##### Consider bandwidth of line is 100 Mbps and sequence

Consider bandwidth of line is 100 Mbps and sequence number of field consists 32 bits on a TCP machine. How much time it will take to cover all possible sequence numbers? ________________ sec (integer value only).

**Answer**

Correct answer: **328 sec**

Explanation will be posted soon.

100MB -> 1 sec

100*(2^20) bits -> 1 sec

(2^32)*8 bits-> (2^32)*8/100*(2^20)=327.68 sec=328 sec

sequence no is not a number given to every bit,

it is given every frame, without frame length (or size) this is an incomplete question

why you have further multiply with 8 ??

we take 10 Mbps as 10*(10^6) bits when we talk ablut bandwidth when we take data size as 10 Mb then we take 10*(2^20)

Answer should br 343

1 sec -----> (100*10^6)/8 seq no.

1 seq no --------> 8/(100*10^6) sec

For 32 bit -------> (2^32 * 8)/(100*10^6) = 343