Network Layer: outline 1 introduction 5 routing algorithms link - PowerPoint PPT Presentation

Datagram forwarding table 4 billion IP addresses, so rather than list individual routing algorithm destination address list range of addresses local forwarding table (aggregate table entries) dest address output link address-range 1 3 address-range 2 2 address-range 3 2 address-range 4 1 IP destination address in arriving packet ’ s header 1 3 2 27

Datagram forwarding table Destination Address Range Link Interface 11001000 00010111 00010000 00000000 through 0 11001000 00010111 00010111 11111111 11001000 00010111 00011000 00000000 through 1 11001000 00010111 00011000 11111111 11001000 00010111 00011001 00000000 through 2 11001000 00010111 00011111 11111111 otherwise 3 Q: but what happens if ranges don ’ t divide up so nicely? 28

Longest prefix matching longest prefix matching when looking for forwarding table entry for given destination address, use longest address prefix that matches destination address. Link interface Destination Address Range 11001000 00010111 00010*** ********* 0 1 11001000 00010111 00011000 ********* 11001000 00010111 00011*** ********* 2 3 otherwise examples: which interface? DA: 11001000 00010111 00010110 10100001 which interface? DA: 11001000 00010111 00011000 10101010 29

Datagram or VC network: why? Internet (datagram) ATM (VC)  evolved from telephony  data exchange among  human conversation: computers  strict timing, reliability  “ elastic ” service, no strict requirements timing req.  need for guaranteed service  many link types  “ dumb ” end systems  different characteristics  telephones  uniform service difficult  complexity inside network  “ smart ” end systems (computers)  can adapt, perform control, error recovery  simple inside network, complexity at “ edge ” 30

Network Layer: outline 1 introduction 5 routing algorithms  link state 2 virtual circuit and datagram networks  distance vector  hierarchical routing 3 what ’ s inside a router 6 routing in the Internet 4 IP: Internet Protocol  RIP  datagram format  OSPF  IPv4 addressing  BGP  ICMP 7 broadcast and multicast  IPv6 routing 31

Router architecture overview two key router functions:  run routing algorithms/protocol (RIP, OSPF, BGP)  forwarding datagrams from incoming to outgoing link forwarding tables computed, routing routing, management pushed to input ports processor control plane (software) forwarding data plane (hardware) high-seed switching fabric router input ports router output ports 32

Input port functions lookup, link forwarding layer line switch protocol fabric termination (receive) queueing physical layer: bit-level reception decentralized switching : data link layer:  given datagram dest., lookup output port e.g., Ethernet using forwarding table in input port memory (“match plus action”)  goal: complete input port processing at ‘ line speed ’  queuing: if datagrams arrive faster than forwarding rate into switch fabric 33

Switching fabrics  transfer packet from input buffer to appropriate output buffer  switching rate: rate at which packets can be transfer from inputs to outputs  often measured as multiple of input/output line rate  N inputs: switching rate N times line rate desirable  three types of switching fabrics memory memory bus crossbar 34

Switching via memory first generation routers:  traditional computers with switching under direct control of CPU  packet copied to system ’ s memory  speed limited by memory bandwidth (2 bus crossings per datagram) input output port port memory (e.g., (e.g., Ethernet) Ethernet) system bus 35

Switching via a bus  datagram from input port memory to output port memory via a shared bus  bus contention: switching speed limited by bus bandwidth  32 Gbps bus, Cisco 5600: sufficient bus speed for access and enterprise routers 36

Switching via interconnection network  overcome bus bandwidth limitations  banyan networks, crossbar, other interconnection nets initially developed to connect processors in multiprocessor  advanced design: fragmenting datagram into fixed length cells, crossbar switch cells through the fabric.  Cisco 12000: switches 60 Gbps through the interconnection network 37

Output ports datagram link switch buffer line layer fabric termination protocol (send) queueing  buffering required when datagrams arrive Datagram (packets) can be lost from fabric faster than the transmission due to congestion, lack of buffers rate  scheduling discipline chooses among queued Priority scheduling – who gets best datagrams for transmission performance, network neutrality 38

Output port queueing switch switch fabric fabric one packet time later at t, packets more from input to output  buffering when arrival rate via switch exceeds output line speed  queueing (delay) and loss due to output port buffer overflow! 39

How much buffering?  RFC 3439 rule of thumb: average buffering equal to “ typical ” RTT (say 250 msec) times link capacity C  e.g., C = 10 Gpbs link: 2.5 Gbit buffer  recent recommendation: with N flows, buffering equal to . RTT C N 40

Input port queuing  fabric slower than input ports combined -> queueing may occur at input queues  queueing delay and loss due to input buffer overflow!  Head-of-the-Line (HOL) blocking: queued datagram at front of queue prevents others in queue from moving forward switch switch fabric fabric output port contention: one packet time later: only one red datagram can be green packet transferred. experiences HOL lower red packet is blocked blocking 41

Network Layer: outline 1 introduction 5 routing algorithms  link state 2 virtual circuit and datagram networks  distance vector  hierarchical routing 3 what ’ s inside a router 6 routing in the Internet 4 IP: Internet Protocol  RIP  datagram format  OSPF  IPv4 addressing  BGP  ICMP 7 broadcast and multicast  IPv6 routing 42

The Internet network layer host, router network layer functions: transport layer: TCP, UDP IP protocol routing protocols • addressing conventions • path selection • datagram format • RIP, OSPF, BGP network • packet handling conventions layer forwarding ICMP protocol table • error reporting • router “ signaling ” link layer physical layer 43

Figure 7.2 IP datagram 44

IP datagram format IP protocol version 32 bits total datagram number length (bytes) header length type of head. ver length (in 4-byte words) len service for “ type ” of data fragment fragmentation/ 16-bit identifier flag offset reassembly max number upper time to header remaining hops live layer checksum (decremented at 32 bit source IP address each router) 32 bit destination IP address upper layer protocol to deliver payload to 0-40 bytes options (if any) e.g. timestamp, how much overhead? data record route  20 - 60 bytes of IP (variable length, taken, specify typically a TCP list of routers Length of data? or UDP segment) to visit.  Total length – header length 45

Service Type 0 0 0 0 x x x x x x x x Precedence x x x x 1 1 interpretation x x x x 0 1 Differential service interpretation 46

• Time to live: control the maximum number of hops (routers) visited • If the value is zero, he router discards the datagram • Protocol: the higher-level protocol that uses the services of the IP layer 47

Example 1 An An IP IP packet et has has arri rive ved wi with the first st 8 bits as as shown wn: The The receiver iver discard ards the packet et. Why? Soluti tion on There is There is an an error error in in th this is pa packet cket. The 4 left left-most ost bi bits ts (0100 0100) sho show the the versi version, on, wh which ich is is corr correct ect. Th The next ext 4 bi bits ts (00 0010 10) show the the wron wr ong hea header er len ength gth (2 × 4 = 8). The The mi minimum imum number umber of of bytes bytes in in the the he head ader er must ust be be 20 20. Th The packet packet has has been been corr corrupted upted in in transmissi nsmission on. 48

Example 2 In an In an IP IP pack acket et, the the va value ue of of HLE HLEN (he (heade ader le length) ngth) is is 1000 1000 in in binary inary. Ho How many many bytes bytes of of op optio ions ns are are being ing carried carried by by th this is packet? acket? Soluti tion on The HLEN The HLEN val value ue is is 8, wh which ich mean means the the total total nu numb mber er of of bytes bytes in in the the he head ader er is is 8 × 4 or or 32 32 bytes bytes. Th The first first 20 20 bytes bytes are are the the base base header, r, the next 12 12 bytes es are the options ons. 49

Example 3 In In an an IP IP pack acket et, the the va value ue of of HLEN is is 5 16 16 and nd the the valu value of of the the total total length length field field is is 00 0028 16 16 . How How ma many ny byte bytes of of da data ta are re being being carri ried ed by by this packet? et? Soluti tion on The The HLEN HLEN val value ue is is 5, wh which ich mean means the the total total nu numb mber er of of bytes bytes in in the the he head ader er is is 5 × 4 or or 20 20 bytes bytes (no (no opti options) ons). The The tot total al length length is is 40 bytes 40 bytes, wh which ich means means the the packet packet is is carr carrying ng 20 20 byte bytes of of data data (40 40 − 20 20). 50

Example 4 An An IP IP packet packet has has ar arrive rived wi with th the the first first few few hex hexad adec ecima imal digits digits as as shown wn belo low: Ho How many any hops ops ca can this this pack packet et tr travel avel bef efor ore be being ing dr dropped opped? The The data belong to to what upper upper layer proto tocol? col? So Solu luti tion on To To fin find the the time time-to to-li live ve fie field, d, we we ski skip 8 byt bytes es (16 16 hexadec hexadecima imal digi gits) ts). The The time time-to to-live ive fie field ld is is th the nin inth th byt byte, e, wh which ich is is 01 01. Th This is me means ans the the packet acket can can tr trave avel on only ly one one ho hop. Th The protoc protocol fi field eld is is the the ne next xt byte byte (02 02), ), which which mean means that at the the upper upper la layer er pr protoco otocol is is IGM GMP. 51

IP fragmentation, reassembly  network links have MTU (max.transfer size) - largest possible link-level fragmentation: frame … in: one large datagram  different link types, out: 3 smaller datagrams different MTUs  large IP datagram divided ( “ fragmented ” ) within net reassembly  one datagram becomes several datagrams  “ reassembled ” only at … final destination  IP header bits used to identify, order related fragments 52

IP datagram format IP protocol version 32 bits total datagram number length (bytes) header length type of head. ver length (in 4-byte words) len service for “ type ” of data fragment fragmentation/ 16-bit identifier flag offset reassembly max number upper time to header remaining hops live layer checksum (decremented at 32 bit source IP address each router) 32 bit destination IP address upper layer protocol to deliver payload to 0-40 bytes options (if any) e.g. timestamp, how much overhead? data record route  20 - 60 bytes of IP (variable length, taken, specify typically a TCP list of routers Length of data? or UDP segment) to visit.  Total length – header length 53

Flags field 1 cannot fragment the datagram D = 0 can be fragmented if necessary 1 the datagram is not the last fragment M = 0 this is the last or only fragment 54

IP fragmentation, reassembly example:  4000 byte datagram  MTU = 1400 bytes Offset = 0000/8 = 0 0000 1399 Offset = 1400/8 = 175 1400 2799 Offset = 2800/8 = 350 2800 3999 55

Detailed fragmentation example 1420 14,567 1 000 820 Bytes 0000 – 1399 14,567 1 175 Fragment 1 4020 14,567 0 000 1420 Bytes 1400 – 2199 14,567 1 175 Fragment 2.1 Bytes 0000 – 3999 Bytes 1400 – 2799 Fragment 2 Original datagram 1220 14,567 0 350 Bytes 2800 – 3999 Fragment 3 56

Example 5 A packet acket has has arr arrived ived with with an an M bit it value value of of 0. Is Is this this the the first first frag fr agment, ment, the la last st fr fragme agment nt, or or a mi middl ddle fragme ragment? nt? Do Do we we know know if if the packet ket wa was fragme agment nted? d? Soluti tion on If If the the M bi bit is is 0, it it me means ns that that th ther ere are are no no more more fr frag agments ments; the the fr fragment agment is is the the la last st on one. Howev However er, we we can cannot not say say if if the the or origi igina nal packet acket wa was frag fragme ment nted or or not not. A non nonfr fragm agmented ented pack packet et is is considere dered the last fragment gment. 57

Example 6 A packet acket has has arr arrived ived with with an an M bit it value value of of 1. Is Is this this the the first first frag fr agment, ment, the la last st fr fragme agment nt, or or a mi middl ddle fragme ragment? nt? Do Do we we know know if if the packet ket wa was fragme agment nted? d? Soluti tion on If If the the M bit it is is 1, it it me mean ans that that there there is is at at least least one more more fr fragment agment. Th This is frag fragme ment nt can can be be the the first first one or or a mid middle le on one, e, but ut not not the the last ast on one. We We don’t know if if it it is is the the first first one or or a mi midd ddle one one; we we nee eed mor more inform informatio ation (the (the valu value of of the the fragme agmentatio ntation offset fset). See also the next examp mple. 58

Example 7 A pack packet et has has arr arrived ived with with an an M bi bit va value ue of of 1 and and a fr frag agmentation mentation of offset fset value lue of of zer zero. Is Is th this is the the first first fr frag agme ment nt, the the last fragme agment nt, or or a middle fra ragment? gment? Soluti tion on Beca Because use th the M bi bit is is 1, it it is is eithe ither the the first first fragmen gment or or a mi middl ddle one one. Becaus use the off ffset set value is is 0, it it is is the first st frag agment ment. 59

Example 8 A pack packet et has as arr arrived ived in in whic which th the of offset fset value value is is 100 100. What What is is the the nu numbe ber of of the the first first byte byte? Do Do we we kno know the the nu number mber of of the the last last byte? e? Soluti tion on To To find find the the nu numb mber er of of the the first first byte, byte, we we mu mult ltip iply ly the the of offset set val value ue by by 8. Thi This me means ans that that the the first first byte yte numb number er is is 800 800. We We cannot annot determ etermine ine the the number umber of of the the last last by byte te unless unless we we kn know the the length th of of the data. 60

Example 9 A pack packet et has as arr arrived ived in in whic which th the of offset fset value value is is 10 100, the the value value of of HLE HLEN (header eader len length) gth) is is 5 and the the val value of of the the tota otal le length gth fie field is is 100 00. What What is is the the num number ber of of the the first first byte byte and and the the last ast byte? e? So Solu luti tion on Th The firs first by byte te number number is is 100 100 × 8 = 800 800. The The tot total al length length is is 100 100 bytes bytes and nd the the he head ader er length ngth is is 20 20 bytes bytes (5 × 4), ), which which me means ans that that ther there ar are 80 80 bytes bytes in in this this datagram datagram. If If the the first first byt byte number number is is 800 800, the last byte te number er must st be be 879 879. 61

Network Layer: outline 1 introduction 5 routing algorithms  link state 2 virtual circuit and datagram networks  distance vector  hierarchical routing 3 what ’ s inside a router 6 routing in the Internet 4 IP: Internet Protocol  RIP  datagram format  OSPF  IPv4 addressing  BGP  ICMP 7 broadcast and multicast  IPv6 routing 62

IP addressing: introduction 223.1.1.1  IP address: 32-bit identifier for host, router 223.1.2.1 interface 223.1.1.2 223.1.1.4 223.1.2.9  interface: connection between host/router and 223.1.3.27 physical link 223.1.1.3 223.1.2.2  router ’ s typically have multiple interfaces  host typically has one or two interfaces (e.g., wired 223.1.3.1 223.1.3.2 Ethernet, wireless 802.11)  IP addresses associated with each interface 223.1.1.1 = 11011111 00000001 00000001 00000001 223 1 1 1 63

Example 1 Find the error, or, if if any, in in the followin wing IPv4 addresses sses: a. 111 111.56 56.045 045.78 78 b. 221 221.34 34.7.8.20 20 c. 75 75.45 45.30 301.14 14 d. 11100010 10.23 23.14 14.67 67 Soluti tion on a. There re should be be no no leading zeroes es (045 045). b. We We may not not have more than 4 bytes es in in an an IPv4 address ss. c. Each byte should be be less than or or equal equal to to 255 255. d. A mixt xture ure of of binary notati ation on and and dotted ted-decim ecimal al notati ation on. 64

Example 2 Find ind the the num number ber of of addresses dresses in in a ran range ge if if the the first first ad address dress is is 146 146.102 102.29 29.0 and and the last address ess is is 146 146.10 102.32 32.255 255. Soluti tion on We We can can sub subtr tract act the the first first ad addr dress ess fr from om the the la last st ad address dress in in ba base se 256 56 (se (see Appe Append ndix ix B) B). The The resu result lt is is 0.0.3.25 255 in in this this base ase. To To fin find the the num umber ber of of address ddresses es in in the the ra range ge (in (in dec decima imal), l), we we convert ert this is number er to to base 10 10 and and add add 1 to to the result lt.. .. 65

Example 3 Th The firs first ad address dress in in a rang range of of addresses dresses is is 14 14.11 11.45 45.96 96. If If the the number umber of of address ddresses es in in the the rang range is is 32 32, what hat is is the the last ast address? ess? So Solu luti tion on We We conver convert the the numb number er of of add address resses es minus minus 1 to to base base 256 256, which wh ich is is 0.0.0.31 31. We We then then add dd it it to to the the first first add address ress to to get get the the la last address ess. Ad Addit ition ion is is in in base 256 256. 66

Figure Bitwise NOT operation 67

Example 4 68

Figure Bitwise AND operation 69

Example 5 70

Figure Bitwise OR operation 71

Example 6 72

Classful Addressing • When IP addressing was first started, it used a concept called “classful addressing”. A newer concept called “classless addressing” is slowly replacing it though. • Regarding “classful addressing”, the address space is divided into five classes: A, B, C, D and E. Class # of addresses Percent of the Space 73

Figure Finding the class of address 1 1 1 1 Start 0 0 0 0 Class: A Class: B Class: C Class: D Class: E 74

Example 7 Find the class s of each address: ss: a. a. 000000 000001 01 00001011 11 00001011 11 11101111 111 b. b. 1100 0000 0001 01 10000011 11 00011011 11 1111111 111 c. c. 10100 0011 111 1 1101101 011 1 10001011 1 01101111 11 d. d. 1111 1100 0011 11 10011011 11 11111011 11 0000111 111 Soluti tion on a. The first st bit is 0. This is a cl class s A ad address. ess. b. b. The first st 2 bi bits s are 1; the third rd bit is 0. This is a cl class s C C address. ess. c. c. The e first rst bit is 1; the second nd bit is 0. This is a class s B address. ess. d. d. The he first st 4 bi bits s are 1s. This is a class s E address. ss. 75

Example 8 Find the class of of each address ss: a. 227 227.12 12.14 14.87 87 b. 193 193.14 14.56 56.22 22 c. 14 14.23 23.12 120.8 d. 252 252.5.15 15.111 11 Soluti tion on a. a. The e first st byte e is 227 7 (bet etween ween 224 4 and d 239); 9); the clas ass s is D. b. The b. e first st byte e is 193 3 (bet etween ween 192 2 and d 223); 3); the clas ass s is C. c. The first st byte is is 14 14 (be between een 0 and and 127 127); the clas ass is is A. d. The first st byte is is 252 252 (bet etween ween 240 240 and and 255 255); the clas ass is is E. 76

Figure Netid and hostid 77

Figure Blocks in Class A • Class A has 128 blocks or network ids • First byte is the same (netid), the remaining 3 bytes can change (hostids) • Network id 0 (first), Net id 127 (last) and Net id 10 are reserved – leaving 125 ids to be assigned to organizations/companies • Each block contains 16,777,216 addresses – this block should be used by large organizations. • The first address in the block is called the “network address” – defines the network of the organization Example • Netid 73 is assigned • Last address is reserved • Recall: routers have addressees 78

Figure Blocks in Class B • Class B is divided into 16,384 blocks (65,536 addresses each) • 16 blocks are reserved • First 2 bytes are the same (netid), the remaining 2 bytes can change (hostids) • For example, Network id 128.0 covers addresses 128.0.0.0 to 128.0.255.255 • Network id 191.225 is the last netid for this block Example • Netid 180.8 is assigned • Last address is reserved • Recall: routers have addresses 79

Figure Blocks in Class C • Class C is divided into 2,097,152 blocks (256 addresses each) • 256 blocks are reserved • First 3 bytes are the same (netid), the remaining 1 byte can change (hostids) • For example, Network id 192.0.0 covers addresses 192.0.0.0 to 192.0.0.255 80

Figure The single block in Class D Class D addresses are made of one block, used for multicasting. 81

Figure The single block in Class E The only block of class E addresses was reserved for future purposes. 82

The range of addresses allocated to an organization in classful addressing was a block of addresses in Class A, B, or C. 83

Figure Two-level addressing in classful addressing 84

Figure Information extraction in classful addressing netid 000 ... 0 First address The network address is the first address. The network address defines the network to the rest of the Internet. Given the network address, we can find the class of the address, the block, and the range of the addresses in the block 85

Example 9 An add An address ress in in a blo lock ck is is gi give ven as as 73 73.22 22.17 17.25 25. Find Find the the number umber of ad of addresse dresses in in the the block, block, th the firs first ad addr dress, ess, and and the the last last address ess. Soluti tion on 32 −n = 2 24 1. The The numb umber er of of addresses addresses in in this this block block is is N = 2 32 24 16,777 16 77,21 216. 2. To To find the first st address, ss, we we keep the leftm tmost ost 8 bits and and set set the the rightm ri ghtmost ost 24 24 bits its all all to to 0s. The The first first add address ress is is 73 73.0.0.0/8, in in wh which 8 is is the value of of n . 3. To To find the last address, ess, we we keep the leftmo tmost st 8 bits and and set set the the ri right ghtmost most 24 24 bits its all all to to 1s. Th The last ast ad address dress is is 73 73.255 55.25 255.255 55/8. 86

Figure Solution to Example 9 87

Example 10 An An add address ress in in a bl block ock is is gi give ven as as 18 180.8.17 17.9. Fi Find the the number number of of ad addresse dresses in in the the block, block, th the firs first ad addr dress, ess, and and the the last last address ess. Soluti tion on 32 −n = 2 16 1. The The numb umber er of of addresses addresses in in this this block block is is N = 2 32 16 65 65,536 36. 2. To To find the first st address, ess, we we keep the leftm tmost ost 16 16 bits and and set set the the ri rightm ghtmost ost 16 16 bits its all all to to 0s. The The first first add address ress is is 18 18.8.0.0/16 16, in in which 16 16 is is the value of of n . 3. To To find the last address, ess, we we keep the leftmo tmost st 16 16 bits and and set set the the ri right ghtmost most 16 16 bits its all all to to 1s. The Th last ast ad address dress is is 18 18.8.25 255.255 55/16 16. 88

Figure Solution to Example 10 89

Example 11 An An add address ress in in a blo lock ck is is gi give ven as as 200 200.11 11.8.45 45. Find Find the the number umber of ad of addresse dresses in in the the block, block, th the firs first ad addr dress, ess, and and the the last last address ess. Soluti tion on 32 −n = 2 8 = 256 1. The The number er of of addresses sses in in this is blo lock is is N = 2 32 256. 2. To To find the first st address, ess, we we keep the leftm tmost ost 24 24 bits and and set set the the ri right ghtmost most 8 bits its all all to to 0s. The The firs first address dress is is 200.11 200 11.8.0/24 24, in in wh which 24 24 is is the value of of n . 3. To To find the last address, ess, we we keep the leftmo tmost st 24 24 bits and and set set the the right rightmost most 8 bits its all all to to 1s. The The last last ad address dress is is 200 200.11 11.8.255 55/24 24. 90

Figure Solution to Example 11 91

Figure Sample Internet 92

Figure Network addresses 93

Network mask • Given the network address, we can easily determine the block and range of addresses • Suppose given the IP address, can we determine the network address (beginning of the block) ? • To route packets to the correct network, a router must extract the network address from the destination IP address • How would we EXTRACT the network address from the IP address? We would use a MASK. A mask is a 32-bit binary number that gives the first address in the block (the network address) when bitwise ANDed with an address in the block. 94

Figure Network mask 95

Figure Finding a network address using the default mask • If bit is ANDed with 1, it’s preserved • If bit is ANDed with 0, it’s changed to a 0. A simple way to determine the netid for un-subnetted cases: (1) if mask byte is 255, retain corresponding byte of the address, (2) if mask byte is 0, set corresponding address byte to 0. 96

Example 12 A router router rec recei eives es a pack packet et with ith the the destin estinatio ation ad address dress 181 81.24 24.67 67.32 32. Sh Show how how the the rout router er fi find nds the the ne network twork ad address dress of of the packet et. Soluti tion on Since Si nce the the class lass of of the the add ddress ress is is B, B, we we assu assume me that that the the router router app pplie ies the the defau default lt mask ask for for cl class ass B, B, 25 255.255 255.0.0 to to find find the the networ work address ss. 97

Recall IP Addresses: Classful Addressing Class # of addresses Percent of the Space 2 31 =2147483648 A 50% 2 30 =1073741824 B 25% 2 29 =536870912 C 12.5% 2 28 =268435456 D 6.25% 2 28 =268435456 E 6.25% 98

5-bit Address Space Illustration 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 No Netid case 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 32 addresses/block 0 1 0 0 0 0 1 0 0 1 Number of blocks: 1 0 1 0 1 0 0 1 0 1 1 Address range per block: 0 to 31 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 Netids: N/A 0 1 1 1 1 1 0 0 0 0 Network Addresses : 00000 1 0 0 0 1 1 0 0 1 0 Broadcast Addresses: 11111 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 99

5-bit Address Space Illustration 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1-bit Netid case 0 0 1 1 0 0 0 1 1 1 16 addresses/block 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 Number of blocks: 2 0 1 0 1 1 0 1 1 0 0 Address range per block: 0 to 15 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 Netids: 0, 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 Network Addresses : 00000, 10000 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 Broadcast Addresses: 01111, 11111 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 100