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https://gateoverflow.in/qa/computer-networks
Powered by Question2AnswerAnswered: Synchronous transmission
https://gateoverflow.in/161671/synchronous-transmission?show=161678#a161678
Total bits in character$s$ = 280 bits.<br />
<br />
280 bits makes 30 chars transferable units. 1 transferable char unit would contain 28/3 bits. Therefore Receiver bit rate which is equal to transfer rate = 4200*3/28 = 450bps.Computer Networkshttps://gateoverflow.in/161671/synchronous-transmission?show=161678#a161678Fri, 20 Oct 2017 18:20:00 +0000Answered: TCP CONGESTION
https://gateoverflow.in/161669/tcp-congestion?show=161674#a161674
<p>Congestion window=$38KB$,</p>
<p>Threshold=$19KB$</p>
<p>If MSS starts from $1KB$, answer is $2800msec$ </p>
<p>$1\,|\,2\,|\,4\,|\,8\,|\,16\,|\,19\,|\,21\,|\,23\,|\,25\,|\,27\,|\,29\,|\,31\,|\,33\,|\,35\,|\,36$</p>
<p>=$100 \times 14 \times 2$msec $=2800$</p>
<p> </p>
<hr>
<p>If MSS starts from $2KB$, answer is $2600msec$ </p>
<p>$2\,|\,4\,|\,8\,|\,16\,|\,19\,|\,21\,|\,23\,|\,25\,|\,27\,|\,29\,|\,31\,|\,33\,|\,35\,|\,36$</p>
<p>=$100 \times 13 \times 2$msec $=2600$</p>
<p>Both are correct-:<a rel="nofollow" href="https://gateoverflow.in/1794/gate2014-1-27">https://gateoverflow.in/1794/gate2014-1-27</a></p>
<p>But In kurose and ross , they are starting MSS from $1KB$</p>Computer Networkshttps://gateoverflow.in/161669/tcp-congestion?show=161674#a161674Fri, 20 Oct 2017 18:02:33 +0000Answered: virtual gate test series
https://gateoverflow.in/161630/virtual-gate-test-series?show=161647#a161647
<p>According to the question :</p>
<p> Transmission time = Round trip delay</p>
<p> = 2 * Propogation delay (as per the normal convention)</p>
<p> ==> Data size / Data rate = 2 * Distance between station / Speed of medium</p>
<p> ==> 2<sup>10</sup> * 2<sup>3</sup> bits / x = 2 * 20 * 10<sup>3</sup> / (3 * 10<sup>8</sup>)</p>
<p> <em><strong> ==> x = 2<sup>10</sup> * 2<sup>3</sup> * 3 * 10<sup>8</sup> / (2 * 20 * 10<sup>3</sup>)</strong></em></p>
<p><em><strong> = 61.44 Mbps</strong></em></p>
<p><em><strong> Actually taking 1 KB = 1000 B = 8000 bits , we get x = 60 Mbps but the thing is that for system oriented things like packet size , memory size etc we take 1 K = 1024 whereas for channel and external to system things like velocity of medium , clock rate etc we take 1 K = 1000 . </strong></em></p>Computer Networkshttps://gateoverflow.in/161630/virtual-gate-test-series?show=161647#a161647Fri, 20 Oct 2017 16:22:56 +0000Answered: Virtual Gate test series
https://gateoverflow.in/161620/virtual-gate-test-series?show=161641#a161641
<p>Here the thing to keep in mind is : </p>
<blockquote>
<p>For every byte in TCP , we have a sequence number assigned .</p>
</blockquote>
<p>Hence for wraparound of all bytes possible , we need to cover maximum number of sequence numbers i.e. we have to start from sequence number 0 and continue till (2<sup>32</sup> - 1) as sequence number field in TCP is of 32 bits . After (2<sup>32</sup> - 1) , the numbering will begin from 0 again which is known as "wraparound of sequence numbers" .</p>
<p>Hence data involved before wraparound of sequence numbers = 2<sup>32</sup> B (as 1 sequence number is for 1 byte)</p>
<p> = 2<sup>35</sup> bits</p>
<p> Given data rate of the network = 200 Mbps</p>
<p> = 200 * 10<sup>6</sup> bps</p>
<p> Hence time taken for wraparound of sequence numbers = Data size / Data rate</p>
<p> = 2<sup>35</sup> / (200 * 10<sup>6</sup>)<sup> </sup> s</p>
<p> <em><strong> = 171.799 s</strong></em></p>Computer Networkshttps://gateoverflow.in/161620/virtual-gate-test-series?show=161641#a161641Fri, 20 Oct 2017 16:11:21 +0000Answered: computer networks
https://gateoverflow.in/161622/computer-networks?show=161636#a161636
<p><strong>Both are not same.</strong></p>
<p>Virtual circuit is a type of Packet Switch Network. Now there is a big difference between circuit switch and packet switch network. I will assume that you know the difference.</p>
<p>In simple words, in packet switch, message is divided into packets of some size and sent independently.<em> Each packet has a header</em>. In Circuit switch, message is not divided into packets, <em>hence no headers are required. </em>No sequence numbers are required in circuit switching.</p>
<p>Therefore, there is an efficient use of bandwidth in packet switching, while the same is not true for circuit switching</p>
<p><strong>Packet Switch network is of two types.</strong></p>
<p><strong>1) Virtual Circuit. </strong></p>
<p><strong>2) Datagram.</strong></p>
<p><strong>Virtual Circuit</strong> is connection oriented. For example, the 2G calls you make are based on VC.<strong> Datagram</strong> is conceptually connectionless. Your 4G calls on Jio are a good examples of it.</p>Computer Networkshttps://gateoverflow.in/161622/computer-networks?show=161636#a161636Fri, 20 Oct 2017 16:04:39 +0000Fragmentation offset doubt.
https://gateoverflow.in/161614/fragmentation-offset-doubt
When calculating fragmentation offset, we try to make the size of frame multiple of 8.<br />
<br />
Can someone tell me when to remove bits and make it multiple of 8 and when to pad bits and make it multiple of 8?<br />
<br />
By default, do we try to remove bits first or pad bits first ?Computer Networkshttps://gateoverflow.in/161614/fragmentation-offset-doubtFri, 20 Oct 2017 14:36:28 +0000Answered: question
https://gateoverflow.in/161206/question?show=161565#a161565
Since it belongs to Class C IP Address, So the subnet mask will be 255.255.255.0 .Computer Networkshttps://gateoverflow.in/161206/question?show=161565#a161565Fri, 20 Oct 2017 09:18:44 +0000Answered: ethernet network
https://gateoverflow.in/161442/ethernet-network?show=161477#a161477
<p>Already explained <a rel="nofollow" href="https://gateoverflow.in/1050/gate2004-54" target="_blank">https://gateoverflow.in/1050/gate2004-54</a>. For another explanation see below</p>
<blockquote>
<p>This is basically the question related to unfairness of exponential back-off algorithm called ‘capture effect’. You can find more info about it here: <a rel="nofollow" href="http://intronetworks.cs.luc.edu/current/html/ethernet.html#capture-effect">http://intronetworks.cs.luc.edu/current/html/ethernet.html#capture-effec</a>t</p>
<p>The solution to the above problem goes like this:</p>
<p>At every attempt to transit a frame, both A and B chooses value of ‘k’ randomly. Based on the value of ‘k’, back-off time is calculated as a multiple of ‘k’. The station or node having the smaller back-off time gets to send the frames earlier.</p>
<p>1st attempt: Value of ‘k’ would be k=0 or k=1 (0 <= k <= 2^n-1; where n=nth attempt). Since A won the first race, A must have chosen k=0 and B must have chosen k=1 (A wins here with probability 0.25). As A won, A will again choose k=0 or k=1 for its 2nd frame, but B will choose k=0,1,2 or 3 as B failed to send its first frame in the first attempt.</p>
<p>2nd attempt: Let kA= value of k chosen by A and kB = value of k chosen by B. We will use notation (kA,kB) to show the possible values. Now the sample space for the 2nd attempt is (kA,kB) = (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2) or (1,3) i.e. 8 possible outcomes. For A to win, kA should be less than kB (kA < kB). Thus, our event space is (kA, kB) = (0,1),(0,2),(0,3),(1,2),(1,3) i.e. 5 possible outcomes.</p>
<p>Thus the probability that A wins the 2nd back-off race = 5/8 = 0.625</p>
</blockquote>
<p>Source:<a rel="nofollow" href="http://www.geeksforgeeks.org/gate-gate-cs-2004-question-54/">http://www.geeksforgeeks.org/gate-gate-cs-2004-question-54/</a></p>Computer Networkshttps://gateoverflow.in/161442/ethernet-network?show=161477#a161477Fri, 20 Oct 2017 04:22:21 +0000Answered: subnet mask
https://gateoverflow.in/161441/subnet-mask?show=161459#a161459
<p>Here we have to find the subnet mask for which both IP addresses belongs to the different network. ie. When we do Bitwise AND of IP addresses and that subnet mask, result should be different network id.</p>
<p>Option a) 143.105.1.113 & 255.255.255.0 = <strong>143.105.1.0</strong> , 143.105.1.91 & 255.255.255.0= <strong>143.105.1.0</strong></p>
<p>Option b) 143.105.1.113 & 255.255.255.128 = <strong>143.105.1.0</strong> , 143.105.1.91 & 255.255.255.128= <strong>143.105.1.0</strong></p>
<p>Option c) 143.105.1.113 & 255.255.255.192 = <strong>143.105.1.64</strong> , 143.105.1.91 & 255.255.255.192= <strong>143.105.1.64</strong></p>
<p>Option d) 143.105.1.113 & 255.255.255.224 = <strong>143.105.1.96</strong> , 143.105.1.91 & 255.255.255.224= <strong>143.105.1.64</strong></p>
<p>So subnet mask 255.255.255.224 should not be chosen, otherwise A and B will be in different network.</p>
<p><strong>Hence Option d) is correct</strong></p>Computer Networkshttps://gateoverflow.in/161441/subnet-mask?show=161459#a161459Thu, 19 Oct 2017 18:47:32 +0000Answered: Test Series
https://gateoverflow.in/161431/test-series?show=161456#a161456
<p>Here the key point is :</p>
<blockquote>
<p>We are focussing on network layer and hence anything till transport layer will be payload for network layer .</p>
</blockquote>
<p>Hence payload(data) size for network layer = User data size + UDP header size</p>
<p> = 1100 + 10</p>
<p> = 1110 B</p>
<p>Now MTU of a router is concerning maximum size of packet a router can handle and transfer and hence is inclusive of IP(network layer) header as well</p>
<p>Hence maximum payload allowed in a single MTU = 206 - 20 (IP header size)</p>
<p> = 186 B</p>
<p>But 186 is not a multiple of 8 and hence not a suitable data size for a fragment as it should be a multiple of 8 as :</p>
<blockquote>
<p> 1 offset represents 8 bytes of payload</p>
</blockquote>
<p>Hence nearest lower size than 186 is 184 which is a multiple of 8 ..</p>
<p>Thus payload size of each fragment = 184</p>
<p>Hence number of fragments possible = ceil ( total payload to network layer / fragment payload size)</p>
<p> = ceil ( 1110 / 184 )</p>
<p> = 7</p>
<p><em><strong>Hence a) option is wrong . </strong></em></p>
<p>Now for offset calculation , as mentioned earlier one offset represents 8 bytes .</p>
<p>So number of offset values needed to represent data of 1 fragment = 1 fragment size / 8</p>
<p> = 184 / 8</p>
<p> = 23</p>
<p>Hence offset values for 1st fragment : (0 - 22)</p>
<p> offset values for 2nd fragment : (23 - 45)</p>
<p> offset values for 3rd fragment : (46 - 68)</p>
<p> offset values for 4th fragment : (69 - 91)</p>
<p>In fragmentation offset field of an IP header , the first offset value of that fragment is stored .</p>
<p>Hence the offset field value of 4th fragment = 69</p>
<p><em><strong>Hence b) is true .</strong></em></p>
<p>Coming to MF flag value which means "more fragment" , this flag suggests if the fragment concerned is last fragment or not .If it is the last fragment , we need not require any more fragments , so MF field value = 0 for the last fragment else the value is 1 suggesting that there are further more fragments after the current fragment.</p>
<p>Now as we have found number of fragments = 7 </p>
<p>So sixth fragment is not the last fragment .Hence MF = 1 for this fragment .</p>
<p><em><strong>Hence c) is also true .</strong></em></p>Computer Networkshttps://gateoverflow.in/161431/test-series?show=161456#a161456Thu, 19 Oct 2017 18:24:45 +0000digital signature
https://gateoverflow.in/161436/digital-signature
A digital signature is required—<br />
(I) To tie an electronic message to the<br />
sender’s identity<br />
(II) For non repudiation of communication<br />
by a sender<br />
(III) To prove that a message was sent by the<br />
sender in a court of law<br />
(IV) In all e-mail transactions<br />
(A) I and II<br />
(B) I, II, III<br />
(C) I, II, III, IV<br />
(D) II, III, IVComputer Networkshttps://gateoverflow.in/161436/digital-signatureThu, 19 Oct 2017 17:55:02 +0000Answered: LOGICAL ADDRESS
https://gateoverflow.in/161382/logical-address?show=161391#a161391
<p>MAC address is the physical address. IP address is the logical address.</p>
<p>Now, logical address should be unque in the World. Whereas, physical address should be unique in a network. That is, no two public host should have the same logical address.</p>
<p>But, MAC is unique for each and every device in this world. Also, IP for public host is also unquie in the entire World. Therefore both MAC and IP are globally unique.</p>
<p><strong>Hence, conceptualy, MAC can be used as logical address. </strong>But we don't do that, becaue the information in the MAC address is not helpful to route the packet. Wheras IP has helpful information like Network ID and Host ID.</p>
<p><strong>Similarly, Logical Address can be used as physical address as well. </strong>But it is not done in all the networks. Like, not in ethernet.</p>Computer Networkshttps://gateoverflow.in/161382/logical-address?show=161391#a161391Thu, 19 Oct 2017 14:31:42 +0000Answered: GATE2007-19
https://gateoverflow.in/1217/gate2007-19?show=161358#a161358
Basically there is a relation between bit rate and baud rate which is:<br />
<br />
BaudRate = Bitrate * c * (1/r)<br />
<br />
c = Case factor usually = 1 (if nothing mentioned in question)<br />
<br />
r = data element per signal element<br />
<br />
In manchester signal or differential manchester a data bit is encoded in two signal element so r = 1/2 because for half duration level is up and for half of bit duration its below so two signal element per data element<br />
<br />
<br />
<br />
hence if we substitue the data of r in above we get Baud rate = 2* Bit rateComputer Networkshttps://gateoverflow.in/1217/gate2007-19?show=161358#a161358Thu, 19 Oct 2017 13:08:20 +0000Answered: At which layer of OSI Model does a router work?
https://gateoverflow.in/50672/at-which-layer-of-osi-model-does-a-router-work?show=161289#a161289
C. network LayerComputer Networkshttps://gateoverflow.in/50672/at-which-layer-of-osi-model-does-a-router-work?show=161289#a161289Thu, 19 Oct 2017 08:56:21 +0000Answered: mtu
https://gateoverflow.in/59997/mtu?show=161271#a161271
<p><strong>Link A-R1</strong></p>
<p>Length= (920Bdata+20Bheader)=940B </p>
<p> </p>
<p><strong>Link R1-R2(here the packet is being fragmented into 2 packets)</strong></p>
<p>Packet1:Length=(504B data+ 8B header)=512B [It is correct bcz data's length i.e 504 is divisible by 8]</p>
<p>Packet 2:Length=(416B data + 8B header)=424B [It is correct bcz data's length i.e 416 is divisible by 8]</p>
<p> </p>
<p><strong>Link R2-B( here packet 1 from link R1-R2 is fragmented into to two packets say packet1,1 and packet 1,2)</strong></p>
<p>Packet1.1 Length=(496 B data+ 12 bytes header)=508 B [It is correct bcz data's length i.e 496 is divisible by 8]</p>
<p>Packet 1.2 Length=(8 B data+ 12 bytes header)=20 B [It is correct bcz data's length i.e 8 is divisible by 8]</p>
<p>Packet 2 Length=(416 B data+ 12 bytes header)=428 B [It is correct bcz data's length i.e 416 is divisible by 8]</p>
<p>So Receiver receives three packets.</p>
<p> </p>
<table border="1" cellpadding="1" cellspacing="1" style="height:105px; width:519px">
<tbody>
<tr>
<td><strong>PACKET</strong></td>
<td><strong>ID</strong></td>
<td><strong>MF</strong></td>
<td><strong>HL=Headersize/4</strong></td>
<td><strong>TL</strong></td>
<td><strong> OFFSET </strong></td>
</tr>
<tr>
<td>PACKET 1,1</td>
<td>X</td>
<td>1</td>
<td>3</td>
<td>508</td>
<td>0</td>
</tr>
<tr>
<td>PACKET 1,2</td>
<td>X</td>
<td>1</td>
<td>3</td>
<td>20</td>
<td>496/8=62</td>
</tr>
<tr>
<td>PACKET 2</td>
<td>X</td>
<td>0</td>
<td>3</td>
<td>428</td>
<td>(496+8)/8=63</td>
</tr>
</tbody>
</table>
<p> </p>Computer Networkshttps://gateoverflow.in/59997/mtu?show=161271#a161271Thu, 19 Oct 2017 07:05:40 +0000Answered: Tanenbaum- Q34 (Network Layer)
https://gateoverflow.in/158602/tanenbaum-q34-network-layer?show=161270#a161270
<p><strong>Link A-R1</strong></p>
<p>Length= (920Bdata+20Bheader)=940B </p>
<p> </p>
<p><strong>Link R1-R2(here the packet is being fragmented into 2 packets)</strong></p>
<p>Packet1:Length=(504B data+ 8B header)=512B [It is correct bcz data's length i.e 504 is divisible by 8]</p>
<p>Packet 2:Length=(416B data + 8B header)=424B [It is correct bcz data's length i.e 416 is divisible by 8]</p>
<p> </p>
<p><strong>Link R2-B( here packet 1 from link R1-R2 is fragmented into to two packets say packet1,1 and packet 1,2)</strong></p>
<p>Packet1.1 Length=(496 B data+ 12 bytes header)=508 B [It is correct bcz data's length i.e 496 is divisible by 8]</p>
<p>Packet 1.2 Length=(8 B data+ 12 bytes header)=20 B [It is correct bcz data's length i.e 8 is divisible by 8]</p>
<p>Packet 2 Length=(416 B data+ 12 bytes header)=428 B [It is correct bcz data's length i.e 416 is divisible by 8]</p>
<p>So Receiver receives three packets.</p>
<p> </p>
<table border="1" cellpadding="1" cellspacing="1" style="height:105px; width:519px">
<tbody>
<tr>
<td><strong>PACKET</strong></td>
<td><strong>ID</strong></td>
<td><strong>MF</strong></td>
<td><strong>HL=Headersize/4</strong></td>
<td><strong>TL</strong></td>
<td><strong> OFFSET </strong></td>
</tr>
<tr>
<td>PACKET 1,1</td>
<td>X</td>
<td>1</td>
<td>3</td>
<td>508</td>
<td>0</td>
</tr>
<tr>
<td>PACKET 1,2</td>
<td>X</td>
<td>1</td>
<td>3</td>
<td>20</td>
<td>496/8=62</td>
</tr>
<tr>
<td>PACKET 2</td>
<td>X</td>
<td>0</td>
<td>3</td>
<td>428</td>
<td>(496+8)/8=63</td>
</tr>
</tbody>
</table>
<p> </p>Computer Networkshttps://gateoverflow.in/158602/tanenbaum-q34-network-layer?show=161270#a161270Thu, 19 Oct 2017 07:05:11 +0000Answered: test-series
https://gateoverflow.in/155630/test-series?show=161162#a161162
Network A has MTU of 1500 bytes<br />
Network B has MTU of 620 bytes<br />
we have to find the size of packets that is received at network C<br />
<br />
Here is the thing , i think the data sent is 1380 , that is .. it does not include the header .<br />
<br />
therefore we have the data as 1380 B.<br />
<br />
now it will have no problem in network A .<br />
<br />
but for network B , MTU = 600+20<br />
<br />
thus , the data is broken into three segments .. (600+20) + (600+20)+(180+20)<br />
<br />
thus the total data received will be 1440 B .<br />
<br />
PS<br />
<br />
made easy solution doesn't make much sense .. i think 1380 is the total data that station -1 is sending , otherwise they would have explicitly mentioned that it contains 20 B header , like they did for MTU of networks .Computer Networkshttps://gateoverflow.in/155630/test-series?show=161162#a161162Wed, 18 Oct 2017 17:48:15 +0000Virtual gate
https://gateoverflow.in/161092/virtual-gate
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=5302622131807981103"></p>Computer Networkshttps://gateoverflow.in/161092/virtual-gateWed, 18 Oct 2017 13:51:42 +0000Virtual gate
https://gateoverflow.in/161083/virtual-gate
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=8651597089807632626"></p>Computer Networkshttps://gateoverflow.in/161083/virtual-gateWed, 18 Oct 2017 13:19:34 +0000Answered: test series
https://gateoverflow.in/160890/test-series?show=161032#a161032
<p>Given Subnet Mask = 255.255.255.224</p>
<p>It is a class C network, so 24 MSB belongs to Network, rest 8 belongs to Host.</p>
<p>Subnet Mask is all 1s for subnet and network ID part. 0's for host ID part.</p>
<p>So we get 3 bit for subnet, Total number of subnet = 2<sup>3</sup> = 8</p>
<p>Now comes the main part.</p>
<p>--</p>
<p><strong>According to <span class="marker">RFC 950</span> specification, we have to subtract the subnet with all 0's and all 1s</strong></p>
<p>Therefore, last subnet's network ID will be = 210.15.131. (110 00000)</p>
<p>Last subnet's Directed Broadcast Address = 210.15.131. (110 11111)</p>
<p>Therefore, last host of last subnet = 210.15.131. (110 11110) = 210.15.131.222</p>
<p>--</p>
<p><strong>According to <span class="marker">RFC 1878</span> specification, we DON'T have to subtract the subnet with all 0's and all 1s</strong></p>
<p>Therefore, last subnet's network ID will be = 210.15.131. (111 00000)</p>
<p>Last subnet's Directed Broadcast Address = 210.15.131. (111 11111)</p>
<p>Therefore, last host of last subnet = 210.15.131. (111 11110) = 210.15.131.254</p>
<p> </p>
<p>Which standard to follow for GATE ?</p>
<p>The new standard, i.e. RFC 1878</p>Computer Networkshttps://gateoverflow.in/160890/test-series?show=161032#a161032Wed, 18 Oct 2017 08:35:11 +0000Answered: hamming distance
https://gateoverflow.in/160941/hamming-distance?show=160947#a160947
<p>Here we need to know : </p>
<blockquote>
<p>a) Minimum Hamming distance needed to detect 'd' errors = d + 1</p>
<p>b) Minimum Hamming distance needed to correct 'd' errors = 2d + 1</p>
<p>c) A simple parity scheme has one bit error detecting capability</p>
</blockquote>
<p> </p>
<p>In the modified parity scheme , we can detect two errors as we are using two separate parity bits ; one for checking bits at odd-numbered positions and other at even numbered positions.</p>
<p>Hence number of errors that can be detected = 2 (corresponding to each parity bit we have one error which can be detected)</p>
<p><em><strong>Hence minimum hamming distance needed for error detection = d + 1 = 2 + 1 = 3</strong></em></p>
<p><em><strong> minimum hamming distance needed for error correction = 2d + 1 = 2(2) + 1 = 5</strong></em></p>Computer Networkshttps://gateoverflow.in/160941/hamming-distance?show=160947#a160947Wed, 18 Oct 2017 03:47:51 +0000Answered: Made easy book
https://gateoverflow.in/160757/made-easy-book?show=160822#a160822
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=18373914284872329426"></p>
<p> </p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=7375376698967946670"></p>
<p> </p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=9859407640601796171"></p>
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=11346031968701400020"></p>
<p> </p>
<p>b=1500 bytes r=1500</p>
<p>N=1000 bytes</p>
<p>T<sub>s</sub>=15ms</p>
<p>T=1500/1000*1500=2=1ms</p>
<p>Rotational Latency= 1/(2*1500)=0.33ms</p>
<p>T<sub>a</sub>= 15+1 + 0.33=16.33ms</p>
<p><strong>Source: William Stallings :)</strong></p>Computer Networkshttps://gateoverflow.in/160757/made-easy-book?show=160822#a160822Tue, 17 Oct 2017 13:01:45 +0000Answered: test series
https://gateoverflow.in/160764/test-series?show=160789#a160789
For any flow control no of sequence numbers >=(window size of receiver +window size of sender)<br />
<br />
for Stop and wait ARQ ws=1 and wr=1 so no of sequence numbers =2 let's suppose {0,1}. So we use modulo 2 arithmetic<br />
<br />
and for GBN it is modulo N+1 and for SR it is modulo 2N..Computer Networkshttps://gateoverflow.in/160764/test-series?show=160789#a160789Tue, 17 Oct 2017 11:11:07 +0000Answered: subnet mask ques
https://gateoverflow.in/160675/subnet-mask-ques?show=160701#a160701
<p>IP address was given 200.1.2.64 2 bits are chosen from host id</p>
<p>as we can see it is of class c(192-223)so 24 bit NID+8 bit HID</p>
<blockquote>
<p>as you say 2 bit for subnet from HID so SUBNET MASK:-255.255.255.192</p>
</blockquote>
<hr>
<p>IP 200.1.2.2 CLASS C (192-223)so 24 bit NID+8 bit HID</p>
<p>because NID+SID=bits in the subnet mask</p>
<blockquote>
<p>firstly tell how many bits are chosen from host id for subnet??</p>
</blockquote>Computer Networkshttps://gateoverflow.in/160675/subnet-mask-ques?show=160701#a160701Tue, 17 Oct 2017 07:13:21 +0000Answered: ISRO2017-30
https://gateoverflow.in/128657/isro2017-30?show=160683#a160683
<p><strong>A) n(n-1)/2 and n-1</strong></p>Computer Networkshttps://gateoverflow.in/128657/isro2017-30?show=160683#a160683Tue, 17 Oct 2017 06:19:51 +0000Answered: ISRO2017-32
https://gateoverflow.in/128661/isro2017-32?show=160680#a160680
<p><strong>B) CSMA/CA is used in IEEE 802.11</strong></p>Computer Networkshttps://gateoverflow.in/128661/isro2017-32?show=160680#a160680Tue, 17 Oct 2017 06:02:40 +0000test series
https://gateoverflow.in/160672/test-series
If directed broadcast address of a subnet is 220.101.92.31. Which of the following will be subnet mask ?<br />
<br />
a)255.255.255.224<br />
<br />
b)255.255.255.192<br />
<br />
c) 255.255.255.198<br />
<br />
<br />
<br />
what will be the answer /? and is this question even valid ? since DBA should have all 1's in the last octet ..Computer Networkshttps://gateoverflow.in/160672/test-seriesTue, 17 Oct 2017 05:42:35 +0000Answered: NYQUIST THEOREM
https://gateoverflow.in/135561/nyquist-theorem?show=160600#a160600
<p>sampling is a process of converting a signal (for example, a function of continuous time and/or space) into a numeric sequence (a function of discrete time and/or space). <a rel="nofollow" href="https://en.wikipedia.org/wiki/Claude_Shannon">Shannon's</a>version of the theorem states:</p>
<blockquote>
<p>If a function x(t) contains no frequencies higher than <em>B</em> <a rel="nofollow" href="https://en.wikipedia.org/wiki/Hertz">hertz</a>, it is completely determined by giving its ordinates at a series of points spaced 1/(2<em>B</em>) seconds apart.</p>
<p>according to nyquist theorem</p>
</blockquote>
<p>A sufficient sample-rate is therefore 2<em>B</em> samples/second, or anything larger. Equivalently, for a given sample rate <em>f</em><sub>s</sub>, perfect reconstruction is guaranteed possible for a bandlimit <em>B</em> < <em>f</em><sub>s</sub>/2.</p>Computer Networkshttps://gateoverflow.in/135561/nyquist-theorem?show=160600#a160600Mon, 16 Oct 2017 17:01:57 +0000Answered: Test series
https://gateoverflow.in/148352/test-series?show=160555#a160555
C) According to shortest path AlgorithmComputer Networkshttps://gateoverflow.in/148352/test-series?show=160555#a160555Mon, 16 Oct 2017 13:42:46 +0000Answered: Please solve this Q
https://gateoverflow.in/149772/please-solve-this-q?show=160554#a160554
<p><strong>C) 89</strong></p>Computer Networkshttps://gateoverflow.in/149772/please-solve-this-q?show=160554#a160554Mon, 16 Oct 2017 13:30:20 +0000Answered: UGCNET-dec2009-ii-26
https://gateoverflow.in/152887/ugcnet-dec2009-ii-26?show=160532#a160532
D) GatewayComputer Networkshttps://gateoverflow.in/152887/ugcnet-dec2009-ii-26?show=160532#a160532Mon, 16 Oct 2017 11:30:02 +0000Answered: UGCNET-dec2009-ii-49
https://gateoverflow.in/152944/ugcnet-dec2009-ii-49?show=160530#a160530
<p><strong><em>C)WIFI</em></strong></p>Computer Networkshttps://gateoverflow.in/152944/ugcnet-dec2009-ii-49?show=160530#a160530Mon, 16 Oct 2017 11:19:52 +0000Answered: UGCNET-june2009-ii-08
https://gateoverflow.in/154201/ugcnet-june2009-ii-08?show=160506#a160506
<p><strong>D), Frequency Band</strong></p>Computer Networkshttps://gateoverflow.in/154201/ugcnet-june2009-ii-08?show=160506#a160506Mon, 16 Oct 2017 09:32:26 +0000Answered: UGCNET-dec2008-ii-16
https://gateoverflow.in/155174/ugcnet-dec2008-ii-16?show=160503#a160503
<p>C)<strong> S=Ge<sup>-G</sup></strong></p>Computer Networkshttps://gateoverflow.in/155174/ugcnet-dec2008-ii-16?show=160503#a160503Mon, 16 Oct 2017 09:25:10 +0000Answered: UGCNET-dec2008-ii-17
https://gateoverflow.in/155176/ugcnet-dec2008-ii-17?show=160502#a160502
<p><strong>A) Congestion Control is done in Network and Transport Layer</strong></p>Computer Networkshttps://gateoverflow.in/155176/ugcnet-dec2008-ii-17?show=160502#a160502Mon, 16 Oct 2017 09:24:26 +0000Answered: UGCNET-dec2008-ii-19
https://gateoverflow.in/155178/ugcnet-dec2008-ii-19?show=160501#a160501
<p><strong>A) shortest path Algorithm</strong></p>Computer Networkshttps://gateoverflow.in/155178/ugcnet-dec2008-ii-19?show=160501#a160501Mon, 16 Oct 2017 09:22:22 +0000Answered: UGCNET-dec2008-ii-20
https://gateoverflow.in/155180/ugcnet-dec2008-ii-20?show=160500#a160500
<p><strong>B) IP address in class B is from 128-191</strong></p>Computer Networkshttps://gateoverflow.in/155180/ugcnet-dec2008-ii-20?show=160500#a160500Mon, 16 Oct 2017 09:20:56 +0000Answered: UGCNET-dec2008-ii-35
https://gateoverflow.in/155359/ugcnet-dec2008-ii-35?show=160499#a160499
A)ATMComputer Networkshttps://gateoverflow.in/155359/ugcnet-dec2008-ii-35?show=160499#a160499Mon, 16 Oct 2017 09:19:39 +0000Answered: UGCNET-dec2008-ii-36
https://gateoverflow.in/155360/ugcnet-dec2008-ii-36?show=160497#a160497
<p><strong>A) layer 1 and 2</strong></p>Computer Networkshttps://gateoverflow.in/155360/ugcnet-dec2008-ii-36?show=160497#a160497Mon, 16 Oct 2017 09:19:02 +0000Answered: UGCNET-june2008-ii-28
https://gateoverflow.in/155612/ugcnet-june2008-ii-28?show=160496#a160496
<p><strong>C) 151000 bps</strong></p>Computer Networkshttps://gateoverflow.in/155612/ugcnet-june2008-ii-28?show=160496#a160496Mon, 16 Oct 2017 09:16:08 +0000Answered: Ethernet
https://gateoverflow.in/112915/ethernet?show=160459#a160459
Efficiency = 1/(1+6.44a)<br />
a=Tp/Tt and Tp=d/v and Tt=L/B<br />
Tp=2500/2*10^8<br />
Tt=1024/2*10^7<br />
You'll get efficiency as 38% approx<br />
The percentage of the time medium is occupied but not used by a station is 62% approxComputer Networkshttps://gateoverflow.in/112915/ethernet?show=160459#a160459Mon, 16 Oct 2017 06:13:52 +0000#GATE
https://gateoverflow.in/160430/%23gate
Packets are being transmitted using GB5 and here every 4th packet is lost.How many packets need to be transmitted to transmit 10 packets?Computer Networkshttps://gateoverflow.in/160430/%23gateMon, 16 Oct 2017 00:21:52 +0000Answered: ace test series
https://gateoverflow.in/155701/ace-test-series?show=160261#a160261
<p><strong>C) N+MlogM</strong></p>Computer Networkshttps://gateoverflow.in/155701/ace-test-series?show=160261#a160261Sun, 15 Oct 2017 12:05:37 +0000Answered: UGCNET-dec2004-27
https://gateoverflow.in/156346/ugcnet-dec2004-27?show=160247#a160247
A) error ControlComputer Networkshttps://gateoverflow.in/156346/ugcnet-dec2004-27?show=160247#a160247Sun, 15 Oct 2017 11:53:10 +0000Answered: computer-networks
https://gateoverflow.in/156713/computer-networks?show=160244#a160244
<p><strong>c) it should divide x<sup>t</sup>+1</strong></p>Computer Networkshttps://gateoverflow.in/156713/computer-networks?show=160244#a160244Sun, 15 Oct 2017 11:43:03 +0000Answered: SWP OPTIMAL WINDOW SIZE
https://gateoverflow.in/157374/swp-optimal-window-size?show=160224#a160224
<p><strong>1+2a </strong></p>
<p><strong>Optimal window size is 6</strong></p>Computer Networkshttps://gateoverflow.in/157374/swp-optimal-window-size?show=160224#a160224Sun, 15 Oct 2017 11:07:56 +0000Answered: checksum
https://gateoverflow.in/158311/checksum?show=160218#a160218
<p><strong>reminder is x<sup>0</sup></strong></p>Computer Networkshttps://gateoverflow.in/158311/checksum?show=160218#a160218Sun, 15 Oct 2017 10:53:34 +0000Answered: hamming code
https://gateoverflow.in/158326/hamming-code?show=160216#a160216
<p><strong>10100110110</strong></p>Computer Networkshttps://gateoverflow.in/158326/hamming-code?show=160216#a160216Sun, 15 Oct 2017 10:52:48 +0000Answered: general conceptual question
https://gateoverflow.in/158380/general-conceptual-question?show=160212#a160212
<p><strong>it prevent from count to infinity problem but it split horizon with poison reverse makes the router alive in the network by placing its distance as infinity.</strong></p>Computer Networkshttps://gateoverflow.in/158380/general-conceptual-question?show=160212#a160212Sun, 15 Oct 2017 10:38:38 +0000Answered: tanenbaum
https://gateoverflow.in/123310/tanenbaum?show=160209#a160209
<p><strong> 2 Times it will send </strong></p>Computer Networkshttps://gateoverflow.in/123310/tanenbaum?show=160209#a160209Sun, 15 Oct 2017 10:33:49 +0000