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# Note for Computer Network - CN By Devesh Sharma

• Computer Network - CN
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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

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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

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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•