Question

1. 1)  Assume that, in a Stop-and-Wait ARQ system, the bandwidth of the line is 64 Mbps, and 1 bit takes 16 ms to make a round trip. What is the bandwidth-delay product? If the system data frames are 1500 bits in length and the utilization percentage of the link 80%. How many frames that can send up before stopping and worrying about the acknowledgments?

   2)  A  Hamming code needs a dataword of at least 24 bits. Calculate the values of k and n that satisfy this requirement?

Answer 1

Ans:

Explanation:

Given that,

The bandwidth of line = 64mbps

1 bit takes 16 ms make round trip

The length of system data frams =1500 bits

Bandwidth- delay product =64×10⁶×16×10^-3

^{= 1024000 bits}

^​​​​

1024000×80/100 = 819.2

during this time it takes the data go from the sender to recipient and then back again the device will send 1024000 bits

The unit However send 1500 bits we can say that the use of the link is only 1500/1024000 for this reason we use stop anand wait arq is wasting the link power of the connection with high bandwidth Or long delay.