1. CSEN 404 Introduction to Networks Amr El Mougy Lamia AlBadrawy

2. Revision 1-2

3. A Day in the Life of a Webpage Request  Bob wishes to retrieve a page from www.google.com  Bob has just booted up the computer Bob 00:16:D3:23:63:8A GUC 68.85.2.0/24 00:22:6B:45:1F:1B 68.85.2.1 ISP 68.80.0.0/13 DNS Server 68.87.71.226 Google 68.233.160.0/19 www.google.com 68.233.169.105

4. 4 A Day in the Life of a Webpage Request GUC 68.85.2.0/24 00:22:6B:45:1F:1B 68.85.2.1 ISP 68.80.0.0/13 DNS Server 68.87.71.226 Google 68.233.160.0/19 www.google.com 68.233.169.105 6867 0.0.0.0255.255.255.255 00:16:D3:23:63:8AFF:FF:FF:FF:FF:FF Yiaddr: 68.85.2.101 6768 68.85.2.1255.255.255.255 00:22:6B:45:1F:1BFF:FF:FF:FF:FF:FF  DHCP ACK also includes IP of default gateway (router in this case) and DNS server Bob 00:16:D3:23:63:8A

5. 5 A Day in the Life of a Webpage Request Bob 00:16:D3:23:63:8A 68.85.2.101 GUC 68.85.2.0/24 00:22:6B:45:1F:1B 68.85.2.1 ISP 68.80.0.0/13 DNS Server 68.87.71.226 Google 68.233.160.0/19 www.google.com 68.233.169.105 ARP Query 68.85.2.10168.85.2.1 00:16:D3:23:63:8AFF:FF:FF:FF:FF:FF ARP Reply 68.85.2.168.85.2.101 00:22:6B:45:1F:1B00:16:D3:23:63:8A  Before sending a DNS query, Bob must first discover the MAC address of the router through ARP

6. 6 A Day in the Life of a Webpage Request GUC 68.85.2.0/24 00:22:6B:45:1F:1B 68.85.2.1 ISP 68.80.0.0/13 DNS Server 68.87.71.226 Google 68.233.160.0/19 www.google.com 68.233.169.105 DNS query: www.google.com 6133453 68.85.2.10168.87.71.226 00:16:D3:23:63:8A00:22:6B:45:1F:1B  To send HTTP request, Bob’s browser has to create a TCP socket. Thus, it needs the IP address of www.google.com Bob 00:16:D3:23:63:8A 68.85.2.101 DNS reply: 68.233.169.105 5361334 68.87.71.22668.85.2.101 MAC of DNSMAC of default R

7. 7 A Day in the Life of a Webpage Request  To establish TCP connection, TCP SYN packet is sent GUC 68.85.2.0/24 00:22:6B:45:1F:1B 68.85.2.1 ISP 68.80.0.0/13 DNS Server 68.87.71.226 Google 68.233.160.0/19 www.google.com 68.233.169.105 Bob 00:16:D3:23:63:8A 68.85.2.101 5212280 68.85.2.10168.233.169.105 00:16:D3:23:63:8A00:22:6B:45:1F:1B 8052122 68.233.169.10568.85.2.101 MAC of web serverMAC of default R

8. 8 A Day in the Life of a Webpage Request  Finally HTTP request can be sent GUC 68.85.2.0/24 00:22:6B:45:1F:1B 68.85.2.1 ISP 68.80.0.0/13 DNS Server 68.87.71.226 Google 68.233.160.0/19 www.google.com 68.233.169.105 HTTP Request: google search 5212280 68.85.2.10168.233.169.105 00:16:D3:23:63:8A00:22:6B:45:1F:1B Bob 00:16:D3:23:63:8A 68.85.2.101 HTTP Reply: search results 8052122 68.233.169.10568.85.2.101 MAC of web serverMAC of default R

9. Solved Problems 1-9 Introduction I

10. Q1: Delays  Host A wishes to send a file of size 4 Mbits to Host B. The path from Host A to Host B has three links of rates R 1 = 500 kbps, R 2 = 1 Mbps, R 3 = 2 Mbps. The propagation speed is 2×10 8 m/s over all links and the lengths of the links are D 1 = 2 km, D 2 = 3 km, D 3 = 2 km. The processing time for each intermediate router is 0.1 × Transmission time. Ignore delays due to queuing and TCP and assume the file is transmitted directly (HTTP is not used).  What is the throughput of the path from Host A to Host B?  What is the total delay of transmitting the file?

11. Q1: Solution Transport Layer 1-11

12. Q2: Transport layer protocols Transport Layer 1-12  A sender, A, and receiver, B, are using the pure selective repeat protocol at the Transport Layer. Both sender and receiver windows can hold a maximum of 4 packets. Initially the sender and receiver windows are synchronized at [0, 1, 2, 3]. Only A is sending data to B. Complete the diagram below by specifying the sender’s window, receiver’s window, the sequence number of the buffered packets and of the ACKs, and any packets that are delivered to the network layer for the following succession of events. Assume NAKs and cumulative ACKs are not used.

13. Transport Layer 1-13

14. 1-14

15. Q3: TCP Transport Layer 1-15  Consider two hosts A and B that are communicating over a TCP connection. Assume that the initial sequence numbers for A and B are 44 and 70, respectively. The figure below shows the exchange of messages between A and B. Assuming that the system is able to buffer packets that are received out of order, fill in the blanks for sequence numbers and acknowledgment numbers.

16. Q3: TCP Transport Layer 1-16

17. Q3: Solution Transport Layer 1-17

18. Q4: Congestion Control 1-18 Consider a host that is sending data over a TCP connection with the following initial values: cwnd = 1 MSS, ssthresh = 50 MSS. Suppose that the receiver buffer size = 20 MSS, and the receiver processing speed is 2 MSS per RTT. Draw a table that shows the sender rate of sending versus RTT (How many MSS per RTT) for the first 15 RTT. Note that three duplicate ACKs are received in the 5th RTT. TCP Reno is used.

19. Q4: Solution 1-19

20. Q5: Routing 1-20  The figure represents a network of one client (A), one server (G), and a series of routers (B  F). Use Dijkstra’s algorithm to determine the best paths from (A) to all other nodes in the network (warning: points will be given for the steps of the algorithm. Simply indicating the best paths without showing the steps will not be rewarded).  Draw the forwarding table at (A).

21. Q5: Solution 1-21

22. Q6: IP Addresses 1-22 An organization has 7 LANs as shown in the figure. The organization has a network address 193.166.124.x. Determine:  The network address for each LAN.  The IP addresses of the hosts (H1  H6) and the routers shown in the figure. Note that LAN 7 has no hosts but requires IP addresses.  The maximum number of hosts that can be connected to each of the first 6 LANs (LAN1  LAN 6).

23. Q6: Solution Transport Layer 1-23 7 networks  3 subnet bits and 5 host bits. Thus, max no. of hosts is 32 Network address LAN 1: 193.66.124.0 LAN 2: 193.66.124.32 LAN 3: 193.66.124.64 LAN 4: 193.66.124.96 LAN 5: 193.66.124.128 LAN 6: 193.66.124.160 LAN 7: 193.66.124.192 Host address LAN 1: 193.66.124.1, 193.66.124.2 LAN 2: 193.66.124.33, 193.66.124.34 LAN 3: 193.66.124.65, 193.66.124.66 LAN 4: 193.66.124.97, 193.66.124. 193.66.124.98 LAN 5: 193.66.124.129, 193.66.124.130 LAN 6: 193.66.124.161, 193.66.124.162 LAN 7: 193.66.124.193, 193.66.124.194 Number of hosts 2^5 = 32 -2 = 30