EC441 Midterm Two Fall 2017

Transcription

1 EC441 Midterm Two Fall 2017 This is an open-book, open-computer, open-notes exam. You may work with a partner, but you must submit one joint answer for each problem. You may not complete any exam with the same partner as any other exam. IP addressing and forwarding a) How many entries will be in a routing table for the network /24? b) Suppose every network in IPv4 is a /20 network. How many different networks would there be? Is there a maximum number of computers in each network? Explain. c) Suppose a typical internet router receives a packet with destination address and a time-to-live field of 5. The IP checksum is correct. The More Fragments bit (MF) is set to 1. What will the router do with this packet? d) Why does ICMPv6 include a Packet Too Big code, and why is this code not present in the ICMP protocol used by IPv4? a) One. b) 20 bits specify the network, so there are 2 20 networks. Each network would have 2 12 addresses. NAT and DHCP would allow for more computers than the number of addresses. c) Discard /8 is a re-usable private IP network. d) Fragmentation is not supported in IPv6, so this ICMP message is used to tell the sender to use a smaller packet. IPv4 just fragments as needed. IP address space Consider the following responses to the Internet address space problem: a) gradually transition computers to IP version 6 1

2 b) take away/buy under-utilized class B networks from their owners and re-assign them to bigger companies c) take away/buy under-utilized class B networks from their owners and re-assign them to CIDR d) use CIDR to improve address usage e) encourage the use of network address translation wherever possible f) encourage the use of DHCP in institutions such as Boston University Explain briefly why (or why not) each of these methods might help the Internet remain functioning. The long-term solution is the transition from IPv4 to IPv6. However, the transition from classful addressing to CIDR has led to more efficient use of the address space. A big problem with classful addresssing is the under-utilization of class A and class B networks. NAT and DHCP are protocols whose primary purpose is elsewhere (simplified operations for ISP, mobility/dynamic support), but both protocols allow for multiple computers to share a single public IP address (by translation or time-sharing). Routing Consider the following network: Figure 1: 2

3 One AS Suppose the network shown is one AS using OSPF and A-H represent routers. Two AS Suppose ABEF is an AS, and CDGH is a different AS. Each AS uses OSPF internally. All AS Now suppose that each node A-H represents an AS, and the link costs shown represent the distance between the participating next-hop routers in each AS. Assume that no AS has local preferences. Link B-C down Explain what happens in each of the three network scenarios above if the link between B and C does down. What messages are exchanged? How are the routing tables updated? Do any of the E->G or F->G path answers change? For one AS: OSPF uses shortest paths. F-B-C-D-H-G E-A-B-C-D-H-G 3

4 For two AS, the AS path is one for both, so the next consideration is next-hop PATH cost to get out of our AS into the other one. This is F-B-C for F, and E-A-B-C for E. Then, in the other AS, OSPF is used. So the answers are F-B-C-D-H-G E-A-B-C-D-H-G For all AS, the shortest path in terms of number of ASes is chosen, regardless of the cost of the router-to-router costs. So: F-C-G E-F-C-G When B-C goes down, the BGP and OSFP protocols will exchange DV packets and LS packets. The routes for parts 1 and 2 will now be F-C-D-H-G E-F-C-D-H-G TCP Operation You are downloading a 2 Gbyte file from a website. After the first minute, your download time estimate is 59 more minutes. After 5 minutes, the download time estimate is 115 more minutes. After 10 minutes, the download time estimate is 5 more minutes, and the download completes after a total of 30 minutes from the start. Explain what might have been happening in the network during these 30 minutes, and how your TCP connection s parameters (such as RTT, timeout, congestion window, receive window) might have evolved. Is it possible you are connected to the network with a 10 Mb/s link? Explain. These numbers are indicative of congestion. This means that the RTT is likely to increase and the congestion window to decrease during those times when the estimate is long. Timeouts are also most likely when the estimate is long, causing the sender to reduce its congestion window and its effective throughput. At one time, the average throughput led to an estimate of 15 minutes, which equates to an average data rate of 28 10ˆ9 / (15*60) which is 17.8 Mb/s. The link speed cannot be 10 Mb/s, it must be at least as high as 17.8 Mb/s. TCP versus UDP Show a complete sequence of packets for a TCP connection in which a single application message is sent from the client to the server (100 bytes long) and the server responds with 4

5 5000 bytes worth of data, after which the connection is closed. Suppose the round trip time is 2 ms for a typical packet and acknowledgment. Estimate the total amount of time taken to complete the TCP session. What would the time have been if UDP was used? About 10 ms for TCP. About 3 ms for UDP. 5