lOMoARcPSD|28579846 PE NWC203c by PH - abc Networking (Trường Đại học FPT) Studocu is not sponsored or endorsed by any college or university Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 PE NWC203c NWC203c Shape Spring 2022 Q1: Explain the difference between connectionless unacknowledged service and connectionless acknowledged service. How do the protocols that provide these services differ? [Answer] Connection Less: Connectionless service comes with a single free-standing data unit for all transmissions. → In this, each unit contains all of the protocols that control information necessary for delivery perspective, but this also contains no provision for sequencing or flow control. + Acknowledged: This is achieved by the use of ACK and NAK control messages. → These types of protocols are well suited for communication over the network, where high layers are very sensitive to loss and can have a significant probability of error in these underlying networks. Example: HDLC, which offers for unnumbered acknowledgment service(setup and release). + Unacknowledge: → This comes with a very simpler version and provides faster communication for networks, which are inherently reliable or provide service to a higher layer, that can tolerate loss in the information, or which has built-in error control/recovery feature. ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q2. Explain the difference between connection-oriented acknowledged service and connectionless acknowledged service. How do the protocols that provide these services differ? [Answer] Less & Oriented: + Connection-oriented: → In this type of service, a setup phase will be initialized between sender and receiver, to establish a context for transferring the information → This connection is provided to the sender for all SDUs. → This service requires a stateful protocol, which is used to keep track of sequence numbers, and timers. + ConnectionLess: → Here, there will be no prior context provided for transferring the information between sender and receiver. → The sender will pass its SDU to an underlying layer without any notice. → And in this, the sender requires an acknowledgment of SDU delivery. → The protocols are very different in these services → this service also does not require transmitting protocols to track the acknowledgment of PDU. → After receiving the PDU, the receiver needs to send acknowledgment, If not received in time, then it will return failure. ========================================================================= Q3: Explain the differences between PPP and HDLC. [Answer] HDLC is a short form of High-level Data Link Control that does the data encapsulation. PPP is an acronym for Point-to-Point Protocol that can be used by different devices without any data format change. A few major differences are as below: + For communication through HDLC, a bit-oriented protocol is used for point-to-point links as well as for multipoint link channels. However, PPP uses a byte-oriented protocol for point-to-point links at the time of communication. + HDLC does the encapsulation for synchronous media only whereas PPP can do the encapsulation for synchronous as well as for asynchronous media. + HDLC can be used only for CISCO devices whereas PPP can be easily used for other devices. Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q4: A 1.5 Mbps communications link is to use HDLC to transmit information to the moon. What is the smallest possible frame size that allows continuous transmission? The distance between earth and the moon is approximately 375,000 km, and the speed of light is 3 x 108 meters/second. [Answer] Go-Back-N Selective Repeat Maximum Send Window Size in Default HDLC Frame Maximum Send Window Size in Extended HDLC Frame 7 4 127 64 The round trip propagation delay 2tprop = 2 = 2 = 2.5s We have: = Go-Back-N: If N = 7 → = 2.5 s → = 535715 bits If N = 127 → = 2.5 s → = 29528 bits Selective Repeat: If N = 4 → = 2.5 s → = 973500 bits If N = 64 → = 2.5 s → = 58594 bits ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q5: Suppose HDLC is used over a 1.5 Mbps geostationary satellite link. Suppose that 250-byte frames are used in the data link control. What is the maximum rate at which information can be transmitted over the link? [Answer] R = 1.5 * 106 bps, and nf =2000 bits. The distance that the information must travel is the earth to satellite distance → d ≈ 36,000 km. The speed of light c: 3 x 108. Tprop = = = 120 ms Tf = = = 1.33 ms We can use either Go-Back-N or Selective Repeat ARQ. The default window size is N = 7 The maximum information rate is achieved with no error, and hence, no retransmission. Tcycle = Tf + 2Tprop = 1.33 + 2 * 120 = 241.33 ms n = N * nf = 7 * 2000 = 14,000 bits Rmax = = = 58 kbps If the extended sequence numbering option (7-bit) is used, the maximum send window size would be N = 127 → n = 127 * 2000 = 254000 bits Rmax = = = 1.052 Mbps ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q6: Suppose that a multiplexer receives constant-length packet from N = 60 data sources. Each data source has a probability p = 0.1 of having a packet in a given T-second period. Suppose that the multiplexer has one line in which it can transmit eight packets every T seconds. It also has a second line where it directs any packets that cannot be transmitted in the first line in a T-second period. Find the average number of packets that are transmitted on the first line and the average number of packets that are transmitted in the second line. [Answer] Find out the probability of the k packets that have reached the T- second. N=60 and p=0.1 The average number for the arrivals of the packets can be given as Np = 6 The average number of packets received through the first line as below: → The average number of packets received through the first line = 4.59 packets The average number of packets transmitted through the second line per T second can be obtained as below: The average number of packets transmitted through the second line = 6 - 4.59 = 1.41 packets Therefore, it will transmit 1.41 packets on average per T second from the second line. ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q7: Consider the transfer of a single real-time telephone voice signal across a packet network. Suppose that each voice sample should not be delayed by more than 20 ms. a, Discuss which of the following adaptation functions are relevant to meeting the requirements of this transfer: handling of arbitrary message size; reliability and sequencing; pacing and flow control; timing; addressing; and privacy, integrity and authentication. b, Compare a hop-by-hop approach to an end-to-end approach to meeting the requirements of the voice signal. [Answer] a, Message size is important because in real-time signals of voice it is necessary to transfer a fixed packet size of that holds no more than 20ms of the speech signal. The handling of arbitrary message size is not as important as long as the desired packet size for voice can be handled. Sequencing is essential because each packet needs to arrive in the same sequence that it was generated. Reliability is moderately important since voice transmission can tolerate a certain level of loss and error. Pacing and flow control are not as important because the synchronous nature of the voice signal implies that the end systems will be matched in speed. Timing, for real-time voice transfer, is important because this adaptation function helps to control the jitter in the delivered signal. Addressing is only during the connection setup phase if we assume some form of virtual circuit packet switching method. Privacy, integrity, and authentication have traditionally not been as important as the other issues discussed above. b, If the underlying network is reliable then the end-to-end approach is better because the probability of error is very low so processing at the edge suffices to provide acceptable performance. If the underlying network is unreliable then the hop-by-hop approach may be required. For example, if the probability of error is very high, as in a wireless channel, then error recovery at each hop may be necessary to make effective communication possible. ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q8: Consider the Stop-and-Wait protocol as described. Suppose that the protocol is modified so that each time a frame is found in error at either the sender or receiver, the last transmitted frame is immediately resent. a, Show that the protocol still operates correctly. b, Does the state transition diagram need to be modified to describe the new operation? c, What is the main effect of introducing the immediate-retransmission feature? [Answer] a, The sender in the stop-and-wait protocol described in the chapter retransmits a frame when an acknowledgment is not received in time. The modified protocol says that the frame is retransmitted every time the sender or receiver sees an error. Therefore, the only difference is that frames are retransmitted sooner. So, the protocol will work correctly. b, No. The state transition diagram will stay the same. c, The error recovery process will be faster with this modified protocol. ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q9: Suppose that two peer-to-peer processes provide a service that involves the transfer of discrete messages. Suppose that the peer processes are allowed to exchange PDUs that have a maximum size of M bytes including H bytes of header. Suppose that a PDU is not allowed to carry information from more than one message. [Answer] Bytes each to be transmitted in several PDUs in order to exchange messages of any size. A single PDU must include all small messages. Peer processes must exchange information that permits messages to be reassembled at the recipient. The message length, for example, could be included in the first PDU. A message end-of-message marker could be included in the last PDU. In connection-oriented networks, sequence numbers can be used to detect loss, while in connectionless networks, they can be used to aid in message reconstruction. Finally, because variable-size PDUs are allowed, the PDU size must be specified in the PDU header. In this instance, each PDU must be identified with a stream ID in addition to all of the header information specified in (b), so that the receiver may treat each stream separately while reassembling messages. Suppose that two peer-to-peer processes provide a service that involves the transfer of discrete messages. Suppose that the peer processes are allowed to exchange PDUs that have a maximum size of M bytes including H bytes of header. Suppose that a PDU is not allowed to carry information from more than one message. [Answer BY PH] a. Develop an approach that allows the peer processes to exchange messages of arbitrary size. To exchange messages of arbitrary size, large messages must be segmented into parts of M-H bytes each in length to be transmitted in multiple PDUs. Small messages must be placed in a single PDU. b. What essential control information needs to be exchanged between the peer processes? The peer processes need to communicate information that allows for the reassembly of messages at the receiver. For example, the first PDU may contain the message length. The last PDU may contain and end-of-message marker. Sequence numbers may also be useful to detect loss in connection oriented networks and to help in reconstruction of the messages in connectionless networks. Lastly, since variable size PDUs are permitted, the size of the PDU must be transmitted in the PDU header. c. Now suppose that the message transfer service provided by the peer processes is shared by several message source-destination pairs. Is additional control information required, and if so, where should it be placed? In this case, in addition to all of the header information mentioned in b), each PDU must be labeled with a stream ID, so that the receiver can treat each stream independently when reassembling messages. Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q10: A 1 Mbyte file is to be transmitted over a 1 Mbps communication line that has a bit error rate of p = 10-6. a, What is the probability that the entire file is transmitted without errorsWe conclude that it is extremely unlikely that the file will arrive error free. b, The file is broken up into N equal-sized blocks that are transmitted separately. What is the probability that all the blocks arrive correctly without error? Does dividing the file into blocks help? c, Suppose the propagation delay is negligible, explain how Stop-and-Wait ARQ can help deliver the file in error-free form. On the average how long does it take to deliver the file if the ARQ transmits the entire file each time? [Answer] a, Note for n large and p very small, (1 − p)n ≈ e − np. P[no error in the entire file] = (1 – p)n ≈ e – np for n > 1, p < 1 = e-8 = 3.35 * 10-4 We conclude that it is extremely unlikely that the file will arrive error free. b, A subblock of length is received without error with probability: P[no error in subblock] = A block has no errors if all subblocks have no errors So P[no error in block] = P[no errors in subblock]N = = (1 – p)n So simply dividing the blocks does not help. c, We assume the following: t0 = tf + tACK = + (tprop ≈ 0). P[nt = i ] = P[one success after i – 1 failure] = (1 – Pf) Pf i – 1 ttotal | i transmissions = i * t0 E[ttotal] = Here, nf = n → na , Thus t0 ≈ tf = E[ttotal] = = = = 23881s = 6.633 hours! The file gets through, but only after many retransmissions. Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q11: In this activity, you are given the network address of 192.168.100.0/24 to subnet and provide the IP addressing for the Packet Tracer network. Each LAN in the network requires at least 25 addresses for end devices, the switch and the router. The connection between R1 to R2 will require an IP address for each end of the link. a, Based on the topology, how many subnets are needed? b, How many bits must be borrowed to support the number of subnets in the topology table? c, How many subnets does this create? d, How many usable hosts does this create per subnet? [Answer] a, Based on the topology, how many subnets are needed? →5 b, How many bits must be borrowed to support the number of subnets in the topology table? →3 c, How many subnets does this create? →8 d, How many usable hosts does this create per subnet? → 25 – 2 = 30 ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q12: Five stations (S1-S5) are connected to an extended LAN through transparent bridges (B1-B2), as shown in the following figure. Initially, the forwarding tables are empty. Suppose the following stations transmit frames: S1 transmits to S5, S3 transmit to S2, S4 transmits to S3, S2 transmits to S1, and S5 transmits to S4. Fill in the forwarding tables with appropriate entries after the frames have been completely transmitted. [Answer] Address S1 S3 S4 S2 S5 Port 1 2 2 1 2 Addres s S5 S2 S3 S1 S4 Port 1 2 2 1 2 ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q13: Consider the network in Figure. a, Use the Dijkstra algorithm to find the set of shortest paths from node 4 to other nodes. Iteration N D1 D2 D3 D5 D6 Initial b, Find the set of associated routing table entries (Destination, Next Hop, Cost) Destinatio n Cost Next Hop [Answer] Iteration Initial 1 2 3 4 5 N {4} { 2, 4 } { 2, 3, 4 } { 2, 3, 4, 5 } { 2, 3, 4, 5, 6 } {1, 2, 3, 4, 5, 6} D1 5 4 4 4 4 4 D2 1 1 1 1 1 1 D3 2 2 2 2 2 2 D5 3 3 3 3 3 3 D6 ∞ 3 3 3 3 3 b, Find the set of associated routing table entries (Destination, Next Hop, Cost) Destinatio n 1 2 3 5 6 Cost Next Hop 2 2 3 5 3 5 1 2 3 3 ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q14: You are a network technician assigned to install a new network for a customer. You must create multiple subnets out of the 192.168.12.0/24 network address space to meet the following requirements: - The first subnet is the LAN-A network. You need a minimum of 50 host IP addresses. - The second subnet is the LAN-B network. You need a minimum of 40 host IP addresses. - You also need at least two additional unused subnets for future network expansion. Note: Variable length subnet masks will not be used. All of the device subnet masks should be the same length. Answer the following questions to help create a subnetting scheme that meets the stated network requirements: a. How many host addresses are needed in the largest required subnet? b. What is the minimum number of subnets required? c. The network that you are tasked to subnet is 192.168.12.0/24. What is the /24 subnet mask in binary? d. The subnet mask is made up of two portions, the network portion, and the host portion. This is represented in the binary by the ones and the zeros in the subnet mask.: In the network mask, what do the ones and zeros represent? e. When you have determined which subnet mask meets all of the stated network requirements, derive each of the subnets. List the subnets from first to last in the table. Remember that the first subnet is 192.168.12.0 with the chosen subnet mask. Subnet Address Prefix Subnet Mask Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 [Answer] a, How many host addresses are needed in the largest required subnet? NNumbeof subnets required = 4 which implies that the network is divided into four parts The given IP address i.e 192.168.12.0 is a Class C IP address. Number of host addresses required = 26 - 2 = 62 → Therefore the max number of hosts possible in each subnet is 62 . b, What is the minimum number of subnets required? Two subnet are required for LAN-A and LAN-B and two subnets are needed to be left for future use. → Therefor the total number of subnets are 4 . c, The network that you are tasked to subnet is 192.168.12.0/24. What is the /24 subnet mask in binary? Subnet mask for any network is obtained by changing the net id bits to 1's and host id bits to 0's. Since the given network is a Class C network so the number of netid bits are 24 and the number of host id bits are 8 and there are two bits reserved for subnet id bits. → The subnet mask for network in binary: 11111111.11111111.11111111.11000000. d, The subnet mask is made up of two portions, the network portion, and the host portion. This is represented in the binary by the ones and the zeros in the subnet mask. 1.) In the network mask, what do the ones and zeros represent? → In the nerwork maslk the ones represent the net id bits and the zeroes represent the host id bits. 2.) When you have determined which subnet mask meets all of the stated network requirements, derive each of the subnets. List the subnets from first to last in the table. Remember that the first subnet is 192.168.12.0 with the chosen subnet mask. Subnets for LAN-A are : 1. 192.168.12.00000000 2. 192.168.12.00000001 3. 192.168.12.00000010 4. 192.168.12.00000011 Last network for the subnet will be 192.168.12.00111111 Subnets for LAN-B are : 1. 192.168.12.01000000 2. 192.168.12.01000001 3. 192.168.12.01000010 4. 192.168.12.01000011 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 ...... Last network for the subnet will be 192.168.12.01111111 e, When you have determined which subnet mask meets all of the stated network requirements, derive each of the subnets. List the subnets from first to last in the table. Remember that the first subnet is 192.168.12.0 with the chosen subnet mask. Subnet address Prefix Subnet Mark 192.168.12.0 /26 255.255.255.192 192.168.12.64 /26 255.255.255.192 192.168.12.128 /26 255.255.255.192 192.168.12.192 /26 255.255.255.192 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Q15: Suppose that Selective Repeat ARQ is modified so that ACK messages contain a list of the next m frames that it expects to receive. Solutions follow questions: a. How does the protocol need to be modified to accommodate this change? b. What is the effect of the change on protocol performance? [Answer] a) 2 things are needed to be changed: The frame header needs to be modified to recieve the list of frames and Since the reciever explicitly indicates which frames are needed to be transmitted. Change in transmitter operation is needed. If the recieved list contains m oldest frames that are yet to be recieved , then it can be used to skip retransmission of frames that have already been received. b) Performance will surely increase if the error rate is high or delay is high. A single frame can ask for the retransmission of several frames. The complexity of the protocol will surely increase relative to the unchanged Selective repeat ARQ Other two subnets are : Starting address of third subnet is : 192.168.12.10000000 Last address of the third subnet is : 192.168.12.10111111 Starting address of forth subnet is : 192.168.12.11000000 Last address of the forth subnet is : 192.168.12.11111111 ========================================================================= Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 More: 0 1 2 3 4 5 6 U U,T U,T,W UTWV UTWVX UTWVXY U 0,U - W ~ 3,U 3,U - V ~ 3,U 3,U 3,U - T ~ 2,U - Y ~ ~ 9,T 9,T 9,T 9,T - X ~ ~ ~ 9,W 6,V - Z ~ ~ ~ ~ ~ 14,X 14,X => Z -> X -> V -> U A A, C A,C,B ACBD ACBDE ACBDEG ACBDEGF A 0,A - B ~ 4,A 4,A - C ~ 3,A - D ~ ~ ~ 9,B - E ~ ~ 10,C 10,C 10,C - F ~ ~ ~ ~ 14,D 14,D 14,D - G ~ ~ ~ ~ 11, G 11,G - Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) z ~ ~ ~ ~ ~ ~ 15, G 15,G step 0 1 2 3 4 5 6 7 lOMoARcPSD|28579846 Các Dạng của Các Kỳ Trước: 1. Let g1(x) = x3 + x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits (1,1,0,1,1,1). a. Find the codeword corresponding to these information bits if g1(x) is used as the generating polynomial. b. Find the codeword corresponding to these information bits if g2(x) is used as the generating polynomial. [Answer] a, G1(x) is used as the generating polynomial G1(x) = x + 1 = 11 → n = 2 Step 1: Add n-1 bit 0 to the information bits → Information bits: 1101110 Step 2: 1101110 11 11 100101 00011 11 0010 11 01 R(x) = 01 Q(x) = 100101 Step 3: Add R(x) to the information bits → Codeword: 1101111 b, G2(x) is used as the generating polynomial G2(x) = x3 + x2 + 1 = 1101 → n=4 Step 01: Add n-1 bit 0 to the information bits → Information bits: 110111000 Step 02: 1101111000 1101 1101 100010 00001100 1101 00010 R(x) = 010 Q(x) = 100010 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Step 3: Add R(x) to the information bits → Codeword: 110111010 2. Let g1(x) = x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits (1,1,0,1,1,0). a. Find the codeword corresponding to these information bits if g1(x) is used as the generating polynomial. b. Find the codeword corresponding to these information bits if g2(x) is used as the generating polynomial. [Answer] a, G1(x) is used as the generating polynomial G1(x) = x + 1 = 11 → n=2 Step 01: Add n-1 bits 0 to the information bits → Information bits: 1101100 Step 02: 1101100 11 11 100100 00011 11 0000 00 0 R(x) = 0 Q(x) = 100100 Step 03: Add R(x) to the information bits → Codeword: 1101100 b, G2(x) is used as the generating polynomial G2(x) = x3 + x2 + 1 = 1101 → n = 4 Step 01: Add n-1 bits 0 to the information bits → Information bits: 110110000 Step 02: 110110000 1101 1101 100011 00001000 1101 01010 1101 0111 R(x) = 111 Q(x) = 100011 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Step 03: Add R(x) to the information bits → Codeword: 110110111 2. Let g1(x) = x + 1 and let g2(x) = x3 + x2 + 1. Consider the information bits (1,1,0,1,1,0). a. Find the codeword corresponding to these information bits if g1(x) is used as the generating polynomial. b. Find the codeword corresponding to these information bits if g2(x) is used as the generating polynomial. [Answer] a, G1(x) is used as the generating polynomial G1(x) = x + 1 = 11 → n = 2 Step 01: Add n-1 bits 0 to the information bits → Information bits: 1101100 Step 02: 1101100 11 11 100100 00011 11 0000 00 0 R(x) = 0 Q(x) = 100100 Step 03: Add R(x) to the information bits → Codeword: 1101100 b, G2(x) is used as the generating polynomial G2(x) = x3 + x2 + 1 = 1101 → n=4 Step 01: Add n-1 bits 0 to the information bits → Information bits: 110110000 Step 02: 110110000 1101 1101 100011 00001000 1101 01010 1101 0111 R(x) = 111 Q(x) = 100011 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Step 03: Add R(x) to the information bits → Codeword: 110110111 3. Consider the 7-bit generator, G=10111, and suppose that D has the value 1010100001. What is the value of R? Show your all steps to have result. [Answer] G = 10111 → The polynomial expression of G: x4 + x2 + x + 1 D = 1010100001 Here, the degree of the expression is 4 → r = 4. Thus, D + r → 10101000010000 10101000010000 10111 10111 1001011001 00010000 10111 0011101 10111 010100 10111 00011000 10111 01111 Therefore, the value of R = 1111 =================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 4. Consider the 7-bit generator, G=10011, and suppose that D has the value 1010101010. What is the value of R? Show your all steps to have result. [Answer] G = 10011 → The polynomial expression of G: x4 + x + 1 D = 1010101010 Here, the degree of the expression is 4 → r = 4. Thus, D + r → 10101010100000 10101010100000 10011 10011 1011011100 0011001 10011 010100 10011 0011110 10011 011010 10011 010010 10011 0000100 Therefore, the value of R = 0100 =================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 8. Let g(x)=x3+x+1. Consider the information sequence 1001. Find the codeword corresponding to the preceding information sequence. Using polynomial arithmetic we obtain [Answer] G(x)=x3+x+1 = 1011 The information sequence: 1001 Here, the degree of the expression is 3 → r = 3. Thus, the information sequence → 1001000 Using polynomial arithmetic we obtain: 1001000 1011 1011 101 001000 1011 00110 → Codeword: 1001110 ==================================================================== 9. Let g(x)=x3+x+1. Consider the information sequence 1011. Find the codeword corresponding to the preceding information sequence. Using polynomial arithmetic we obtain [Answer] G(x) = x3+x+1 = 1011 The information sequence: 1011 Here, the degree of the expression is 3 → r = 3. Thus, the information sequence → 1011000 Using polynomial arithmetic we obtain: 1011000 1011 1011 1000 0000000 → Codeword: 1011000 ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 14. Sender A wants to send 100111010011110 to receiver B. This transmission uses CRC algorithm for error detection with generator polynomial bits string is 10110. What is bits string will be transmitted on the medium. Show your all steps to have result [Answer] D: 100111010011110 Generator polynomial: 10110 → n = 5 Step 1: Add n-1 bits 0 to the bit stream → D: 1001110100111100000 Step 2: 1001110100111100000 10110 10110 101000010100010 0010110 10110 0000010011 10110 0010111 10110 000010000 10110 001100 R(x) = 1100 Step 3: Add R(x) to the information bits → Codeword: 1001110100111101100 Bits string will be transmitted on the medium is: 1001110100111101100 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 15. A bit stream 10011101 is transmitted using the standard CRC method. The generator polynomial is x3 + 1. Show the actual bit string transmitted. Suppose the third bit from the left is inverted during transmission. Show that this error is detected at the receivers end. [Answer] Bit stream: 10011101 Generator polynomial: x3+ 1 = 1001 → n = 4 Step 1: Add n-1 bits 0 to the bit stream → bit stream: 10011101000 Step 2: 10011101000 1001 1001 10001100 00001101 1001 01000 1001 000100 (Remainder) R(x) = 100 Step 3: Actual frame transmitted: 10011101000 – 100 = 10011101100 (modulo 2 subtraction) Now suppose the third bit from the left is garbled and frame is received as 10111101100. Hence on dividing this by the polynomial generator we get a remainder of 100 which shows that an error has occurred. Had the received frame been error free we would have got a remainder of zero. 10111101100 1001 1001 10101000 001011 1001 001001 1001 0000100 → Remainder indicating error = 100 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Sender A wants to send 100111010011011 to receiver B. This transmission uses CRC algorithm for error detection with generator polynomial bits string is 10111. What is bits string will be transmitted on the medium. Show your all steps to have result. [Answer] Generator polynomial is 10111 = x4 + x2 + x1 + x0 → r = 4 Step 1: Add 0000 to data bits string. It will be 1001110100110110000 Step 2: 1001110100110110000 10111 10111 101010001000110 0010010 10111 0010110 10111 000010110 10111 000011100 10111 010110 10111 000010 R(x) = 0010 Step 03: add R(x) to data bits string → bits string: 1001110100110110010 Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 5. Consider the following network Figure 1. With the indicated link costs, use Dijkstra’s shortest-path algorithm to compute the shortest path from u to all network nodes. Show how the algorithm works by computing a table. [Answer] u u, x u, x, v u, x, v, y u, x, v, y, w u - v 2, u 2, u 2,U - w 5, u 4, x 4, x 4, x - x 1, x - U -> X y ~ 2, x - z ~ ~ 5, y 5, y 5, y -> Y -> Z ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 6. A router has the following CIDR entries in its routing table: 135.46.56.0/22 Interface 0 135.46.60.0/22 Interface 1 192.53.40.0 /23 Router 1 default Router 2 a, What does the router do if a packet with an IP address 135.46.63.10 arrives? b, What does the router do if a packet with an IP address 135.46.57.14 arrives? [Answer] a, The bit pattern of 135.46.63.10 is: 10000111.00101110.00111111.00001010 Taking the first 22 bits of the above IP as network address (it is equivalent of making the last 10 bit zero), we have: 10000111.00101110.00111100.00000000 = 135.46.60.10 → It matches the network address of the 2rd row. The packet will be forwarded to interface 1. b, The bit pattern of 135.46.57.14 is 10000111.00101110.00111001.00001110 Taking the first 22 bits of the above IP as network address (it is equivalent making the last 10 bits zero), we have 10000111.00101110.00111000.00000000 = 135.46.56.0 → It matches the network address of the first row. The packet will be forwarded to interface 0. c, The bit pattern of 135.46.52.2 is 10000111.00101110.00111001.00001110 Taking the first 22 bits of the above IP address as network address, we have 135.45.52.0. It dose not matche the network addresses of the first three rows. → The packet will be forwarded to default gateway which is Router 2. d, The bit pattern of 192.53.40.7 is 11000000.00110101.00101000.00000111 Taking the first 23 bits of the above IP address as network address, we have 192.53.40.0. → It matches the network address of the third row. The packet will be forwarded to Router 1. e, The bit pattern of 192.53.56.7 is 11000000.00110101.00111000. 00000111 Taking the first 23 bits of the above IP address as network address, we have 192.53.56.0. → It does not matche the network addresses of the first three rows. The packet will be forwarded to default gateway which is Router 2. ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 7. Suppose two hosts, A and B, are separated by 30,000 kilometers and are connected by a direct link of R = 3 Mbps. Suppose the propagation speed over the link is 2.5 x 108 meters/sec. a. Calculate the bandwidth-delay product, R * dprop. b. Consider sending a file of 900,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time? [Answer] a. The distance between two hosts A and B = 30,000 km = 3 * 107 m Transmission rate(R) of the direct link between A and B = 3Mbps = 3 * 106 bps Propagation Speed(S) of the link between A and B = 2.5 * 108 meters/sec Propagation delay: dprog = band-width delay product: R * dprog = 3 * 106 * 0.12 = 360000 bits Therefore, band-with delay product is: 360000 bits b. Size of the file = 900 000 bits = 9 * 105 bits Transmission rate(R) of the direct link between A and B = 3Mbps = 3 * 106 bps The band-width delay product: R * dprog = 3 * 106 * 0.12 = 360000 bits Therefore, the maximum number of bits at a given time will be 360000 bits ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 10. A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 2,500 bytes and the link rate is 3 Mbps. What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue? [Answer] Packet length = L Transmission rate = R Currently transmitted packet = x bits Waiting queue = n packets Formula for Queuing delay: Queuing delay = Given data: L = 2500 bytes = 8 * 2500 = 20000 bits R = 3 Mbps. = 3 * 106 bps/s X = bytes = 8 * 1250 = 10000 bits n=4 Queuing delay = Thus, the queuing delay: 0.03s ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 11. Suppose a header consists of four 16-bit words: (11111111 11111110, 11111111 00000000, 11110000 11110000, 11000000 11000001). Find the Internet checksum for this code [Answer] A0 = 11111111 11111110 = 215 → 21 = 65534 A1 = 11111111 00000000 = 215 → 28 = 65280 A2 = 11110000 11110000 = 215 + 214 + 213 + 212 + 27 +26 + 25 + 24 = 61680 A3 = 11000000 11000001 = 215 + 214 + 27 + 26 + 20 = 49345 X = (A0 + A1 + A2 + A3) % 65534 = 241839 % 65534 = 45237 A4 = -X % 65534 = -45237 % 65534 = 20297 So, 20297 is 0100111101001001 binary → The Internet checksum = 01001111 01001001 ==================================================================== 8. Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000, 11110000 11110000, 11000000 11000000). Find the Internet checksum for this code [Answer] A0 = 11111111 11111111 = 215 → 20 = 65535 A1 = 11111111 00000000 = 215 → 28 = 65280 A2 = 11110000 11110000 = 215 + 214 + 213 + 212 + 27 +26 + 25 + 24 = 61680 A3 = 11000000 11000000 = 215 + 214 + 27 + 26 = 49344 X = (A0 + A1 + A2 + A3) % 65535 = 241839 % 65535 = 45234 A4 = -x % 65535 = -45234 % 65535 = 20301 So, 20301 is 0100111101001101 binary → The Internet checksum = 01001111 01001101 ==================================================================== 8. Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000, 11110000 11110000, 11000000 11000001). Find the Internet checksum for this code [Answer] A0 = 11111111 11111111 = 215 → 20 = 65535 A1 = 11111111 00000000 = 215 → 28 = 65280 A2 = 11110000 11110000 = 215 + 214 + 213 + 212 + 27 +26 + 25 + 24 = 61680 A3 = 11000000 11000001 = 215 + 214 + 27 + 26 + 20 = 49345 X = (A0 + A1 + A2 + A3) % 65535 = 241840 % 65535 = 45235 A4 = -x % 65535 = -45234 % 65535 = 20300 So, 20300 is 0100111101001100 binary → The Internet checksum = 100111101001100 9. Consider a packet of length 2,000 bytes that propagates over a link of distance 3,500 km with propagation speed of 2,5 · 108 m/s, and transmission rate 2 Mbps? Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 a. How long does the packet propagation take? b. Does this propagation delay depend on the packet length? c. Does this propagation delay depend on the transmission rate? [Answer] Length: 2000 bytes = 16000 bits Distance: 3,500 km = 35 * 105 m Propagation Speed: 2,5 * 108 m/s Rate: 2Mbps = 2 * 106 bit/s a. Transmission delay = = = 8 ms Propagation delay = = = 14 ms Therefore, the total time = 8ms + 14ms = 22ms b. No, the propagation delay only depend on D,S not L,R → the delay depend on packet length is not true. c. No, the propagation delay only depend on D,S not L,R → the delay depends on transmission rate is not true. ==================================================================== 9. Consider a packet of length 1,000 bytes that propagates over a link of distance 2,500 km with propagation speed of 2,5 · 108 m/s, and transmission rate 2 Mbps? a. How long does the packet propagation take? b. Does this propagation delay depend on the packet length? c. Does this propagation delay depend on the transmission rate? [Answer] a. Length: 1000 bytes = 8000 bits Distance: 2,500 km = 25 * 105 m Propagation Speed: 2,5 * 108 m/s Rate: 2Mbps = 2 * 106 bit/s Transmission delay = = = 4 ms Propagation delay = = = 10 ms Therefore, the total time = 4ms + 10ms =14ms b. No, the propagation delay only depend on D,S not L,R → the delay depend on packet length is not true. c. No, the propagation delay only depend on D,S not L,R → the delay depends on transmission rate is not true. 10. Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1 = 250 kbps, R2 = 3 Mbps, and R3 = 2 Mbps. a. Assuming no other traffic in the network, what is the throughput for the file transfer? Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 b. Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take to transfer the file to Host B? [Answer] a, Consider given data: R1 = 250 kbps R2 = 3 Mbps R3 = 2 Mbps The throughput for the file transfer = min {R1, R2, R3} min {250 kbps, 3 Mbps, 2 Mbps} = 250 kbps Thus, the throughput for the file transfer = 250 kbps b, Consider given data: The file size = 4 million bytes = 32000000 bits - From (a) Throughput for the file transfer=250 Kbps = 250 000 bps - Time to transfer the file to Host B = = 128 s Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 10. Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links of rates R1 = 500 kbps, R2 = 2 Mbps, and R3 = 1 Mbps. a. Assuming no other traffic in the network, what is the throughput for the file transfer? b. Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take to transfer the file to Host B? c. Repeat (a) and (b), but now with R2 reduced to 100 kbps. [Answer] a) Consider given data: R1 = 500 kbps R2 = 2 Mbps, R3 = 1 Mbps The throughput for the file transfer = min {R1, R2, R3} min {500 kbps, 2 Mbps, 1 Mbps} = 500 kbps So, the throughput for the file transfer = 500 kbps b) Consider given data: The file size = 4 million bytes = 32000000 bits - From (a), Throughput for the file transfer = 500 Kbps = 500000 bps - Time to transfer the file to Host B = = 64 s c) Repeat (a) and (b), but now with R2 reduced to 100 kbps. Consider given data: R1 = 500 kbps, R2=100 kbps, and R3 = 1 Mbps The throughput for the file transfer = min {R1, R2, R3} min {500 kbps, 100 kbps, 1 Mbps} = 100 kbps So, the throughput for the file transfer = 100 kbps - From (a), Throughput for the file transfer = 100 Kbps = 100000 bps - Time to transfer the file to Host B = = 320 s Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 11. Suppose an application layer entity wants to send an L-byte message to its peer process, using an existing TCP connection. The TCP segment consists of the message plus 20 bytes of header. The segment is encapsulated into an IP packet that has an additional 20 bytes of header. The IP packet in turn goes inside an Ethernet frame that has 18 bytes of header and trailer. What percentage of the transmitted bits in the physical layer correspond to message information, if L = 200 bytes, 1000 bytes, 2000 bytes. [Answer] TCP/IP over Ethernet allows data frames with a payload size up to 1460 bytes. Therefore, L = 200, L=1000 and L= 2000 are within this limit The message overhead includes: • TCP: 20 bytes of header • IP: 20 bytes of header • Ethernet: total 18 bytes of header and trailer. Therefore, the total message overhead is 58 bytes Therefore L = 200 bytes, ( ) * 100 = 77,5% L = 1000 byte, () * 100 = 94.5% L = 2000 bytes, () * 100 = 97,1% ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 11. Suppose an application layer entity wants to send an L-byte message to its peer process, using an existing TCP connection. The TCP segment consists of the message plus 20 bytes of header. The segment is encapsulated into an IP packet that has an additional 20 bytes of header. The IP packet in turn goes inside an Ethernet frame that has 18 bytes of header and trailer. What percentage of the transmitted bits in the physical layer correspond to message information, if L = 100 bytes, 500 bytes, 1000 bytes [Answer] TCP/IP over Ethernet allows data frames with a payload size up to 1460 bytes. Therefore, L = 100, 500 and 1000 bytes are within this limit. The message overhead includes: • TCP: 20 bytes of header • IP: 20 bytes of header • Ethernet: total 18 bytes of header and trailer The total message overhead is 58 bytes Therefore: L = 100 bytes, ( ) * 100 = 63,3% efficiency L = 500 byte, ( ) * 100 = 89,6% efficiency L = 1000 bytes, ( ) * 100 = 94,5% efficiency Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 12. Suppose the size of an uncompressed text file is 1 megabyte a. How long does it take to download the file over a 35 kilobit/second modem? b. How long does it take to take to download the file over a 1 megabit/second modem? c. Suppose data compression is applied to the text file. How much do the transmission times in parts (a) and (b) change? If we assume a maximum compression ratio of 1:6, then we have the following times for the 35 kilobit and 1 megabit lines respectively: [Answer] a, Size text file = 1 megabyte = 1 *1024 * 1024 * 8 = 8388608 bits Speed = 35 kilobit/second = 35 * 1000 = 35000 bits/s Time to download the file over = = 239.67 s b, Size text file = 1 megabyte = 1 *1024 * 1024 * 8 =8388608 bits Speed = 1 megabit/second = 1000000 bits/s Time to download the file over = = 8.38 s c, T(35k) = = 39.9 s T(1M) = = 1.39 s Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 12. Suppose the size of an uncompressed text file is 1 megabyte a. How long does it take to download the file over a 35 kilobit/second modem? b. How long does it take to take to download the file over a 1 megabit/second modem? c. Suppose data compression is applied to the text file. How much do the transmission times in parts (a) and (b) change? If we assume a maximum compression ratio of 1:6, then we have the following times for the 35 kilobit and 1 megabit lines respectively: [Answer] a, Size text file = 1 megabyte = 1 *1024 * 1024 * 8 = 8388608 bits Speed = 35 kilobit/second = 35 * 1000 = 35000 bits/s Time to download the file over = = 239.67 s b, Size text file = 1 megabyte = 1 *1024 * 1024 * 8 =8388608 bits Speed = 1 megabit/second = 1000000 bits/s Time to download the file over = = 8.38 s c, T(35k) = = 39.9 s T(1M) = = 1.39 s Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 13. Consider the three-way handshake in TCP connection setup. (a) Suppose that an old SYN segment from station A arrives at station B, requesting a TCP connection. Explain how the three-way handshake procedure ensures that the connection is rejected. (b) Now suppose that an old SYN segment from station A arrives at station B, followed a bit later by an old ACK segment from A to a SYN segment from B. Is this connection request also rejected? [Answer] a. In a three-way handshake procedure, one must ensure the selection of the initial sequence number is always unique. If station B receives an old SYN segment from A, B will acknowledge the request based on the old sequence number. When A receives the acknowledgment segment from B, A will find out that B received a wrong sequence number. A will discard the acknowledgment packet and reset the connection b. If an old SYN segment from A arrives at B, followed by an old ACK segment from A to a SYN segment from B, the connection will also be rejected. Initially, when B receives an old SYN segment, B will send a SYN segment with its own distinct sequence number set by itself. If B receives the old ACK from A, B will notify A that the connection is invalid since the old ACK sequence number does not match the sequence number previously defined by B. Therefore, the connection is rejected ==================================================================== A university has 150 LANs with 100 hosts in each LAN. (a) Suppose the university has one Class B address. Design an appropriate subnet addressing scheme. (b) Design an appropriate CIDR addressing scheme. [Answer] a) Class B's address has 14 bits for the network ID and 16 bits for the host ID. We need to decide how many bits to allocate to the subnet id versus the host id for designing an appropriate subnet addressing scheme. We can choose either 7 bits or 8 bits to identify the hosts.ideTCP b) The CIDR addressing scheme involves generating a prefix length that indicates the length of the network mask. Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Suppose that a group of computers is connected to an Ethernet LAN. If the computers communicate only with each other, does it make sense to use IP protocol in the computers? Should the computers run TCP directly over Ethernet? How is addressing handled? [Answer] IP convention can be utilized since the IP convention is a lot of necessities for tending to also as directing information on the Internet. Since the computer can't run principles-based TCP without IP, TCP uses IP addresses. There is a need to utilize an individual custom streaming convention. that depends on Ethernet or another link layer as a lower layer. =================================================================== (1) The figures below show the TCP/UDP communication pattern diagrams. Which diagram works for TCP? Why? (2) Fill the missing steps (blank boxes) in both diagrams for TCP/UDP correspondingly. [Answer] 1) Diagram (b) will work for TCP since it is a network protocol that shows the details of how data is sent as well as received. 2) Missing steps: (a) bind, connect, (b) bind>listen>accept, connect. ==================================================================== Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Suppose that two peer-to-peer processes provide a service that involves the transfer of discrete messages. Suppose that the peer processes are allowed to exchange PDUs that have a maximum size of M bytes including H bytes of header. Suppose that a PDU is not allowed to carry information from more than one message. a. Develop an approach that allows the peer processes to exchange messages of arbitrary size. b. What essential control information needs to be exchanged between the peer processes? c. Now suppose that the message transfer service provided by the peer processes is shared by several message source-destination pairs. Is additional control information required, and if so, where should it be placed? [Answer] a) To convert arbitrary sizes, large contents must be split into bytes of each length that will be transmitted in multiple PDUs. b) Peer-to-peer processes move information allowing messages to be aggregated. c) A PDU must be attached to the sequence ID, to process each sequence independently when a message is selected. This stream ID can be dodged if the source and target work for them to process a content transfer at a certain time. ==================================================================== Consider a network link that has distance of 100 meters, and signal traverses at the speed of light in cable 2.5 x 10^8 meters per second. The link has transmission bandwidth of 100 megabits/second (100 x 10^6 bits per second). The packet size is 400 bits. What is the signal propagation delay? [Answer] d = 100 megabits / second (Cách đổi đơn vị: 100 megabits / second = 100*103 kilobits / seconds, đổi ngược lại thì chia cho 1000) v = 2.5 * 108 meters / second Signal Propagation = = = 4 *10-7 s Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com) lOMoARcPSD|28579846 Consider a network link that has distance of 100 meters, and signal traverses at the speed of light in cable 2.5 x 108 meters per second. The link has transmission bandwidth of 100 megabits/second (100 x 106 bits per second). The packet size is 400 bits. What is the packet transmission delay? [Answer] L = 400 bits R = 100 *106 bit/s Packet transmission = = = 4 *10-6 s ==================================================================== Three users are sharing a common link of 1 Mb/s. User A is downloading large files and is connected to the common link via a slow access link at x Mb/s, user B is connected via a 100 Mb/s link, but the application she is using requires at most x Mb/s, and finally user C is connected via a 1 Gb/s link and is downloading a movie that can take up any amount of bandwidth available to it. What is the max-min fair allocation for these three flows at the common link? [Answer] Two Cases: Case 1: If x < Mb/s then A and B get x Mb/s, C gets (1-2x) Mb/s. Otherwise: A, B, and C get Mb/s Downloaded by Tùng Lâm Nguy?n (kudolam2k3@gmail.com)