds) is ___________

Answer 1

In stop-and-wait protocol, once a packet is sent, the sender waits until ACK comes back which takes one Round Trip Time (RTT).

RTT = Transmission Time for frame + Propagation Delay for frame + Transmission Time for ACK  +Propagation Delay for Ack
    
 $\quad \dfrac{2000 \times 8}{5 \times 10^5}+ 2PD = 0.032 + 2PD$
    
Efficiency = Amount of Data Sent/Amount of data that could be sent.
    
$ \qquad = \dfrac{2000 \times 8 }{5 \times 10^5 \times RTT} = 0.35$
    
So, we get $RTT = 0.0.0914 \implies 0.032 + 2PD = 0.0914$
    
  $\qquad \implies PD = 0.0297s= 29.7\;\text{ms}$

Comments

2000?

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yes

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Where is it missed?