Note: Simplified values are used in the calculation: 1g ≈; 1M≈； 1k ≈ 3
1. The application layer data with a length of 5 bits is submitted to the transport layer for processing, and a 2-byte TCP header is required. Submit it to the network layer for processing, and add a 2-byte IP header. Finally, the Ethernet transmission submitted to the data link layer needs to add 18 bytes of header and trailer. Assuming no other overhead, the transmission efficiency of this data is about (? ） 。
answer: 51.9%
analysis: data length: 5/8=62.5 bytes
transmission efficiency = 62.5/(62.5+2+2+18) = 62.5/12.5 ≈ 51.9%
2. The network prefix corresponding to the mask has (28) bits.
answer: 21.23.65.112 28
resolution IP: 21.23.65.1111
255.255.1111
21.23.65.111 = 21.23. The network prefix ***24+4=28 bits. In addition, the default mask of class C address is 24 bits, and 4 bits are borrowed, so it can be divided into 16 subnets. There are 2 4-2 = 14 valid hosts
Note: A 1... to 126... Valid .. and 127... are reserved
B 128.1.. to 191.254. Effects 192... and 223.255.255. reserve
d 224.. to 239.255.255 for multicast
e 24.. to 255.255.255. The meaning of the confirmation serial number at each end is an indication ().
answer: 3 the sequence number that should be sent next
analysis
solution: TCP three-way handshake TCP has six kinds of flags:
SYN(synchronous connection)?
ACK(acknowledgement)
? PSH(push transfer) FIN(finish)?
RST(reset)?
urge
Sequence number?
Acknowledge number
We commonly use the following three flag bits: SYN-to create a connection, FIN-to terminate a connection, and ACK- to acknowledge the received data Three-way Handshake, which means that when establishing a TCP connection, the client and the server need to send three packets in total.
4. it is assumed that TCP protocol is used to transfer files. The segment size of TCP is 1kbytes (assuming no congestion and no lost packets), and the notification window of the receiver is 1M bytes. When the sending window of slow startup reaches 32kbytes, it takes () round-trip delay (RTT).
Answer 5
Parsing slow start is a congestion control mechanism used by transmission control protocols. Slow start is also called exponential growth period. Slow start means that the TCP receive window will grow every time it receives an acknowledgement. The sender starts by sending a message segment and then waits for an ACK. When the ACK is received, the congestion window is increased from 1 to 2, that is, two segments can be sent. When the ACK of these two segments is received, the congestion window increases to 4, so after the nth time, it is? =32, then n=5, and the relationship increases exponentially.
note: the relationship between TCP buffer and window Conclusion 1 The sending window is not the window advertised by the sender, but the window advertised by the receiver. 2 Window size notified by the receiver = window size of the sender. 3 The window size notified by the receiver is also equal to its own window size (that is, the receiver's window size) 4 Sending cache >; Sender window > Sent Byte 5 Receive Buffer >: Receiving window > In the process of receiving 6TCP communication with unacknowledged byte 6, the sizes of sending buffer and receiving buffer remain unchanged, and the sending window and receiving window may change.
2. Multiple choice questions (1 point for each small question, ***5 points)
1. The Ethernet adapter of host A intercepted a frame on the line and submitted it to host A for processing. In the following statement (d), the process cannot be explained.
a. The network adapter works in promiscuous mode
b. The destination MAC address of the frame is in the same network segment as host A
c. The destination MAC address of the frame is a broadcast address
d. The destination MAC address of the frame is the multicast group address where host A is locateD
Answer d
Analyzing IP multicast (also called multicast or multicast) technology is All hosts that use the same IP multicast address to receive multicast packets form a host group, also known as multicast group. The members of a multicast group change at any time. A host can join or leave the multicast group at any time, and the number and geographical location of multicast group members are not limited. A host can also belong to several multicast groups. In addition, hosts that do not belong to a multicast group can also send packets to the multicast group.
2. use the hub for network interconnection, in the following statement (? ) is correct.
a. the data transmission rate can be different and the data link layer protocol is the same
b. the data transmission rate is the same and the data link layer protocol can be different
c. the data transmission rate and the data link layer protocol are the same
d. the data transmission rate and the data link layer protocol can be different
the answer C
the analytical bridge can interconnect two networks with different link layer protocols, different transmission media and different transmission rates.
3. in order to realize transparent transmission, PPP protocol uses (? ) method.
a. character padding
b. bit padding
c. using bit padding in asynchronous transmission; Use character padding in synchronous transmission
D. Use character padding in asynchronous transmission; Using bit padding in synchronous transmission
Answer C
The transmission of analytic frames needs to be transparent, that is, transparent transmission. Transparent transmission means that no matter what bit combination the transmitted data is, it should be able to transmit on the link. When the bit combination in the transmitted data happens to be exactly the same as some control information, appropriate measures must be taken so that the receiver will not mistake such data for some control information. This is to ensure that the data link layer transmission is transparent. Transparent transmission uses zero-bit padding method in HDLC protocol, and
the synchronous transmission link of PPP protocol also uses zero-bit padding method;
when PPP protocol is transmitted asynchronously, the BSC protocol uses the character padding method.
(1) Zero-bit padding method: six consecutive 1s will not appear between two control fields in a frame; When the sender has five consecutive 1s, it immediately fills in a ; When there are five consecutive 1s at the receiver, delete the one after them.
(2) Character filling method (end-to-end delimiter method): In this frame synchronization method, in order not to misjudge the same characters in the data information bits as the end-to-end delimiters of the frame, an escape control character (DLE STX) can be filled in the frame head of this data frame, and it ends with DLE ETX () to show the difference, thus achieving data transparency. If the DLE character appears in the data of the frame, the sender will insert a "DLE" character, and the receiver will delete the DLE character.
4. A company has assigned a Class B address, and plans to divide the internal network into 2 subnets, and 18 subnets will be added in the future, with the number of hosts in each subnet approaching 7. A feasible masking scheme is ().
A．255．255．128．?
B．255．255．248．
C．255．255．252．?
d.255.255.254.
answer C
analysis 255.255.11111. = 255.255.252.
2? ≥38 2^? ≥7? m+n=16 ? M = 6
5. In the system with digital certificate mechanism, public key cryptosystem is adopted to provide security services, in which the user's public key can be used for ().
A. Encryption and authentication
B. Decryption and authentication
C. Encryption and signature
D. Decryption and signature
Answer A
Analysis of public key function: Encryption and authentication of private key function: decryption and signature
III. Noun explanation (2 points for each small question, 4 points for * *) < p This parameter of maximum transmission unit is usually related to communication interface (network interface card, serial port, etc.).
2. Spanning tree algorithm
Answer: In Ethernet, on the one hand, a spanning tree rooted at a port of a switch can be created to avoid loops (that is, to avoid forwarding frames constantly circling in the network). Second, when the topology of Ethernet network changes, the purpose of convergence protection is achieved through spanning tree protocol. (Note: Convergence means that the topology changes and these switches recalculate a new stable tree process).
iv. Question and answer and calculation questions (***15 points) Description: Simplified values are used in the calculation: 1g ≈ 1 9; 1M≈1^6； 1K ≈ 1 3
1. (4 points) Host A transmits data frames to Host B through a network link with a bandwidth of 1Mbps, assuming that the data carried in each frame is 1kbytes and the one-way delay of the link is 15ms. If a sliding window protocol is designed so that the size of the sending window and the receiving window are the same, how many bits are needed to represent the serial number?
a: the time required to send a frame is 1 * 8 * 1 3/1 * 1 6 = .8 ms
the RTT is 15*2=3ms
if the sliding window protocol is used, the number of frames that can be sent continuously is 1+3/.8 = 376 frames 2. 376> 2 9, therefore, at least use 9-bit serial number
2. (5 points) Consider a metropolitan area network. If the average distance between the source and the destination host is 1km, the signal propagation rate in the transmission medium is 2× m/s. Try to answer the question:
① When the data transmission rate is what, the transmission delay of 2kbyte packets is equal to the round-trip propagation delay of the link?
② If the line bandwidth is 1Gbps and the line length is 2km, and the data transmission adopts stop-and-wait protocol, and a 1K-kilobyte file is transmitted, can the time required for successful transmission of the file be significantly shortened by increasing the bandwidth? Try to explain the reasons briefly.
Answer
3. (6 points) In the network shown in Figure 1, the IP address and MAC address of each port of host H_A, host H_B, router R_1 and router R_2 are expressed as (MAC address, IP address) respectively, that is, H _ a (MAC _ a, IP _ a). IP _ 11), port 12(MAC_12, IP _ 12) of R _ 1, port 21(MAC_21, IP_21) of R_2, and port 22(MAC_22, IP_22) of R _ 2. Try to answer the question:
① How many times does the host H_A need to use ARP protocol to send data to the host H_B at most? Briefly explain the reasons.
② host H_A sends data to host H_B, and tries to write out the address fields of the data link layer protocol header and the network layer protocol header in the data unit received by port 11 of router R_1 and forwarded by port 22 of router R_2 respectively.
answer: (1)*** It takes 3 times. Host A first gets the MAC of the first router through arp, the first router gets the MAC of the second router through arp, and the second router still needs to get the MAC of host B through ARP before sending IP packets to host B, ***3 times.
(2) During the whole transmission, the source and destination IP addresses of the IP datagram header will not change, as long as the source and destination MAC addresses must change across network segments (across broadcast domains), and the destination MAC address is the MAC address of the next hop.