Serial : BS_CS_A_Computer Network_040 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: E-mail: info@madeeasy.in Ph: 0-45242 CLASS TEST 20-9 COMPUTER SCIENCE & IT Subject : Computer Network Date of test : 04/0/20 Answer Key. (b) 7. (c) 3. (c) 9. (b) 25. (b) 2. (c). (b) 4. (b) 20. (a) 2. (c) 3. (c) 9. (b) 5. (d) 2. (a) 27. (a) 4. (d). (c). (b) 22. (c) 2. (a) 5. (b). (b) 7. (b) 23. (c) 29. (b). (c) 2. (a). (d) 24. (c) 30. (a)
Computer Science & IT Detailed Explanations. (b) In redirection packet is not discarded but it is redirected to a network as the host doesn t belong to this network. In source quench packet is discarded due to congestion in the network. Destination unreachable means host is not present in the network or the host is not responding to the request, then the packet is discarded. 2. (c) Repeater are used at physical layer. Bridge are used at data link layer. Gateway are used at all layers. Router are used at network layer. 3. (c) Options (a), (b) and (d) are correct statements. HTTP server maintains no information about clients, such HTTP server is said to be Stateless. Option (c) is incorrect statement. 4. (d) Option (a) : Error Checking is only for header part. Option (b) : There is no acknowledgment for packets reaching the destination. Option (c) : IP has minimal error control and there is no concept of error correction for IP datagram. All the options are correct. 5. (b) 27.0.0. Loop back address for testing. 255.255.255.255 Limited broadcast for local network. 239.27.255.255 Class D address for multicast. 0.0.0.0 Computer address for bootstrap.. (c) For shared channel average requests for,000 stations 4 3 requests/ sec 0 0 0 requests/sec Slot time 0 µsec slot 0 sec 4 0 0 slots sec Number of slots,000 slots/sec G Channel load No. of requests/sec No. of slots/sec 0 00 0
CT-20 CS Computer Network 7 7. (c) Data link control is part of data link layer used for services like flow control, error detection and error correction. So, more appropriate answer is data link control.. (b) 000 bit 30 bits/sec 2 5 sec 3 5 km/sec 2 L L 3 km 2 L 5 3 km/sec 9. (b) Each frame has a chance of 0. of getting through, the chance of whole message getting through is (0.) 20 which is about 0.05. The mean number of transmission P 0.05.95. (c) UDP and TCP are transport layer protocol. TCP supports electronic mail.. (b) 242.20.5.200/2 255.255.255.240 240 0000 200 0 00 Network Id First IP 0 000. Last IP 0 i.e. last IP is 242.20.5.20 20 is the fourth octet of the last IP address. 2. (a) One way delay msec One window can be sent 2 PT RTT 40 msec One window 20 millisec One window 40 3 sec 3 40 sec 25 windows/sec 250 5535 bytes/sec 3375 bytes/sec Maximum data rate.3375 Mega bits/sec.3375 Mega bits/sec [ byte bits]
Computer Science & IT Line efficiency 3.7 Mbps 0.0255 52 Mbps in % 0.0255 0 2.5 3. (c) For station A (at network layer) : Total data is 00 bytes At network 2 MTU Data + Header or, Data MTU Header 40 20 40 Bytes As 40 is not divisible by. So, add 4 byte in it. [ 40 + 4 44/ 5] Total data size Number of fragments (n) Actual data size used n 00 Bytes 3.44 44 Bytes Take it as 4 fragments st fragment data 44 Bytes 20 44 st packet 2 nd fragment data 44 Bytes 20 44 2 nd packet 3 rd fragment data 44 Bytes 20 44 3 rd packet 4 th fragment data 20 Bytes 20 20 4 th packet Total data size transmitted to destination 0 Bytes. 4. (b) Net outflow from the token bucket is: Mbps Mbps As a result the time it takes for the full bucket of Mb to empty is: Mb/5 Mbps 0.2 sec During first 0.2 sec, it transmits with Mbps, then switches to Mbps. 5. (d) Transmission time of A for putting packet on to the ethernet, 00 2 µs 2+ 2+ 2+ 2+ B C D A Switch Switch 2 Switch 3 E The time needed for last bit of packet to propagate to the first switch is µs. The time needed for first switch to transmit the packet to second switch is (2 + ) µs and the same happens for remaining switches, each segment introduces a 2 µs T delay, µs P delay. Thus, total latency (2 + ) + (2 + ) + (2 + ) + (2 + ) 57 µs.
CT-20 CS Computer Network 9. (b) X 2 3 4 5 7 9 2 3 4 5 7 2 3 4 5 7 9 7 9 2 2 Y Lost Lost 7. (b) The network mask of class C 255.255.255.0. To satisfied the need of 5 host we have to borrow bit from network portion i.e. 255.255.254.0. The DBA of this network is: 255.255.255.254 i.e., 255.255.... (d) Efficiency for a sliding window protocol is N η + 2 a where N is the window size. N 3 2 Utilization is given as 0.50 0.50 2 + 2a where a t p td... () t d x bytes Mbps 3 t p 2 000 3 0.4 sec a 0.4 Mbps / x byte +2(0.4 Mbps/x byte) 24...() 2(0.4 Mbps/x byte) 23 0. Mbps 23 x byte x 30 byte
Computer Science & IT 9. (b) If LSP is with less recent data than the data stored in database is received, then new LSP is updated with database data and is sent back only over the link from which the first LSP was received. So, option (a) is wrong. 20. (a) The functionality of network layer is source to destination delivery of packets. 2. (a) Burst length of computer is given by Bucket capacity S C M e Output rate Bucket fill rate S 5 4 3.75 seconds 22. (c) IP Addr : 54.33.7.220 Mask : 255.255.255.92 {apply BITWISE AND} ----------------------------------------------------------- 0. 00000. 00000. 00000 220 : 00 Host bits in IP Destination address is of class B network. Therefore Network ID is 54.33 and as per mask we can say that bits are used for subnetting and remaining are host bits. Subnet ID bits are 00000 and Host ID bits are 00 23. (c) Number of subnets 204 Bits required for subnet Network mask 255.255.255.224 Number of hosts / subnet 2 5 2 Ranges are: 4.7.0.0/27 to 4.7.0.3/27 st subnet 4.7.0.32/27 to 4.7.0.3/27 2 nd subnet 4.7.0.4/27 to 4.7.0.95/27 3 rd subnet 4.7.255.224/27 to 4.7.255.255/27 4 th subnet 24. (c) CRC polynomial is the divisor and the message is dividend. The remainder is added to the message and then it is send. CRC is always bit < divisor
CT-20 CS Computer Network 0000 00 000 0 0 00 0000 Message 0 25. (b) 2. (c) Attempt Sender's Congestion Window (kb) Threshold (kb) 0 2 4 4 4 2 4 3 4 4 32 4 5 2 4 7 9 20 During attempt number 9, senders congestion window size Transmission Time 2 Propagation Time Data size B.W. d 2 v Velocity is same when media is same, bandwidth for fast ethernet is 0 Mbps. In order to maintain the same frame size since bandwidth is increased from to 0 Mbps the distance will be reduced from L to L/. Here length of cable is given 9 m. So, length of cable to support frame size 4 byte 9 9.. 27. (a) I.R.T.T 30 ms N.R.T.T 2 ms (α 0.9) E.R.T.T α I.R.T.T + ( α) N.R.T.T 0.9 29 + 0. 2 E.R.T.T 2. + 2. 2.7 ms I.R.T.T 2.7 ms N.R.T.T 32 ms
2 Computer Science & IT E.R.T.T 0.9 2.7 + 0. 32 25.3 + 3.2 29.03 ms I.R.T.T 29.03 ms N.R.T.T 24 ms E.R.T.T 0.9 29.03 + 0. 24 2.27 + 2.4 2.527 ms 2. (a) Propagation delay to travel from S to R L L S R 2 L R 3 2 0 km 0 km 0 km R 3 D 0 km 5 Distance Link Speed ms Total propagation delay to travel from S to D 4 ms 4 ms Total transmission delay for packet (Number of bits) 4 Bandwidth (00) 4 4ms So the first packet will take ms to reach D. While first packet was reaching D, other packets must have been processing in parallel. So D will receive remaining packets packet per ms from R 3. So remaining 999 packets will take 999 ms and total time will be 999 + 07 ms. 29. (b) For pure aloha maximum throughput is.4% For slotted aloha maximum throughput is 3.% Pure aloha.4 90.5 Kbps 0 L 4 For slotted Aloha 3. 90 33.2 Kbps 0 Difference Slotted aloha Pure aloha 33.2.5.5 30. (a) Node throughput 50 frames/sec System throughput 0 * (node throughput) 0 * 50 5000 frames/sec Maximum system rate bps transmission rate 2500bits avg frame length 40000 frames/sec 5000 System throughput Efficiency 0.25 40000 Max system rate 2.5%