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Two peer processes A(sender) and B(receiver) use 'stop-and—wait ARQ to send packets over a Single link with capacity C. All packets have the same length of

100 bits. The round-trip time (which is the time until A receives an acknowledgment for a sent packet) is equal to 2 seconds. Assume that no packets or Ack's are dropped and that all packets and ACK's arrive error-free. Furthermore,assume that the capacity C is

equal to 100,000 bits per second.

Find the average (transmission) rate (in bits per

seconds) with which process A sends data to

process B? ‘

(a) 34.47 bps (b) 49.97 bps

c) 51.45 bps (d) 67.75 bps

100 bits. The round-trip time (which is the time until A receives an acknowledgment for a sent packet) is equal to 2 seconds. Assume that no packets or Ack's are dropped and that all packets and ACK's arrive error-free. Furthermore,assume that the capacity C is

equal to 100,000 bits per second.

Find the average (transmission) rate (in bits per

seconds) with which process A sends data to

process B? ‘

(a) 34.47 bps (b) 49.97 bps

c) 51.45 bps (d) 67.75 bps

+3 votes

Best answer

0 votes

dont get confused by the terms...here by transmission rate means the throughput of system= **datasize / total time**

100 / (TT+ 2PT)

**TT=data size / bandwidth**

= 100/ 100000= 1ms=0.001 sec

**2PT=round trip time=2 sec**

**transmission rate=** 100/(2+0.001) =** 49.97 bps**

**this means even the bandwidth provided is 100000 bps, but we can transmit with the rate of 49.97 bps.**

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