Computer Networks ( Classroom Practice Booklet Solutions)

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2 Computer Networks ( Classroom Practice Booklet Solutions). Concept Of Layering 0. Ans: (b) Sol: Data Link Layer is responsible for decoding bit stream into frames. 0. Ans: (c) Sol: Network Layer has the functionality of determining which route through the subnet to use. 0. Ans: (c) Sol: Given: At each layer, n bits of information is added/appended. = nh Total message = original message+overhead m + nh nh % of overhead 00 m nh 04. Ans: (b) Sol: packet NPDU Frame DPDU 05. Ans: (b) Sol: Network Layer 4 times Data Link Layer times H H H T Layer visited Layer 7 times Layer times Layer 5 times Layer 4 times Layer 4 times Network Layer Layer times Data Link Layer Layer times 0. Ans: (c) Sol: Transport Layer is responsible for the End to End delivery of the entire message. 07. Ans: (a) Sol: Data link layer ensures reliable transport of data over a Physical point to point link. Network layer routes data from one network node to next. Transport layer allows end to end communication between two processes. 0. Ans: (c) Sol: Fragment: Network layer (fragmentation) Segment: Transport layer (segmentation) There is a restriction on the message length in the subnet, so breaking the lengthy message starts at transport layer R R 5 4

3 : : Computer Networks. LAN Technologies 0. Ans: (c) Sol: When the transmission delay is high and propagation delay is low the number of collisions decreases. When the collisions decreases throughput increases. 0. Ans: (a) Sol: Ethernet uses Manchester encoding in which is bit has two signal segments, so 0Mbps = 0M signal segments per seconds = 0 mega baud. 0. Ans: (c) Sol: B = Gbps d = km v = km/sec; L =? L d * B v 9 L = 0000 bits or 50 bytes 04. Ans: 00 Sol: L =? B = 0 Mbps T p = 40 micro sec T x = L/B = 00 ms T x = T p 05. Ans: (b) Sol: Collision number for A is, and for B it is. Possible numbers for A from backoff algorithm is (0,), for B they are (0,,, ) Going by the Combinations, A will have 5 chances and B has chance out of. Rest of the two is Undecided. n =, A = (0,), B = (0,) A B Remark 0 0 Collision 0 A = ¼ 0 B = /4 Collision n =, A = (0,), B = (0,,,) A B Remark 0 0 Collision 0 A 0 A 0 A 0 B Collision A A 5 A 0. 5, B 0. 5 Hence Probability for A in 5/ = 0.5. L = Tp B L min = T p B = = (40) (0) bits = 00 bits = 00 Bytes 0. Ans: (c) Sol: Frame Transmission time = 000/00 = 00s At time t = 0 both A & B transmit At time t =.5s a detects collision

4 : : CSIT Postal Coaching Solutions At time t = 5 s Last bit of B s aborted transmission arise at A. At t = 7.5s first bit of A s retransmissions arrives at B. At 7.5s A s packet is completely arrives B = Ans: Sol: All k-stations For a stations P( P) k For some stations among k-station = k.p( P) k S S S S 4 P P P P For S (0.) (0.) (0.7) (0.) = 0.0 For S (0.9) (0.) (0.7) (0.) = For S (0.9) (0.) (0.) (0.) = 0.9 For S 4 (0.9) (0.) (0.7) (0.4) = Probability for any one station among S, S, S, S 4 to send a frame without collision = Ans: (b) Sol: B = 0 Mbps Slot time = 5. sec L = 5 bytes Number of slots =.7 L Transmission time = B 5 bits 00 = Contention width = no. of slots slot time =.7 5. = 7.5 sec L B contention width = seconds L B =.%

5 : 4 : Computer Networks. Data Link Layer 0. Ans: (c) Sol: 0) ( CRC 0. Ans: (a) =+ Even parity 00 5=+4 = =++4 9=+ 0=+ = Hamming code = Ans: 4.7 L = 000 bits d = 00 0 m V = 0 m/sec B = 0 Mbps = 0 0 bps L 000 bits 5 TX 50 sec B 00 bps d 000 m 4 TP 50 sec v 0 m / sec 4 TP 50 a 0 5 TX 50 Efficiency () a 0 = = 4.7% 04. Ans: 0 bits Sol: B = 4 Kbps Propagation delay = 0 msec = 50% RTT = Propagation = 40 msec L = BR N = 50 then L = BR = = 0 bits

6 05. Ans: 0. Sol: B =.5 Mbps RTT (Round Trip Time) = 45ms L = KB L Link utilization L BR 04 (04) % 0. Ans: (c) Sol: Propagation delay = 00 sec d = 0 KM L = Kb = 04 bits B =? RTT = Transmission delay RTT = Propagation delay RTT = 00 sec Transmission delay = L B 04 B = 40 Mbps Ans: 500 Sol: B = 0 kbps L = 000 bytes T p = 00 ms T x = L/B = 00 ms Tax = ack size/ bandwidth = 00 ms 00 Efficiency = tx/(tx +tp+tax) = Throughput = efficiency * bandwidth = 0.5 *0 4 bytes = 500 bytes : 5 : CSIT Postal Coaching Solutions 0. Ans: (c) Sol: L = 000 bits frame BER = d = 00 km = 00 0 m B = 0 Mbps = 0 0 bps v = 0 m/sec L 000 TX 0.50 B d TP 0.50 v 0 GBN w = 0 w LP a LPw = 0. =.% 4% SR w = 0 T P = T X = TP 0.50 a 0 4 TX a = 0 So, + a = + (0) = Here (w) < ( + a) so smaller window w LP Efficiency a 0 LP %

7 : : Computer Networks 09. Ans: (d) Sol: 5 bytes x bits/b = 409 bits per frame 409/4000bps= 4 msec to send one frame Round trip delay = 540 msec Window size : send 409 bits per 540msec 409bits/540msec = bps throughput Window size 7: 755 x 7 = 509 bps Window size 9 and greater: 755 x 9 = 5 bps but the maximum capacity is 4 kbps so for window sizes greater than 9 the maximum throughput is 4 kbps. Ans: (d) Sol: B = Mbps Latency delay (or) Propagation delay =.5 sec L = KB () RTT =.5 =.5 sec () sec = 0 bits.5 sec =? bits () p pkt size.5 0 = (4) sequence no. = p = 05 k = 05 k = 9 bits for GBN for SR W p = 0 so k = 0 bits. Ans: (c) Sol: d = 000 km B =.5 Mbps L = 4 bytes Propagation speed = sec/km Propagation delay for 000 km 000 sec () RTT = 000 sec = = msec () sec.5 0 bits ms? bits () p pktsize = 0 (4) Sequence num p = 0 k k (5) 0 p k 7 k = 7 0. Ans: Sol: w = Total 9 packets Every fifth packet lost w = w = w = Packets Attempts Total attempts

8 : 7 : CSIT Postal Coaching Solutions. Ans: % Sol: Calculation for 00 frames = = 9000 Overhead 00 0 = 4000 NAK 40 Retransfers = Total = % of Bandwith that is wasted = [040/( )] * 00 =.99 % 4. Ans: (b) Sol: Instantaneously all 4 frames arrive at router. Time line t = 0 Remark frame 0 is sent out [0] f 0 0 t = f 0 is B ack is sent out [0] f ack 0 receiver slide down for next frame f is sent out [0 ] t = f is B ack is sent out receiver slide down for next frame f is sent out previous ack is arrived, next round Frame is instantly joined in window [0] ack f outstanding frames are there in the window =, Now onwards

9 : : Computer Networks 5. Ans: 4 Sol: 5 step problem. Calculate RTT = (T p ). Calculate BR, window size in bits. Calculate W = window in packets = BR/L 4. For selective repeat, ASN is set to W 5. Sequence number, k Bandwidth (B) = 0 bps Propagation delay (T P ) = 50 msec Packet size(l) = kilobyte L Transmission delay (T t ) = B 0 0 Tt T t sec T t = 4 msec W S = sender window size WS Tt T T t W S p bits bps 4 W S 4 Ws = 5.75 W S + W R = Available sequence numbers for SR W S = W R ASN = W S ASN = 5.75 ASN =.75 No. of bits in the sequence number = log ASN = log.75 = 4. Ans: (d) Sol: Given: B = 0 bps Distance = 0000 km T P = 0 m/s L = B L p = T x = 4 B = = = 400 msec d q = = v = ms = 50 ms = 7. Ans: 9. Sol: B = Mbps T p = 0.75 ms T proc = 0.5 ms Payload = 90 B Ack = 0 B OH = 0 B L = Payload + OH = = 000 Bytes L 000 T x = = B = ms 0 0 Tax = 0 sec 0 = 0. msec Total time = T x +T p + T proc + T ax + T p + T aproc = ms ms + 0.5ms+ 0.ms+0.75ms = 7.9 ms = = T x Total Time 7.9 = 9.%

10 : 9 : CSIT Postal Coaching Solutions 4. Switching(Circuit, Packet) 0. Ans: (a) Sol: Given data Circuit setup time = S sec Bandwidth = bit rate = b bps Path = K -hop Propagation delay = d sec per hop Connection release = not given Packet size = p bits Message size = x bits K = K hop path (hop means jump) T P d v m m /s sec Total delay = I + II + III I. Circuit setup time = S II. T X L B messagesize bit rate III. T P =one hop propagation time= d sec x b 0. Ans: (d) Sol: Pkt Pkt Pkt Pkt Pkt Pkt Pkt Pkt Pkt The last packet is getting retransmitted at k hops so the delay is (k ) b p. There is no set of time (NO S) Transmission delay is x/b p p.. pn = b Message For k hop propagation time? = k d p Total time = x/b + k.d + (k ) b For k hop propagation time? = k d Total delay = S + x/b + k.d () () () S d d d

11 : 0 : Computer Networks 5. Network Layer 0. Ans: (b) Sol: C F 5 C F = 94 = 47 = = 0 From NID, bits are reserved for prefix of class C address therefore number of network all allowed under class C address are 4 =. 04. Ans : (b) NID Class B = Ans (c) Sol: In given problem network part is of 0 bits. NID 0 Class B 0 = Among 0 NID bits we are not going to use bits which are fixed for class B prefix so number of network possible are 0 = and number of hosts possible are. Host per subnet = Ans: (c) Sol: NID + Class C Ans: (c) Sol: NID Class C x = 7 x = SID 5 + = subnets can be formed 5 = 0 host per subnet id 0 =

12 : : CSIT Postal Coaching Solutions (or) Given 5 host per subnet x = 5 x = 5 host per subnet NID + Class C SID 5 0. Sol: SM = = NID 4 Class C SID Subnet mask /7 0. Ans: (d) Sol: Class B NID + 4 SID 4 departments = / Ans : Sol: 00 LAN s x = 00 hosts Class B x = 7 NID + 4 SID 9 7 Class C network has 4-bits NID and bit (a) bits are borrowed from (b) no. of subnets = = 4 (c) no. of system per subnet = = 4 = 09. Ans: 5 Sol: /7 clearly indicates that first bits (, 4, ) in the last octet are borrowed for subnet, 5 bits for Host ID. and mask is If you perform AND operation between IP ( ) and Subnet mask ( ) then we get So subnet ID is and network ID is We have 5 bits for host ID. We cannot have all s in host ID, therefore we will have 0 (last 5 bits) for the last IP address. Therefore in last octet we will have 000, it is 5

13 0. Ans : 4 Subnet mask / Given LANS = 50 x = 50 x = NID + = 4. Ans: (a) Sol: (b) / and / same can not be given to two subnets (c) / and / same can not be given to two subnets (d) /4 and / same /4 will not be required. Ans: (c) Sol:.5.4.0/ /4.5..0/ / Ans: Sol: For each hop TTL is reduced by (minimum) and there are hops here hence =. + 4 Class B SID Change after bit Change / : : Computer Networks 4. Ans: 00 bytes Sol: Offset 00 means there are 00 fragments before this, bytes for each fragment 00 bytes. 5. Ans: (c) Sol: For last fragment always M = 0. If HLEN is 0 then header length is 40 bytes (We use scale factor of 4 in HLEN).Therefore total data in fragment is = 0 bytes. Since offset is 00 total bytes ahead of this fragment is 00 = 400 bytes (we use scale factor of in offset). Therefore it is last fragment, starting byte is 400 and ending byte is 759 (Actually = 70 bytes but byte number starts with zero, so it is from 400 to 759). Ans: (c) Sol: For the first network the maximum allowed payload size =00 bytes per frame and for the second network the maximum allowed payload size= 400 bytes per frame. Per packet IP overhead is given as 0 bytes. So first we will calculate the total number of packets formed. Note: If first network consider: For first network 00 bytes will be divided into packets of size 00 and 900 bytes. So IP overhead of st network = (* 0=40 bytes) But given is second network. For second network 00 bytes will be divided into packets 5 of 400 bytes and of 00 bytes. So, IP overhead of the nd network = (* 0 = 0 bytes) Thus, the maximum IP overhead for the nd network = 0 bytes

14 : : CSIT Postal Coaching Solutions 0. Routing Algorithms 0. Ans: (c) Sol: Going via B gives (,, 4,,, ). Going via D gives (9, 5, 9,, 9, 0). Going via E gives (,,, 4, 5, 9). Taking the minimum for each destination except C gives (,, 0,, 5, ). The outgoing lines are (B, B, -, D, E, B). 0. Ans: (c) Sol: RIP uses distance vector routing RIP packets are sent using UDP OSPF doesn t use UDP or TCP and sends directly via IP OSPF operation is based on LSR 0. Ans: (a) Sol: Perform AND operation Given IP address and net mask, and compare results with network number if it matches with network number, then forward packet through that interface. If not matched with any entry then use default route. Ex: for (i) AND = Hence packet must be transferred through Interface 0. Sometimes result matches with multiple network number, if so use interface that has longest length subnet mask. So similarly for i : a ii : c iii : e iv: d 04. (i) Ans: (a) Sol: is one of the IP address in the same subnet, it uses interface Subnet Host it uses R 4 as the next because this IP address is under default subnet number. 04. (ii) Ans: (d) Sol: uses R as the next hop because it falls under Uses R as the next hop because it falls under

15 : 4 : Computer Networks 7. TCP, UDP and Congestion Control 05. Ans: (c) Sol: 0. Ans: (c) T KB Sol: TCP pseudo Header Format KB Source IP Destination Pseudo header 4KB TCP Header KB 0. Ans: (b) Sol: Each socket is binded with a port 4 9KB 0. Ans: (d) 04. Ans: (b) Sol: RTT = 0 msec Scap = 4 KB Lcap = KB ACK ACK ACK ACK KB 4KB KB KB 4KB After 40 ms a full window is transmitted c KB 4 KB = ( )ms = 40 ms s When timeout occurs thresh hold 9 Minimum (Congestion Window, Receiver Window). Minimum (, ). Minimum (, ). Minimum (4, ) 4. Minimum (, ) 9 Since it is crossing threshold, instead of KB it sent 9 KB 0. Ans : 7.4 Sol: Transport data unit Total numbers available = = 5 and they should be consumed in 0 sec. 5 Data rate per connection 0

16 07. Ans: (b) Sol: Given M = max burst = Mbps = const rate = token arrive rate Mbps C = Mbps S =? C Mbits S M Mbits / sec sec 5 =. sec 0. Ans: (c) Sol: Given L = 000 byte M = 50 million bytes/sec = 0 million byte/sec C = 0 bytes S =? C 0 M = 5 msec S 09. Ans :. Sol: C= MB M = 0 MB parsec Arrive rate = 0 MB per sec Actual file size = 0 bytes S =? C S M 0 0 0sec 0.sec 40 The computer runs with bursty rate for the duration of 5 sec. the amount of data outlet = equation () : 5 : CSIT Postal Coaching Solutions 0MB 0.sec 5 = MB data is outlet.. step () Current file size = MB = already outlet data with M rate = MB MB Remaining data = 0 MB This remaining data 0 MB goes as with constant rate. 0 MB sec Remaining data 0 MB? 0MB sec step () 0MB Total time taken = S + S = 0. + =. sec 0. Ans: (d) Sol: Data in st segment is from byte number 0 to byte number 9, that is 0 bytes As st is lost so, TCP will send ACK for the next in-order segment receiver is expecting. So it will be for 0.. Ans: (b) SYNTAX: int connect(int sockfd, const struct sockaddr *addr, socklen_t addrlen); The connect() system call connects the socket referred to by the file descriptor sockfd to the address specified by addr. The addrlen argument specifies the size of addr. The format of the address in addr is determined by the address space of the socket sockfd;

17 : : Computer Networks If the socket sockfd is of type SOCK_DGRAM, then addr is the address to which datagrams are sent by default, and the only address from which datagrams are received. If the socket is of type SOCK_STREAM or SOCK_SEQPACKET, this call attempts to make a connection to the socket that is bound to the address specified by addr. Generally, connection-based protocol sockets may successfully connect() only once; connectionless protocol sockets may use connect() multiple times to change their association. Connectionless sockets may dissolve the association by connecting to an address with the sa_family member of sockaddr set to AF_UNSPEC (supported on Linux since kernel.). So A process can successfully call connect function again for an already connected UDP socket for the above reason. Hence statement II is correct. Statement I is wrong as the UDP is a connection less service and basically used for query and response purpose and there is no meaning of concurrent service (simultaneous) 0.Application Layer Protocols 0. Ans: (b) Sol: Refer page 00 for the concept of base 4 encoding 0. Ans: (c) Sol: The concept to be followed. Step : The client(browser) initiates a DNS query for remote server. It may be that they already have this server in their DNS cache, in which case the client may simply send a TCP SYN directly to the application server. Step : The client will next send a connection request to the application server. This will be a TCP SYN packet, the first in the TCP three-way handshake. Step : Next, after the TCP connection has been established, the client will request data from the server. In the web-based application, the client performs an HTTP GET.. Ans: 9.5 Sol: RTT = 0 msec = 0.9 NRTT = Basic algorithm = (IRTT)+(-) (NRRT) = (-0.9) () = 9. msec nd round = 9.4 msec rd round = 9.5 msec 09. Basics Of Wi-Fi 0. Ans: (b) Sol: RTS and CTS mechanism is used for collision avoidance, not collision detection.

18 : 7 : CSIT Postal Coaching Solutions 0. Network Security Chapter (a) Private key 0. Sol: Symmetric in nature 0 (b) Cipher Modes 0. Ans: (b) Sol: Bit error causes its impact on two blocks only (i, i+). 04. Ans: (c) as per concept 05. Ans: (b) as per concept 0. Ans: (c) as per concept 07. Ans: (c) as per concept 0 (c) Key Management Ans: (d) Sol: Given n = 47, g =, x = n, g, g x mod n 47,, mod 47 x = A B Both same Both are in symmetric nature. Hence IP = IP mod 47= (47,, ) 0. Ans: (c) Sol: nodes nodes 09. Ans: (b) Sol: Given y = 0, n = 47, g =. key 0 mod 47 = 7 A 0 mod 47 B 4 nodes keys + 0. Ans: (a) Sol: n, g, g x mod n ++ = keys 4 nn n nodes +.+(n-) = keys A B y = 0 0 mod 47 = 7 Session key = 7 mod 47 = 4

19 : : Computer Networks. Ans: (a) & (d) Sol: Property for good candidate n Choose n in such a way that n, both should be prime. 7 (a) 7, (7, ) (b) is not prime 7 (c) 7, (7, ) 47 4 (d) 47, (47, ) Option (a) & (d) is correct Chapter 4. Sol: a =, b = (a) p = 7 q = n = p q = 7 = 77 z = (p ) (q ) = (7 ) ( ) = 0 = 0 GCD (d, z) = GCD (d, 0) =. Ans: (c) Sol: x =? Session key = 9 5 % 79 = mod 79 e = d = Ans: (b) Sol: n = 7, g =, x =, y = 5 A A 79,, % 79 = 54 7,, mod 7 B B 5 mod 7 5 % 7 = %7 = 4 y = Five legal values for d is 7,,, 7, 9 (b) p =, q =, d = 7. Find e P =, q =, n = p q z = (p ) (q ) = 0 = 0 d = 7 GCD (d, z) = (e d) mod z = k u = {e, n} (e d) = mod Z k r = {d, n} Encryption p e mod n = c Decryption c d mod n = p (e d) = mod 0 (e 0) = mod 0 (multiple of 0 + ) % 0 = i = 0 = 0 + = % 0 = e 7 = ; e = fraction 7 i = 0 = 70 + = 7% 0 = e 7 = 7 7 e 0 7

20 (c) p = 5; q = ; d = 7 p = 5 q = n = 5 = 55 z = 4 0 = 40 (e d) = mod z (e 7) % 40 = Multiple of 4 + i = 40 (40 + ) % 40 = e 7 = 4, e = fraction i = 40 (0 +) % 40 = e 7 =, e = Encrypted abcdefghij P P e mod n P mod 55 a = mod 55 = b = mod 55 = c = 7 mod 55 = 7 d = 4 4 mod 55 = 9 e = 5 5 mod 55 = 5 f = mod 55 = 5 g = 7 4 mod 55 = h = 5 mod 55 = 7 i = 9 79 mod 55 = 4 j = mod 55 = 0 5. Ans: (c) Sol: e = 5, p = 7, q = 7 Z = = 9 (e d) = mod z = e = 5 (e d) = multiple of 9 + i = 9 + = 97% 9 = e d = 97;d = 97/5 = fraction : 9 : CSIT Postal Coaching Solutions i = 9 = 9 + = 9 e d = 9 fraction i = 9 = + = 9 e d = 9 fraction i = = = 5 e d = 5 5 d Ans: (d) Sol: A =, B =, C =, D = 4, E = 5, F = M = F = Given p = 7, q = 7 n = 7 7 = 9 C = M e mod n = 5 mod 9 = 4 7. Ans: (d) Sol: p = 97, q = 40, e = 4, d =? z = = 5400 So, d = 007 n Q e T = A B * Q A = B B = T R = (n) % e (n) = Extended Euclidean Algorithm Q A (n) B e

21 : 0 : Computer Networks. Ans: (a) Sol: encrypted p = 4 p e mod n e = 4 n = n = 59, mod mod d= 4, X 999 X 49 X Chapter 9. Ans: (d) Sol: Definition of Digital sign and PKC.