c) With the selective repeat protocol, it is possible for the sender to receive an ACK for a packet that falls outside of its current window.

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1 Part 1 Question 1 [0.5 Marks] Suppose an application generates chunks of 40 bytes of data every 20 msec, and each chunk gets encapsulated by a TCP segment and then an IP datagram. What percentage of each datagram will be overhead, and what percentage will be application data given that both TCP and IP have 20 byte headers each and there is no fragmentation? a) 80% b) 75% c) 60% d) 50% Answer: As TCP and IP are 20 bytes each, it adds 40 bytes to the data which makes is 80 total bytes. The overhead is thus 40 of the 80 total, so that's 50% Question 2 [2 Marks] TCP provides reliable transfer through a mixture of sequence number, receiver buffer, cumulative acknowledgement, and fast retransmission. Indicate whether the following statements are True or False. a) Suppose host A sends host B one segment with sequence number 38 and 4 bytes of data. Then in the same segment the acknowledgement number is necessarily 42. False. The acknowledgement number has nothing to do with the sequence number. The ack. number indicates the next sequence number A is expecting from B. b) Suppose that the last samplertt in a TCP connection is equal to 1 second. Then timeout for the connection will necessarily be set to a value >= 1 second. False. Next_RTT = alpha * last_estimated_rtt + (1- alpha)*newly_collected_rtt_sample. In this case even though the last samplertt which is the newly_collected_rtt_sample is 1sec, the next_rtt still depends on alpha and last_estimated_rtt. Therefore, the next_rtt is not necessarily greater than 1sec. c) With the selective repeat protocol, it is possible for the sender to receive an ACK for a packet that falls outside of its current window. False. SR uses selective acknowledgement. The ack. number has no way to fall outside the current window.

2 d) With the Go-Back-N, it is possible for the sender to receive an ACK for a packet that falls outside of its current window. True. GBN uses cumulative acknowledgement. Imagine a scenario where ACK1 arrives AFTER ACK2. Once the sender receives ACK2, it would know that both packet1 and 2 were received correctly. So it can remove packet1 and 2 from its window. Now if ACK1 arrives, then ACK1 actually falls outside the current window. Question 3 [0.5 Marks] Is it necessary that every autonomous system use the same intra-as routing algorithm? a) Yes b) No c) No need for an intra-as routing algorithm Answer No: each AS has administrative autonomy for routing within an AS. Question 4 [0.5 Marks] What is the difference between routing and forwarding? a) There is no difference b) Routing determines which outport an incoming packet should be sent to and forwarding is transmitting the packet on the link c) Forwarding is moving a packet from a router s input link to the appropriate output link. Routing is determines the end-to-routes between sources and destinations Answer: Forwarding is about moving a packet from a router s input link to the appropriate output link. Routing is about determining the end-to-routes between sources and destinations. Question 5 [0.5 Marks] How many IP addresses does a router with 4 interfaces have? a) Zero - Routers do not have IP addresses b) One IP address c) Two IP addresses d) Four IP Addresses Answer: Each interface needs an IP address Question 6 [0.5 Marks] If the transmission rate is 1 Mbps and the RTT is 5ms, what is the maximum data rate that can be achieved when transmitting 1 kbyte packets, if the end protocols uses a simple stop and wait protocol? Assume no errors. a)111 kps b) 1 Mbps

3 c) 500 Kbps Answer: Time to transmit 1 kbyte = (8 x10 3 )/(1x 10 6 ) = 8 x Therefore, to transmit 8 x10 3 bits it takes 13 ms. Thus the throughput = (8/13) x 10-6 = 615 kbps Question 7 [0.5 Marks] Specify the IP subnet mask for a subnet that can contain up to 30 clients? a) w.x.y.z/28 b) w.x.y.z/27 c) w.x.y.z/32 d) w.x.y.z/16 Answer: Assume address starting at ; Host range: ; Mask bits: 27; Part 2 Question 1 [1 Mark] Suppose the TCP congestion window is set to 18KB and a timeout occurs. List the three things that happen? And explain how big the congestion window will be if the next four transmission bursts are all successful? Assume that the m Answer: When a timeout occurs, three things happens. First, slow start will be initiated. Second, the congestion window would start at 1. Third, the threshold will be reset to 18KB/2=9KB. If the next four transmissions are all successful, then 1st transmission: 1 segment, 1KB 2nd transmission: 2 segments, 2KB 3rd transmission: 4 segments, 4KB 4th transmission: 8 segments, 8KB After these four successful transmissions, the window size is supposed to be 16. However, since the threshold is 9KB, the window size can only be 9KB. Question 2 [1 Mark] Consider the network shown below, and assume that each node initially knows the cost to each of its neighbours. Consider the distance-vector algorithm and show the distance table entries at node z.

4 Question 3 [1 Mark] Assume you request a webpage consisting of one document and five images. The document size is 1 kbyte, all images have the same size of 50 kbytes, the download rate is 1 Mbps and the RTT is 100ms. How long does it take to obtain the whole webpage under the following conditions? (Assume no DNS name query is needed and the impact of the request line and the headers in the HTTP messages are negligible) a) Non-persistent HTTP with serial connections b) Persistent Connection with one connection. a) 2 x 100 ms bits / 10 6 bits/s + 5 (2 x 100 ms + 4 x 10 5 bits / 10 6 bits/s) = s b) 2 x 100 ms + 8 x 10 3 bits / 10 6 bits/s + 5 (4 x 10 5 bits / 10 6 bits/s) = 2.208s

5 Question 4 [1 Mark] Consider the effect of using slow start on a line with a 10-msec round-trip time and no congestion. The receive window is 24KB and the maximum segment size is 2KB. How long does it take before the first full window can be sent? With slow start, the first RTT sends out 1 segment (or 2KB), the 2nd RTT sends out 2 segments (or 4KB), the 3rd 4 segments (or 8KB), the 4th 8 segments (or 16KB). The 5th RTT would have sent out 16 segments (or 32KB), however, it'll exceed the receiver's window. Therefore, the amount of time it takes BEFORE the 5th RTT (or full window, that is, 24KB) is 4*10=40msec. Question 5 [1 Mark] What four pieces of information does a DHCP client obtain from the DHCP server after a successful communication between the two? Answer: IP address, IP address of first hop router, name and IP address of DNS server, network mask.