EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Chapter 7, Problem P7P
Program Plan Intro
IEEE 802.11:
- IEEE 802.11 stands for wireless LAN that was ratified by the Institute of Electrical and Electronics Engineering.
- There are many 802.11 standards for wireless LAN technology in the IEEE 802.11. The different 802.11 standards all share some common characteristics.
- Basic service set (BSS) is a fundamental building block of 802.11 architecture. The BSS also known as Access Point (AP).
- Every wireless station in 802.11 wants to associate with an access point before it can send or receive network layer data.
DIFS:
DIFS stands for DCF Interframe Spacing. It is a time delay for which sender wait after finishing its back off, previously sending RTS package.
SIFS:
SIFS stands for Shortest Interframe Spacing and it is consider as shortest among above mention networking technology.
Request to send (RTS)/ clear to send (CTS):
- The optimal
mechanism used by the IEEE 802.11 is RTS. This is used to decreasing the frame collisions which was introduced by the hidden node problem. - In order to avoid the collisions problem, the IEEE 802.11 protocol allows a station to use Request to Send (RTS) and Clear to Send (CTS) control frames to reserve access to the channel.
- When a sender want to send a data frame, it can initial send an RTS frame to the AP. After AP receives the RTS frame, it will responds by broadcasting a CTS frame.
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
Suppose that a SW ARQ system with 1-bit sequence numbering has a time-out value
that is less than the time required to receive an ACK. Station A sends 5 frames to station
B. No errors occur during transmission of all 5 frames. One ACK is lost when it is
transmitted the first time (see the following table), and no
transmission of that ACK. No errors occur during transmission of the other 4 ACKS.
Sketch the sequence of frame exchanges that transpire between stations A and B.
(а)
error occurs in re-
Last digit of your student ID number
0, 1
2, 3
Lost ACK
1
4, 5
6, 7
8, 9
2
3
4
Compute the fraction of the bandwidth
that is wasted on overhead (headers
and retransmissions) for sliding
window protocol using selective
repeat on a heavily loaded 50 kbps
satellite channel with data frames
consisting of 40 header and 3960 data
bits. ACK frames never occur. NAK
frames are 40-bits. The error rate for
data frames is 1 percent and the error
rate for NAK frames is negligible. The
sequence numbers are 8-bits.
Computer A uses the Go-back-N ARQ protocol to send a 110 Mbytes file to computer B with a window size of 15. Given each frame carries 100K bytes data. How long does it take to send the whole file (the total time taken from A sending the first bit of the file until A receiving the last acknowledgment)? Given that the transmission rate of the link is 500 Mbps and the propagation time between A and B is 15ms. Assume no data or control frame is lost or damaged and ignore the overhead due to header and trailer.
Chapter 7 Solutions
EBK COMPUTER NETWORKING
Ch. 7 - Prob. R1RQCh. 7 - Prob. R2RQCh. 7 - Prob. R3RQCh. 7 - Prob. R4RQCh. 7 - Prob. R5RQCh. 7 - Prob. R6RQCh. 7 - Prob. R7RQCh. 7 - Prob. R8RQCh. 7 - Prob. R9RQCh. 7 - Prob. R10RQ
Ch. 7 - Prob. R11RQCh. 7 - Prob. R12RQCh. 7 - Prob. R13RQCh. 7 - Prob. R14RQCh. 7 - Prob. R15RQCh. 7 - Prob. R16RQCh. 7 - Prob. R17RQCh. 7 - Prob. R18RQCh. 7 - Prob. R19RQCh. 7 - Prob. R20RQCh. 7 - Prob. R21RQCh. 7 - Prob. R22RQCh. 7 - Prob. R23RQCh. 7 - Prob. P1PCh. 7 - Prob. P2PCh. 7 - Prob. P3PCh. 7 - Prob. P5PCh. 7 - Prob. P6PCh. 7 - Prob. P7PCh. 7 - Prob. P8PCh. 7 - Prob. P10PCh. 7 - Prob. P11PCh. 7 - Prob. P12PCh. 7 - Prob. P13PCh. 7 - Prob. P14PCh. 7 - Prob. P15PCh. 7 - Prob. P16P
Knowledge Booster
Similar questions
- Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25 ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). 1.1 What is the minimum number of bits (1) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.arrow_forward4. In this problem, you will derive the efficiency of a CSMA/CD-like multiple access protocol. In this protocol, time is slotted and all adapters are synchronized to the slots. Unlike slotted ALOHA, however, the length of a slot (in seconds) is much less than a frame time (the time to transmit a frame). Let S be the length of a slot. Suppose all frames are ofarrow_forwardConsider a half-duplex point-to-point link using a stop-and-wait scheme, in which a series of messages is sent, with each message segmented into a number of frames. Ignore errors and frame overhead. A. What is the effect on line utilization of increasing the message size so that fewer messages will be required? Other factors remain constant. B. What is the effect on line utilization of increasing the number of frames for a constant message size? C. What is the effect on line utilization of increasing frame size?arrow_forward
- Plagiarism and wrong answers will downvotes Compute the fraction of the bandwidth that is wasted on overhead (headers and retransmissions) for sliding window protocol using selective repeat on a heavily loaded 50 kbps satellite channel with data frames consisting of 40 header and 3960 data bits. ACK frames never occur. NAK frames are 40-bits. The error rate for data frames is 1 percent and the error rate for NAK frames is negligible. The sequence numbers are 8-bits.arrow_forwardThe following frame should be transmitted over a given network (the frame is shown in hexadecimal): 7E DB DD 7E C0 7D DB DC FF C0. a) If the SLIP framing protocol is used; what are the bytes that are sent on the wire (after framing)? Show your workarrow_forwardSuppose that a binary message – either 0 or 1 – must be transmitted by wire from location Ato location B. However, the data sent over the wire are subject to a channel noise disturbance,so to reduce the possibility of error, the value 2 is sent over the wire when the message is1, and the value −2 is sent when the message is 0. If X, X = {−2,2}, is the value sent atlocation A, the value received at location B, denoted as R, is given byR = X + N ,where N is the channel noise disturbance, which is independent of X. When the message isreceived at location B, the receiver decodes it according to the following rule:if R ≥.5, then conclude that message 1 was sentif R < .5, then conclude that message 0 was sentAssuming that the channel noise, N, is a unit normal random variable and that the message0 or 1 is sent with equal probability, what is the probability that we conclude that the wrongmessage was sent? This is the probability of error for this communication channelarrow_forward
- Consider the following scenario: Frames of 12000 bits are sent over a 400kbps channel using a satellite whose propagation delay is 270ms. Acknowledgements are always piggybacked onto data frames. (That is, in the case of stop-and-wait, the next frame cannot be transmitted until the entire frame containing the acknowledgement is received.) Five-bit sequence numbers are used. What is the maximum achievable channel utilization for (a) Stop-and-wait. (b) Protocol 5 (i.e., Go Back N). (c) Protocol 6 (i.e., Selective Repeat)arrow_forwardIn a byte-oriented link layer protocol, the receiver adds all the bytes between the start and end marker bytes (not including those markers) modulo 239, and expects to get a result of 0 if there have been no errors. The byte immediately before the end marker is a checksum, chosen at the transmitter to make this possible. The probability of an error in any received bit is 0.00075 . The total length of the frame is 205 bytes (including start and end markers). Estimate the probability of errors occurring in the received frame but not being detected. You need to consider what combinations of bit errors could cause the error detection system to fail (i.e, conclude that there are no errors). As you are asked for an estimate, you need only consider the most likely scenarios - it is safe to ignore events that could only occur with much lower probability. Round your answer to three significant figures. As the probability will be very low in some cases, you may have to enter up to 7 digits after…arrow_forwardConsider a full-duplex 256 Kbps satellite link with a 240 millisecond end-to-end delay. The data frames carry a useful payload of size 12,000 bits. Assume that both ACK and data frames have 320 bits of header information, and that ACK frames carry no data. What is the effective data throughput When using sliding windows with a sender window size of W = 5 data frames?arrow_forward
- b) The distance between a sender and a receiver of a data communication system is 1400 km and the propagation speed is 2.5 x 10° m/s. The system uses 5-bit sequence numbers whenever possible. Ignore waiting and processing delays. Also ignore the overhead due to the header and the trailer and assume no data or control frame is lost or damages during transmission. If each packet carries 1000 bits of data how long does it take to send 7 million bits of data for the following cases: If Go-Back-N ARQ protocol is used. If Selective-Repeat ARQ protocol is used. ii.arrow_forwardb) The distance between a sender and a receiver of a data communication system is 1400 km and the propagation speed is 2.5 x 10° m/s. The system uses 5-bit sequence numbers whenever possible. Ignore waiting and processing delays. Also ignore the overhead due to the header and the trailer and assume no data or control frame is lost or damages during transmission. If each packet carries 1000 bits of data how long does it take to send 7 million bits of data for the following cases: i. If Go-Back-N ARQ protocol is used. ii. If Selective-Repeat ARQ protocol is used. c) Compare and contrast byte-stuffing and bit-stuffing, then demonstrate bit-stuffing the following frame payload shown in Figure 3(c). 111110 1011| 1 1 1 1|o0 1 10|O Figure 3(c)arrow_forwardThe following character encoding is used in a data link protocol: A: 01000111; B: 11100011; FLAG: 01111110; ESC: 11100000 Show the bit sequence transmitted in binary for the four character frame: A B ESC FLAG when each of the following framing methods is used: a) Character count b) Flag bytes with byte stuffing c) Starting and ending flag bytes, with bit stuffing. Please solve within 30 minutes. Asap.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Operations Research : Applications and AlgorithmsComputer ScienceISBN:9780534380588Author:Wayne L. WinstonPublisher:Brooks Cole
Systems ArchitectureComputer ScienceISBN:9781305080195Author:Stephen D. BurdPublisher:Cengage Learning
Operations Research : Applications and Algorithms
Computer Science
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Brooks Cole
Systems Architecture
Computer Science
ISBN:9781305080195
Author:Stephen D. Burd
Publisher:Cengage Learning