Wireless and Mobile Networks EECS 489 Computer Networks

Wireless and Mobile Networks EECS 489 Computer Networks http: //www. eecs. umich. edu/courses/eecs 489/w 07 Z. Morley Mao Monday March 19, 2007 Acknowledgement: Some slides taken from Kurose&Ross 1

Mobile IP r RFC 3220 r has many features we’ve seen: m home agents, foreign-agent registration, care-of-addresses, encapsulation (packet-within-a-packet) r three components to standard: m indirect routing of datagrams m agent discovery m registration with home agent 2

Mobile IP: indirect routing foreign-agent-to-mobile packet sent by home agent to foreign agent: a packet within a packet dest: 79. 129. 13. 2 dest: 128. 119. 40. 186 Permanent address: 128. 119. 40. 186 dest: 128. 119. 40. 186 Care-of address: 79. 129. 13. 2 packet sent by correspondent 3

Mobile IP: agent discovery r agent advertisement: foreign/home agents advertise service by broadcasting ICMP messages (typefield = 9) H, F bits: home and/or foreign agent R bit: registration required 4

Mobile IP: registration example 5

Components of cellular network architecture recall: correspondent wired public telephone network MSC MSC MSC different cellular networks, operated by different providers 6

Handling mobility in cellular networks r home network: network of cellular provider you subscribe to (e. g. , Sprint PCS, Verizon) m home location register (HLR): database in home network containing permanent cell phone #, profile information (services, preferences, billing), information about current location (could be in another network) r visited network: network in which mobile currently resides m visitor location register (VLR): database with entry for each user currently in network m could be home network 7

GSM: indirect routing to mobile home network HLR 2 home MSC consults HLR, gets roaming number of mobile in visited network correspondent home Mobile Switching Center 1 VLR 3 Mobile Switching Center 4 Public switched telephone network call routed to home network home MSC sets up 2 nd leg of call to MSC in visited network mobile user visited network MSC in visited network completes call through base station to mobile 8

GSM: handoff with common MSC r Handoff goal: route call via new base station (without interruption) r reasons for handoff: VLR Mobile Switching Center old routing old BS m new routing m new BS m stronger signal to/from new BS (continuing connectivity, less battery drain) load balance: free up channel in current BS GSM doesn’t mandate why to perform handoff (policy), only how (mechanism) r handoff initiated by old BS 9

GSM: handoff with common MSC VLR Mobile Switching Center 2 4 1 8 old BS 5 7 3 6 new BS 1. old BS informs MSC of impending handoff, provides list of 1+ new BSs 2. MSC sets up path (allocates resources) to new BS 3. new BS allocates radio channel for use by mobile 4. new BS signals MSC, old BS: ready 5. old BS tells mobile: perform handoff to new BS 6. mobile, new BS signal to activate new channel 7. mobile signals via new BS to MSC: handoff complete. MSC reroutes call 8 MSC-old-BS resources released 10

GSM: handoff between MSCs r anchor MSC: first MSC visited during call home network correspondent Home MSC anchor MSC m call remains routed through anchor MSC r new MSCs add on to end PSTN MSC (a) before handoff of MSC chain as mobile moves to new MSC r IS-41 allows optional path minimization step to shorten multi-MSC chain 11

GSM: handoff between MSCs r anchor MSC: first MSC visited during cal home network correspondent Home MSC anchor MSC m call remains routed through anchor MSC r new MSCs add on to end PSTN MSC (b) after handoff of MSC chain as mobile moves to new MSC r IS-41 allows optional path minimization step to shorten multi-MSC chain 12

Mobility: GSM versus Mobile IP GSM element Comment on GSM element Mobile IP element Home system Network to which the mobile user’s permanent phone number belongs Home network Gateway Mobile Switching Center, or “home MSC”. Home Location Register (HLR) Home MSC: point of contact to obtain routable address of mobile user. HLR: database in home system containing permanent phone number, profile information, current location of mobile user, subscription information Home agent Visited System Network other than home system where mobile user is currently residing Visited network Visited Mobile services Switching Center. Visitor Location Record (VLR) Visited MSC: responsible for setting up calls to/from mobile nodes in cells associated with MSC. VLR: temporary database entry in visited system, containing subscription information for each visiting mobile user Foreign agent Mobile Station Roaming Number (MSRN), or “roaming number” Routable address for telephone call segment between home MSC and visited MSC, visible to neither the mobile nor the correspondent. Care-ofaddress 13

Wireless, mobility: impact on higher layer protocols r logically, impact should be minimal … m best effort service model remains unchanged m TCP and UDP can (and do) run over wireless, mobile r … but performance-wise: m packet loss/delay due to bit-errors (discarded packets, delays for link-layer retransmissions), and handoff m TCP interprets loss as congestion, will decrease congestion window un-necessarily m delay impairments for real-time traffic m limited bandwidth of wireless links 14

Chapter 6 Summary Wireless r wireless links: m m m capacity, distance channel impairments CDMA r IEEE 802. 11 (“wi-fi”) m CSMA/CA reflects wireless channel characteristics r cellular access m architecture m standards (e. g. , GSM, CDMA-2000, UMTS) Mobility r principles: addressing, routing to mobile users m m m home, visited networks direct, indirect routing care-of-addresses r case studies m mobile IP m mobility in GSM r impact on higher-layer protocols 15

MPLS is used today! r % traceroute -q 1 -w 2 www. att. net r r r r r r traceroute to www. att. net (204. 127. 135), 30 hops max, 46 byte packets 1 eecscomp 2 -4 (141. 213. 4. 1) 0. 291 ms 2 141. 213. 127. 37 (141. 213. 127. 37) 0. 305 ms 3 ge-caen-bin-seb. r-bin-seb. umnet. umich. edu (192. 122. 183. 53) 0. 515 ms 4 pc-bin-arb-seb. r-bin-arb. umnet. umich. edu (192. 122. 183. 193) 0. 621 ms 5 ge-1 -1 -0 x 984. aa 1. mich. net (192. 122. 183. 45) 0. 660 ms 6 65. 77. 89. 177 (65. 77. 89. 177) 2. 002 ms 7 brvwil 1 wcx 2 -pos 14 -1. wcg. net (64. 200. 240. 33) 8. 129 ms 8 64. 200. 249. 186 (64. 200. 249. 186) 7. 904 ms 9 te-4 -4. car 2. Chicago 1. Level 3. net (4. 68. 110. 37) 8. 259 ms 10 ae-14 -53. car 4. Chicago 1. Level 3. net (4. 68. 101. 72) 215. 627 ms 11 ggr 2 -p 3110. cgcil. ip. att. net (192. 205. 33. 185) 9. 075 ms 12 tbr 2 -p 033301. cgcil. ip. att. net (12. 123. 6. 26) 32. 783 ms MPLS Label=32719 Co. S=0 TTL=0 S=1 13 tbr 2 -cl 3641. phlpa. ip. att. net (12. 122. 10. 94) 33. 031 ms MPLS Label=31344 Co. S=0 TTL=0 S=1 14 tbr 1 -cl 9. wswdc. ip. att. net (12. 122. 2. 85) 33. 738 ms MPLS Label=31295 Co. S=0 TTL=0 S=1 15 gbr 5 -p 10. wswdc. ip. att. net (12. 122. 11. 170) 29. 483 ms MPLS Label=317 Co. S=0 TTL=0 S=1 Co. S: 16 gbr 1 -s 83. mdtva. ip. att. net (12. 122. 4. 81) 32. 068 ms Class of Service 17 12. 122. 247. 254 (12. 122. 247. 254) 32. 438 ms S: Stacking bit 16 18 *

Boeing Connexion Mobility Service 17

Example problem 1 r If all the links in the Internet were to provide the reliable delivery service, would the TCP reliable delivery service be redundant? Why or why not? 18

Example problem 1 r If all the links in the Internet were to provide the reliable delivery service, would the TCP reliable delivery service be redundant? Why or why not? r Ans: TCP is still needed to ensure end-toend reliability. Packets may be lost due to routing loops or equipment failures. TCP also provide in-order delivery. 19

Example problem 2 r Suppose two nodes start to transmit at the same time a packet of length L over a broadcast channel of rate R. Denote the propagation delay between the two nodes as tprop. Will there be collision if tprop < L/R? Why or why not? 20

Example problem 2 r Suppose two nodes start to transmit at the same time a packet of length L over a broadcast channel of rate R. Denote the propagation delay between the two nodes as tprop. Will there be collision if tprop < L/R? Why or why not? r Ans: Yes, there will be collision, because while a node is transmitting it will start to receive a packet from the other node. 21

Review: Ethernet CSMA/CD algorithm 1. Adaptor receives datagram 4. If adapter detects from net layer & creates another transmission while frame transmitting, aborts and sends jam signal 2. If adapter senses channel idle (for 96 bit times), it 5. After aborting, adapter starts to transmit frame. enters exponential If it senses channel busy, backoff: after the mth waits until channel idle (for collision, adapter chooses 96 bit times), and then a K at random from transmits {0, 1, 2, …, 2 m-1}. Adapter 3. If adapter transmits entire waits K·512 bit times and returns to Step 2 frame without detecting another transmission, the adapter is done with frame! 22

Review: Ethernet’s CSMA/CD (more) Jam Signal: make sure all other transmitters are aware of collision; 48 bits Bit time: . 1 microsec for 10 Mbps Ethernet ; for K=1023, wait time is about 50 msec Exponential Backoff: r Goal: adapt retransmission attempts to estimated current load m heavy load: random wait will be longer r first collision: choose K from {0, 1}; delay is K· 512 bit transmission times r after second collision: choose K from {0, 1, 2, 3}… r after ten collisions, choose K from {0, 1, 2, 3, 4, …, 1023} 23

Example problem 3 r In CSMA/CD, after the fifth collision, what is the probability that a node chooses K=4? The result K=4 corresponds to a delay of how many seconds on a 10 Mbps Ethernet? 24

Example problem 3 r In CSMA/CD, after the fifth collision, what is the probability that a node chooses K=4? The result K=4 corresponds to a delay of how many seconds on a 10 Mbps Ethernet? r Ans: After the fifth collision, the adapter chooses from {0, 1, 2, …. , 2^5 -1}. The probability that it chooses 4 is 1/32. r It waits K· 512 bit transmission times m 4 x 512 bit/(10 Mbps) = 204. 8 microseconds. 25

Example problem 4 r Suppose nodes A and B are on the same 10 Mbps Ethernet segment and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? 26

r Suppose nodes A and B are on the same 10 Mbps Ethernet segment and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? r Ans: Assuming A chooses K=0, B chooses K=1, basically we need to find out whether A’s retransmission will reach B before B’s scheduled retransmission time. (It’s unlikely that A can finish retransmission before B retransmits. ) 27

r Suppose nodes A and B are on the same 10 Mbps Ethernet segment and the propagation delay btw. the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? r Draw a time line: m t=0 m t=225+48=273 m 273+225=498 A and B begin transmission A and B detect collision A and B finish sending Jam signal B’s last bit arrives at A; A detects an idle channel m 498+96=594 A starts transmitting m 273+512=785 B returns to step 2, B must sense idle channel for 96 bit times: 785+96=881 m 594+225=819 A’s transmission reaches B 28 m 819<881: A’s retransmission reaches B before B can send.

Example problem 5 r Why are link layer acknowledgements used in 802. 11 but not in wired Ethernet? 29

Example problem 5 r Why are link layer acknowledgements used in 802. 11 but not in wired Ethernet? r Ans: Two reasons why 802. 11 uses ACKs: (1) wireless channel’s bit error rate is much higher. (2) in wired Ethernet, a transmitting station can detect when there has and hasn’t been a collision, but in 802. 11 a station cannot detect a collision due to hidden terminal problem. 30

Example problem 6 r Suppose the IEEE 802. 11 RTS and CTS frames were as long as the standard DATA frames. Would there be any advantage to using the CTS and RTS frames? Why and why not? 31

Example problem 6 r Suppose the IEEE 802. 11 RTS and CTS frames were as long as the standard DATA frames. Would there be any advantage to using the CTS and RTS frames? Why and why not? r Ans: No, there wouldn’t be any advantage. Suppose there are two stations that want to transmit at the same time, and they both use RTS/CTS. If the RTS frame is as long as a DATA frame, the channel would be wasted for as long as it would have been wasted for colliding DATA frames. Thus, the RTS/CTS exchange is only useful when they are significantly smaller than DATA frames. 32

Example problem 7 r Consider three LANs interconnected C A E B by two routers r Assign IPs to all the interfaces. Subnet 1 uses addresses of the form 111. xxx, subnet 2 uses Subnet 1 addresses of the form 122. 222. xxx, and subnet 3 uses addresses of the form 133. 333. xxx r Assign MAC addresses (arbitrarily). r Consider sending an IP datagram from Host A to Host F, suppose all the ARP tables are up-to-date. Enumerate all the steps. F D Subnet 3 Subnet 2 33

Example problem 7 122. 222. 001 44 -44 -44 -44 111. 001 00 -00 -00 -00 A C 111. 002 22 -22 -22 -22 133. 333. 001 77 -77 -77 -77 B 122. 222. 002 33 -33 -33 -33 111. 003 11 -11 -11 -11 Subnet 1 122. 222. 003 55 -55 -55 -55 E F 133. 333. 002 88 -88 -88 -88 D 122. 222. 004 66 -66 -66 -66 133. 333. 003 99 -99 -99 -99 Subnet 3 Subnet 2 34

122. 222. 001 44 -44 -44 -44 A B 111. 001 C 00 -00 -00 -00 111. 003 11 -11 -11 -11 Subnet 3 111. 002 22 -22 -22 -22 133. 333. 001 77 -77 -77 -77 122. 222. 002 33 -33 -33 -33 122. 222. 003 55 -55 -55 -55 DSubnet 2 122. 222. 004 66 -66 -66 -66 E F 133. 333. 002 88 -88 -88 -88 133. 333. 003 99 -99 -99 -99 1. Forwarding table in A determines that the datagram should be routed to interface 111. 002 2. Host A uses ARP to determine the LAN address for 111. 002, namely 22 -22 -22 -22. 3. The adapter in A creates an Ethernet packet with destination address of 22 -22 -22 4. The first router receives the packet and extracts the datagram. The forwarding table in this router indicates that the datagram is to be routed to 122. 222. 003. 5. The first router then uses ARP to obtain the associated Ethernet address: 55 -55 -55 -55. 6. The process continues until the packet has reached host F. 35