TCP Window size

Question

Answer 1

1)  T_t <= RTT (because we cant transmit at more speed at given bandwidth. You cant have window size considering max segment lifetime)

Thus, window size = 16Mb

Thus, bits to represent this window  = 21 bits

2) We need atleast 22 bits as we do in GBN because we need to differentiate between the sequnce nos for the 1st window and next consecutive window. If we take 21 bits, consider the foolowing situation:

Host A transmits with seq no 0.

Host B receives it and now, as 2²¹ bytes are received, Host B sends ack no of next byte it expects. But it needs to send ACK 0 as sequence space is over for host B.

Now, host A wouldnt understand if this is proper ACK or if the previous packet was lost.

So, 21 bits are insufficient.

Comments

@ 

Sushant Gokhale  how did you get 16Mb  ?

Answer 2

a) Maximum segment life time =80s

Data transmitted in 80 s=(100/8)*10 6 *80 = 10 ⁸ bytes

window size in bits = log₂ (10 ⁸)= 8 log ₂ 10~ 24

b) Propagation delay is 80 ms. Hence round trip time is 160 ms.

One complete window is sent in this time by sender

No of bytes send in RTT= (100/8)*10 ⁶ *160 *10 ^-3 = 2 x 10 ⁶

No of bits = log ₂(2 x 10 ⁶) = log₂ 2+ log₂ 10 ⁶ = 1+6 log₂ 10 ~21 bits

Comments

@Shiva. 80 sec is the max packet lifetime considering congestion on the network. Lets consider ideal situation that there is no congestion at all. So, if you transfer $10^{8}$Bytes in just 2*80ms(i.e RTT ), the receiver would be overwhelmed. Also, this not possible at given bandwidth.

We must always have T_t <= RTT.  

Answer 3

1 sec=100MB

10^8 byte in =1 sec

1 seq number=1/10^8

so in 80sec=10^8*80/8

taking log on base 2 of 10^8*80/8=~27

So none of these will be option.