Conversion of Bandwidths Convert the bandwidths given below then provide on i Given bandwidth Convert into Converted Unit 1. 1,000 kbps 2. 6,056,389 bps 3. 8.380 Gbps 4.7,890,389,560 kbps 5:5 Tbps mbps gbps mbps tbps kbps

Conversion of Bandwidths Convert the bandwidths given below then provide on i Given bandwidth Convert into Converted Unit 1. 1,000 kbps 2. 6,056,389 bps 3. 8.380 Gbps 4.7,890,389,560 kbps 5:5 Tbps mbps gbps mbps tbps kbps

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Suppose that a 20-Mbps 802.11 LAN is transmitting 64-byte frames back-to-back over a radio channel with a bit error rate of 10−7 (i.e., the probability of a bit to flip duringthe transmission). How many frames per second will be damaged on average?(*hint: you would need to calculate the probability that a frame does not suffer any bit errors, thisis equivalent to the fraction of frames that do not suffer any bit errors on average.)
Dear Writer could oyu please kindly explin why i get this soltuion and it is different that your solution? The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.025ms 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 * 0.025ms + 8ms = 0.05ms + 8ms = 8.05ms The transmission time of the frame from A to D is 8.05 ms. b) To calculate the efficiency of the CSMA/CD protocol in this LAN, we can use the formula: Efficiency = (1 / (1 + 5tprop / tframe)) tprop is the propagation delay tframe is the time it takes to transmit a frame. To calculate tprop, we need to determine the time it takes for a signal to propagate through the 5 Km distance between Station A and Station D. The speed of propagation is 200,000 Km/sec The propagation delay is: tprop = distance /…
Why CDMA can make users communicate at the same time using the same frequency band without interference? Explain the advantages and disadvantages of CDMA. Why ADSL can achieve the transmission rate of several Mbps using the bandwidth less than 1MHz? Each question is a judgment question. Write down your answer with T for True, F for False, and give the explanation if the answer is False. Consider a system of bandwidth 10Mbps, with multiple users, each is active (i.e., have data to transmit at rate 1Mbps) with a low probability, e.g. 0.1. From the capacity perspective, packet switching is better than circuit switching because the former can support more users. Denial of Service (DoS) attack can be generated by a large number of (legitimate) TCP connection requests to the same targeted server at the same time. These TCP requests can be originated from different hosts in a botnet, which is controlled by the attacker. In VSAT systems the micro-stations do not have sufficient…
Dear Writer this is my calculation solution.could you please double check it and let me know it if it soccrect or should i proceed with your solution: Please provide justification if correction is need and correct answer.Thank you in advance. 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…
What is the total utilization of a circuit-switched network, accommodating five users with equal bandwidth share, and the following properties: Two users each using 91% of their bandwidth share Two users each using 43% of their bandwidth share One user using 16% of their bandwidth share Give answer in percent, with one decimal place (normal rounding) and no percentage sign (e.g. for 49.15% you would enter "49.2" without the quotes).
1. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps.What is the throughput if the system(All stations together)produces 100 frame per second.(Hints:44.5; 10.5) 2. what is the vulnerable time frame size of pure ALOHA?If the time unit is P then find it out.[Hints: 2P;0.2P]
A pipelined reliable transfer protocol has a window size of N segments and runs over a link with bandwidth R=100Mbps. The link has a propagation delay t_prop=300ms. Each data packet sent over the link is of size L_D=5kBytes, whereas acknowledgements are of size L_A=50Bytes. Answer the following questions and show your work where indicated. A. What is the achieved throughput of the above pipelined protocol for N=4, 8 and 16? B. What is the link utilization for N=4, 8 and 16? Note that in class, we calculated the link utilization in terms of the time the sender is busy sending. We can also do this in terms of the achieved vs. available throughput. C. Will the achieved throughput of the above pipelined protocol grow or shrink if the packet size was increased? D. Will the achieved throughput of the above pipelined protocol grow or shrink if the propagation time was increased? E. Will the achieved throughput of the above pipelined protocol grow or shrink if the transmission rate of the link…
For the experiment, let’s assume that the concrete distance between Canterbury and London is 60miles (or ca. 100km) and our pigeons fly at a speed of 60 miles per hour (or ca. 100km/h). (ii) With current technology and considering the weight a pigeon can carry, we are able to send 2TB of data on a USB stick with a pigeon. On a 4G mobile network, we can likely send about 2MB/s. Which of the two transport media would have the shorter message transmission time for a 2TB message? (We again assume pigeons to be perfectly reliable).Briefly explain your reasoning.
Networking question: Consider the following bit stream-a. 00000000 b. 11111111 c. 01010101 d. 00110011 Draw the graph of thei. NRZ-Lii. NRZ-Iiii. Manchesteriv. differential Manchester schemeusing each of the given data streams, assuming that the last signal level has been positive. From the graphs, guess the bandwidth for this scheme using the average number of changes in the signal level.
Consider the following Aloha systems.(a) A group of N users share a 56 kbps pure Aloha channel. Each user generates at a Poisson rateof one 1000-bit packet every 100 sec, even if the previous one has not yet been sent. What is themaximum value of N (Hint: The pure Aloha maximum channel utilization is 18.4%)? (b) Ten thousand airline reservation stations are competing for the use of a single slotted Alohachannel. The average station makes 18 requests/hour. A slot is 125 μsec. What is the approximatetotal channel load (requests per slot)?
suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 108 meters/sec. 1. Calculate the bandwidth-delay product, R x dprop. 2. Consider sending a file of 800, 000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time? 3. Provide an interpretation of the bandwidth-delay product?
Que.a) Consider a CSMA/CD network that transmits data at a rate of 400 x 106 bits per second. The maximum signal propagation time from one node to another node is 20 microseconds. The minimum size of a frame in the network is(. bytes).