×

Close

- Computer Network - CN
- Note
**9 Views**- Uploaded 3 months ago

P a g e |2 COMPUTER NETWORKS Delay Calculations: •••••• ! #$•%!&!' () *$•+! ,$ -• .!• · Propagation delay = · Transmission delay = /$,-•0••,• •-!!' 1,.,*(••• •, •$•••*!$ #$•••+••••,• •-!!' ,$ (••'3•'•4(•• (-•) Channel Utilization: 1. For Ethernet, channel utilization, u = 7 789• , where a = -$,-•0•••,• '!&•) •$•••+••••,• '!&•) 1.1. for 1- persistent CSMA/CD LAN 10-1000Mbps speed, simplified channel efficiency is, [:ℎ<==>? >@@ABA>=BC] = #E FG H #E 8 , where A =IJ(1 − J).M7 , k stations each with probability p to transmit in a contention slot. N' is the time to transmit a frame. O is the worst case one way propagation delay. 1.2. Efficiency of 802.3 Ethernet at 10Mbps with 512 bit connection: (:ℎ<==>? P@@ABA>=BC) = 7 FQRS 78 TU , where, B = network bandwidth. L = Cable Length. C = speed of signal propagation. E = optimal number if contention slots per frame. F = frame size 2. For Token ring (release after transmission or early token release), 2.1. Channel utilization, u = 7 V , where N is the number of stations in the ring. 78W 3. For Token ring (release after reception or delayed token release), 3.1. Channel utilization, u = 7 78• , where N is the number of stations in the ring. 4. For unrestricted simplex protocol: If a frame has d data bits and h overhead bits and channel bandwidth = b bits/sec then, •••• X•Y! ' = '84 4.1. Maximum channel utilization = Z$•+! X•Y! ' •••• X•Y! × \<=^_A^`ℎ = ×a 4.2. Maximum data throughput = Z$•+! X•Y! '84 5. For stop-and-wait, 7M-$,-•0•••,• '!&•) 5.1. Channel utilization, u = and p is the probability that a , where a = `c<=deAddAf= '!&•) 78b• frame is in error. #•+! •, •$•••+•• • *$•+! ' ' = 5.2. Also Maximum channel utilization = × g,h•' •$•- ••+! (g) (×g '84

P a g e |3 • ×" = • ×! ! 6. For Simplex positive acknowledgement with retransmission (PAR) protocol: 6.1. Maximum channel utilization and throughput is similar to stop-and-wait protocol when the effect of errors is ignored. 7. For Sliding Window Protocols with window size of w, 7.1. Go-Back-N, 5.3. Maximum data throughput = u 1−% , '( *'-./* ('003 4ℎ6 %'%6 '. 6. , * ≥ 2& + 1 1+2&% =# *(1−%) , '( *'-./* ./63 -/4 ('003 4ℎ6 %'%6 '. 6. , * < (1+2&)(1−%+*%) 7.1.1.Channel utilization, 7.2. Selective reject, 7.2.1.Channel utilization 2& + 1 (1 − %), '( *'-./* ('003 4ℎ6 %'%6 '. 6. , * ≥ 2& + 1 ,u =;*(1−%) , '( *'-./* ./63 -/4 ('003 4ℎ6 %'%6 '. 6. , * < 2& + 1 (1+2&) [='>6 4/ 4?&-3>'4 @ (?&>63] ≥ [A/B-. =?'% ='>6] 7.3. Condition for maximum utilization or throughput is: Throughput Calculations: Throughput = Channel Utilization × Channel Bandwidth Signal and Noise Calculations: 1. Signal to Noise Ratio (in decibels, dB) = 10logDE F , G a. where S= Signal strength and N = noise strength. HIJKLMNOOP• QRSPI , 2. Signal Attenuation (in decibels, dB) = 10logDE !PTPNUP• QRSPI Data Rate and Channel Capacity Calculations: 1. Nyquist Theorem: Maximum data rate = 2Hlog V W bits/sec, where H is bandwidth in hertz (Hz) and V is number of levels. Z 2. Shannon’s theorem: Channel capacity = X log2 Y1 + ^ bits/sec, where H is bandwidth in hertz \ (Hz). (Note: here F G is not in decibels). Baud rate: A baud is the number of changes per second in the signal. · For Manchester encoding, baud rate = 2 × bit-rate MAC Sub layer: Static channel allocation in LANs and MANs. If C = channel capacity in bps _ = arrival rate of frames (frames/sec) D ` = no. of bits per frame, then

P a g e |4 Mean time to delay, T = • !−# , Now, if the channel is divided into N sub channels each with capacity each of the N channels is T’(fdm) = • − ' ( ) ( = & % $ % and arrival rate or input rate on then, % !−# Dynamic Channel Allocation: + ,-./0/12134 .5 0 5-067 = = #Δ3 , where # is the arrival rate of frames. /7189 9787-037: 18 0 ;7-1.: .5 27893ℎ Δ3 Multiple access protocol: Pure ALLOHA protocol · Infinite senders assumed. · Poisson distribution with mean rate S frames per frame time · Combined frame rate with retransmission G frames per frame time. · ‘t’ is the time required to transmit a frame. · In multiple access protocol, a frame is successful if no other frames are transmitted in the vulnerable period. · Probability of k frames being generated during a frame transmission time: ,? = · · @ A B CD ?! Hence, probability of zero frames in 2 frame periods is, ,F = 7 GH@ Therefore, for pure ALLOHA, - Mean rate I = J,0 = J7−2J which becomes maximum at G = ½, Max(S) = · • HB = 0.184 = 18.4% throughput. Vulnerability period in pure ALLOHA: For successful frame transmission, no other frame should be on the channel for vulnerability period equal to twice the time to transmit one frame. That is, NO287-0/12134 ;7-1.: Y = 23 , where t is the time to transmit one frame. M 18 P0Q7 .5 ,RST UVVWXU Slotted ALLOHA protocol · Time is divided into discrete frame slots. · A station is required to wait for the beginning of the next slot to transmit a frame. · Vulnerability period is halved as opposed to pure ALLOHA protocol. That is, · NO287-0/12134 ;7-1.: Y = 3 , where t is the time to transmit one frame. M 18 P0Q7 .5 IVWZZT[ UVVWXU Probability of k frames being generated during a frame transmission time: ,? = 7 G@ (1 − 7 G@ )?G•

P a g e |5 · Hence, probability of zero frames in 1 frame period is, •• = • •• - Mean rate = !"0 = !•−! which becomes maximum at G = 1, Max(S) = · PPP · $ % = 0.368 = 36.8% throughput. Expected number of retransmission,& = •• . In Point to Point Protocol (PPP), number of channels grows as square of the number of ()*+•, -. /ℎ233•45 -, 46375 <(<•$) ;= @ .-, 3 8-*9):•,5 computers. That is, ' Binary Exponential Backoff Algorithm: · After i collisions wait a random number of slots between 0 and 2B − 1 with a maximum of 1023. (Note: After 16 collisions, failure is reported to the higher layers.) Minimum frame size for IEEE 802.3 (Ethernet) frame = 64 bytes. ROUTING ALGORITHMS Shortest Past algorithm Flooding Algorithm Flow Based Routing Algorithm, It uses the formula, STATE Non-Adaptive (or Static) Non-Adaptive (or Static) Delay time, T = Non-Adaptive (or Static) $ DE•F , where C= channel capacity, 1/G = mean packet size in bits, and H = mean number of arrivals in packets/sec Distance Vector Routing (DVR) · Based on Bellman-Ford Algorithm and the Ford-Fulkerson Algorithm. · It suffers from the “Count to infinity problem” · Exchange information of the entire network with its neighbors. Link State Routing Algorithm (LSR) · Discovers its neighbors and construct a packet telling all it has just learned and send this packet to all other routers. Hierarchical Routing Algorithm · For N router subnet, the total number of levels = ln ( · And each router will have e × ln ( number of entries in their routing tables. Broadcast Routing Algorithm Adaptive Algorithm (or non-static) Adaptive algorithm (or non-static) Adaptive Algorithm (or non-static) Adaptive Algorithm (or non-static) · Congestions deals with wires and routers while flow deals with hosts. · Traffic Shaping: · § Leaky Bucket Algorithm (If the bucket overflows, then packets are discarded). § Token Bucket Algorithm (causes a token to be generated periodically). Congestion control through Load Shedding may lead to deadlock and unfairness.

## Leave your Comments