Question

Computer A uses Stop and Wait ARQ to send packets to computer B. If the distance between A and B is 40000 km, the packet size is 5000 bytes and the
bandwidth is 10Mbps. Assume that the propagation speed is 2.4x108m/s
a) How long does it take computer A to receive acknowledgment for a packet?
b) How long does it take for computer A to send out a packet?

Answer 1

The time that it takes the computer to receive acknowledgment for a packet is 0.1667 seconds. The time it takes to send out a packet is  4 x 10⁻³seconds

1 The acknowledgment time for the packet

speed =  2.4x108m/s

Distance = 40000 km,

Time = distance/ speed

= 40000 x10³/ 2.4x10⁸m/s

= 0.1667

The time that it take is 0.1667 seconds.

b. Number of bytes = 5000

5000x 8 = 40000bits

10 mbps = 10000 kbps

10000 kbps = 10000000

packet size / bit rate = 40000/10000000

= 4 x 10⁻³seconds to send a packet out

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