Selective Repeat ARQ MCQ in Data link layer

1. In Selective Repeat ARQ if sequence number is range is 0-7 than size of sending window is

• A. 7
• B. 8
• C. 4
• D. 1
• E. None of the above

Ans-C. 4

2. In Selective Repeat ARQ if sequence number size is 5 bit than size of sender window is

• A.32
• B. 16
• C. 8
• D. 31
• E. None of the above

Ans-B. 16

3. In Selective Repeat ARQ if sequence number size is k bit than size of sender window is…

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

Ans-A. 2 ^k-1

4. In Selective Repeat 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 sender window size(w_s) 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 Selective Repeat ARQ Efficiency = w_s* Useful Time/Total Time

Sender window size = w_s
Useful Time= Transmission time
Total Time = Transmission time + Propagation time

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

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

Total Time = Transmission time + Propagation time or Total time = Round trip time(RTT)
Throughput = Length of Frame * w_s / 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 * w_s / T_T+2T_p
Throughput = 10*256*10³/32+8 =>64 Kbps

Medium Access Control-MAC MCQ in Data link Layer

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5. In Selective Repeat ARQ if sequence number size is 4 bit than size of receiver window is…

• A. 2 ^4-1
• B. 2 ⁴-1
• C. 2 ⁴
• D. 2 ⁴ +1
• E. None of the above

Ans-A. 2 ^4-1

6. In Selective Repeat ARQ if sequence number size is 3 bit than size of sender window is…

• A. 2 ^3-1
• B. 2 ³-1
• C. 2 ³
• D. 2 ³+1
• E. None of the above

Ans-A. 2 ^3-1

7. In Selective Repeat ARQ if sender send a packet with sequence number 5 than acknowledgement number will be…

• A. 6
• B. 4
• C. 5
• D. 7
• E. None of the above

Ans-C. 5

8. In Selective Repeat ARQ if sender window size is 6 than range of sequence number is…

• A. 0-6
• B. 0-11
• C. 0-10
• D. 0-5
• E. None of the above

Ans-B. 0-11

9. In Selective Repeat ARQ if sequence number is range is 0-7 than size of receiver window is

• A. 7
• B. 8
• C. 4
• D. 1
• E. None of the above

Ans-C. 4

10. In Selective Repeat ARQ if receiver window size is 4 than range of sequence number is…

• A. range is 0-8
• B. range is 0-7
• C. range is 0-6
• D. range is 0-9
• E. None of the above

Ans-B. range is 0-7

Medium Access Control-MAC MCQ in Data link Layer

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11. In Selective Repeat ARQ if receiver window size is 4 than minimum sequence number is…

• A. 8
• B. 6
• C. 7
• D. 10
• E. None of the above

Ans-A. 8

12. In Selective Repeat ARQ if maximum sender window size is 10 to achieve maximum link utilization. Instead of Selective Repeat ARQ if we are using Stop and Wait ARQ than efficiency is(in Percent)…

• A. 20
• B. 5
• C. 10
• D. 100
• E. None of the above

Ans-C. 10

Solution:- Efficiency_{(Selective Repeat ARQ)}= W_s * Efficiency_{(Stop and Wait ARQ)}

As per given question Maximum link utilization means Efficiency_{(Selective Repeat ARQ)}=1

Sender window size W_s=10

Put the value in formula Efficiency_{(Selective Repeat ARQ)}= W_s * Efficiency_{(Stop and Wait ARQ)}

Efficiency_{(Selective Repeat ARQ)}= 10 * Efficiency_{(Stop and Wait ARQ)}

1= 10 * Efficiency_{(Stop and Wait ARQ)}

Efficiency_{(Stop and Wait ARQ)}=1/10=.1

Efficiency_{(Stop and Wait ARQ)}=1*100=>10

12. In Selective Repeat ARQ the acknowledgement is…

• A. Commutative
• B. Independent
• C. Repetitive
• D. Same
• E. None of the above

Ans-B. Independent

Medium Access Control-MAC MCQ in Data link Layer

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