Go back N ARQ MCQ in Data Link Layer

1. In Go back N ARQ Duplicate packet problem can be solved by

• A. Increasing receiving window size
• B. Increasing sending window size
• C. Increasing sequence number
• D. Decreasing sequence number
• E. None of the above

Ans- C. Increasing sequence number

Solution:- Duplicate packet problem can be solved by two ways:-

Ws+Wr=< Available Sequence Number

1. Increasing the size of sequence number
2. Decreasing the sending window size

2. In Go back N ARQ if sending window size is 3 and sender wants to send 8 packet. If every 4^th packet lost than How many transmission is required to send all these packets.

• A.19
• B. 18
• C. 20
• D. 22
• E. None of the above

Ans- A.19

Total Number of transaction we get 19

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3. In Go Back N ARQ receiver window size is always…

• A. 5
• B. 6
• C. 2
• D. 1
• E. None of the above

Ans- D. 1

3. In Go Back N ARQ receiver window size is always…

• A. 5
• B. 6
• C. 2
• D. 1
• E. None of the above

Ans- D. 1

4. In Go Back 5 ARQ sender window size is…

• A. 5
• B. 6
• C. 2
• D. 1
• E. None of the above

Ans- A. 5

5. In Go Back 5 ARQ available sequence number range is…

• A. 0-6
• B. 0-4
• C. 0-5
• D. 0-1
• E. None of the above

Ans- C. 0-5

6. In Go Back N ARQ if sender window size is N than available sequence number range is…

• A. 0- N -1
• B. 0- N+1
• C. 0 – N
• D. 0-N/2
• E. None of the above

Ans-C. 0 – N

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7. In Go Back N ARQ if highest sequence number is N than sender window size is …

• A. N -1
• B. N+1
• C. N
• D. N/2
• E. None of the above

Ans-C. N

8. In Go Back N ARQ if size of sequence number is 6 bit than total sequence number is…

• A. 62
• B. 64
• C. 66
• D. 63
• E. None of the above

Ans-B. 64

9. In Go Back N ARQ if size of sequence number is K bit than total sequence number is…

• A. 2^k
• B. 2^k-1
• C. 2^k+1
• D. 2^k+2
• E. None of the above

Ans-A. 2^k

10. In Go Back N ARQ if size of sequence number is K bit than window sender size is…

• A. 2^k
• B. 2^k-1
• C. 2^k+1
• D. 2^k+2
• E. None of the above

Ans-A. 2^k-1

11. In Go Back N ARQ if sender window size is 6 than min sequence number required is…

• A. 8
• B. 6
• C. 5
• D. 7
• E. None of the above

Ans-D. 7

12. In Go Back N ARQ if 256 bit size packet is to be send on a link with 8 Kbps bandwidth. if one way propagation delay is 10msec and value of n is 10 than find efficiency and data rate of this link.

• A. 8 and 64 Kbps
• B. 8 and 128 Kbps
• C. 16 and 256 Kbps
• D. 24 and 32 Kbps
• E. None of the above

Ans-A. 8 and 64 Kbps

Solution:-In Go back n ARQ Efficiency = N* Useful Time/Total Time

Useful Time= Transmission time
Total Time = Transmission time + Propagation time

Transmission time T_T=256/8*103=> 32 msec
Efficiency = N*T_T/T_T+T_p =>10*32/40=> 8

Throughput = Length of Frame * N /Total Time(Round Trip Time)

Total Time = Transmission time + Propagation time or Total time = Round trip time(RTT)
Throughput = Length of Frame * N / T_T+2T_p
now first we calculate Transmission time which is length of frame/link bandwidth

T_T=L/B_w
T_T=256/8*10³=> 32 msec
Now put the value of transmission time into Throughput formula

Throughput = Length of Frame * N / T_T+2T_p
Throughput = 10*256*10³/32+8 =>64 Kbps

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