QUESTION 13 Which ONE of the following statements satisfies the requirement of multiplexing? Bandwidth is one of the most valuable tools we have. An effective system like multiplexin a. resources. O b. To achieve goals like privacy and anti-jamming, we don't need to increase the bandwidth Oc. Multiplexing is one of the bandwidth consumption types, and it is used to decrease perf d. To use multiple channels with greater bandwidth, we must merge multiple low-bandwid ck Save and Submit to save and submit. Click Save All Answers to save all answers.

QUESTION 13 Which ONE of the following statements satisfies the requirement of multiplexing? Bandwidth is one of the most valuable tools we have. An effective system like multiplexin a. resources. O b. To achieve goals like privacy and anti-jamming, we don't need to increase the bandwidth Oc. Multiplexing is one of the bandwidth consumption types, and it is used to decrease perf d. To use multiple channels with greater bandwidth, we must merge multiple low-bandwid ck Save and Submit to save and submit. Click Save All Answers to save all answers.

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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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 /…
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