EKSAMEN / EXAM TTM4100 18 05 2007



Similar documents
Note! The problem set consists of two parts: Part I: The problem specifications pages Part II: The answer pages

CSE 473 Introduction to Computer Networks. Exam 2 Solutions. Your name: 10/31/2013

Attenuation (amplitude of the wave loses strength thereby the signal power) Refraction Reflection Shadowing Scattering Diffraction

Stop And Wait. ACK received; transmit frame 2 CS 455 3

EINDHOVEN UNIVERSITY OF TECHNOLOGY Department of Mathematics and Computer Science

CSE331: Introduction to Networks and Security. Lecture 9 Fall 2006

Homework 2 assignment for ECE374 Posted: 02/21/14 Due: 02/28/14

Midterm Exam CMPSCI 453: Computer Networks Fall 2011 Prof. Jim Kurose

Note! The problem set consists of two parts: Part I: The problem specifications pages Part II: The answer pages

Internet Packets. Forwarding Datagrams

TCP in Wireless Mobile Networks

High-Level Data Link Control

First Midterm for ECE374 02/25/15 Solution!!

CH.1. Lecture # 2. Computer Networks and the Internet. Eng. Wafaa Audah. Islamic University of Gaza. Faculty of Engineering

Transport Layer Protocols

ECE 358: Computer Networks. Homework #3. Chapter 5 and 6 Review Questions 1

EECS 489 Winter 2010 Midterm Exam

Data Link Layer. Flow Control. Flow Control

CSE 123: Computer Networks Fall Quarter, 2014 MIDTERM EXAM

Computer Networks. Chapter 5 Transport Protocols

Final Exam. Route Computation: One reason why link state routing is preferable to distance vector style routing.

TCP over Multi-hop Wireless Networks * Overview of Transmission Control Protocol / Internet Protocol (TCP/IP) Internet Protocol (IP)

Ring Local Area Network. Ring LANs

First Midterm for ECE374 03/24/11 Solution!!

Frame Burst Adjusting for Transmitting Video Conference in Gigabit Ethernet

First Midterm for ECE374 03/09/12 Solution!!

Final for ECE374 05/06/13 Solution!!

Based on Computer Networking, 4 th Edition by Kurose and Ross

CS 5480/6480: Computer Networks Spring 2012 Homework 4 Solutions Due by 1:25 PM on April 11 th 2012

COMP 361 Computer Communications Networks. Fall Semester Midterm Examination

RESOURCE ALLOCATION FOR INTERACTIVE TRAFFIC CLASS OVER GPRS

LAN TCP/IP and DHCP Setup

First Semester Examinations 2011/12 INTERNET PRINCIPLES

Chapter 6 Wireless and Mobile Networks

Answer: that dprop equals dtrans. seconds. a) d prop. b) d trans

EXAMPLES AND PROBLEMS. Competence Based Education Internet Protocols

The Internet. Reminders. Internet Protocol Routing: Simplified. Part 2: Networks of Networks - Internet Workings

>>> SOLUTIONS <<< c) The OSI Reference Model has two additional layers. Where are these layers in the stack and what services do they provide?

CS6956: Wireless and Mobile Networks Lecture Notes: 2/11/2015. IEEE Wireless Local Area Networks (WLANs)

Module 15: Network Structures

Simulation-Based Comparisons of Solutions for TCP Packet Reordering in Wireless Network

ECE 333: Introduction to Communication Networks Fall 2002

Data Networks Summer 2007 Homework #3

CSE331: Introduction to Networks and Security. Lecture 6 Fall 2006

CS268 Exam Solutions. 1) End-to-End (20 pts)

Computer Networks - CS132/EECS148 - Spring

Operating System Concepts. Operating System 資 訊 工 程 學 系 袁 賢 銘 老 師

AE64 TELECOMMUNICATION SWITCHING SYSTEMS

The Quality of Internet Service: AT&T s Global IP Network Performance Measurements

CS335 Sample Questions for Exam #2

Feature and Technical

Transport and Network Layer

ESSENTIALS. Understanding Ethernet Switches and Routers. April 2011 VOLUME 3 ISSUE 1 A TECHNICAL SUPPLEMENT TO CONTROL NETWORK

TCP and Wireless Networks Classical Approaches Optimizations TCP for 2.5G/3G Systems. Lehrstuhl für Informatik 4 Kommunikation und verteilte Systeme

Issue 1 EN. Nokia and Nokia Connecting People are registered trademarks of Nokia Corporation

2. What is the maximum value of each octet in an IP address? A. 28 B. 255 C. 256 D. None of the above

Chapter 1 Configuring Internet Connectivity

Objectives. The Role of Redundancy in a Switched Network. Layer 2 Loops. Broadcast Storms. More problems with Layer 2 loops

Chapter 14: Distributed Operating Systems

NOVEL PRIORITISED EGPRS MEDIUM ACCESS REGIME FOR REDUCED FILE TRANSFER DELAY DURING CONGESTED PERIODS

Computer Networks. Data Link Layer

Figure 1: cellular system architecture

Issue 2EN. Nokia and Nokia Connecting People are registered trademarks of Nokia Corporation

How To Make A Multi-User Communication Efficient

How To Solve A Network Problem In An Ipa (Ipa 2) (Ipam) (Networking) (For Acedo) (Ipta) (Powerline) (Network) (Nem

Names & Addresses. Names & Addresses. Hop-by-Hop Packet Forwarding. Longest-Prefix-Match Forwarding. Longest-Prefix-Match Forwarding

Mobile Computing/ Mobile Networks

Local Area Networks transmission system private speedy and secure kilometres shared transmission medium hardware & software

Mobile Communications Chapter 9: Mobile Transport Layer

TCP for Wireless Networks

Solving complex performance problems in TCP/IP and SNA environments.

Random Access Protocols

Chapter 16: Distributed Operating Systems

Load Balancing. Final Network Exam LSNAT. Sommaire. How works a "traditional" NAT? Un article de Le wiki des TPs RSM.

WAN Data Link Protocols

EXPLORER. TFT Filter CONFIGURATION

GATE CS Topic wise Questions Computer Network

Chapter 1 Configuring Basic Connectivity

Chapter 3: Sample Questions, Problems and Solutions Bölüm 3: Örnek Sorular, Problemler ve Çözümleri

Improving the Performance of TCP Using Window Adjustment Procedure and Bandwidth Estimation

The OSI and TCP/IP Models. Lesson 2

ECE 358: Computer Networks. Solutions to Homework #4. Chapter 4 - The Network Layer

ICOM : Computer Networks Chapter 6: The Transport Layer. By Dr Yi Qian Department of Electronic and Computer Engineering Fall 2006 UPRM

QoS issues in Voice over IP

Wireless Networks. Reading: Sec5on 2.8. COS 461: Computer Networks Spring Mike Freedman

15-441: Computer Networks Homework 2 Solution

RTT 60.5 msec receiver window size: 32 KB

Avaya IP Office 8.1 Configuration Guide

Prefix AggregaNon. Company X and Company Y connect to the same ISP, and they are assigned the prefixes:

Data Link Layer(1) Principal service: Transferring data from the network layer of the source machine to the one of the destination machine

SIPSTATION User Guide. Schmooze Com Inc.

How To Set Up Virtual Fax On Zcoo (U20/50/60/100) On A Pc Or Macbook Or Ipod (U21/50) On An Ipod Or Ipo (U19/50 Or 60

10. Wireless Networks

Agenda. Distributed System Structures. Why Distributed Systems? Motivation

Referring to the above question, the end-to-end delay (transmission delay plus propagation delay) is

Unlicensed Mobile Access (UMA) Handover and Packet Data Performance Analysis

EPL 657 Wireless Networks

Exam 1 Review Questions

Transcription:

1.1 1.1.1...... 1.1.2...... 1.1.3...... 1.1.4...... 1.1.5...... 1.1.6...... 1.1.7...... 1.1.8...... 1.1.9...... 1.1.10.... 1.1.11... 1.1.16.... 1.1.12... 1.1.17.... 1.1.13... 1.1.18.... 1.1.14... 1.1.19.... 1.1.15... 1.1.20.... 1.2 1.2.1...... 1.2.2...... 1.2.3...... 1.2.4...... 1.2.5...... 1.2.9..... 1.2.13.... 1.2.6...... 1.2.10... 1.2.14.... 1.2.7...... 1.2.11... 1.2.15.... 1.2.8...... 1.2.12... 1.2.16.... 1.3 1.3.1...... 1.3.2...... 1.3.3...... 1.3.4...... 1.3.5...... 1.3.6...... 1.3.7...... 1.3.8...... 1.3.9...... 1.3.10.... 1.3.11... 1.3.12... 1.3.13... 1.3.14... 1.3.15... 1.3.16.... 1.3.17.... 1.3.18.... 1.3.19.... 1.3.20.... Kontroller: Studentnr. på alle sider Samme studentnr. over alt Eksamensvaktens signature / Invigilator's signature 1/14

1.4 1.4.1...... 1.4.5...... 1.4.9..... 1.4.13.... 1.4.2...... 1.4.6...... 1.4.10... 1.4.14.... 1.4.3...... 1.4.7...... 1.4.11... 1.4.15.... 1.4.4...... 1.4.8...... 1.4.12... 1.4.16.... 1.5 1.5.1...... 1.5.5...... 1.5.9..... 1.5.2...... 1.5.6...... 1.5.10... 1.5.3...... 1.5.7...... 1.5.11... 1.5.4...... 1.5.8...... 1.5.12... 1.5.13.... 1.5.14.... 1.5.15.... 1.5.16.... 1.6 1.6.1...... 1.6.2...... 1.6.3...... 1.6.4...... 1.6.5...... 1.6.6...... 1.6.11... 1.6.16.... 1.6.7...... 1.6.12... 1.6.17.... 1.6.8...... 1.6.13... 1.6.18.... 1.6.9...... 1.6.14... 1.6.19.... 1.6.10.... 1.6.15... 1.6.20.... 2/14

2.1. Flow control is the receiver controls the data flow (sending rate) from the sender. 2.2. Flow control is commonly used in Transport Layer and Data Link Layer. It may also be used in Application Layer. The reason of having flow control in these layers is that, due to limited processing capacity, limited storage space and/or other reasons, the receiver may not be able to handle the incoming data as they arrive and will lose them, if the sender sends the data too fast. This scenario can happen in Transport Layer, Data Link Layer and Application Layer. 3/14

2.3. In stop-and-wait flow control, the sender sends one packet and then waits for an acknowledgment from the receiver before proceeding to send the next packet. In particular, after sending a packet, the sender keeps waiting and checking if there is an acknowledgement for this packet. If a pre-defined timeout time has passed for the waiting, the sender re-sends the packet. If the acknowledgement is received, the sender sends the next packet. At the receiver side, it sends back to the sender an acknowledgement for the packet it receives correctly 4/14

2.4. One problem/disadvantage is that the stop-and-wait flow control is inefficient in utilizing the communication channel between the sender and the receiver: The following equation can be found from the channel utilization slide of stop-and-wait flow control: U = (X/C)/[(X/C)+2τ] Where U denotes the maximum channel utilization: X packet size; C the channel capacity; τ one way propagation delay. This equation implies that, if the propagation cannot be ignored, the stop-and-wait flow control can never fully utilize the channel capacity. Another problem is that duplicate packets may be sent and received. If packets are not associated with some sequence number, the receiver side will not notice such duplication, and this will cause problem. 2.5. Under stop-and-wait flow control, it takes at least one round-trip time to send one data packet. This round trip time, in this question, is at least twice the one way propagation delay. In other words, at most one packet is sent every 2ms. Suppose the maximum size of the packet is X. Then, the maximum data rate is X/(2ms). 5/14

3.1. In the Internet, textual domain names are used. However, the Internet protocol is based on numerical IP addresses. Hence, DNS is needed to map textual domain names and e-mail destinations/addresses to numerical IP-addresses. WWW and E-mail are two sample Internet applications that use DNS. 3.2. DNS is generally unchanged over short time. So, operations can be repeated without harm. When a process makes a DNS request, it typically starts a timer. If the timer expires, it just makes the request again. So, no harm is done. 6/14

3.3. Normally, each zone has a primary DNS name server and one or several secondary DNS name servers. Secondary name servers are used to increase the reliability of the system in cases where some name server(s) are down. 4.1. MAC sub-layer is needed for a network with multi-access channels. It is used to determine who goes next on a multi-access channel. 7/14

4.2. In CSMA/CD, the time is divided into slots. The algorithm works as follows: Before a node wants to send a data frame, it senses the channel: If the channel is idle, the node starts to transmit the frame. At the same time, it senses the channel for collision. If there is collision, the node stops transmitting and wait for some random time before taking another try of transmitting. If no collision is detected, the node finishes the transmission. If the channel is busy, the node keeps sensing the channel until the channel becomes idle. Then, the above step is repeated. 8/14

5. The router will route the packets as follows: Packet (a) to Interface 1 Packet (b) to Interface 0 Packet (c) to Router 2 Packet (d) to Router 1 Packet (e) to Router 2 9/14

6.1. Frequencies are re-used in different cells to increase the system capacity, or to increase the number of users that can be supported at the same time in the system. 6.2. Each cell has six neighbors. If the central cell uses frequencies Group 1, its six neighbors can use Group 2, 3, 2, 3, 2, and 3 respectively, as illustrated in the figure below. Accordingly, only three unique cells are needed. So, each cell can have 280 frequencies. 10/14

7.1. WRONG The first and-clause is true. The second and-clause is false: this leg is called 'forwarding leg' and is charged to ZN. This leg goes from fixed to mobile in this case from fixed to fixed in other cases, it is and is called 'forwarding leg'. (>From this we also see that 'ZN pays nothing' is false.) Conclusion: The combined statement is false (wrong). Explanation: The major billing principle from POTS is also used in GSM, and is as follows: P1: A subscriber shall be charged for a call based on information available for him. (Other good explanations are also acceptable as well. They may be quite short.) For the calling party this principle P1 leads to: Caller A shall be charged for a call based on information available for him, i.e. based on the dialed number, numbering plans and charging info related to these numbers Additional info: This is in PSTN typically based on a numbering plan i.e. on the called number. Previous local fixed calls were cheaper than regional and national fixed calls, and the different international charging applied (to be looked up in the phonebook). Today we have separate number series for mobile calls, hence A knows is he places a mobile-to-mobile or a mobile-to-fixed call. We also still have separate international prefix For B the same principle P1 applies as follows: CFU service is known to B, (but may not be known to A, A does not know if CFU is active or not). Hence forwarding is charged to the subscriber of the forwarding service 'forwarding leg'. Since CFU in not known to A, A is not charged for this. Hence B is charged for this, (since the service is not for free). In the case in the exam: A = X1N is charged for the 'calling leg' (from mobile to fixed) B = ZN is charged for the for the 'forwarding leg' from fixed to mobile Note also that there is nothing called 'mobile leg', and this also makes the 2nd and-clause false: X2N is not charged, (only receiving a call without roaming). There is something called 'roaming leg' but does not apply to this case since X2N is not roaming, while only utilizing regional mobility within OpM2N's network (as explained in the text in the exam) Note: we said that the same person Ola is behind both the subscriptions X2N and ZN, hence the same person will get the bill, but the charging still goes to ZN and X2N as two different subscriptions. Tvilling etc. was excluded, such services also ties several numbers to a common bill. 11/14

KOMMENTARER COMMENTS 1.1.16 False. This is because the connection set up packets may follow different routes from data packets. So, it is possible that the setup time is shorter even though there need several rounds of info. packet exchanging. 1.3.5 False. The maximum number of errors that can be corrected is 8. 1.4.4 False. The statement is too strong to hold. If always converges is changed to may converge, then the statement is true. 1.5.9. True (based on the English version); False (based on the Norwegian version). Everyone will get the corresponding point from this statement due to the mismatch between the English version and the Norwegian version. 1.6.5 The statement is a bit ambiguous. For a new domain, permission is required from the domain where it will be included. However, for any subdomain, the permission is not required. (Book P.582) Everyone will get the corresponding point from this statement. 1.6.6 False. The correct statement is: An e-mail system is made up of two subsystems that are user agents and message transfer agents. 1.6.10. The statement is a bit ambiguous. Even though, by default, POP3 removes messages from the mail server, it can be configured such that the messages can still be left on the mail server. Everyone gets the corresponding point for this statement. 1.6.15. The terms Webmail and IMAP have commonly been used in a way broader than what is seen from this textbook. As a consequence, this statement becomes a bit ambiguous. Everyone gets the corresponding point from this statement. 1.6.18. Due to the ambiguous definition of an en IP telephone, everyone gets the corresponding point for this statement. 12/14

KOMMENTARER COMMENTS 13/14

KOMMENTARER COMMENTS 14/14