Internet Technology 4/5/2016

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1 Question 1 Compute the 8-bit Internet checksum of the following values: Internet Technology 11. Spring 2016 Exam 2 Review Paul Krzyzanowski Rutgers University Spring ~= Overflow (carry) Complement this value April 5, Paul Krzyzanowski 1 April 5, Paul Krzyzanowski 6 Question 2a Question 2b Apply Dijkstra s Link-State algorithm to the graph of nodes and costs to find the least-cost paths from node a. Express your answer as a list of { D(v), p(v) } tuples, where D(v) is the cost of the least-cost path from node a to some vertex v and p(v) is the previous node of v along that path. Nʹ D(b), p(b) D(c), p(c) D(d), p(d) D(e), p(e) a 1, a 3, a 3, a ab 1, a 3, a 2, b abd 1, a 3, a 2, b abdc 1, a 3, a 2, b 5, c abdce 1, a 3, a 2, b 5, c N = set of tested vertices 1. Pick the untested vertex with the least-cost path 2. Compute cost to each node through the new vertex 3. If it s lower then it displaces the previous least-cost path e 2 2 c 3 bd = 1, D(d) = D(b) + 1 = 2 d a b What is the forwarding table at node a? Nʹ D(b), p(b) D(c), p(c) D(d), p(d) D(e), p(e) abdce 1, a 3, a 2, b 5, c Create a path by following the previous nodes. Each entry of the forwarding table is the link that a packet must take when leaving node a Destination Path Next hop b b a b c c a c d d b a b e e c a c April 5, Paul Krzyzanowski 7 April 5, Paul Krzyzanowski 8 Question 3 Which statement is true about UDP? (a) UDP requires a connection setup between the client and server but the connection does not guarantee reliable delivery. (b) Message boundaries are not preserved: the application simply sees a stream of bytes. (c) Messages from different computers sent to the same port will be delivered to the same socket. (d) For efficiency, the kernel waits until it receives some minimum number of bytes before sending a UDP datagram (a) No connection setup in UDP (b) Message boundaries are preserved: # bytes sent = # bytes read (d) No. Data is sent immediately. Question 4 The Go-Back-N protocol is an example of: (a) Alternating bit protocol. (b) Negative acknowledgements. (c) Stop-and-wait. (d) Pipelining. (a) No. Alternating bit is only useful for stop-and-wait, where we have one at most one outstanding message. alternating bit = one-bit sequence # (b) No. GBN uses sequence #s, not negative ACKs (c) No. GBN allows multiple segments to be in transit at a time (= pipelining) April 5, Paul Krzyzanowski 9 April 5, Paul Krzyzanowski 10 Paul Krzyzanowski 1

2 Question 5 Question 6 A drawback of the Go-Back-N protocol is: (a) A failure to receive one packet may require retransmitting many packets. (b) Only N packets may be in transit at any one time. (c) It does not support pipelining. (d) It uses a timer. Selective Repeat does not use: (a) Timeouts. (b) A sliding window. (c) A receive window. (d) Cumulative acknowledgements. Drawback = undesirable consequence (a) If the sender times out (does not receive a cumulative ACK), it will retransmit ALL segments in its window. Worst case = first segment is lost but all others are received. Selective Repeat fixes this (at a cost of non-cumulative ACKs). (b) Yes but this is not a drawback. Only N segments (the window size) can be in transit. This allows control of effective network utilization (c) Yes it does this is not a drawback. It improves network utilization. (d) Yes it does this is not a drawback. It enables detection of lost segments. SR was designed to address the drawback of GBN: having to re-transmit segments that may have been successfully received. It does this by sending individual ACKs instead of cumulative ACKs. (a) Timeouts are used on a per-segment basis to detect lost ACKs or lost segments (b) Sliding window: set of sent but unacknowledged segments (c) Receive window used as a buffer to assemble out-of-order segments Since each segment is acknowledged individually it will not be re-transmitted. (d) No. SR uses individual ACKs. April 5, Paul Krzyzanowski 11 April 5, Paul Krzyzanowski 12 Question 7 The goal of Path MTU Discovery is to: (a) Determine the most efficient route to the destination. (b) Avoid having IP datagrams get fragmented. (c) Define the largest transmit window size. (d) Define the largest receive window size. Question 8 Which of these is not a field in the TCP header? (a) Checksum. (b) Acknowledgement number. (c) Destination IP address. (d) Source port number. MTU = Maximum Transmission Unit = largest acceptable payload size that a link can carry Path MTU discovery discovers the smallest MTU along the entire path. This allows the formation of datagrams that will not get fragmented It does not report the path. (a) IP, UDP, and TCP headers all have checksums (b) TCP headers have an ACK number (c) The source and destination IP addresses are in the IP header, not TCP or UDP (d) TCP and UDP headers have source & destination port numbers Windows size segment size April 5, Paul Krzyzanowski 13 April 5, Paul Krzyzanowski 14 Question 9 The receive window in the TCP header: (a) Defines the maximum number of bytes that the receiver can accept. (b) Identifies the address of the first free byte in the receive buffer. (c) Is the next sequence number needed by the receiver. (d) Contains the number of bytes that have been successfully received so far. Question 10 Which value is doubled when a timeout occurs in TCP? (a) Round-trip time variation. (b) Smoothed round-trip time. (c) Timeout interval. (d) Exponentially weighted moving average (a, b, c): these are measured values. April 5, Paul Krzyzanowski 15 April 5, Paul Krzyzanowski 16 Paul Krzyzanowski 2

3 Question 11 Question 12 You sent five segments starting at 1460 and received the following TCP acknowledgements: 2400, 3860, 3860, 3860 What is this most likely to indicate? (a) One segment was lost. (b) Two segments were lost. (c) The receiver got three duplicate segments. (d) All segments have been received Each duplicate ack means: here is what I need I still didn t get it ACK 2400: I got everything up to 2399 ACK 3860: I got everything up to 3859 ACK 3860: I got a segment but still need bytes starting at 3860 ACK 3860: I get another segment but still need bytes starting at n n LOST ack: 2400 ack: 3860 ack: 3860 ack: 3860 Piggybacking acknowledgements means that: (a) Multiple acknowledgements can be sent in the same segment. (b) An acknowledgement is sent in the same segment as return data. (c) A cumulative acknowledgement is sent. (d) The receiver is sending a duplicate acknowledgement April 5, Paul Krzyzanowski 17 April 5, Paul Krzyzanowski 18 Question 13 One way of dealing with a SYN flooding attack is to: (a) Not allocate TCP buffers until the initial handshake is complete. (b) Limit the total number of TCP connections at a host. (c) Limit the number of TCP connections from each client. (d) Authenticate the identity of each client SYN floods often sends a SYN packet with an unreachable return address. The OS allocates buffers for the connection but the handshake never completes, leaving resources allocated until a timeout period (typically >30 sec) (a) SYN cookies: Don t allow any single packet to cause the OS to allocate buffers. Instead, allocate when the connection handshake is complete: when the final ACK is sent, which will contain a sequence # that is f(addresses, ports, server_secret). (b) Every OS does this this does not prevent SYN flooding. (c) This does not prevent SYN flooding either since source addresses are usually random unreachable addresses. (d) The server does not perform any client authentication; it simply checks that the ACK is a legitimately related to the SYN-ACK that the server sent. Question 14 When a host receives a TCP segment with a FIN (finish) flag, it means that the sending host: (a) May send data until it receives a FIN as a response. (b) Will not send any more data. (c) Will not receive any more data. (d) Will neither send nor receive any more data FIN means that you re done communicating and want to close the connection. You will not send more data but there still may be data coming your way that you may receive. April 5, Paul Krzyzanowski 19 April 5, Paul Krzyzanowski 20 Question 15 TCP s fast retransmit occurs when a segment is retransmitted: (a) Before a timeout occurs. (b) Immediately after a timeout. (c) At a faster transmission rate than normal. (d) Upon receiving a negative acknowledgement (NACK). Three received segments with duplicate ACK #s assume sent segment was lost without Retransmit the missing segment without waiting for the timer to expire (b) No we beat the timer (c) No we transmit at whatever the transmission rate of the interface is (d) No TCP doesn t use NACKs Question 16 You sent five segments of 1000 bytes with a starting sequence number of 1000 and received the following TCP acknowledgements: 2000, 2000, 2000, What is this most likely to indicate? ack: 2000 (a) Two segments are lost (b) Three duplicate segments have been received. (c) One segment was lost. (d) All segments have been received. We send segments with seq # 1000, 2000, 30000, 4000, 5000 If ALL are received, our cumulative ACK will be for 6000 (we got up to byte #5999) Looks like we lost a segment because we received duplicate ACKs. But then we get an ACK for DELAYED or LOST 5000 ack: 2000 ack: 2000 ack: 4000 ack: 6000 April 5, Paul Krzyzanowski 21 April 5, Paul Krzyzanowski 22 Paul Krzyzanowski 3

4 Question 17 Question 18 TCP differs from a Go-Back-N protocol in that: (a) Go-Back-N does not support reliable delivery. (b) Go-Back-N does not support cumulative acknowledgements. (c) TCP does not use a sliding window. (d) TCP will only retransmit one segment on a timeout TCP s congestion window (cwnd): (a) Is a backlog queue for storing data when the network is congested. (b) Increases when the network appears to be less congested. (c) Is negotiated with the receiver to maximize the segment size. (d) Is a time interval that determines when a segment can be retransmitted (a) No: Go-Back-N is designed for reliable delivery (b) No: Go-Back-N does use cumulative acknowledgements (c) No: TCP does use a sliding window (d) Yes: GBN retransmits the entire window on timeout TCP only must retransmit the first lost segment on a timeout The congestion window is the transmit window size based on a guess of network congestion (packet loss congestion). Larger cwnd more pipelining increased network utilization cwnd increases during slow start until a threshold is reached cwnd increases during congestion avoidance until a timeout occurs (or 3 duplicate ACKs are received) April 5, Paul Krzyzanowski 23 April 5, Paul Krzyzanowski 24 Question 19 TCP s slow start means that the transmission rate: (a) Increases exponentially by one MSS with each received acknowledgement. (b) Increases linearly by one MSS every RTT. (c) Increases sub-linearly by less than one MSS every RTT. (d) Decreases gradually by one less MSS every RTT Question 20 In TCP s congestion avoidance state: (a) The window size remains constant. (b) The window size gradually increases. (c) The window size gradually decreases. (d) Transmission is temporarily stopped to relieve network congestion Slow start = cwnd size increases by one MSS with each received ACK +MSS per segment Congestion avoidance = cwnd increases by one MSS each RTT cwnd MSS segments per window Congestion avoidance = cwnd increases by one MSS each RTT cwnd MSS segments per window cwnd increases until: 3 duplicate ACKs received cwnd = cwnd/2 + 3 MSS timeout cwnd = 1 April 5, Paul Krzyzanowski 25 April 5, Paul Krzyzanowski 26 Question 21 The control plane of a router: (a) Checks and recomputes IP checksums. (b) Looks up the destination for an incoming IP datagram. (c) Manages output queues. (d) Updates the router s forwarding table Any decisions having to do with real-time routing are handled at the data plane. Question 22 What address belongs in the /22 network? (a) bits 8 bits 8 bits 8 bits (b) (c) (d) We need to match the first 6 bits of the 3 rd byte 8 = (a) 11 = (b) 12 = (c) 7 = (d) 5 = bits April 5, Paul Krzyzanowski 27 April 5, Paul Krzyzanowski 28 Paul Krzyzanowski 4

5 Question 23 Question 24 How does a DHCP client receive an offer from a server if it does not yet have an IP address? (a) The DHCP server uses a broadcast to send the response. (b) The client acquires a temporary address from which it contacts the server. (c) The response is sent via a link layer protocol. (d) The client must be configured with an address; DHCP just provides DNS servers and the gateway address. The client sends a request via a limited broadcast ( ). Since it does not have an address, the server sends a reply to everyone via a limited broadcast. Unlike IPv4, an IPv6 header does not contain a: (a) Checksum. (b) Protocol ID. (c) Source address. (d) Destination port number IPv6 got rid of checksums: At the link layer, Ethernet & Wi-Fi have them At the transport layer, TCP & UDP have them It s redundant for IP April 5, Paul Krzyzanowski 29 April 5, Paul Krzyzanowski 30 Question 25 Network Address Translation, NAT, translates: (a) Between IPv4 and IPv6 addresses. (b) The destination address of a packet to the address of the next router at each hop in the path. (c) Internal IP addresses in a packet to external ones. (d) User friendly domain names to IP addresses Question 26 An autonomous system is a set of hosts and routers that: (a) Expose a single routing policy. (b) Are hidden behind a NAT gateway. (c) Are fully configured and do not need to use protocols such as BGP to determine external routes. (d) Are disconnected from the rest of the Internet Autonomous Systems are an administrative collection of systems that provide a uniform policy for what traffic they permit to be routed through them. (b) No they have nothing to do with NAT and are not hidden (c) No they rely on BGP to advertise their routes and discover routes (d) No they are a core part of the Internet April 5, Paul Krzyzanowski 31 April 5, Paul Krzyzanowski 32 Question 27 TCP does not use negative acknowledgements. Question 28 Fragmentation requires recomputing TCP checksums since data is now split across multiple packets. TCP uses cumulative acknowledgements and timeouts No. The packets must be reassembled into one datagram before the segment is extracted and sent to the transport layer. April 5, Paul Krzyzanowski 33 April 5, Paul Krzyzanowski 34 Paul Krzyzanowski 5

6 Question 29 Question 30 Route aggregation reduces the number of entries in a forwarding table. An alternating bit protocol is a form of a stop-and-wait protocol that identifies duplicate packets. It allows you to use a single prefix to specify multiple IP addresses Alternating bit = 1 bit sequence number We added this to distinguish new data from duplicate data due to a retransmission April 5, Paul Krzyzanowski 35 April 5, Paul Krzyzanowski 36 Question 31 A NAT-aware router may have to recompute TCP and UDP checksums in addition to the IP checksum. Question 32 The traceroute program shows the route to a remote node by querying each router for its next hop. Yes. If it changes port numbers, which are in TCP and UDP headers, those checksums will need to be recomputed No the router will not state its next hop. Traceroute relies on TTL timeouts. A router sends an ICMP TTL Exceeded error to the sender. The sender looks at the source address to identify the router. April 5, Paul Krzyzanowski 37 April 5, Paul Krzyzanowski 38 Question 33 Poison reverse helps avoid the count-to-infinity problem with unreachable networks. Question 34 BGP uses a distance vector algorithm to establish least-cost paths. A router will not advertise routes to a router X if any of those routes go through X. BGP peers exchange information by sending route advertisements. It tells its peers which other routers it can reach. April 5, Paul Krzyzanowski 39 April 5, Paul Krzyzanowski 40 Paul Krzyzanowski 6

7 Question 35 A Stub AS (autonomous system) is connected only to one other AS: The end A stub autonomous system cannot provide transit service between ASes. April 5, Paul Krzyzanowski 41 April 5, Paul Krzyzanowski 42 Paul Krzyzanowski 7