In GBN

Ws +Wr=sequence no

Wr= 1

Ws=1+2a= 40000 bit

Sequence no = 40001

No of bit in sequence no=log2(40001)=15.28=16 bit

Ws +Wr=sequence no

Wr= 1

Ws=1+2a= 40000 bit

Sequence no = 40001

No of bit in sequence no=log2(40001)=15.28=16 bit

The Gateway to Computer Science Excellence

0 votes

Bandwidth(B)=100 Mbps,Distance(d)=10000 km Packet size(M)=100000 bits, Prop speed (v)= 2 * 10^8 m/s

we know in ideal case to send 1 packet we need Tt time.So,in Tt +2*Tp time,we can send Tt + 2*tp/Tt=1+ 2a packets (a=Tp/Tt).

therefore (a)sender window size(Ws) =1+ 2a and since it is GBN receiver window size(Wr)=1 .

(b)We know Ws +Wr<=available sequence numbers. Number of bits in sequence number field = ceil (log (2+ 2a)).

(c)In TCP, the timeout value is not static and so we need to use algorithms like Jacobson Algorithm for changing the timerout timer value after each transmission cycle but in data link layer it is fixed and its value is 2*RTT=2*2Tp=4Tp.

Putting the values,we get Ws=40000 bits,Wr=1,No of bits in sequence number =16,Timer out timer value=0.20 sec.

we know in ideal case to send 1 packet we need Tt time.So,in Tt +2*Tp time,we can send Tt + 2*tp/Tt=1+ 2a packets (a=Tp/Tt).

therefore (a)sender window size(Ws) =1+ 2a and since it is GBN receiver window size(Wr)=1 .

(b)We know Ws +Wr<=available sequence numbers. Number of bits in sequence number field = ceil (log (2+ 2a)).

(c)In TCP, the timeout value is not static and so we need to use algorithms like Jacobson Algorithm for changing the timerout timer value after each transmission cycle but in data link layer it is fixed and its value is 2*RTT=2*2Tp=4Tp.

Putting the values,we get Ws=40000 bits,Wr=1,No of bits in sequence number =16,Timer out timer value=0.20 sec.

0 votes

For maximum efficiency, no. of packets that can be sent = 1 + 2a where a = Tp/Tt

Tp = Distance/propagation speed = 10^7/(2*10^8) = .05 sec

Tt = Packet size/Bandwidth = 10^5/10^8 = 0.001 sec

Therefore, window size = 1 + 2a = 1 + 2*.05/.001 = 101 packets.

Total no. of sequences >= Ws + Wr = 101 + 1 = 102.

No. of bits needed to represent 102 sequences = logbase2(102) = 7.

Tp = Distance/propagation speed = 10^7/(2*10^8) = .05 sec

Tt = Packet size/Bandwidth = 10^5/10^8 = 0.001 sec

Therefore, window size = 1 + 2a = 1 + 2*.05/.001 = 101 packets.

Total no. of sequences >= Ws + Wr = 101 + 1 = 102.

No. of bits needed to represent 102 sequences = logbase2(102) = 7.

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.5k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.3k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.1k
- Non GATE 1.5k
- Others 1.5k
- Admissions 595
- Exam Queries 576
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 17

50,647 questions

56,497 answers

195,490 comments

100,816 users