Consider the window size is 10, bandwidth is 1500 bps, transmission delay is 2 ms atpropagation delay is 50 ms. What is the throughput using Go Back N protocol? a) 294.11 b) 140.23 c) 96 d) 154.21
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Consider the window size is 10, bandwidth is 1500 bps, transmission delay is 2 ms at
propagation delay is 50 ms. What is the throughput using Go Back N protocol?
a) 294.11
b) 140.23
c) 96
d) 154.21
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- Computer Science A Go-Back-N (GBN) ARQ scheme has been designed with window-size W = 10 such that it is fully efficient when there are no errors on the channel. Assume that frame size is 20000 bits and the overhead bits are negligible. What is the bit probability error (as seen at Layer 2) that makes the Go-Back-N ARQ 50% efficient?Consider the window size is 10, bandwidth is 1500 bps, transmission delay is 2 ms and propagation delay is 50 ms. What is the throughput using Go Back N protocol? a) 294.11 b) 140.23 c) 96 d) 154.21Topic: Network Delay Assume two hosts, A and B, that are linked by a single link with a rate of R bps and are spaced m meters away from one another. A packet of size L bits is being sent from Host A to Host B. Suppose that the propagation through the connection is occurring at s meters/second. (Answer the following question while showing work) 1) When d-prop is bigger than d-trans, where is the packet's first bit at time t = d-trans? 2) At time t = d-trans, where d-prop is less than d-trans, where is the packet's first bit?
- 28 A paging scheme uses a TLB. A TLB access takes 10ns and main memory access takes 50ns. What is the effective access time (in ns) if the TLB hit ratio is 90%? 50 65 70 55Topic: Network Delay Assume two hosts, A and B, that are linked by a single link with a rate of R bps and are spaced m meters away from one another. A packet of size L bits is being sent from Host A to Host B. Suppose that the propagation through the connection is occurring at s meters/second. (Answer the following question while showing work) 1) Give an explanation of the propagation delay, d prop, in terms of the separation between the two hosts and the link's propagation speed. 2) Give examples of how the packet size and link speed affect the transmission time of the packet, d trans.Assume a shared connection of 1 Mbps (for example, an Ethernet bus). How long does it take (in microseconds) to transmit a 1000-bit frame over this connection?
- Consider a bus network with two computers connected shown below: -d m- -X m- A В Rate: R1 Mbps Rate: R2 Mbps Velocity of Propagation: v2 m/s Velocity of Propagation: v, m/s The two computers A and B transmit at two different bandwidths R1 and R2 respectively. The signals transmitted from each computers travel over the bus with different propagation speeds vi m/s and v2 m/s respectively. (Note: This is a hypothetical scenario) Each computer transmits only one bit at the same instance of time. a) Show the distance X m from Computer A where the collision of the two bits occur is R2 vid + Viv2 X = \Vị + v2, R1R2 Vị + v2. b) If, both the computers transmit with the bandwidth R and both signals travel with a propagation speed v, determine X.Efficiency Principle:Ethernet V1 access protocol was designed to run at 10 Mbps over 2.5 Km using 1500 Byte packets. This same protocol needs to be used at 100 Mbps at the same efficiency (utilization). What distance can it cover if the frame size is not changed? (Note: be careful with the units.1.3 10 packets are continuously sent over a 1 Mbps link. Each packet is of 1,000 bits long and RTT is 10 ms. What is the throughput of the link (i.e. on average, how many bits transmitted per second)? A. 511.856 bps B. 511.856 Kbps C. 500 bps D. 500 Kbps E. 666.667 Kbps rej res
- Consider the following Promela model for the Alternating Bit Protocol /* Alternating Bit Protocol, cf. Holzmann 1993, Tutorial */ #define N 2 #define MAX 8 #define FETCH mt = (mt+1)%MAX #define ACCEPT printf("ACCEPT %d\n", mr); assert(mr==(last_mr+1)%MAX) mtype = {data, ack, error} proctype lower_layer(chan fromS, toS, fromR, toR) { byte d; bit b; do ::fromS?data(d,b) -> progress0: if :: toR!data(d,b) /* correct */ :: toR!error(0,0) /* distorted */ fi ::fromR?ack(b) -> progress1: if :: toS!ack(b) :: toS!error(0) fi od } proctype Sender(chan in, out) { byte mt; /* message data */ bit at; /* alternation bit transmitted */ bit ar; /* alternation bit received */ FETCH; /* get a new message */ out!data(mt,at); /* send it */ do ::in?ack(ar) -> /* await response */ if ::(ar == at) -> /* correct send */ FETCH;…timeout timeout Sender Consider the figure below (simplex communication scenario). 23 Seq1, 10 bytes of data Seq2, 10 bytes Seq3, 10 bytes ACK1 ACK2 XA ACK4 Seq4 ACK3 The initial SEQ Number (Seq1) is 567. 1. 123 2. 3. What is Seq2 and Seq3 What is ACK1, ACK2, Ack3 What is Seq4 and ACK4 Receiver 3Suppose you are designing a sliding window protocol for a 1-Mbps point-to-point link to a stationary satellite revolving around the earth at 3 x 104 km altitude. Assuming that each frame carries 1 KB of data, what is the minimum number of bits you need for the sequence number in the following cases? Assume the speed of light is 3 x 108 m/s. (a) RWS=1 (b) RWS=SWS