If 'B' is the network bandwidth, 'L’ is the cable length and 'C' is the speed of the signal propagation, 'e' is the number of contention. slots per frame and 'F’is the frame length, then channel efficiency 'i' is:
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A: Please upvote. I am providing you the correct answer below. Please please please.
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A: Channel capacity is determined by the most efficient component of the communication process.
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Q: In asynchronous transmission, the gap time between bytes is _______. a. fixed b. variable c. a…
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- A 4000 octet user data is to be transmitted over a network which supports a maximum user data size of 536 octets. Assuming the header in each IP data gram requires 20 octets, derive the number of datagrams (fragments) required and the contents of the following fields: Identification Total length Fragment offset More Fragments flagGiven that you are to transfer a total datagram of size 28939 bytes over a network that has an MTU of 3038 bytes. The given header size is 38 bytes. (a) Calculate the total number of packets required to send this data. (b) What is the data size of the last packet? (c) What would be the header size of the 8th packet and the offset value of the 4th packet?Suppose that R = 1 Gbps and Rc is 300 Mbps and Rs is 200 Mbps. Assuming that the servers are sending at their maximum rate possible, enter the link utilization of the shared link, whose rate is R, below. Enter your answer as a decimal, of the form 1.00 (if the utilization is 1, or 0.xx if the utilization is less than 1, rounded to the closest xx).
- 186. The parameter that refers to uneven delay of data packets in the delivery is a. Jitter b. Timelessness c. Accuracy d. Transmission medium(a) In the second segment sent from Host A to B, what are the sequence number, source port number, and destination port number? (b) If the first segment arrives before the second segment, in the acknowledgment of the first arriving segments, what is the ACK number, the source port number. and the destination port number?Unless specified otherwise, use the free space propagation speed of light 3x108 m/s to calculate the propagation delay. For numerical calculations, the results must show the appropriate unit. Show a complete solution for each problem5. An access multiplexer (AM) installed in the roadside box of the NBN (National Broadband Network). Assume that 200 homes are connected to the AM. The average packet interarrival arrival time at the AM is 200 μs. The average packet length is 2,500 bytes where the packet length is exponentially distributed.a. Calculate the number of packets arriving at the AM every hour.b. Calculate the incoming traffic rate (in bits/sec) at the input of the AM.c. Assume that the output of the AM is connected to the NBN core network via a 2-Gbps link. Calculate the total normalised load ρ at the AM.d. Assume that the packet arrival process is exponentially distributed. Calculate the average queuing delay introduced by the AM.
- In asynchronous transmission, the gap time between bytes is variable why?The End-to-End Delay consists of a combination of which delays? Do not exclude negligible delays? transmission + propagation queuing + processing + transmission processing + transmission + propagation processing + transmission + propagation + queuingA large number of consecutive IP address are available starting at 192.32.0.0. Suppose that three organizations, A, B and C, request a000 (a thousand), b000 (b thousand) and c000 (c thousand) addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the w.x.y.z/s notation. The student should write details of calculations. (Note: a, b and c are defined as your last three numbers of student ID. For example, if your student ID is 18290102 then a=1, b=0 and c=2. So, organization A requests 1000, B requests 0 and C requests 2000 addresses). MY NUMBER:16290025
- This is my solution, Please check the solution and let me know If my solution is correct:Please adjust your decisiona nd provide corrected version if it is needed. a) The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.0025ms Transmission time (Tt) = frame size/bandwidth = 10000 * 8bits / (1Mbps * 10 ^ 6) = 0.008sec = 8ms Total time for transmission = 2 * Tp + Tt(since the frame has to travel from A to D and then D to A) = 2 * 0.0025ms + 8ms = 0.005ms + 8ms = 8.005ms The transmission time of the frame from A to D is 8.005 ms. b) The efficiency of the CSMA/CD protocol is given by the formula: Tt is the transmission time of a frame C is the number of collisions, Tp is the propagation time of a signal from one end of the segment to the other. Efficiency = Tt / (C * 2 * Tp + Tt + Tp) Assuming that there are no other stations transmitting or attempting to transmit on the segment, the transmission from…Computer Science In Python, I am trying to make a client and server code that passes the payload with the requirements. The client has to specify the length of the data packet, up to max, specify how many times to pass the data packet, and specify the server IPv4 address. Both the client and the server need timer functions and compute the time needed to perform the number of sends both the Client and the server need to display the local and remote IPv4 addresses, total time, and bytes per second. Please help, Thanks.Calculate the transmission time of a packet, which is transmitted by a station where the length of the packet is 2 million bytes and the bandwidth of the channel is 100 Kilobits per second?
