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+1 vote

Part 1)

The 15 bit wide clock counter can produce 2^{15 }sequence numbers, producing 1 every 100 milliseconds. (1 msec = 10^{-3 }sec)

$\Rightarrow$ 1 sequence number is produced in 100 x 10^{-3 }sec,

$\Rightarrow$ 2^{15} sequence numbers can be produced in 2^{15 }x 100 x 10^{-3 }seconds = 3276.8 sec $\Rightarrow$ 54.61 minutes.

As the sequence numbers will start repeating after the expiry of lifetime (which is given as 60 sec, we dont have any issue)

Part 2)

240 sequence numbers in 60 sec,

$\Rightarrow$ 1 sequence number in $\Rightarrow \frac{60}{240}$ sec

$\Rightarrow$ 2^{15 }sequence numbers in $\Rightarrow \frac{60}{240} \times 2^{15}$ = 8192 sec $\Rightarrow$ 136.53 minutes

0 votes

i think this should be the answer

15 bit counter can count 2^15 sequence numbers . so total sequence number will be 2^15. total time to blow up all sequence number =100msec*2^15

=3276.800 seconds (worst case) we have to reset the counter. well as life time of packet is far less than the time to reset counter no issue will occur .

2-

if we use 240 sequence number per minute then.

total time to eat up all sequence number will be

2^15/240=136.5 minutes =137 minutes

15 bit counter can count 2^15 sequence numbers . so total sequence number will be 2^15. total time to blow up all sequence number =100msec*2^15

=3276.800 seconds (worst case) we have to reset the counter. well as life time of packet is far less than the time to reset counter no issue will occur .

2-

if we use 240 sequence number per minute then.

total time to eat up all sequence number will be

2^15/240=136.5 minutes =137 minutes

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