GATE CSE Flow Control Computer Network
| Q1➡ | GATE 2021 Set-1 Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following: • The time taken for processing the data frame by the receiver is negligible. • The time taken for processing the acknowledgement frame by the sender is negligible. • The sender has an infinite number of frames available for transmission. • The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits. • The link data rate in each direction is 1 Mbps (=106bits per second). • One way propagation delay of the link is 100 milliseconds. The minimum value of the sender’s window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is _____. |
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Answer: 50 to 52
Explanation:
Given,
Processing delay of data frame = 0
Processing delay of Acknowledge frame = 0
Size of Data frame = 2000 bits
Size of Acknowledge frame = 10 bits
Data rate (or Bandwidth) = 106 bits per second
Propogation Delay (Tp) = 100 ms
Efficiency (or Link utilization) = 50% or 1/2
Ask,
Sender window sizr (N) = ?
Formula,
where,
N= window size
Tt (data) = Transmission Delay for Data Frame
Tt (Ack) = Transmission Delay for Acknowledge Frame
Tp = Propogation Delay
Let’s Calculate,
Note: 1 second = 1000 millisecond
So, The minimum value of the sender’s window size is 51.
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| Q2➡ | GATE 2017 Set-1 The value of parameters for the Stop-and-Wait ARQ protocol are as given below: Bit rate of the transmission channel = 1 Mbps. Propagation delay from sender to receiver = 0.75 ms. Time to process a frame = 0.25 ms. Number of bytes in the information frame = 1980. Number of bytes in the acknowledge frame = 20. Number of overhead bytes in the information frame = 20. Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is ________. (correct to 2 decimal places). |
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Answer: 86.5 to 89.5
Explanation:
Given Data,
Propagation Delay (Tp) = 0.75 ms
Bit rate = 1Mbps = 106 bits per sec
Number of bytes in the information frame = 1980
Number of overhead bytes in the information frame = 20
Number of bytes in the acknowledge frame = 20
Time to process a frame = 0.25 ms.
Formula,
Calculation,
Useful Data = information frame + overhead = 1980+20 = 2000 B
Processing Delay( Tprocess ) = Tprocess (Data) + Tprocess (Acknowledge) = 0.25 + 0.25 = 0.5 ms
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GATE CSE Flow Control Computer Network
| Q3➡ | GATE 2017 Set-2 Consider two hosts X and Y, connected by a single direct link of rate 106bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2×108m/sec. Host X sends a file of 50,000 bytes as one large message to host Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the values of p and q are |
| i ➥ p=400 and q=50 |
| ii ➥ p=100 and q=50 |
| iii ➥ p=50 and q=400 |
| iv ➥ p=50 and q=100 |
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Answer: I
Explanation:
Given,
Data rate (or Bandwidth) = 106 bits/sec
Distance = 10,000 km
Speed = 2×108 m/sec
Data Length = 50,000 Bytes
Ask,
Transmission Delay = ?
Propagation Delay = ?
Formula,
Calculation,
So,Option(I) is correct.
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| Q4➡ | GATE 2016 Set-2 Consider a 128 × 103 bits/ second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is _________. |
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Answer: 4
Explanation:
Given,
Bandwidth = 128 x 103 bps = 128 Kbps
Propagation Delay = 150 milliseconds
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1
Ask,
The minimum number of bits required to represent sequence number field =?
Formula,
For selective repeat protocol,
Efficiency (η) = 
where,
N= Sender window size = Receiver window size
Tp = Propogation Delay
Tt = Transmission Delay = 
Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required for the sequence number = ceil of(2N)
Calculation,
Efficiency (η) =
=> 1 =
=> N = 1+2a
=> N = 1+2*2.4 = 5.8 = 6(approximate)
Sequence Number = sender window size + receiver window size = N+N = 6+6 = 12
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (12) = 4bits
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| Q5➡ | GATE 2015 Set-1 Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________. |
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Answer: 320
Explanation:
Given,
Bit rate (or Bandwidth) = 64 kilobits =64 x 103 bps
Processing Delay (or Tp) = 20ms
Link utilization (or Efficiency) = 50% = 1/2
Ask,
Frame size = ?
Formula,
For stop-wait protocol,
Efficiency (η) = 
where,
Tp = Propogation Delay
Tt = Transmission Delay =
Calculation,
η =
=> 
=> 1+2a = 2
=> Tt =2*Tp
=> 
=> L = 2 * Tp * Bw
=> L = 2* 20 ms * 64 x 103 bps
=> L = 2* 20* 10-3 s * 64 x 103 bps
=> L = 40 * 64 / 8 Byte = 320 Byte
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| Q6➡ | GATE 2015 Set-2 A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at the nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ________. |
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Answer: 12
Explanation:
Given,
Bandwidth = 106 bps
Frame size = 1000 B = 1000* 8 bits
Efficiency = 25% = 1/4
Ask,
Propogation Delay = ?
Formula,
For stop-wait protocol,
Efficiency (η) =
where,
Tp = Propogation Delay
Tt = Transmission Delay =
Calculation,
=> η =
=> 1/4 =
=> 1+2a = 4
=> Tp =3*Tt /2
=> Tp = 3*8ms/2
=> Tp = 12ms
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| Q7➡ | GATE 2015 Set-3 Consider a network connected two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 106 bits per second. The propagation speed of the media is 4 × 106 meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 107 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then the minimum size in bits of the sequence number field has to be _________. |
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Answer: 8
Explanation:
Given,
Distance = 8000 km = 8 * 106 meters
Bandwidth = 500 * 106 bps
Propagation Speed= 4 * 106 meters per sec
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1
Ask,
The minimum number of bits required to represent the sequence number field =?
Formula,
For selective repeat protocol,
Efficiency (η) =
where,
N= Sender window size = Receiver window size
Tp = Propogation Delay
Tt = Transmission Delay =
Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required to represent the sequence number = ceil of(2N)
Calculation,
Efficiency (η) =
=> 1 =
=> N = 1+2a
=> N = 1+2*2.4 = 5.8 = 6(approximate)
Sequence Number = sender window size + receiver window size = N+N = 6+6 = 12
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (12) = 4bits
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GATE CSE Flow Control Computer Network
| Q8➡ | GATE 2014 Set-1 Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________. |
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Answer: 5
Explanation:
Given,
Frame size = 1 KB = 8 x 210 bits = 8*1024 bits
Bandwidth = 1.5 Mbps = 1.5 x 106 bps
Propagation Delay(Tp) = 50 milliseconds
Link utilization (or Efficiency)= 60% = 0.6
Ask,
The minimum number of bits required to represent the sequence number field =?
Formula,
For selective repeat protocol,
Efficiency (η) =
where,
N= Sender window size = Receiver window size
Tp = Propogation Delay
Tt = Transmission Delay =
Sequence Number = sender window size + receiver window size = N+N =2N
Minimum number of bits required for the sequence number = ceil of(2N)
Calculation,
Efficiency (η) =
=> 0.6 =
=> N = 0.6(1+2a)
=> N = 0.6(1+2*9.15) = 11.58 = 12(approximate)
Sequence Number = sender window size + receiver window size = N+N = 12+12 = 24
Minimum number of bits required for the sequence number = ceil of(2N) =ceil of (24) = 5bits
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| Q9➡ | GATE 2013 Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s. |
| i ➥ 1 |
| ii ➥ 2 |
| iii ➥ 2.5 |
| iv ➥ 5 |
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Answer: II
Explanation:
Given,
Data rate (or Bandwidth) = 500 Mbps = 500 x 106 bps
Frame size = 10,000 bits
Speed = 200,000 km/sec
Ask,
Distance =?
Formula,
Length = 2 * Tp * Bandwidth
Calculation,
=> L = 2 * Tp * Bandwidth
=> L = 2 *(Distance/Speed) * Bandwidth
So, Option(II) is correct.
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GATE CSE Flow Control Computer Network