Question

sender must send a 1 MB file to receiver using TCP. RTTT is 100 ms, if advertised window of the receiver is 64KB, and each segment is 1 KB, and there is no congestion and loss in the network. How long does it take to transmit the file ?

Answer 1

Step 1 :- Data given is

File size is = 1MB

RTT (Round Trip Time ) = 2Tp = 100ms

Where Tp is Propagation Delay

-> Advertised window of receiver is 64KB = wr

->And segment size is 1KB which is Congestion capacity = Wc

So,sender window will be Ws= min(wc,wr) = min(1kB,64kb) = 1KB

->Total number of segments required to send 1MB file is = 1MB/1KB = 2^10 segments

Step 2 :-

According to TCP congestion algorithm ,

Threshold value will be = Wr/2 = 64kB/2 = 32KB

So firstly sender will send packet in exponential manner upto threshold value then after that in linear manner upto 64KB like this

1MSS,

2MSS,

4MSS,

8MSS,1

6MSS,

32MSS,

33MSS,

34MSS,

35MSS,

36MSS,

37MsS,

38MSS,

39MSS,

40MSS,

41MSS,

42MSS,

43MSS,

44MSS,

45MSS,

46MSS,

47MSS,

48MSS,

49MSS,

50MSS,

51MSS,

52MSS,

53MSS,

54MSS,

4MSS

Total transmissions = 29

Last segment is 4MSS because upto 54MSS we have already sent 1020 segments and remaining 4 segments will be send at last

Where MSS is Maximum Segment Size

Step 3:- Total time required to send 1MB file is = Total number of transmission which is 29 transmission when you will add 1,2,4 like this all will be counted as single transmission = 29*2*Tp = 58*100ms = 5800ms