GATE CSE Flow Control Computer Network

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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

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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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 GATE CSE Flow Control Computer Network

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Explanation:
Given,
Data rate (or Bandwidth) = 106 bits/sec
Distance = 10,000 km
Speed = 2×108 m/sec
Data Length = 50,000 Bytes

Transmission Delay = ?
Propagation Delay = ?

Formula,  Calculation,  So,Option(I) is correct.

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Explanation:
Given,
Bandwidth = 128 x 103 bps = 128 Kbps
Propagation Delay = 150 milliseconds
Frame size = 1 kilobyte
Link utilization (or Efficiency)= 100% = 1

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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Explanation:
Given,
Bit rate (or Bandwidth) = 64 kilobits =64 x 103 bps
Processing Delay (or Tp) = 20ms
Link utilization (or Efficiency) = 50% = 1/2

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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Explanation:
Given,
Bandwidth = 106 bps
Frame size = 1000 B = 1000* 8 bits
Efficiency = 25% = 1/4

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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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

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

GATE CSE Flow Control Computer Network

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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

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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Explanation:
Given,
Data rate (or Bandwidth) = 500 Mbps = 500 x 106 bps
Frame size = 10,000 bits
Speed = 200,000 km/sec

Distance =?
Formula,
Length = 2 * Tp * Bandwidth Calculation,
=> L = 2 * Tp * Bandwidth
=> L = 2 *(Distance/Speed) * Bandwidth So, Option(II) is correct.

GATE CSE Flow Control Computer Network

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