In The Name of God, The Merciful, The Compassionate

Transcription

1 In The Name of God, The Merciful, The Compassionate Mini-Exam#1 Computer Networks 1, Department of Computer Engineering, Sharif University of Technology, Fall 2012 CE40443 Total Points: 110 / Time: 2 hours 1) General. Give the technical term that best fits the following descriptions. (5 points) a) The service model of the current Internet. Best Effort b) The basic transmission unit at the Network layer. Datagram c) A mechanism by which the receiver of data throttles the transmission rate of the sender, so that data will not arrive too quickly to be processed. Flow Control d) A text string used to identify the location of Internet resources in the Web. URL e) A network based on Internet protocols that is accessible only by a specific organization s members, employees, or others with authorization. Intranet (I accepted VPN too) 2) Packet-switching vs. Circuit switching (15 points) We studied two types of approach to transfer data in network core: packet-switching and circuit switching. Outline the main differences between them? In the following scenario, how many simultaneous users can be supported in each of these approaches: Each user sends 100 kbps when active. Moreover, each user is active p =10% of time Refer to the slides #32 to #46 of Introduction chapter. The main differences between them (5 points) Simultaneous users in circuit switching (3 points) Simultaneous users in packet switching (7 points) 3) Internet structure (13 points) Explain the following concepts? Show their relations in a figure: ISP, IXP, PoP, Content Provider Network ISP (4 points): Regardless of the type of device an individual or business uses to connect to the Internet, the device must connect through an Internet service provider (ISP). An ISP is a company or organization through which a subscriber obtains Internet access. A subscriber can be a business, a private consumer, a government body, or even another ISP. ISPs are classified into 1 1

2 different tiers according to how they access the Internet backbone. Three tiers of ISPs are identified: Tier 1, Tier 2 (Regional), and Tier 3 (access) ISPs. PoP (2 points): Users connect to the ISP through a Point of Presence (POP) using a variety of access technologies. IXP (2 points): The ISP POPs connect to an Internet Exchange Point (IXP). In some countries, this is called a Network Access Point (NAP). An IXP or NAP is where multiple ISPs are joined to gain access to each others networks and exchange information. Currently more than 100 major exchange points exist worldwide. Content provider networks (2 points): a content provider network (e.g., Google, Microsoft, Akamai) is a private network that connects its data centers to Internet, often bypassing tier-1, regional ISPs. (It may run their own network, to bring services, content close to end users.) Figure (3 points) 4) (12 points) In the context of Electronic Mail, Explain the following major components? Show their relations in a figure (e.g., a scenario about User1 sends message to User2): user agents, mail servers, SMTP, IMAP, PoP3 Refer to the slides #58 to #73 of Application chapter. 5) DNS (10 points) Suppose that a given host (HostA: ee.someschool.edu) wants to learn the IP address of another host (HostB: cs.anotherschool.edu). The local DNS server of HostA is dns.someschool.edu and the authoritative DNS server of HostB is dns.anotherschool.edu. Ignore DNS caching. In both (a) and (b), you must show the order of the queries by numbering them. (a) Assume that HostA makes a recursive query, and the rest of the queries between DNS servers are iterative. Draw the interaction of DNS servers and HostA. Refer to the slide #101 of Application chapter. 2

3 (b) Assume that all queries are recursive. Draw the interaction of DNS servers and HostA. Refer to the slide #100 of Application chapter. 6) Content Delivery (15 points) There exist some problems for content delivery in Internet such as congestion in routers and high latency for users. Explain (in brief) three general ideas (architecture) to solve these problems. Ideas: Caching (HTTP, web proxy and server farms) Content Delivery Networks (CDNs) Peer-to-Peer Networks (P2P) 7) TCP fairness (14 points) What is meant by TCP fairness? How AIMD (Additive increase/multiplicative decrease) algorithm can leads to fairness. Fairness (4 points): The idea is that if two TCP flows share a common bottleneck link and have similar round-trip times, than they should attain (approximately) equal throughputs. However, this is not true when one TCP flow has a larger RTT than the other. In this case, the flow with the shorter RTT gets higher throughput. AIMD (Adaptive Increase, Multiplicative-Decrease) (5 points) AI: increase cwnd by 1 MSS every RTT in the absence of loss events. Additive increase gives slope of 1, as throughout increases MD: cut cwnd in half after loss event. multiplicative decrease decreases throughput proportionally Convergence to optimal point and related figure (5 points) 8) Transmission Control Protocol (TCP) (16 points) Consider the following plot of TCP window size as a function of time. 3

4 Assuming TCP Reno is the protocol experiencing the behavior shown above, answer the following questions. In all cases, you should provide a short discussion justifying your answer. (a) Identify the intervals of time when TCP slow start is operating. TCP slow start is operating in the intervals [1,6] and [23,26] (b) Identify the intervals of time when TCP congestion avoidance is operating. TCP congestion avoidance is operating in the intervals [6,16] and [17,22] (c) After the 16th transmission round, is segment loss detected by a triple duplicate ACK or by a timeout? After the 16th transmission round, packet loss is recognized by a triple duplicate ACK. If there was a timeout, the congestion window size would have dropped to 1. (d) After the 22nd transmission round, is segment loss detected by a triple duplicate ACK or by a timeout? After the 22nd transmission round, segment loss is detected due to timeout, and hence the congestion window size is set to 1. (e) What is the initial value of Threshold at the First transmission round? Answer Threshold is initially 32, since it is at this window size that slow start stops and congestion avoidance begins. (f) What is the value of Threshold at the 18th transmission round? Threshold is set to half the value of the congestion window when packet loss is detected. When loss is detected during transmission round 16, the congestion windows size is 42. Hence the threshold is 21 during the 18th transmission round. (g) During what transmission round is the 70th segment sent? During the 1st transmission round, packet 1 is sent; packet 2-3 are sent in the 2nd transmission round; packets 4-7 are sent in the 3rd transmission round; packets 8-15 are sent in the 4th transmission round; packets are sent in the 5th transmission round; packets are sent in the 6th transmission round; packets are sent in the 7th transmission round. Thus packet 70 is sent in the 7th transmission round. 4

5 (h) Assuming a packet loss is detected after the 26th round by the receipt of a triple duplicated ACK, what will be the values of the congestion control window size and of Threshold? The congestion window and Threshold will be set to half the current value of the congestion window (8) when the loss occurred. Thus the new values of the threshold and window is 4. 9) Network Delay (Extra Point -10 points) Consider a server sending a 64 MB audio file to a receiver over a 1Mbps connection using packets of size 1 MB. After a packet is sent, the sender waits until an ACK packet of size 8 bytes is received before a new packet can be sent (no pipelining). Find the latency of the connection if the data transfer lasts 10 minutes in total. Assume that the packet processing delays (t proc ) at the sender and the receiver are negligible. Answer: M = 512 Mbits; R = 1 Mbps; P = 8 Mbits; T = 600 s; A = 64bits; Total transmission time is: T = (M/P) (t tran1 + t prop + t proc-rec + t tran2 + t prop + t proc-tran ) = (M/P) (P/R + 2 t prop + t proc-tran + A/R + t proc-rec ) Neglecting both processing delays (t proc-tran = t proc-rec =0), latency = 2t prop = (PT / M) - P/R - A/R sec total latency = (M/P) latency 88 sec 5