* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Piggybacking (Internet access) wikipedia , lookup
Remote Desktop Services wikipedia , lookup
Distributed firewall wikipedia , lookup
Multiprotocol Label Switching wikipedia , lookup
Asynchronous Transfer Mode wikipedia , lookup
Point-to-Point Protocol over Ethernet wikipedia , lookup
Deep packet inspection wikipedia , lookup
Computer network wikipedia , lookup
Zero-configuration networking wikipedia , lookup
Network tap wikipedia , lookup
Airborne Networking wikipedia , lookup
Wake-on-LAN wikipedia , lookup
TCP congestion control wikipedia , lookup
Cracking of wireless networks wikipedia , lookup
Real-Time Messaging Protocol wikipedia , lookup
Recursive InterNetwork Architecture (RINA) wikipedia , lookup
Service time evaluation (transfer) for a message sent over a LAN through TCP protocol Network Routers service time PROTOCOL TCP/IP STRATIFICATION HTTP transport layer network layer data link layer FTP NFS RPC TELNET TCP DNS UDP IP SNMP DATAGRAM IP FORWARDING The router holds a routing table, used to find the following router or host which forward the datagram Host A TCP Host B IP TCP IP IP Network layer Network layer Data layer router 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. Layer n Layer n Layer n -1 Layer n -1 network 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 NETWORK PROTOCOLS Protocol TCP UDP IPv4 IPv6 ATM Ethernet Ieee 802.3 Ieee 802.5 TR FDDI PDU name Segment Datagram Datagram Datagram Cell Frame Frame Frame Frame Max PDU Overhead Max data size area 65535 Ltd by IP 65535 65535 53 1518 1518 4472 4500 20 8 20 40 5 18 21 28 28 65515 Ltd by IP 65515 65495 48 1500 1497 4444 4472 TCP • 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 Host B Host A syn syn ack data data fin ack • 3 segments are required to establish a connection • 4 segments are required to end it in both directions data fin ack 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 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. 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. 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 = Example WITHOUT fragmentation 300 byte message sent from a Client to a Server 300 20 300 client Server TCP TCP IP IP Network layer Network layer 20 20 300 20 300 20 20 300 T.R. 18 20 20 300 28 20 20 300 ethernet FDDI router2 router1 28 20 20 300 300 LAN service time 300 byte message sent from a Client to a Server 28 20 20 4404 Service Time Eth = 10.000.000 358 x 8 = = 10.000.000 368 x 8 28 20 20 300 Service Time FDDI = 100.000.000 = 16.000.000 0.00002944 sec. = 100.000.000 368 x 8 28 20 20 300 Service Time TR = 0.000286 sec. = = 16.000.000 0.000184 sec. Example WITH fragmentation The Server sends a 10.000 byte reply to Client Hypothesis: TCP does not know the local network MTU client Server TCP TCP IP IP Network layer Network layer 10000 20 4404 4424 1172 20 20 4404 20 4424 20 1172 28 20 20 4404 T.R. 28 20 4424 ethernet 28 20 1172 FDDI 28 20 20 4404 28 20 4424 28 20 1172 - 28 - 28 - 28 FDDI =20 4424 =20 4424 = 1192 ET data 18 1192 FDDI TR ET IP data 18 20 1480 18 20 1480 18 20 1464 28 20 20 4404 28 20 4424 28 20 1172 - 28 28 28 + + + 28 28 28 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 ( 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 ) x8 Example WITH fragmentation The Server sends a 10.000 byte reply to Client Hypothesis: TCP knows the local network MTU client Server TCP TCP IP IP Network layer Network layer 10000 20 4404 20 4404 20 1192 20 20 4404 20 20 4404 20 20 1192 28 20 20 4404 T.R. 28 20 20 4404 ethernet 28 20 20 1192 FDDI - 28 28 20 20 4404 - 28 28 20 20 1192 - 28 28 20 20 4404 FDDI = 20 4424 = 20 4424 = 1232 ET data 18 1232 FDDI TR ET IP TCP data 18 20 20 1460 18 20 1480 18 20 1464 28 20 20 4404 28 20 20 4404 28 20 20 1192 - 28 28 28 + + + 28 28 28 LAN Service Time The Server sends a 10.000 byte reply to Client Hypothesis: TCP knows the local network MTU Case of Token Ring ( 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. 0.005082 sec. 20 msec different ) x8 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 Average service time Without fragmentation TCP does not know LAN MTU NSegments = MessageSize 65495 NDatagrams = TCP knows LAN MTU NSegments = NDatagrams MessageSize + NSegments x TCPOvhd minn MTUn - IPOvhd (rough estimate) Overheadn = NSegments x TCPOvhd + Ndatagrams x (IPOvhd + FrameOvhdn) ServiceTimen = 8 x (MessageSize + Overheadn ) 106 x Bandwidth Exercise - text Client Ethernet 10 Mbps MTU: 1518 bytes DB server router1 FDDI 100 Mbps MTU: 4472 bytes T.R. router2 Token RingI 16 Mbps MTU: 4444 bytes • 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. Exercise - solution • Using NDatagrams = MessageSize + Nsegment x TCPOvhd 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 (8092+20)/(1500-20) = 6 (100000+40)/(1500-20) = 68 request short reply long reply Exercise - solution • Using 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 ServiceTimen = 8 x (MessageSize + Overheadn ) 106 x Bandwidth We can evaluate the Service Time (ethernet) OverheadEth = 8 x (400 + 58) / 10.000.000 = 0.366 msec OverheadEth = 8 x (8092 + 248) / 10.000.000 = 6.67 msec OverheadEth = 8 x (100.000 + 2604) / 10.000.000 = 82.1 msec request short reply long reply Service times evaluation Eth FDDI TR Router Ndatagrams Overhead (byte) ServiceTime(msec) Ndatagrams Overhead (byte) ServiceTime(msec) Ndatagrams Overhead (byte) ServiceTime(msec) Latency (134 usec/packet) Request Short reply Long reply 1 58 0.366 1 68 0.0374 1 68 0.234 134 6 248 6.67 6 308 0.672 6 08 4.2 804 68 2604 82.1 68 3284 8.26 68 3284 51.6 9.112 Network Router Router queues Service times at Routers Router latency (msec per packet): time spent by the router to process a datagram, given by manufacturer. RouterServiceTime: Ndatagrams * RouterLatency Where MessageSize + TCPOvhd NDatagrams = minn MTUn - IPOvhd Exercise Client Ethernet 10 Mbps MTU: 1518 bytes • DB server router1 FDDI 100 Mbps MTU: 4472 bytes T.R. router2 Token Ring 16 Mbps MTU: 4444 bytes Router 1 and 2 process 400,000 packets/sec Service time: 2.5 msec (=1/400,000) Service demand at router Client Request Short Reply Long Request 1 x 2.5 = 2.5 msec 6 x 2.5 = 15 msec 68 x 2.5 = 170 msec