Network Switching and Layered Models: Circuit Switching vs. Packet Switching, Lecture notes of Data Communication Systems and Computer Networks

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Chapter 3 —- End-to-End Switching and Layered Models  3.1 – Network Switching • Few computers and small distance  direct connection (LAN) • Many computers or large distance  switching • Two fundamental approaches: – Circuit Switching – Packet SWitching • Message Switching: – No physical path is established in advance – Store and forward, in its entirety • Packet Switching: – Messages are broken into discrete units called packets – Each packet contains data and headers – Store and forward, on a packet basis – Two approaches: • Datagram– mainly used in the Network Layer • Virtual Circuit– a Data Link Layer technology – Datagram packet switching • Connectionless  no connection established before transmission of packet • Each packet may be individually routed • May take different routes and arrive out of order • Reassembled at the destination • Advantages: – Efficient for bursty data – Easy to provide bandwidth on demand with variable rates • Disadvantages: – Variable delays – Difficult to provide QoS assurances (Best-effort service) – Packets can arrive out-of-order – Virtual Circuit packet switching • It is connection-oriented • Merges datagram packet switching and circuit switching to extract advantages of both • Packets flow on logical circuits; no physical resources are allocated • Each packet carries a circuit identifier which is local to a link and updated by each switch on the path of the packet from its source to its destination • A virtual circuit: sequence of mappings (established during connection setup) between a link taken by packets and the circuit identifier packets carry on this link. • Packet Switching Vs Circuit Switching – Packet-switching is not suitable for real-time services – However, Packet switching : • offers better sharing of bandwidth, and • is simpler, more efficient, and less costly to implement than circuit switching • is great for bursty data. Examples • 1 Mbps link, each user sends data at 100kbps when busy, busy for 10% of time – Circuit switching with TDM: • Only 10 users can connect at a time – Packet switching with statistical multiplexing: • With 35 users, probability that more than 10 users are active at the same time is less than .0004; thus 35 users can connect at a time Given L = 7.5 Mbits and R = 1.5 Mbps • For message switching: – Delay = 3L/R =3(7.5)/1.5 = 15 sec • For packet switching with 5000 packets: smessagethetransmittorequiredtimeTotal smslinkslastthethroughpacketlastthetransmittoTime ssxlinkfirstthethroughpacketsalltransmitTo mssmbpsbitsRLlinkoneonpacketonetransmittoTime bitsbitsxpacketeachofSize 002.5 002.022 5105000 11015001500/ 15005000105.7 3 3 6        • Group of cars analogy – Given: a group of 10 cars, moving at 100 km/hr. A petrol station takes 12 sec to service a car (transmission time). • Analogy: car ~ bit; group ~ packet. – How long will it take until the group of cars is lined up before the 2nd petrol station? – Solution • Time to “push” the entire group from 1st petrol station onto highway = 12*10 = 120 sec • Time for last car to propagate from 1st to 2nd petrol station: 100km/(100km/hr)= 1 hr • Total time = 62 minutes (1 hr + 120sec) • Suppose the cars “propagate” at the rate of 1000 km/hr and petrol station takes 1 min to service a car. Will cars arrive at 2nd station before all cars are serviced at 1st station? Solution • After 7 min ( ), 1st car will be at 2nd station and 3 cars are still at 1st station. hrhkm km 1 min60 * /1000 100 min1  Delay Analysis Circuit Switching: Assume: Number of hops = M Per-hop processing delay = P Link propagation delay = L Transmission speed = R bit/s Message size = S bits Total Delay = total propagation + total transmission + total processing = 4ML + S/R + (M-1)P Total Delay excluding time for acknowledgment: = 3ML + S/R + (M-1)P Q1. Almaz and Belay are 4 hops apart on a datagram packet-switched network where each link is 160 km long. Per-hop processing delay is 5s. Packets are 1500 bytes long. All links have a transmission speed of 56kbps (kbit/s). The speed of light in the wire is approximately 200,000 km/s. If Belay sends a 10-packet message to Almaz, how long will it take Almaz to receive the message up to the last bit (measured from the time Belay starts sending)? Solution: We know the following: – Number of hops M=4, – Number of packets N=10, – Per-hop processing delay P=5s=0.000005s, – Link propagation delay L = distance/speed of light = 160/200,000 = 0.0008s, – Packet size = 1500 bytes = 1500*8=12,000 bits, – Packet transmission delay T = packet size/transmission speed = 12,000/56000 =0.214s. Delay = ML + NT + (M-1)T + (M-1)P =0.0032 + 2.14 + 0.642 + 0.000015 = 2.785s. Note that the total delay is dominated by the transmission delay which depends on link speed. A link with a higher transmission speed can reduce the delay dramatically. Q2. Now, suppose the link transmission speed is 1Gbps (Gbit/s). How long will it take Almaz to receive the message up to the last bit (measured from the time Belayb starts sending)? Answer: As before, we know the following: – Number of hops M=4, – Number of packets N=10, – Per-hop processing delay P=5ms, – Link propagation delay L = distance/speed of light = 160/200,000 = 800ms, – Packet size = 1500 bytes = 1500*8=12,000 bits, – Packet transmission delay T = packet size/transmission speed = 12,000/109 =12s. Delay = ML + NT + (M-1)T + (M-1)P =3200 + 120 + 36 + 15 = 3371s = 3.371ms. Note that the total delay is now dominated by the propagation delay which cannot be improved because it is constrained by the speed of light. Hence, it is unlikely that future technologies will significantly reduce the delay of Belay’s message at this point. Q3. Repeat Q1 and Q2, assuming that the network uses circuit switching instead of datagram packet switching. Belay’s message is the same length as before. Answer: Year 1989: – Number of hops M=4, – Message size S = 10 * 1500 * 8 =120,000 bits (it is not packetized) – Link transmission speed R = 56kbit/s, – Per-hop processing delay P=0.000005s, – Link propagation delay L = distance/speed of light = 100/125,000 = 0.0008s, Delay = 3ML + S/W + (M-1)P =0.0096 + 2.14 + 0.000015 = 2.1496s Note that the delay improved over the case of datagram packet switching for the same link speed. Why? Now, let link transmission speed be R= 1Gbp Delay = 3ML + B/W + (M-1)P =9600 + 120 + 15 = 9735ms = 9.735ms Note that the delay is worse than in the case of datagram packet switching. Why? Network Models (Cont’d) • In a computer network, each layer may perform one or more of the following tasks: – Error control – Flow control – Segmentation and reassembly – Multiplexing – Connection setup – Routing Network Models (Cont’d) • Two Network Models: OSI and TCP/IP • OSI (Open System Interconnect) consists of 7 layers: • Application Set – Layer 7: Application – here, communication partners are identified, quality of service is identified, user authentication and privacy are considered, and any constraints on data syntax are identified. – Layer 6: Presentation – usually part of an operating system, it converts incoming and outgoing data from one presentation format to another (for example, from a text stream into a popup window with the newly arrived text). – Layer 5: Session – coordinates, and terminates conversations, exchanges, and dialogs between the applications at each end. It deals with session and connection coordination. Network Models (Cont’d) • Transport Set – Layer 4: Transport - provides transparent transfer of data between end systems, and is responsible for end- to-end error recovery and flow control It ensures complete data transfer. – Layer 3: Network - handles the routing of the data, from source to destination. – Layer 2: Data Link – defines procedures for operating the communication links; frames packets and detects and corrects packets transmit errors. – Layer 1: Physical - conveys the bit stream through the network at the electrical and mechanical level. It provides the hardware means of sending and receiving data on a carrier.