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Communications and Networks. Lecture 10 Instructor: Rina Zviel-Girshin. Lecture overview. Routing Problems. Network is represented as a graph . Graph has nodes – computers and edges (links). Graph abstraction for routing algorithms: graph nodes are routers
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Communications and Networks Lecture 10 Instructor: Rina Zviel-Girshin
Lecture overview • Routing • Problems
Network is represented as a graph. Graph has nodes – computers and edges (links). Graph abstraction for routing algorithms: graph nodes are routers graph edges are physical links link cost: delay, $ cost or congestion level 5 3 5 2 2 1 3 1 2 1 A D E B F C Routing protocol Routing Goal: determine “good” path (sequence of routers) through network from source to destination • “good” path: • typically means minimum cost path • other definitions can be provided (min distance, min time)
Keeping table up-to-date • There are lots of protocols that are used to keep routing tables in the routers and hosts up-to-date. • Example protocols include: • RIP: • HELLO: • OSPF: • GGP: • EGP: • BGP: • All of them try to find a “good” path • minimum cost path
Fragmentation • Maximum Packet Size • Break Large Packets into Segments • Re-assembly • At exit gateway • At the receiving end
Network links have MTU (max.transfer size) - largest possible link-level frame. different link types, different MTUs Large IP datagram divided (“fragmented”) within network one datagram becomes several datagrams “reassembled” only at final destination IP header bits used to identify, order related fragments IP Fragmentation & Reassembly FRAGMENTATION: in: one large datagram out: 3 smaller datagrams reassembly
length =4000 length =1500 length =1040 length =1500 ID =x ID =x ID =x ID =x fragflag =0 fragflag =1 fragflag =1 fragflag =0 offset =1480 offset =2960 offset =0 offset =0 IP Fragmentation and Reassembly Example • 4000 byte datagram • MTU = 1500 bytes One large datagram becomes several smaller datagrams
Firewalls • Security Issue • Information leaking out • Information leaking in • Viruses • Types • Packet-level filter • Source and destination IP • Application-level filter • Mail contents Firewall
IPv6 • Problems in Current Internet • Too few addresses • IPv6 • Long addresses • 128 bit • Multiple addresses per node • Better security • Better quality of service
1 27 Problem 1 • Your company has 27 computers. • You wish to connect all company’s computers to the internet. • How many bits should be given for subnet mask? • What should be your subnet mask address? • How many computers can be connected to your network without changing subnet mask?
1 27 Answer • Answer: • 27 computers – need 5 bits for encoding, 25=32. • 32 bits address – 5 bits = 27 bits. • 11111111.11111111.11111111.11100000 • Subnet mask address: 255.255.255.224 • Number of additional IP addresses your company holds: 32-27 –(2 special addresses)=3 computers
Problem 2 • Your company has several sub-networks. • Your company wants to connect those sub-networks to the Internet. • Your ISP gave you the following IP address: 156.161.130.0/20. • A.What is a class of your network? • Answer: class B (ip address starts with 10) • B. How many computers may be connected to your network? • Answer: 212-2=4094
Subnet B RB Subnet X RA RX Subnet A Internet Problem 2 (cont.) • Lets say you wish to connect your computers as described in the picture ( subnet X does not have any host connected to it). • C. What should be minimal subnet mask (max number of bits) for each subnetwork to have 150 computers in it (except subnet X)? • D. What is IP address of the requires subnet mask?
Subnet B RB Subnet X RA RX Subnet A Internet Solution 2 • 2 sub-networks with 150 computers: 150+150=300 computers. • 9 bits for IP addresses: 29=512. • Therefore: • C. minimal subnet mask is 32 –9 = 23 bits. • D. IP address is:255.255.254.0
Subnet B RB Subnet X RA RX Subnet A Internet Problem 2 (cont). • E. How many computers can be connected to this network in the future? • Answer: • All together 29-2=510 • 300 already in use • 210 can be connected in the future.
Subnet B RB Subnet X RA RX Subnet A Internet Problem 2 (cont). • F. How many sub-networks can be in your architecture (keeping in mind C structure)? • Answer: • Your ISP company holds a prefix of 20 bits. • Your current computers require additional 9 bits. • 3 bits are left for intra networking: 23=8 sub-networks
Problem 3 • Which address fits 11.0.0.0/8 • 11.150.0.34 • 11.155.280.98 • 8.85.29.189 • 14.50.97.103
Solution 3 11.150.0.34 11.155.280.98 8.85.29.189 14.50.97.103 • Which address fits 11.0.0.0/8? • Subnet mask is only 8 bits so only address that starts with 11 can be used. • 11.150.0.34 • 11.155.280.98 • However 11.155.280.98 is not a correct IP address. • Therefore the answer is: 11.150.0.34.
Problem 4 • To which network belongs the following IP address: 12.174.56.16 • 12.174.56.18/30 • 12.174.56.166/28 • 12.174.57.0/24 • 12.174.50.16/26
Solution 4 12.174.56.18/30 12.174.56.166/28 12.174.57.0/24 12.174.50.16/26 • Given 12.147.56.16 and 4 masks: 24,26,28,30 bits. • 24 bits is the smallest mask. Therefore in all 4 cases prefix of 24 bits remains the same and it should remain 12.147.56. • Therefore 12.174.57.0/24 and 12.174.50.16/26 are incorrect addresses. • Binary representation of 16 is: 00010000. • In case of 12.174.56.18/30 binary representation of 18 is 00010010 and if the mask is 30 bits then both representation (00010000) have the same prefix 000100 and can belong to the same network. • In case of 12.174.56.166/28 binary representation of 166 is 10100110 and if the mask is 28 bits then both IP’s should have the same prefix 1010. However 16 starts with 0001.
