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Service time evaluation (transfer) for a message sent over a LAN through TCP protocol

Service time evaluation (transfer) for a message sent over a LAN through TCP protocol Network Routers service time. PROTOCOL TCP/IP STRATIFICATION. NFS. HTTP. FTP. TELNET. DNS. SNMP. RPC. transport layer. TCP. UDP. network layer. IP. data link layer. DATAGRAM IP FORWARDING.

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Service time evaluation (transfer) for a message sent over a LAN through TCP protocol

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  1. Service time evaluation (transfer) for a message sent over a LAN through TCP protocol Network Routers service time

  2. PROTOCOL TCP/IP STRATIFICATION NFS HTTP FTP TELNET DNS SNMP RPC transport layer TCP UDP network layer IP data link layer

  3. DATAGRAM IP FORWARDING The router holds a routing table, used to find the following router or host which forward the datagram Host A Host B TCP TCP IP IP IP Network layer Network layer Data layer router

  4. Layer n Layer n Layer n -1 Layer n -1 network PROTOCOLS Communication protocol design is based on a layer architecture. Every layer n entity communicates, exchanging Physical Data Unit (PDU), only with layer n remote entities, using the services provided by layer n-1 and offers services to layer n+1. When data are transferred an header is added by each layer Data layer n -1 Header layer n -1 Header layer n Data layer n

  5. NETWORK PROTOCOLS

  6. TCP Host B Host A syn syn • TCP provides a service: connection oriented, reliable, end-to-end, with flow control. It assures the data delivery in the same sending order, without losses. • TCP implements a connection reliable mechanism called Three way handshake • 3 segments are required to establish a connection • 4 segments are required to end it in both directions ack data data fin ack data fin ack

  7. IP • IP defines the format of the packets sent in Internet • The service is unreliable, isn’t end-to-end like, and the datagrams can be lost

  8. FRAGMENTATION • Protocol entities at each layer communicate with each other by exchanging PDUs composed of a header and a data area. • PDUs have a maximum dimension for the data area (named, for the network layer, Maximum Transmission Unit ((MTU)) • Being MTUs lengths variable for different protocols, router has to be able to fragment datagrams, that will be reassembled at the IP level by the destination host.

  9. Fragmentation disadvantages • The router has to be able to divide PDU • Destination node must reassemblate the fragments • WARNING: IP standard tells to fragment from source node, in anticipation of the path to make.

  10. Service times Service Time of a message Time to transmit the message over the network = = # of bytes (including protocol header-trailer, and fragmentation if there is) = bandwidth

  11. Example WITHOUT fragmentation 300 byte message sent from a Client to a Server 300 client Server 300 TCP TCP 20 300 20 300 IP IP 20 20 300 20 20 300 Network layer Network layer T.R. 28 20 20 300 18 20 20 300 ethernet FDDI router2 router1 28 20 20 300

  12. LAN service time 300 byte message sent from a Client to a Server 28 20 20 4404 358 x 8 = = 0.000286 sec. Service Time Eth = 10.000.000 10.000.000 368 x 8 0.00002944 sec. 28 20 20 300 = = Service Time FDDI = 100.000.000 100.000.000 368 x 8 28 20 20 300 = = 0.000184 sec. Service Time TR = 16.000.000 16.000.000

  13. Example WITH fragmentation The Server sends a 10.000 byte reply to Client Hypothesis: TCP does not know the local network MTU client Server 10000 TCP TCP 20 4404 4424 1172 IP IP 20 20 4404 20 4424 Network layer Network layer 20 1172 28 20 20 4404 T.R. 28 20 4424 ethernet 28 20 1172 FDDI FDDI - 28 20 20 4404 28 = 20 4424 - = FDDI TR 28 20 4424 28 20 4424 - ET IP data - = + 28 20 1172 28 1192 28 20 20 4404 28 28 18 20 1480 - + 28 20 4424 28 28 ET data 18 20 1480 - + 28 20 1172 28 28 18 1192 18 20 1464

  14. LAN Service Time The Server sends a 10.000 byte reply to Client Hypothesis: TCP does not know the local network MTU Case of Token Ring ( ) x 8 28 20 20 4404 + 28 20 4424 + 28 20 1172 Service Time = 16.000.000 (4472+4472+1220) x 8 Service Time = = 0.005082 sec. 16.000.000

  15. Example WITH fragmentation The Server sends a 10.000 byte reply to Client Hypothesis: TCP knows the local network MTU client Server 10000 TCP TCP 20 4404 20 4404 20 1192 IP IP 20 20 4404 20 20 4404 Network layer Network layer 20 20 1192 28 20 20 4404 T.R. 28 20 20 4404 ethernet 28 20 20 1192 FDDI FDDI - 28 20 20 4404 28 = 20 4424 - = FDDI TR 28 20 20 4404 28 20 4424 - ET IP TCP data - = + 28 20 20 1192 28 1232 28 20 20 4404 28 28 18 20 20 1460 - + 28 20 20 4404 28 28 ET data 18 20 1480 - + 28 20 20 1192 28 28 18 1232 18 20 1464

  16. LAN Service Time The Server sends a 10.000 byte reply to Client Hypothesis: TCP knows the local network MTU Case of Token Ring ( ) x 8 28 20 20 4404 + 28 20 20 4404 + 28 20 20 1192 Service Time = 16.000.000 (4472+4472+1260) x 8 Service Time = = 0.005102 sec. 16.000.000 0.005102 sec. 20 msec different 0.005082 sec.

  17. Average service time Evaluation without fragmentation Let’s specify: MTUn: network n MTU (byte) XOvhd: X protocol overhead (byte) FrameOvhdn: network n overhead (byte) Overheadn: network n total overhead (TCP+IP+frame) (byte) Bandwidthn: network n bandwidth (Mbps) Ndatagrams: number of IP datagrams required Nsegments: number of TCP segments required (< or = Ndatagrams) N: number of networks

  18. Average service time Without fragmentation TCP does not know LAN MTU TCP knows LAN MTU MessageSize NSegments = NSegments = NDatagrams 65495 MessageSize + NSegments x TCPOvhd NDatagrams = (rough estimate) minn MTUn - IPOvhd Overheadn = NSegments x TCPOvhd + Ndatagrams x (IPOvhd + FrameOvhdn) 8 x (MessageSize + Overheadn ) ServiceTimen = 106 x Bandwidth

  19. Exercise - text Client DB server • The client sends 3 transactions per minute (0.05 tps), whose messages average lenght is 400 bytes. The 80% of replies are 8092 bytes lenght and the 20% are 100.000 bytes • Assuming that no fragmentation exists and the TCP layer does not know the netwotk MTU, evaluate the average service time for requests and replies for each network. FDDI 100 Mbps MTU: 4472 bytes T.R. Ethernet 10 Mbps MTU: 1518 bytes router1 router2 Token RingI 16 Mbps MTU: 4444 bytes

  20. Exercise - solution MessageSize + Nsegment x TCPOvhd NDatagrams = • Using minn MTUn - IPOvhd • We can evaluate the numbers of Datagrams required in each case (request, short reply and long reply) • (400+20)/(1500-20) = 1 request • (8092+20)/(1500-20) = 6 short reply • (100000+40)/(1500-20) = 68 long reply

  21. Exercise - solution Overheadn = TCPOvhd+Ndatagrams x (IPOvhd + FrameOvhdn) We can evaluate the network overhead (ethernet) OverheadEth = 20 + 1 (20 + 18) = 58 request OverheadEth = 20 + 6 (20 + 18) = 248 short reply OverheadEth = 20 + 68 (20 + 18) = 2604 long reply • Using • Using 8 x (MessageSize + Overheadn ) ServiceTimen = 106 x Bandwidth We can evaluate the Service Time (ethernet) OverheadEth = 8 x (400 + 58) / 10.000.000 = 0.366 msec request OverheadEth = 8 x (8092 + 248) / 10.000.000 = 6.67 msec short reply OverheadEth = 8 x (100.000 + 2604) / 10.000.000 = 82.1 msec long reply

  22. Service times evaluation

  23. Network Router Router queues

  24. Service times at Routers Router latency (sec per packet): time spent by the router to process a datagram, given by manufacturer. RouterServiceTime: Ndatagrams * RouterLatency Where MessageSize + TCPOvhd NDatagrams = minn MTUn - IPOvhd

  25. Exercise Client DB server • Router 1 and 2 process 400,000 packets/sec • Service time: 2.5 sec (=1/400,000) • Service demand at router FDDI 100 Mbps MTU: 4472 bytes T.R. Ethernet 10 Mbps MTU: 1518 bytes router1 router2 Token Ring 16 Mbps MTU: 4444 bytes

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