TCP slow start right ans????????????

Question

Answer 1

Im not sure whether question is wrong.
Here it is how i did it,
1st rtt 1packets
2nd rtt 2packets
3rd rtt 4packets

nth rtt 2^n-1 packets

Hence, Summation of packets will be (2^n+1 - 2)

To transmit 2000MB , 1000 packets are needed.

1000=2^n+1 - 2
log 998 = n+1
n= 9.96-1
n= 8.96 for 1000 packets

thus n transmission each with 100sec = 8.96*100 = 896/60 = 14.93 minutes

You can also calculate direct but for 9th(summation will be 1023) packet you will get 900/60 = 15 minutes

Comments

you are absolutely right,i forgot to mention that ,if we do direct 15 should be answer..I forgot to mention that part also.I did it as per qs,but ur 1st approach is perfect

Answer 2

the qs is wrong if 4 packets during 3 rd RTT,means 2^2 packets during 3 rd RTT,2^3 packets during 4 th RTT,so How 2^n packets during n th RTT,it should be n+1 RTT.

Though here we need ,no of packets 2000/2 =1000 ,almost 2^10.

so as per qs during 10 th RTT.

total 10*100 = 1000 sec = 1000/60 mint.

 

but originally,it should be 2^n-1 packets in nth RTT.

2^n-1 = 1000

so,n =9.

so total time =9*100= 900/60 = 15 minutes