Q2. (a) Two towns 20km apart are connected by a microwave data link. Data is sent across the link in 800 byte frames and the link operates at a data rate of 120Mbps. Assuming the link has a bit error rate of 4 x 10-5, and a velocity of propagation v = 3 x 108 m/s, determine the link utilisation efficiency using the following protocols: (i) Idle RQ (ii) Selective repeat RQ with a send window buffer k=2. (iii) Suggest a way to increase the link utilisation of the Selective repeat RQ scheme. Discuss potential disadvantages of such an approach.

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A: Manchester encoding (first published in 1949) is a synchronous  clock-encoding technique used by the…

Q: Consider a selective repeat sliding window protocol that uses a frame size of 2 KB to send data on a…

A: Given the Frame Size as 2KB, Bandwidth=4Mbps Latency =60ms, which is also known as Propagation delay…

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Q: a) In one of the conferences held in North America, to investigate performance of a radio network,…

A: - We need to talk about the mechanisms shown about packet transferred.

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Q: erkeley Computer Science example (CS-168): Picture this: there are two hosts (A and B). The two…

A: Given Data : Tp = 50 msec Data rate = 10^6 Bps W = 4 Packet size = 1000 Bytes

Q: sts are conne mission spe its per secon d wait protoc

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Q: part 1:…

A: As multiple questions are posted, only first can be solved and kindly repost the questions that are…

Q: Suppose you are designing a sliding window GBN protocol for a 5 Mbps point to point link, which has…

A: Find the required answer with calculation given as below :

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A: Given Bandwidth B=10Mbps=10*106bps=107bps Distance L=200m Propagation speed P=2*108m/sec

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A: Actually, given data transmission rate is 2000 bits/sec.. propagation rate 1000m/sec..

Q: 17. A 1 KM 1long CSMA/CD with a propagation 10Mbps delay 5usec has bandwidth. Repeaters are not…

A: Here is the solution with an explanation:-

Q: Suppose you are designing a sliding window protocol for a 2-Mbps point-to-point link to the moon,…

A: Given- Bandwidth = 2 Mbps Propagation delay (Tp) = 1.25 sec Packet size = 1 KB = 210B= 210*8 bits

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Q: Consider a network which uses a frame size oT RRo send data on a 1.5 Mbps link with a one-way…

A: Given: Frame Size: 1KB Bandwidth: 1.5Mbps Utilization: 60% Solution:- Transmission delay = Frame…

Q: 17. A 1 KM long CSMA/CD with a propagatr delay 5usec has 10Mbps bandwid Repeaters are not allowed in…

A: Effective data rate = data (excluding Ack) / total time Total time = transmitter seizes cable +…

Q: Consider a CDMA system with two users A and B. and their spreading sequences are a-[1,1, 1] and b-…

A: Lets see the solution.

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A: According to collision detection in half duplex ethernet.  Tt >= 2* TP  Here we have given the…

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A: Q

Q: d) Assume that, in a Stop-and-Wait ARQ system, the bandwidth of the line is 5 Mbps, and 1 bit takes…

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Q: 1.) There is a data that has a file size of 26Mb with an ethernet overhead of 8 Mb. Find the Total…

A: Answer is

Q: data transmission

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A: Answer: I have given answer in the handwritten format.

Q: Q6) Assuming the router's buffer is infinite, the queuing delay is 1.2957 ms, and 1476 packets…

A: Given,Router buffer size = infinitequeuing delay = 1.2957 mspackets arriving = 1476formula for…

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A:

Q: Consider a CSMA/CD network that transmits data at a rate of 400 x 106 bits per second. The maximum…

A: Introduction : Given  ,A CSMA/CD network data rate = 400 x 106 bits /second.Tp = 20 microsecondsWe…

Q: multiples of 512 bits, and each repeater will insert a store-and-forward delay equivalent to 20-bit

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Q: Consider a forwarding station A on a SONET STS-1 line, receiving frames from the downstream end B…

A: Introduction  SONET is b Synchronous Optical Network and depends on clock-based framing.   SONET,…

Q: Consider a full-duplex 256 Kbps satellite link with a 240 millisecond end-to-end delay. The data…

A: The answer is