A reliable data transport protocol is to be implemented for an occasionally erro

Question

A reliable data transport protocol is to be implemented for an occasionally error-prone but loss-free link with round-trip-time (RTT) of 10 msecs and a link speed of 1 Mbyte per sccond as pictured below. Each packet is to carry an S-bit sequence number field and be 1000 bytes long (including all headers). te/sec Host packet format: 1000 byte packets, including an & bit sequence number (4) 9. The largest raw throughput in packets/sec that can be achieved if the stop-and-wait protocol is implemented is (a) 0.01 packets/sec (b) 0.1 packets/sec (c) 1 packet/sec (d) 10 packeta/sec (e) 100 packets/sec (0) 1000 peckets/sec 000000 1000 pa 1 000 0 ) 10. When the Go-Back-N (GBN) protocol is implemented, tbe minimum transmit window required to maximize raw throughput, i.e., to "fill the pipe", is (a) 1 packet (b) 10 packets (c) 100 packets (d) lk packets (e) 10k packets (f) 100k packets quence numbers , anl successively faster and faster links (with the transmit buffer scaled up accordingly) the raw throughput increases until, at a certain link speed, the protocol suddenly becomes unreliable. The link speed at which the protocol becomes-unreliable is (a) 1.275 Mbytes/sec 0-256 (b) 2.55 Mbytes/sec (e) 3.825 Mbytes/sec (d) 12.75 Mbytes/sec (e)25.5 Mbytes/sec 38.25 Mbytes/sec ,01

Explanation / Answer

9)

      We know in stop n wait protocol throghput is = 1 Packet / RTT

      Given RTT is = 10 msec = 0.01 sec

      Hence throghput is = 1 Packet / 0.01

                                  = 100 packets

      Hence throghput of stop n wait protocol is = 100 packets.

10)

        Given link speed is = 1 Mbyte / sec

        implies in 1sec we can send 1 Mbyte of data

        Now in the RTT of 10 msec we can send 10,000 bytes of data.

        Here 1 packet size is = 1000 bytes.

        Hence total number of packets is = 10,000 bytes / 1000 bytes = 10

        Hence window size to Max throughput is = 10.

11)

      Now using 8 bit sequence number we can have a total of

      256 sequence numbers in which 255 is for sender and 1

      is for receiver. Here 255 packets are the max that we can send

     Given 1 packet size is = 1000 bytes

     Hence for 255 packets it will be 255 * 1000 bytes = 255000 bytes

     Now in 1 RTT 0.01 sec the link speed is = 255000 bytes / 0.01

                                                               = 25.5 Mbytes /sec

     Hence the link speed after which protocol is unreliable is = 25.5 Mbytes /sec

     

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