通信网基本概念与主体结构(第二版)答案Ch2
Solutions to Chapter 2 (Note: solution to Problem 62 to be added)
1. Explain how the notion of layering and internetworking make the rapid growth of applications such as the World Wide Web possible.
Solution:
Internetworking allows many component networks each with different underlying technology and operation to work together and form one large network. As new network technologies are introduced they can be readily incorporated into the Internet. This provides the ubiquitous connectivity for applications like WWW.
The layering concept hides the specific underlying network technology from the upper layers and provides a common networking platform. Using the communication service provided by the layers below, new applications can be introduced independently and at a rapid rate.
2. (a) What universal set of communication services is provided by TCP/IP?
Solution:
The TCP/IP protocol stack provides two basic types of communications services through its two transport layer protocols: TCP provides reliable connection-oriented transfer of a byte stream; UDP provides for best-effort connectionless transfer of individual messages. TCP/IP provides with
globally unique logical addressing that enables machines connected to the Internet to access these two services. The IP addressing scheme is very scalable because of its hierarchical structure.
2. (b)How is independence from underlying network technologies achieved?
Solution:
The two basic communications services provided by TCP and UDP are built on the connectionless packet transfer service provided by the Internet Protocol (IP). Many network interfaces are defined to
support IP. The salient part of the above figure is that all of the higher layer protocols access the network interfaces through IP. This is what provides the ability to operate over multiple networks.
2. (c) What economies of scale result from (a) and (b)?
Solution:
Once a network interface for IP is defined for a given network technology, then hosts connected using the given network technology can connect to the Internet. This allows the reach of the Internet to grow rapidly, leveraging multiple coexisting networks technologies. Thus investment in new network technologies extends the reach of the Internet.
3. What difference does it make to the network layer if the underlying data link layer provides a connection-oriented service versus a connectionless service?
Solution:
If the data link layer provides a connection-oriented service to the network layer, then the network layer must precede all transfer of information with a connection setup procedure. If the connection-oriented service includes assurances that frames of information are transferred correctly and in
sequence by the data link layer, the network layer can then assume that the packets it sends to its neighbor traverse an error-free pipe.
On the other hand, if the data link layer is connectionless, then each frame is sent independently through the data link, probably in unconfirmed manner (without acknowledgments or
retransmissions). In this case the network layer cannot make assumptions about the sequencing or correctness of the packets it exchanges with its neighbors.
The Ethernet local area network provides an example of connectionless transfer of data link frames.
The transfer of frames using "Type 2" service in Logical Link Control (discussed in Chapter 6)
provides a connection-oriented data link control example.
4. Suppose transmission channels become virtually error-free. Is the data link layer still needed?
Solution:
The data link layer is still needed for framing the data and for flow control over the transmission
channel. In a multiple access medium such as a LAN, the data link layer is required to coordinate access to the shared medium among the multiple users.
5. Why is the transport layer not present inside the network?
Solution:
Some of the functions provided by the transport layer can be provided inside the networks, but other functions cannot. For example, the transport layer provides functions at the end-system to
compensate for the limitations and impairments of the network layer, in order to meet requirements
(e.g. QoS) of the upper layer. For example in TCP/IP, IP provides only best effort service. To provide
the reliable service required by some applications - that is, to compensate for the shortcomings of best effort service - TCP establishes connections and implements error control on an end-to-end basis. One can imagine that a service provider could incorporate this error control function at the edge of its network. On the other hand, one of the main purposes of the transport layer is to allow multiple processes in the end systems to share a network service. This cannot be achieved inside the network.
6. Which OSI layer is responsible for the following?
(a)Determining the best path to route packets.
The network layer is concerned with the selection of paths across the network.
(b)Providing end-to-end communications with reliable service.
The transport layer is concerned with providing reliable service on an end-to-end basis across the network.
(c)Providing node-to-node communications with reliable service.
The data link layer provides for the reliable transfer of information between adjacent nodes in a
network.
7. Should connection establishment be a confirmed service or an unconfirmed service? What about data transfer in a connection-oriented service? Connection release?
Solution:
In general, the establishment of a connection needs to be confirmed before information transfer can commence across a connection. Therefore connection establishment should be a confirmed service.
A connection-oriented service is usually reliable so confirmation of data delivery is not necessary. In
certain situations, however, it is possible that the transfer across a connection is not reliable; in this case confirmation of correct data transfer may be required.
In general it is desirable that the release of a connection be confirmed by the parties involved. We will see in Chapter 8, section 5, that sometimes it is not easy to confirm that a connection has been
closed. Consequently, many protocols attempt to confirm the closing of a connection several times, and then give up and simply stop transmitting.
8. Does it make sense for a network to provide a confirmed, connectionless packet transfer service?
Solution:
Yes. Connectionless packet transfer is often unreliable, that is, packets may be lost or discarded inside a network. Certain applications, for example, signaling in connection setup, require
confirmation to acknowledge the receipt of packets.
9. Explain how the notion of multiplexing can be applied at the data link, network, and transport layers. Draw a
figure that shows the flow of PDUs in each multiplexing scheme.
Solution:
Transport Layer: Multiple application layers processes can share the service provided by UDP.
When a UDP PDU arrives from the network layer, the destination port number in the PDU is used to deliver the SDU to the appropriate application layer process. Multiple application layer processes also share the service provided by TCP. In this case, when a TCP segment arrives, the TCP
connection ID, consisting of (source port #, source IP address, destination port #, destination IP address), is used to determine which application process to deliver the SDU to.
Network Layer: The packet transfer service provided by IP can be used by all transport layers
operating in a machine. Each transfer layer passes SDUs to the IP layer which prepares IP packets with appropriate source and destination IP addresses for transfer across the Internet. Upon receiving an IP packet, a machine examines the protocol type field to determine which transport layer service to deliver the SDU to. We can also view all transport layer PDUs as sharing the IP packet transfer service between a source machine and a destination machine.
Data Link Layer: Network layer packets from different protocols (IP, IPX, Appletalk, etc) can share a data link (such as PPP or Ethernet). We can also view packet flows that traverse a data link
between two routers as sharing the link.
10. Give two features that the data link layer and transport layer have in common. Give two features in which they differ. Hint: Compare what can go wrong to the PDUs that are handled by these layers.
Solution:
Features they have in common:
?Both layers can provide recovery from transmission errors.
?Both layers can provide flow control.
?Both layers can support multiplexing.
Features in which they differ:
?The transport layer is end to end and involves the interaction of peer processes across the network. The data link layer involves the interaction of peer-to-peer processes that are
connected directly. In general, the time that elapses in traversing a data link is much smaller than the time traversing a network, where packets can become trapped in temporary routing loops.
Consequently, transport layer protocols must be able to deal with out-of-sequence PDUs and a
much larger backlog of PDUs than data link layers.
?The data link layer is concerned with framing and the transport layer is not.
?The data link layer may be concerned with medium access control, the transport layer does not have this concern.
11(a). Can a connection-oriented, reliable message transfer service be provided across a connectionless packet network? Explain.
Solution:
Yes. To provide a connection-oriented service, the transport layer can establish a logical connection across the connectionless packet network by setting up state information (for example, packet
sequence number) at the end systems. During the connection setup, the message is broken into separate packets, and each packet is assigned a sequence number.
Using the sequence numbers, the end-system transport-layer entities can acknowledge received packets, determine and retransmit lost packets, delete duplicate packets, and rearrange out-of-order packets. The original message is reassembled as packets arrive at the receiving end.
For example, TCP provides a connection-oriented reliable transfer service over IP, a connectionless packet transfer service.
11b. Can a connectionless datagram transfer service be provided across a connection-oriented network?
Solution:
Yes. The connectionless datagram transfer service can be implemented by simply setting up a
connection across the network each time a datagram needs to be transferred. Alternatively, all nodes can have permanent connections to a “connectionless server” that has the function of relaying
datagrams in connectionless fashion.
12. An internet path between two hosts involves a hop across network A, a packet-switching network, to a router and then another hop across packet-switching network B. Suppose that packet switching network A carries the packet between the first host and the router over a two-hop path involving one intermediate packet switch. Suppose also that the second network is an Ethernet LAN. Sketch the sequence of IP and non-IP packets and frames that are generated as an IP packet goes from host 1 to host 2.
Solution:
The IP layer in Host 1 generates an IP packet addressed to the destination host on the destination network and sends it to the router. The network interface in the host encapsulates the IP packet into the packet PDU used by network A. This packet is encapsulated in a frame that traverses data link 1 to the packet switch. The packet is recovered and then forwarded inside a frame along data link 2.
The data link at the router recovers the Network A packet, and the IP network interface at the router recovers the IP packet and determines that the next hop is on Network B. The router encapsulates the IP packet into an Ethernet frame, puts the host 2 Ethernet physical address in the frame and sends it to the LAN. The Ethernet card on the host captures the frame and extracts the IP packet and passes it to the host.
13. Does Ethernet provide connection-oriented or connectionless service?
Solution:
Ethernet provides connectionless transfer service of information frames.
14. Ethernet is a LAN so it is placed in the data link layer of the OSI reference model.
(a)How is the transfer of frames in Ethernet similar to the transfer of frames across a wire? How is it
different?
The transfer of frames in Ethernet occurs directly over a transmission medium and in this sense is similar to direct transmission over a wire. The sequence of frames into Ethernet arrive in the
same order they are transmitted. However multiple stations can transmit in Ethernet which differs from direct transmission over a wire.
(b)How is the transfer of frames in Ethernet similar to the transfer of frames in a packet-switching network?
How is it different?
Ethernet supports the transfer of frames among multiple end systems and in this sense is similar to a packet switching network. Ethernet does not involve routing which is a feature of packet
switching. Ethernet depends on broadcasting and/or bridging which differs from packet networks.
15. Suppose that a group of workstations is connected to an Ethernet LAN. If the workstations communicate only with each other, does it make sense to use IP in the workstations? Should the workstations run TCP directly over Ethernet? How is addressing handled?
Solution:
Ethernet supports the exchange of frames between stations and can support the direct exchange of information. Using Ethernet without IP would result in an inflexible and difficult to manage
system. Ethernet addresses are fixed and tables need to be changed whenever a machine is
moved, while IP addresses are logical and can be changed whenever a machine is moved. A
TCP connection uses the IP addresses in its connection ID so Ethernet addresses could not be
used.
16. Suppose two Ethernet LANs are interconnected by a box that operates as follows. The box has a table that tells it the physical addresses of the machines in each LAN. The box listens to frame transmissions on each LAN. If a frame is destined to a station at the other LAN, the box retransmits the frame onto the other LAN, otherwise the box does nothing.
Solutions follow questions:
a.Is the resulting network still a LAN? Does it belong in the data link layer or the network layer?
The resulting network is a local area network that has been extended. The extended LAN transfers frames, and so it still belongs in the data link layer.
b.Can the approach be extended to connect more than two LANs? If so, what problems arise as the number of
LANs becomes large?
Yes, more than two LANs can be connected using the above approach to form an extended LAN. As the number of LANs becomes large, the number of physical addresses stored in the bridge grows and becomes unmanageable. Each time a machine is added the addresses in all the boxes need to be updated. Serious problems arise if boxes are connected so that loops can occur.
17. Suppose all laptops in a large city are to communicate using radio transmissions from a high antenna tower. Is the data link layer or network layer more appropriate for this situation?
Solution:
The data link layer is concerned with the transfer of frames of information across a single hop. The network layer involves the transfer of information across a network using multiple hops per path in general. The connection from a radio antenna to the laptops is direct, and thus a data link layer
protocol is more suitable for this situation.
Now suppose the city is covered by a large number of small antennas covering smaller areas. Which layer is more appropriate?
A number of areas each covered by small antennas can be interconnected using the "bridging"
approach of problem 16, which remains in the data link layer. However, the network layer may be more appropriate because it provides for the transfer of data in the form of packets across the
communication network. A key aspect of this transfer is the routing of the packets from the source machine to the destination machine, typically traversing a number of transmission link and network nodes where routing is carried out.
18. Suppose that a host is connected to a connection-oriented packet-switching network and that it transmits a packet to a server along a path that traverses two packet switches. Suppose that each hop in the path involves a point-to-point link, that is, a wire. Show the sequence of network layer and data link layer PDUs that are generated as the packet travels from the host to the server.
Solution:
Assume that a network connection has already been set up between the host machine and the
network machine. When the host generates an IP packet for transfer to the server, the IP packet will be transferred using the network connection as follows.
?The IP packet is encapsulated into a network packet that has a connection ID in its header.
The packet may then be encapsulated into a frame that traverses data link 1 and arrives at
switch 1.
?The network packet is recovered from the data link 1 frame. The connection ID in the packet is used to determine the outgoing port from switch 1. The connection ID may need to be
mapped into a corresponding connection ID over data link 2. The packet is encapsulated into
a frame that traverses data link 2.
?The network packet is recovered from the data link 2 frame. The connection ID in the packet determines the outgoing port from switch 1 and the next connection ID. The packet is
encapsulated into a frame that traverses data link 3.
?The network packet is recovered from the data link 3 frame. The connection ID in the arriving packet indicates that this is the destination node. The IP packet is recovered.
The connection-oriented network in this example could correspond to ATM or to frame relay.
19. 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?
Solution:
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.
Therefore
L = 100 bytes, 100/158 = 63% efficiency.
L = 500 bytes, 500/558 = 90% efficiency.
L = 1000 bytes, 1000/1058 = 95% efficiency.
20. Suppose that the TCP entity receives a 1.5 megabyte file from the application layer and that the IP layer is willing to carry blocks of maximum size 1500 bytes. Calculate the amount of overhead incurred from segmenting the file into packet-sized units.
Solution:
1500 - 20 -20 = 1460 bytes
1.5 Mbyte / 1460 byte = 1027.4, therefore 1028 blocks are needed to transfer the file.
Overhead = ((1028 x 1500 - 1.5M)/1.5M) x 100 = 2.8%
21. Suppose a TCP entity receives a digital voice stream from the application layer. The voice stream arrives at a rate of 8000 bytes/second. Suppose that TCP arranges bytes into block sizes that result in a total TCP and IP header overhead of 50 percent. How much delay is incurred by the first byte in each block?
Solution:
Assume the stream is segmented as shown below, where the white cells represent data and the shaded cells represent the TCP header overhead.
Therefore, block size = 80 bytes and the payload size = 40 bytes.
Assume zero processing delay due to data arrangement and segmenting.
The delay incurred by the first byte of each block = 40/8000 = 0.5 ms.
22. How does the network layer in a connection-oriented packet-switching network differ from the network layer in
a connectionless packet-switching network?
Solution:
The network layer in connection-oriented networks maintains state information about every
connection. It can allocate resources at the switches through admission control. The network layer in connectionless networks has no knowledge of "connections", and instead deals independently with each packet.
The network layer in connection-oriented networks performs routing on a per connection basis. Each packet is routed based on a connection identifier of some sort and packets of the same connection have the same identifier value. In a connectionless network, routing is performed on per packet basis;
each packet is routed independently based on information carried in the packet header, for example, the destination address.
In connection-oriented networks, the network layer forwarding table is set up by a signaling procedure during the connection establishment. In connectionless networks, the routers may execute a
distributed algorithm to share network state information and dynamically calculate the routing table continuously.
In case of failure, the connection must be re-established in connection-oriented networks, whereas in connectionless networks, the packets are re-routed. The network layer in connectionless networks is more robust against failures.
Summary of differences:
Connection-oriented Connectionless Maintain state information about every connection No knowledge of the "connection"
Allocate resources to connections at switches No resource allocation
Admission control No admission control
Per connection routing Per packet routing
Route packet based on identifier Route packet based on destination address.
Forwarding table specifies the output port and outgoing identifier value as function of the incoming identifier value Routing table specifies the output port depending on the destination address
Forwarding table set up by signaling during connection establishment.Router executes distributed algorithm to share network state information and dynamically calculate the routing table
Connection must be re-established in cases of failure Packets are rerouted around failures, robust against failures
23. Identify session layer and presentation layer functions in the HTTP protocol.
Solution:
Presentation layer functions:
The request message and the response message headers include information about the content type of the documents (e.g. text/html, image/gif).
Session layer functions:
The HTTP protocol defines the client/server interaction in three steps:
1. Client sends the request for a file
2. Server replies with the file or error message if file is not found.
3. Server closes the TCP connection after some timeout period.
24. Suppose we need a communication service to transmit real-time voice over the Internet. What features of TCP and what features of UDP are appropriate?
Solution:
TCP is desirable in that it provides a connection for the transfer of a stream of information, which characterizes a digital voice stream. However, to provide reliable service TCP uses acknowledgments and retransmissions that result in packet delay and jitter that can not be tolerated by real-time traffic.
UDP provides connectionless service and delivers packets quickly. In case of packet loss, UDP does not provide retransmission, but some degree of packet loss can be tolerated by voice.
25. Consider the end-to-end IP packet transfer examples in Figure 2.15. Sketch the sequences of IP packets and Ethernet and PPP frames that are generated by the three examples of packet transfers: from the workstation to the server; from the server to the PC, and from the PC to the server. Include all relevant header information in the sketch.
Solution:
Workstation to Server:
IP packet header
Ethernet Frame
The Ethernet frame is broadcast over the LAN. The server's NIC card recognizes that the frame is intended for its host, so it captures the frame and examines it. It finds that the protocol type is set to IP, so it passes the IP datagram up to the IP entity.
Server to PC:
IP packet header
Ethernet Frame
The Ethernet frame is broadcast over the LAN. The router examines frame and passes IP
datagram to its IP entity which discover that the IP datagram is not for itself, but is to be routed on. The routing tables at the router show that the machine with address (2,2) is connected
directly on the other side of the point-to-point link. The router encapsulates the IP datagram in a PPP frame.
IP packet header
PPP Frame
The PPP receiver at the PC receives the frame, checks the protocol type field and passes the IP datagram to its IP entity.
PC to Server:
The PC IP entity generates the IP packet shown below. The PPP transmitter at the PC encapsulates the IP packet into a PPP frame sends it to the point-to-point link. There's no need for a physical address specification
IP datagram
IP packet header
PPP Frame
The router examines the PPP frame and passes the IP datagram to its IP entity which discovers
that the IP datagram is not for itself, but is to be routed on. The routing table at the router shows
that the machine with address (1,1) is connected in the other side of the Ethernet network. The
router then encapsulates the IP datagram into an Ethernet frame that is broadcast in the LAN.
The server's NIC card recognizes that the frame is intended for its host, so it captures the frame
and examines it. It finds that the protocol type is set to IP, so it passes the IP datagram up to the IP entity.
26. Suppose a user has two browser applications active at the same time, and suppose that the two applications are accessing the same server to retrieve HTTP documents at the same time. How does the server tell the difference between the two applications?
Solution:
A client application generates an ephemeral port number for every TCP connection it sets up. An
HTTP request connection is uniquely specified by the five parameters: (TCP, client IP address,
ephemeral port #, server IP address, 80). The two applications in the above situations will have
different ephemeral port #s and will thus be distinguishable to the server.
27. Consider the operation of non-persistent HTTP and persistent HTTP.
(a)In non-persistent HTTP (version 1.0): Each client-server interaction involves setting up a TCP connection,
carrying out the HTTP exchange, and closing the TCP connection. Let T be the time that elapses from
when a packet is sent from client to server to when the response is received. Find the rate at which HTTP
exchanges can be made using non-persistent HTTP.
(b)In persistent HTTP (version 1.1) the TCP connection is kept alive. Find the rate at which HTTP exchanges
can be made if the client cannot send an additional request until it receives a response for each request.
(c)Repeat part (b) if the client is allowed to pipeline requests, that is, it does not have to wait for a response
before sending a new request.
Solution:
(a) Each HTTP exchange involves: 1. a three-way handshake to set up the TCP connection; 2. an
HTTP request-response interaction; and 3. a TCP close. The client can send its request after the first two handshakes in part 1 (which takes up T seconds). The request and response then take an additional T second. A new request can be initiated with an associated new TCP connection even while the previous TCP connection is being closes. Thus a maximum of one HTTP exchange per 2T seconds is possible.
(b) Since each exchange is completed in T seconds, after the connection is setup, the exchange rate
is 1/T.
(c) The rate depends on how long it takes to send a request and how late it takes to compose a
response. Considering the maximum of these to be t seconds, exchange rate can be up to 1/t.
28. What is the difference between a physical address, a network address, and a domain name?
Solution:
The physical address is the unique hardware address that identifies an interface of a machine on a physical network such as a LAN. Physical addresses are used in the data link layer.
A network address is a machine's logical address on a network. The network address is used in the
network layer. The network address used on the Internet is the IP address.
Domain names are used as an aid to identify hosts and networks in the Internet, since names are easier to remember than numbers. The DNS system is used to translate between domain names and IP addresses. The domain name for the network address 128.100.132.30 is https://www.sodocs.net/doc/8010749395.html,.
29. Explain how a DNS query proceeds if the local name server does not have the IP address for a given host when the following approaches are used. Assume an example where four machines are involved in ultimately resolving a given query.
(a)When a machine B cannot resolve an address in response to a query from A, machine B sends the query to
another machine in the chain. When B receives the response, it forwards the result to B.
(b)When a machine B cannot resolve an address in response to a query from A, machine B sends a DNS reply
to A with the IP address of the next machine in the chain, and machine A contacts that machine.
Solution:
(a) Host A sends a query to a name server B. B cannot resolve an address, therefore sends the query
to C. C cannot resolve an address either, and send the query to D. Similarly, D cannot resolve an address and sends the query to E, where finally an address is resolved and returned to D. D replies the address to C, C replies it to B, and finally B passes it to the host. In this scenario each server should remember the state of the query and its source.
(b) Host A sends a query to name server B. B cannot resolve an address, replies to A with the IP
address of C. Host A send a query to C this time. C cannot resolve an address, and replies to A with the IP address of D. A sends a query to D. D cannot resolve an address, and replies with the IP address of E. A sends a query to E, E finally resolves an address and returns it to A. In this scenario
the servers do not need to remember the queries.
30. Suppose that the DNS system used a single centralized database to handle all queries. Compare this centralized approach to the distributed approach in terms of reliability, throughput (volume of queries/second that can be processed), query response delay, and maintainability.
Solution:
The centralized approach suffers from scalability, reliability, and security issues. Given the central role of DNS, a centralized system needs to be ultra-reliable whereas reliability is provided by
redundancy in the distributed approach. Maintainability is easier in the distributed approach because individual servers can be brought down without interrupting service. The response delay will increase dramatically in the centralized approach beyond a certain level of traffic whereas the distributed approach can be scaled up by adding more servers. Security is a serious issue in both approaches but the centralized system is more vulnerable to complete service interruption.
31. What is wrong with the following methods of assigning host id addresses?
Solutions follow questions:
(a)Copy the address from the machine in the next office.
There is an address conflict. The host id must be unique to each machine.
(b)Modify the address from the machine in the next office.
The resulting address may be an existing address and result in address conflict, or the address may not be recognizable by the routers.
(c)Use an example from the vendor's brochure.
The address has different network and subnetwork ids, and is not recognized by the routers.
32. Suppose a machine is attached to several physical networks. Why does it need a different IP address for each attachment?
Solution:
The IP address dictates through which network the packets are sent to and from the machine.
Therefore each network connection must have a different address. The use of the IP address for the two attachments could also confuse nearby routers and introduce routing loops.
33. Suppose a computer is moved from one department to another. Does the physical address need to change? Does the IP address need to change? Does it make a difference if the computer is a laptop?
Solution:
The physical address does not change. It is globally unique to the computer's NIC card.
The IP address may need to be changed to reflect a new subnetwork id and host id.
The situation is the same for laptops.
34. Suppose the population of the world is 6 billion, and that there is an average of 1000 communicating devices per person. How many bits are required to assign a unique host address to each communicating device? Suppose that each device attaches to a single network and that each network on average has 10000 devices. How many bits are required to provide unique network ids to each network?
Solution:
log2 (6 x 109 x 103) = 42.44
? 43 bits are required to assign a unique host address to each communicating device.
log2 ((6 x 109 x 103) / 10,000) = 29.2
? 30 bits are required to provide unique network ids to each network.
35. Can IP be used to run a homogeneous packet-switching network, that is, a network with identical packet switches interconnected with point-to-point links?
Solution:
Yes. For a homogeneous packet-switching network, the network interface function in each switch will be the same and will operate over the point-to-point links.
36. Is it possible to build a homogeneous packet-switching network with Ethernet LANs interconnecting the packet switches? If so, can connection-oriented service be provided over such a network?
Solution:
A homogeneous packet-switching network can be built where Ethernet LANs are used to interconnect
packet switches. In the most common example the packet switches are routers running IP.
A connection-oriented service can be provided over such a packet-switching network in several ways.
If the packet-switching network operates in connectionless manner, then additional functions can be added at the ingress and egress to the network to provide a connection-oriented transfer service.
Alternatively, the packet-switching network itself could be designed to operate in connection-oriented fashion. In this case the packet switches might use a layer above Ethernet to ensure reliable and sequenced transfer of frames between packet switches. Such a packet-switching network can provide connection-oriented service.
37. In telephone networks one basic network is used to provide worldwide communications. In the Internet a multiplicity of networks are interconnected to provide global connectivity. Compare these two approaches, namely, a single network versus an internetwork, in terms of the range of services that can be provided and the cost of establishing a worldwide network.
Solution:
At a national level, the telephone network has a hierarchical structure. Each new telephone line
should be connected to a local telephone center which is hierarchically connected to other telephone centers in different levels such that full connectivity can be achieved. At an international level
national networks are interconnected through gateways. Scaling this structure is costly, primarily because the network must keep track of every single connection.
In the case of internetworking, new networks can be connected to existing networks through routers.
The Internet can be scaled up by building larger routers that use higher speed links. The
connectionless nature of the Internet keeps the operation of these routers simpler and hence more readily scalable.
The Internet can support existing and new services through the well-defined TCP and UDP
communications services. The telephone network on the other hand, provides a limited number of services and new services cannot easily be introduced.
38. Consider an internetwork architecture that is defined using gateways/routers to communicate across networks but that uses a connection-oriented approach to packet switching. What functionality is required in the routers? Are there any additional constraints imposed on the underlying networks?
Solution:
The routers must be able to setup and release connections across the internetwork. A connection must be established so that routers can forward packets along a path in the network. The underlying networks may or may not operate in connection-oriented fashion. Therefore it is still possible that packets may get out of sequence while traversing a given network. If we require that packets always traverse the end-to-end path in order, then either the underlying networks must be connection-
oriented or protocols must operate above each network to ensure sequenced transfer of information.
39. The internet below consists of three LANs interconnected by two routers. Assume that the hosts and routers have the IP addresses as shown.
network 3
(a)Suppose that all traffic from network 3 that is destined to H1 is to be routed directly through router R2, and
all other traffic from network 3 is to go to network 2. What routing table entries should be present in the
network 3 hosts and in R2?
H5H6R2
Destination Next hop Destination Next hop Destination Next hop
Default(3,1)default(3,1)(1,2)(1,4)
(1,0)(2,1)
(2,0)(2,4)
(3,0)(3,1)
(b)Suppose that all traffic from network 1 to network 3 is to be routed directly through R2. What routing table
entries should be present in the network 1 hosts and in R2?
R2R1H1H2
Destination Next hop Destination Next hop Destination Next hop Destination Next hop (1,0) (2,1)(1,0)(1,1)(1,0)(1,2)(1,0)(1,3)
(2,0)(2,4)(2,0)(2,1)(2,0)(1,1)(2,0)(1,1)
(3,0)(3,1)(3,0)(2,4)(3,0)(1,4)(3,0)(1,4)
40. Explain why it is useful for application layer programs to have a "well-known" TCP port number?
Solution:
The TCP layer entity uses the port number to determine which application program the packets
belong to. In the TCP connection setup process it is very convenient to have a unique well-known port number, otherwise some protocol or procedure would be required to find the desired number.
41. Use a web browser to connect to https://www.sodocs.net/doc/8010749395.html,. Explain what layers in the protocol stack are involved in the delivery of the video newscast.
Solution:
The delivery of a video newscast over the Internet involves the transfer of a long stream of
information without assurance of delivery or protection from data loss. The current practice is to use TCP to send the video stream. This can be checked by doing an Ethereal packet capture while accessing a video clip. By observing the video display window it is apparent that some sort of
protocol particular to video streaming is in operation. After the connection request, the video display application buffers a certain amount of information before initiating display. This buffering is done in an attempt to ensure a steady supply of information to feed the audio and video decoder. Running out of information would result in a freezing of the picture image and loss of the audio signal. The
protocols used in video streaming are discussed in Chapter 10.
42. Use a web browser to connect to an audio program, say https://www.sodocs.net/doc/8010749395.html,/radio/ (Rolling Stone Radio) or www.cbc.ca (CBC Radio). Explain what layers in the protocol stack are involved here. How does this situation differ from the delivery of video in problem 41?
Solution:
The delivery of audio information is quite similar to that of video information. A significant difference is that the volume of information that has to be transferred for audio is much less than that required by video. In addition the video application must be concerned with the synchronization of the display of audio and video information, otherwise "lip synch" will not be achieved. For this reason, many
newspaper web sites prefer to combine images of a speaker along with audio commentary instead of full-blown video.
43. Which of the TCP/IP transport protocol (UDP or TCP) would you select for the following applications: packet voice, file transfer, remote login, multicast communication (i.e., multiple destinations).
Solution:
Packet Voice - This example involves the transfer of a stream of information in real time across the network. At first, it may appear that TCP is suitable because of its connection orientation. However the acknowledgment and retransmission mechanisms in TCP introduce too much delay in the transfer of packets, and so UDP is the preferred approach to transferring a real-time voice stream across the network.
File Transfer - In general, file transfer requires reliable transfer and so TCP is preferred.
Remote Login - TCP is preferable because it provides for the reliable transfer of the stream of
keystrokes that forms the basis for a remote login application.
Multicast Communication - In multicast services, a source sends information to a subset of
destinations attached to the network. It is easy to imagine multicast applications that require reliable transfer of a stream of information to a set of destinations, and multicast applications that require only best effort transfer of individual messages. Therefore neither TCP nor UDP is preferred. A more pertinent point is that providing reliable multicast stream transfer service is quite difficult to implement, and TCP is not designed for this.
44. (a) Use the Telnet program to send an e-mail by directly interacting with your local mail server. The SMTP server has port 25. You can find the list of commands for the SMTP protocol in RFC 2821, which can be downloaded from https://www.sodocs.net/doc/8010749395.html,.
Solution:
telnet
(follow Table 2.3 to send the e-mail)
(b) Use Ethereal to capture and analyze the sequence of messages exchanged. Identify the various types of
addresses for Ethernet, IP, and TCP PDUs. Examine the data in the Telnet messages to determine whether the login name and password are encrypted.
Solution:
The following screen capture shows some of the SMTP messages exchanged in the sending of email. The sequence begins with the TCP connection setup to the SMTP server. This is followed by authentication, and then by a command to send an email.
45. (a)Use the Telnet program to retrieve an e-mail by directly from your local mail server. The POP3 server has port 110. You can find the list of commands for the POP3 protocol in RFC 1939, which can be downloaded from https://www.sodocs.net/doc/8010749395.html,.
Solution:
Many servers do not allow access to email using telnet. The Ethereal screen capture below shows the first few lines in the interaction with a POP3 server to retrieve email. The sequence of frames is
followed by an authentication phase with an exchange of user name and password.
《通信网》作业答案
思考题一 1（ok）构成现代通信网的结构和要素有哪些？它们各自完成的功能有哪些？ 它们之间的相互协调通信通过什么机制来实现？ 现代通信网络的三大组成部分：传输、交换和终端设备，其发展是和这些通信设备、电子器件、计算机技术的发展紧密相关的。 通信网构成要素 实际的通信网是由软件和硬件按特定的方式构成的通信系统，从硬件构成来看：通信网由终端节点、交换节点、业务节点和传输系统构成，完成接入、交换和传输；软件设施包括了信令、协议、控制、管理、计费等，完成网络的控制、管理、运营和维护、实现通信网的智能化。 上述的网络在传输信息的类型、方式、所提供的服务的种类等方面各不相同，但它们在网络结构、基本功能、实现原理上都是相似的，它们都实现了以下四种功能： （1）信息传送 它是通信网的基本任务，传送的信息有三大类：用户信息、信令信息、管理信息，信息传输主要由交换节点、传输系统来完成。 （2）信息处理 网络对信息的处理方式对最终用户是不可见的，主要目的是增强通信的有效性、可靠性和安全性。 （3）信令机制 它是通信网上任意两个通信实体间为实现某一通信任务，进行控制信息交换的机制，如NO.7信令、TCP/IP协议等。 （4）网络管理 它负责网络的运营管理、维护管理、资源管理，以保证网络在正常和故障情况下的服务质量。是整个网络中最具有智能的部分，已形成的网络管理标准有：电信管理网标准TMN系列，计算机网络管理标准SNMP等。
2（ok）在通信网中，交换节点主要完成哪些功能？分组交换与电路交换的各自方式和特点？ （1）电路交换（Circuit Switching） ITU定义为：“根据请求，从一套入口和出口中，建立起一条为传输信息而从指定入口到指定出口的连接”。电路交换是一种电路间的实时交换，所谓实时，是指任意用户呼叫另一用户时，应立即在两用户之间建立通信电路的连接，这时通信网内的相关设备和线路都被这一对用户占用着，不能再为其他用户服务，这种在一次呼叫中由通信网根据用户要求在指定的呼叫路由上固定分配设备的交换方式，称之为电路交换方式。 电路交换的主要特点：话音或数据的传输时延小且无抖动，“透明”传输。无需存储、分析和处理、传输效率比较高；但是，电路的接续时间较长，电路资源被通信双方独占，电路利用率低。 （2）分组交换（Packet Switching） 分组交换也称包交换，它将用户的一整份报文分割成若干数据块，即分组。 分组交换是一种综合电路交换和报文交换的优点而又尽量避免两者的缺点的第三种交换方式。它的基本原理是“存储——转发”，是以更短的、被规格化了的“分组”为单位进行交换、传输。 分组交换相对于电路交换的方式来说，具有高效、灵活、迅速、可靠等特点。
主体结构试题及答案
试题 姓名：分数 一、填空题（每空5分，共25分） 1.粘贴、喷抹质量检验的取样，应符合下列规定：梁、柱类构件以同规格、同型号的构件为一个检验批。每批构件随机抽取的受检构件应按该批构件总数的 10%确定，但不得少于3根；以每根受检构件为一检验组；每组3个检验点。板、墙类构件应以同种类、同规格个构件为一个检验批，每批按实际粘贴、喷抹的加固材料表面积（不论粘贴的层数）均划分为若干区，每区 100m2，且每一楼层不得少于1区；以每区为一个检验组，每组 3个检验点。 2.植筋锚固件质量的非破损检验，对重要结构构件及生命线工程的非结构构件，应取每一检验批植筋总数的 3%且不少于5件进行检验。对一般结构构件，应取每一检验批植筋总数的 1%且不少于3件进行检验；对非生命线工程的非结构构件，应取每一检验批锚固件总数的 0.1%且不得少于3件进行检验。 3.采用水泥基胶粘贴外墙饰面时，可按胶粘剂使用说明书的规定时间或在粘贴外墙饰面砖14d及以后进行饰面砖粘结强度检验。粘贴后28d以内达不到标准或有争议时，应以28～60d内约定时间检验的粘结强度为准。 二、选择（每题5分，共15分） 1.后锚固件应进行抗拔承载力现场非破损检验，满足下列条件（ B ）时，还应进行破坏性检验： A、安全等级为甲级的后锚固构件 B、悬挑结构和构件 C、重要结构构件 D、非结构构件
2.后锚固承载力检测，进行非破损检验时，荷载检验值应取 A、0.9?yk A s B、0.8N RK,* C、0.9?yk D、0.9?yk A s和0.8N RK,*的较小值 3、监理单位应从粘贴外墙饰面砖的施工人员中随机抽选一人，在每种类型的基层上应各粘结至少（ D ）饰面砖样板件，每种类型的样板件应各制取一组（）饰面砖粘结强度试样。 A、 3㎡、1个 B、1㎡、1个 C、3㎡、3个 D、1㎡、3个 三、解答题（每题20分，共60分） 1.混凝土后锚固承载力抗拉拔试验的适用范围？ 适用于以钢筋混凝土、预应力混凝土以及素混凝土为基材的后锚固连接的设计、施工与验收；不适用于以砌体、轻骨料混凝土及特种混凝土为基材的后锚固连接。 2.加固材料粘贴、喷抹质量按组检验结果的合格评定，应符合什么规定？ 1）当组内每一试样的正拉粘结强度fti均达到fti≥1.5MPa，且为混凝土内聚破坏的要求时，应评定该组为检验合格组； 2）若组内仅一个试样达不到上述要求，允许以加倍试样重新作一组检验，如检验结果全数达到要求，则仍可评定该组为检验合格组； 3）若重作试验中，仍有一个试样达不到要求，则应评定该组为检验合格组。 3.简述外墙饰面砖粘结强度现场检测的取样数量。 答：现场粘贴饰面砖粘结强度检验应以每1000mm2同类墙体饰面砖为一个检验批，不足1000mm2应按1000mm2计，每批应取一组3个试样，每相邻的三个楼层应至少取一组试样，试样应随机抽取，取样间距不得小于500mm。
外贸函电翻译参考答案修订稿
外贸函电翻译参考答案 WEIHUA system office room 【WEIHUA 16H-WEIHUA WEIHUA8Q8-
Lesson 2 1. We avail ourselves of this opportunity to introduce to you as a foreign-invested corporation specializing in arts and crafts. 2. We want to acquaint ourselves with the supply position of steel products。 3 We are enclosing a cope of pricelist. have 28 distributors across the world. 5. We are one of the leading exporters of Chinese industrial products and are desirous of entering into business relations you. Lesson 3 are a dealer in Egypt.。 2. I have?2 years of?follow-up purchase orders and shipping experience 3. We have pictures of sports shoes selling well in UK. 4. We are sending some samples and brochures under separate cover for your reference. 5. We are interested in the electric appliance in your sample. Please kindly send your quotations to us. 6. We’d like to inform that you’ll find our new products are at Stand 16. Lesson 4 1. We are looking for a reliable supplier who can provide us the laptop. 2. I would like to buy computers and computer parts. Please send me a detailed pricelist with min. Order and shipping costs.
主体结构检测测试题
主体结构检测试题 一、单项选择题（在所选项中，只有一个选项正确，每题1分） 1．《回弹法检测混凝土抗压强度技术规程》JGJ/T 23-2011规程适用于普通混凝土的（ B ）检测。A．表面硬度 B．抗压强度 C．表面强度 D．抗折强度 2．《回弹法检测混凝土抗压强度技术规程》JGJ/T 23-2011中指的普通混凝土系主要由水泥、砂、石、外加剂、掺合料和水配置的密度为（ D ）Kg/m3的混凝土。 A．2000～2400 B．2200～2800 C．2400～2800 D．2000～2800 3．回弹法检测构件混凝土强度时的一个检测单元称为（ B ）。 A．测点 B．测区 C．测域 D．以上均不对 4．由测区的（）值和（）值通过测强曲线或强度换算表得到的测区现龄期混凝土强度值称为测区混凝土强度换算值。( D ) A．最大，最小 B．平均，最小 C．最大，碳化深度 D．平均，碳化深度 5．混凝土强度推定值相应于强度换算值总体分布中保证率不低于（ B ）的构建中的混凝土强度值。A．90% B．95% C．99% D．% 6．《回弹法检测混凝土抗压强度技术规程》JGJ/T 23-2011中一般用（ D ）符号表示碳化深度值。 A．R B．S C．f D．d 7．《回弹法检测混凝土抗压强度技术规程》JGJ/T 23-2011中一般用符号S表示下列（ C ）参数。A．碳化深度 B．测区强度修正量 C．标准差 D．回弹值 8．回弹仪除应符合《回弹仪》GB/T9138的规定外，在弹击时，弹击锤脱钩瞬间，回弹仪的标称能量应为（ A ）J。 A． B．3.207 C． D． 9．用于率定回弹仪的钢砧的洛氏硬度HRC应为（ B ）。 A．50±2 B．60±2 C．70±2 D．80±2 10．在洛氏硬度HRC为60±2的钢砧上，回弹仪的率定值应为（ C ）。 A．80±1 B．60±1 C．80±2 D．60±2 11．数字式回弹仪的数显回弹值与指针直读示值相差不应超过（ A ）。 A．1 B．1.5 C．2 D．
主体结构试题答案
主体结构试题一答案 一、填空题（10分） 1、当混凝土有下列情况之一时，不能按《回弹法检测混凝土抗压强度技术规程》（DBJ14-026-2004）进行混凝土强度的检测：（1）测试部位表层与内部的质量有明显差异或内部存在缺陷。 （2）遭受、、、。冻害、化学侵蚀、火灾、高温损伤。 2、预制构件结构重要性系数，当无专门要求时取。1.0 3、检测时应确保所使用的仪器设备在或周期内，并处于。仪器设备的精度应满足的要求。检定、校准、正常状态、检测项目 4、《钻芯法检测混凝土抗压强度技术规程》DBJ14-029-2004适用于抗压强度为的普通混凝土抗压强度的检测。10~80MPa 5、后装拔出法检测混凝土抗压强度施加拔出力应，速度应控制在kN/s。每一结构或构件至少均匀布置个测点。在构件的及部位应布置测点，相邻两测点的间距不应小于mm，测点距构件边缘不应小于mm。 连续均匀、0.5～1.0kN/s、3、受力较大、薄弱部位、250、100 二、判断题（10分） 1、混凝土回弹仪在钢砧上的率定值在80±2时，则说明回弹仪完全正常，不需要检定。（×） 2、采用回弹法检测砌筑砂浆强度不用考虑砂浆原材料品种。（×） 3、砂浆试块贯入深度—砂浆试块抗压强度曲线比砌体灰缝砂浆贯入深度—砂浆试块抗压强度曲线，更能够准确、科学、合理地反映出测钉贯入砂浆的深度和砌筑砂浆抗压强度间的相关关系。（×） 4、超声仪声时显示是否正确，可用空气声速标定值与实测空气声速比较的方法进行校验。（√） 5、荷载效应组合是指按极限状态设计时，为保证结构的可靠性而对同时出现的各种荷载效应设计值规定的组合。（√） 三、选择题（10） 1、预制混凝土构件承载能力极限状态检验，受力主筋拉断对应检验系数允许值是（C ）。 A、1.20 B、1.30 C、1.50 D、1.55 2、回弹法检测混凝土抗压强度，按批抽样检测时，应根据结构或构件类型和受力特征布置测区，测区数量不得少于（ C ）个。 A、5 B、10 C、3 D、1 3、《钻芯法检测混凝土抗压强度技术规程》DBJ14-029-2004规定，芯样试件的直径应为（ A ）mm。 A、150，100，75 B、100，75，50 C、200，150，100 D、200，100，75
外贸函电课后练习答案
Unit 2 Establishing Business Relations & Inquiring Credit Reference I. Translate the following sentences into English. 1. We are one of the leading importers dealing in electronic products in the area, and take this opportunity to approach you, in the hoping of establishing business relations. 2. We have been engaged in handling importing and exporting of machinery and equipment for many years, and our products have enjoyed great popularity in many countries. 3. We owe your name and address to the Commercial Counselor’s office of our Embassy in Beijing. 4. We are given to understand that you are a manufacturer of daily chemicals. One of our clients intends to buy cosmetics from your country. We will appreciate it highly if you airmail the catalogue and price list of the goods available at present. 5. For our credit standing, please refer to the Bank of China, Shanghai Branch. II. Write a letter Dear Sirs, We have obtained your name and address from the Commercial Counsel of your Embassy in Beijing and are pleased to write to establish business relations with you. We are informed that you are in the market for Chinese Cotton Piece Goods, and this item falls with the scope of our business activities. To acquaint you with our goods available for export, we are enclosing a catalogue and price list. We are looking forward to your early reply. Faithfully yours, Unit 3 Enquiries and Replies II. Translate the following sentences into English. 1. Please quote us your best price CIF Shanghai, including our 3% commission. 2. Should your price be found competitive, we intend to place an order with you for 300,000 yards of Cotton Cloth. 3. One of our customers is now interested in the Qingdao Haier Refrigerator made in your country. Please offer CIF London for 400 sets to be delivered in April.
主体结构检测考试试题
一.填空 1、工程质量检测的主要内容包括:建筑材料检测、地基及基础检测、主体结构检测、室内环境检测、 建筑节能检测、钢结构检测、建筑幕墙检测、建筑门窗检测、建筑智能系统检测等。 2、检测机构应对出具的检测数据与结论的真实性、规范性与准确性负法律责任 3、环境条件记录应包括环境参数测量值、记录次数、记录时间、监控仪器编号、记录人签名等、 4、用于工程检测的回弹仪必须具有产品合格证及检定单位的检定合格证。 5、回弹仪累计弹击次数超过 6000 次时应送检定单位检定。 6、回弹法测混凝土强度时,混凝土测区的面积不宜大于0、04㎡。 7、检测混凝土强度就是,如果采用钻芯修正,直径100㎜的混凝土试件,钻取芯样数量不应少于6个,若采用直径小于70㎜的芯样,则数量不应少于9个。 8、碳化深度值测量时,采用浓度为 1% 的酚酞酒精溶液。 9、非金属超声检测仪如在较长时间内停用,每月应通电一次,每次不少于 l h。 10、平面换能器的工作频率宜在50～100kHz范围以内。 11、测量回弹值应在构件测区内超声波的发射与接收面各弹击8个点。超声波单面平测时,可在超声波的发射与接收测点之间弹击16个点,每一测点的回弹值测读精确至1。 12、钻芯法对结构混凝土造成局部损伤,就是一种微(半)破损的现场检测手段。 13、钻芯机应具有足够的刚度、操作灵活、固定与移动方便,并应有水冷却系统。 14、抗压试验的芯样试件宜使用标准芯样试件,其公称直径不宜小于骨料最大粒径的3倍,也可采用小直径芯样试件,但其公称直径不应小于70mm且不得小于骨料最大粒径的2倍。 15、当用钻芯法确定单个构件的混凝土强度推定值时,有效芯样试件的数量不应少于3个;对于较小的构件,有效芯样构件的数量不得少于2个;当用于钻芯修正时,标准芯样试件的数量不应少于6个,小直径芯样试件数量宜适当增加。 16.抗压芯样试件的高度与直径之比(H/d)宜为1、00。 17、芯样试件内不宜含有钢筋,不能满足此项要求时,标准芯样试件,每个试件内最多只允许有2根直径小于10mm的钢筋。 18、比较成熟的拔出法分为预埋或先装拔出法与后装拔出法两种。 19、拔出仪显示仪表的正常使用环境为4～40℃。 20、后装拔出法强度计算时,当构件3个拔出力中的最大与最小拔出力与中间值之差均小于中间值的15％时取最小值作为该构件拔出力计算值。 21、构件的结构性能载荷试验,按其在被测构件或结构上作用载荷特性的不同,可分为静荷载试验
通信网的基本要素功能
通信网的基本要素功能（基本要素：传输、交换、终端） (1) 传输：传输系统指完成信号传输的介质和设备的总称，其在终端设备与交换设备之间以及交换系统相互之间链接起来形成网络。按传输介质分为有限传输和无线传输系统。（2）交换设备以节点的形式与邻接的传输链路构成各种拓扑结构的通信网，是现代通信网的核心。 （3）终端设备是通信网中的源点和终点。终端设备的主要功能是将输入信息变换为易于在信道中传送的信号；用于发送和接收用户信息；与网络交换控制信息；通过网络实现呼叫和接入服务。如：电话机、传真机、计算机、智能多媒体终端设备等。 简述电信网的组成及作用 1.业务网：用于向公众提供诸如话音、视频、数据、多媒体等业务。 (1)传送网：指在不同地点的各点之间完成信息传递功能的网络； (2)交换网：交换设备是核心，由交换节点和通信链路组成，功能是完成对接入交换节 点的传输链路的汇集、转接接续和分配。 2.支撑网 (1)信令网：信令的功能是控制电信网中各种通信连接的建立和拆除，并维护通信网的 正常运行。 (2)数字同步网：保证数字交换局之间、数字交换局与数字传输设备之间的信号时钟同 步，并使通信网中所有数字交换系统和数字传输系统工作在同一时钟频率下。 (3)电信管理网：各种不同应用的管理系统按照标准接口互连，在有限点上与电信网接 口及电信网络互通，达到控制和管理目的。 通信网常用拓扑结构有哪些？试分析各种拓扑结构的特点。 简述模拟信号的数字化过程 抽样—量化---编码 抽样：每隔一定时间间隔T，抽取语音信号的一个瞬时幅度值，抽样后所得到的一条列在时间上离散的抽样值称为样值序列
量化：对抽样后的信号进行离散化处理，对幅值进行化零取整处理 编码：抽样、量化后的信号还不是数字信号，需将此信号转换成数字编码脉冲。 什么是基带传输？数字信号传输的主要技术内容有哪些？ 基带是由消息转换而来的原始信号所固有的频带，不搬移基带信号的频谱而直接进行传输的方式称为基带传输。从数字通信终端送出的数字信号，称为基带信号。 1. 再生中继技术 再生中继的作用是对基带信号进行放大和均衡，对已失真的信号进行判决，再生出与发送信号相同的标准波形。在传输通路的适当地点设置再生中继器，使信号在传输过程中的衰减得到补偿，并消除干扰的影响。再生后的信号与未受干扰的信号一样，继续往前传，从而延长通信距离。 2. 均衡技术 定义：对传输系统中的线性失真进行补偿或者校正的过程称为均衡。 频域均衡：是使整个传输系统（包括均衡器在内）满足无失真传输条件。基本思想是分别校正幅频特性和群时延特性，利用可调滤波器的频率特性去补偿基带系统的频率特性。 时域均衡：以传输信号的时域脉冲响应为出发点，力求传输系统（包括其本身在内）所形成的接收波形接近于无失真信号波形，目的是消除取样点上的码间干扰（而不要求整个信号波形无失真）。 时域均衡关注取样点的瞬时值，使该点上的码间干扰和噪声对判决的影响达到最小，从而提高取样判决的正确率。 什么是调制？什么是解调？简述调制的作用和分类。 调制是在发送端把基带信号的频谱搬移到传输信道通带内的过程。 解调：在接收端把已调制信号还原成基带信号的过程，是调制的逆过程。 模拟调制：基带信号是连续变化的模拟量。——幅度调制、频率调制、相位调制。 数字调制：用数字基带信号对载波进行调制，使基带信号的频谱搬移到载波频率上。——幅度键控、频移键控、绝对相移键控、相对（差分）相移键控。 信道中的差错主要包括哪两类？常用的差错控制方式有哪些？ 2. 差错的分类 ①随机差错：由随机噪声导致，表现为独立、稀疏和互不相关发生的差错。 ②突发差错：相对集中出现，即在短时段内有很多错码出现，而在其间有较长的无错码时间段，例如有脉冲干扰引起的错码。 差错控制方式： ①前向纠错方式 发送端对信息码元进行编码处理，使发送的码组具有纠错能力。接收端收到该码组后，通过译码能自动发现并纠正传输中出现的错误。不需反向通道，系统实时性好。 ②检错重发方式 发送端经过编码后发出能够检错的码组，接收端收到后，若检测出错误，则通过反向信道通
2017主体结构试题及标准答案7
主体结构试题七 一、概念题(2０分) １、结构性能检验 针对结构构件的承载力、挠度、裂缝控制性能等各项指标所进行的检验。 2、混凝土内部钢筋保护层厚度 混凝土内部钢筋外边缘到混凝土表面的距离 3、回弹法检测混凝土抗压强度 是指通过检测结构或构件混凝土的回弹值和碳化深度值来推定该结构或构件混凝土抗压强度的方法。 4、混凝土强度检测总体修正量 ,与非破损全部测区混凝土抗压强度换算值的平均用芯样试件混凝土抗压强度换算值的平均值f c ｃor,m 值f c cu,mz ,进行比较，确定修正量。 ５、抽样检测 从检测批中抽取样本,通过对样本的测试确定检测批质量的检测方法。 二、填空题（1０分) 1、《回弹法检测砌筑砂浆强度技术规程》（DＢJ１4-030-２004),每一测区应记取个回弹值,相邻测点的净距一般不小于,读数精确至。检测过程中，回弹仪应始终处于,且不得移位。１ 2、240mm、1、水平状态 2、检测人员必须经过培训取得,对特殊的检测项目,检测人员应有相应的。现场检测工作应由或检测人员承担。上岗资格、检测资格证书、2名、2名以上 3、混凝土内部钢筋保护层厚度检测对于具有饰面层的构件，其饰面层应、，并与基体混凝土结合。饰面层主体材料以及夹层均不得含有。对于含有金属材质的饰面层，应进行。对于厚度超过的饰面层,宜清除后进行检测,或者钻孔验证。不得在饰面层上进行检测。清洁、平整、良好、金属、清除、50mm、架空 ４、结构性能检测，单跨简支结构构件和连续梁的支座除一端支座应为支座外，其它支座应为支座;安装时,各支座轴线应彼此并垂直于被检测结构构件的,各支座轴线间的距离取为结构构件的。固定铰、滚动铰、平行、纵轴线、检测跨度 三、判断题(１0分） １、在建工程，砌筑砂浆强度的检测与评定可按《回弹法检测砌筑砂浆强度技术规程》（DBＪ１4-030－２00４)执行。（)× 2、回弹法检测结构测区混凝土强度换算值相当于施工现场预留混凝土试块２8天标养强度。( ）× 3、对选定的梁类构件,应抽取不少于５０%纵向受力钢筋的保护层厚度进行检验（) × ４、超声回弹综合法检测混凝土抗压强度,非同一测区的回弹值、超声声速值及碳化深度值，在计算混凝土强度换算值时不得混用。( ）√ ５、超声仪声时显示是否正确,可用空气声速标定值与实测空气声速比较的方法进行校验。(）√ 四、选择题（1０） 1、混凝土构件结构性能检验，在荷载标准值作用下,应持续( )。B A、15min Ｂ、30mｉｎC、1h 2、混凝土预制板结构性能检验裂缝测量，对正截面裂缝，应量测（)处的最大裂缝宽度。B A、板底 B、受拉主筋 C、板侧 3、回弹法检测混凝土抗压强度强度推定值相应于强度换算值总体分布中保证率不低于( ）的强度值。C A、９0% B、85% C、95% 4、混凝土回弹仪校准有效期为( )。C
主体结构试题
主体结构试题二 一、概念题（20分） 1、局部破损检测方法 2、超声回弹综合法检测混凝土强度 3、回弹法检测砌筑砂浆强度 4、混凝土内部钢筋保护层厚度 5、抽样检测 二、填空题（10分） 1、检测单位应有固定的、健全的和相应的。 2、按检测批检测的项目，应进行，且宜符合GB/T50344-2004标准规定。 3、《钻芯法检测混凝土抗压强度技术规程》DBJ14-029-2004规定，芯样试件的直径应为 mm、 mm或 mm，且不应小于骨料最大粒径的倍。如结构或构件工作条件比较干燥，芯样试件应以状态进行试验；如结构工作条件比较潮湿，芯样试件应在状态进行试验。 4、《回弹法检测砌筑砂浆强度技术规程》DBJ14-030-2004不适用于下列情况砌筑砂浆的检测：（1）测试部位与的质量有明显差异或内部存在。（2）遭受、或。 5、钢筋混凝土构件和允许出现裂缝的预应力混凝土构件结构性能检验的检验内容包括：、和。 三、判断题（10分） 1、回弹法检测结构测区混凝土强度换算值相当于施工现场预留混凝土试块28天标养强度。（） 2、后装拔出法检测结构混凝土强度，每一结构或构件上至少均匀布置3个测点。（） 3、贯入深度测量表由百分表改制而成，应按计量仪器要求进行管理。（） 4、原位轴压法属原位测试砌体抗压强度方法，测试结果除能反映砖和砂浆的强度外，还反映了砌筑质量对砌体抗压强度的影响，因此原位轴压法测试结果综合反映了砌体的抗压、抗剪等性能。（） 5、混凝土构件结构性能检验裂缝宽度可用精度为0.2mm的刻度放大镜等进行观测。( )
四、选择题（10） 1、后置埋件锚固质量检测，对于重要结构构件及生命线工程非结构构件，应采用（ ）检验。 A 、破坏性 B 、非破坏性 C 、抗压承载力 2、混凝土构件挠度检验，应量测构件（ ）。 A 、跨中位移 B 、跨中位移和支座沉陷 C 、构件自重产生挠度和跨中挠度 3、超声回弹综合法检测混凝土抗压强度时，超声测距量测应精确至（ ）。 A 、0.1mm B 、1mm C 、0.5mm 4、回弹法检测混凝土抗压强度，被测混凝土龄期应为（ ） A 、14~1000天 B 、28~1100天 C 、14~1100天 5、砌体的受压工作应力检测应采用（ ）法。 A 、原位轴压 B 、扁顶 C 、点荷 五、简答题（20分） 1、预制混凝土构件抗裂检验，开裂荷载实测值如何确定？ 2、如何检测混凝土的碳化深度值？ 3、混凝土回弹仪在什么情况下应送进行校准或技术性能测试？ 4、按《后装拔出法检测混凝土抗压强度技术规程》DBJ14-028-2004检测混凝土抗压强度时，测点布置有哪些要求？ 六、论述题（10分） 回弹仪应如何保养？ 七、计算题（20分） 1、某烧结砖砌体结构住宅楼主体施工完毕，三层砌筑砂浆试块抗压强度未达到设计要求，建设单位要求检测三层砌筑砂浆强度，现场调查，砌筑砂浆为混合砂浆，设计强度等级为M7.5，采用符合国家标准的R32.5普通硅酸盐水泥，中砂，自然养护，龄期3个月，为减少对结构损伤，确定采用回弹法检测，原始记录见下表1，计算三层砌筑砂浆强度推定值。 注：回弹法检测砌筑砂浆强度山东地区测强曲线 1、水泥砂浆测强曲线: ) 0156.0(4280 .2,,10 003542.0i m d i m i R f -= 2、混合砂浆测强曲线： ) 0132.0(9249 .2,,10 0006384.0i m d i m i R f -=
通信网基本概念与主体结构(第二版)答案Ch1
Solutions to Chapter 1 1a. Describe the step-by-step procedure that is involved from the time you deposit a letter in a mailbox to the time the letter is delivered to its destination. What role do names, addresses and mail codes (such as ZIP codes or postal codes) play? How might the letter be routed to its destination? To what extent can the process be automated? Solution: The steps involved in mailing a letter are: 1. The letter is deposited in mailbox. 2. The letter is picked up by postal employee and placed in sack. 3. The letter is taken to a sorting station, where it is sorted according to destination, as determined by the mail code and grouped with other letters with the same destination mail code. (If there is no mail code, then it is determined by the largest geographical unit, for example, country (if specified), otherwise state (if specified), otherwise city (if specified).) 4. The letter is shipped to the post office that handles the mail for the specific mail code (or country or city). 5. The letter is then sorted by street address. 6. The letter is picked up at the post office by the postal worker responsible for delivering to the specified address. 7. The letter is delivered according to the number and street. The name is not really used, unless the street address is missing or incorrect. The name is at the destination to determine who the letter belongs to. (Unless of course the letter is being sent to a small town, where most inhabitants are known to the postal worker.) The mail delivery process can be automated by using optical recognition on the mail code. The letter can then be sorted and routed to the destination postal station, and even to the destination neighborhood, depending on the amount of geographical detail built into the mail code. 1b. Repeat part (a) for an e-mail message. At this point, you may have to conjecture different approaches about what goes on inside the computer network. Solution: The steps involved in e-mailing a message are: 1. The message is sent electronically by clicking 'Send'. (In Chapter 2 we see that the Simple Mail Transfer Protocol (SMTP) is used to do this.) 2. The mail provider of the sender sends a request to a name server for the network address of the mail provider of the recipient. The mail provider is determined by the information following the @ symbol. 3. If the mail provider finds the network address of the recipient's mail provider, then it sends the message to that address.
主体结构工程检测与地基基础检测考试题修订版
主体结构工程检测与地基基础检测考试题修订 版 IBMT standardization office【IBMT5AB-IBMT08-IBMT2C-ZZT18】
2018年主体结构工程检测与地基基础检测考试题 判断题 1.同一根钢梁，用于锚桩横梁的反力装置和用于压重平台的反力装置，其允许使用最大试荷载是不同的。（） 2.灰土挤密桩用于处理含水量大于24%、饱和度大于65%的粉土、黏性土、素填土、杂填土和湿陷性黄土等地基。（） 3.一般而言，土压力仪测得的是土的有效应力。（） 4.单桩水平静载试验主要检测目的是确定单桩竖向抗拔极限承载力；判定竖向抗拔极限承载力是否满足设计要求；通过桩身内力及变形测试，测定桩的抗拔摩阻力。（） 5.强夯法加固地基的机理的中，预压变形作用指的是各种颗粒成份在结构上的重新排列，还包括颗粒结构或型态的改变。（√） 6.单桩竖向抗压静载试验主要检测目的是确定单桩竖向抗压极限承载力；判定竖向抗压承载力是否满足设计要求；通过桩身内力及变形测试，测定桩侧、桩端阻力；验证高应变法的单桩竖向抗压极限承载力检测结果。（） 7.单桩竖向静载试验中，当工程桩作锚桩时，锚桩上拔量已达到允许值时应终止加载。（）
8.人工地基检测应在竖向增强体满足设计要求及地基施工后周围土体达到休止稳定后进行，其中稳定时间对黏性土地基不宜少于28d，对于粉土地基不宜少于14d天，其他地基不应少于7d。（） 9.单桩竖向抗拔静载试验主要检测目的是确定单桩平临界和极限承载力，推定土抗力系数；判定水平承载力是否满足设计要求；通过桩身内力及变形测试测定桩身弯矩。（） 10.标准贯入试验（SPT）是用质量为63.5kg的穿心锤，以76cm的落距自由下落，将标准规格的贯入器自钻孔孔底预打15cm，测记再打入30cm的锤击数的原位测试。（） 11.回弹法检测混凝土抗压强度碳化深度测量，每次读数精确至1mm。（） 12.建筑结构检测对于通用的检测项目，应选用国家标准或行业标准；对于有地区特点的检测项目，可选用地方标准。（） 13.动力触探试验时，每贯入0.1m所需锤击数连续三次超过50击时，可停止试验。（） 14.排水固结法加固地基，可采用钻探取样试验，原位测试等方法进行土体强度检查。规范要求每加一级荷载后，都应进行一次检查。（） 15.单桩竖向静载试验中，当荷载沉降曲线呈缓变型时，可加载至桩顶总沉降量 60~80mm时终止加载。（）
通信网概念
第一章通信网组网结构：星，网状，环，树，总线，复合型。衡量通信网质量的三个目标：接通的任意性与快速性，信号传输的透明性与传输的一致性，网络的可靠性与经济合理性。Osi七层模型：应用表示会话传输网络链路物理。 第二章本地网又称市话网，设置有两个等级的交换中心，分别为汇接局tm,端局c5。长途网包括1234级交换中心，分别用c1234表示，国际局是对外的出入口，通过国际电路与其他国家的国际局连通。电话长途网正有四级向两级过度。C1和C2间直达电路的增多，C2的转接功能随之减弱，C3形成扩大的本地网，C4失去原有作用几乎消失。C1和C2之间的长途交换中心合并成DC1，构成长途网的高平面网，即省际平面，C3被称为DC2，构成长途网的地平面网，即省内平面。然后主簿向无级网和动态网过渡。话路子系统由交换网络、信令设备、中继器、接口电路。控制子系统功能：对呼叫进行处理，对整个交换机的运行进行管理、监测维护。硬件由存储器、处理机、输入输出设备。处理机是整个系统的核心部分，主要的运行，管理，监测，维护都由他来完成，存储器负责存储交换机工作程序和数据。输入输出设备包括键盘，打印机，显示器及远端接口等。控制系统是整个交换机的核心，集中控制，分散控制两种方式。多台处理器之间的分工方式有功能分担方式，负荷分担方式和容量分担方式。数字程控交换机的服务功能：呼叫转移，呼出限制，呼叫等待，自动振铃回叫，缩位拨号，热线服务，三方通话，免打扰，闹钟叫醒。PCM：R0=8*8000b/s=64kb/s R1=N*64kb/s 时隙交换：在交换网络的一侧，某条电路上的某个时隙内的8比特话音信号，通过交换网络的交换，转移到交换网络的另一侧的某条电路上的某个时隙的位置。通过这种时隙交换来的实现话音电路的交换时分交换原理，顺序存入，控制读或控制存入，顺序读出话务理论源—请求服务的用户，服务器—被请求对象。话务量A T=nh av（h）话务流量A1=A T=Nnh av/T=N λ1h av=λh av空间接线器由电子交叉矩阵和控制存储器构成。第一级t型接线器采用顺序存入、控制读出，第三级t型采用控制存入、顺序读出。帧同步就是从接受的数据流中搜索并识别这一同步码字，并以该时隙作为一帧的排头，使接收端的帧结构和发送端完全一致，从而保证两个交换机能够同步工作。这样才能实现数字信息的正确接收和交换。复帧同步是使接收端的复帧结构和排列与发送端一致。信令就是用户信息以外的各种控制命令。信令按工作区域可分用户线信令和局间信令，按传送通道分随路信令和共路信令。信令按功能分为线路路由管理信令。两交换机的信令设备之间没有直接相连的信令通道，信令是通过话路传达的。两交换机的信令设备之间有一条直接相连的专用通道，信令的传送是与话路分开且无关的信令传送方式端到端、逐段转发、混合。信令控制方式非互控，半互控，全互控。信令消息中消息信令单元，链路状态信令单元，填充信令单元，SIO指明MSU的类型。七号信令由信令点，信令转接点，信令链组成。智能网部件独立于现有的固定电话网络，是一个附加的网络结构。信令体系结构第四级用户部分，第一级数据链路，第二级链路控制，第三级网功能层信令单元结构消息信令单元，链路状态信令单元和填充信令单元智能网：业务交换点SSP，业务控制点SCP，业务数据点SDP，智能外设IP，业务生成环境SEC和业务管理系统SMS被叫集中计费业务（800）将含有该业务特服号码的呼叫，经智能网送到数据库里检索取得真正的被叫号码，然后建立呼叫连续，并允许呼叫的费用集中记录在预先登记的被叫号码上好处：免费快速易用，无需等待话务员接线，只记同一个号码即可任何地方使用，较高的可靠性；增加销售机会，减少花费，提高效率。滑码：如果每个交换系统接收到的数字比特流与其内部时钟位置的偏移和错位，造成帧同步的丢失，这就会产生帧失步主从同步方式是指在通信网内某一个主交换局设置高精度高稳定度的时钟源，并以其作为主基准时钟的频率。连接方式可采用星型树形结构，我国主要采用主从同步 第三章移动通信的发展第一阶段20世纪20年代至40年代初第二阶段40年代到60年代初第三阶段60年代到70年代中期第四阶段70年代中期到目前移动通信系统工作方式分为单工半双工双工，组成：移动台基站移动业务交换中心多址技术频分多址FDMA，时分多址
《建筑构造》试题及答案 精华!!!!!!!!!!!!!!
《建筑构造》试题及答案精华！！！！！！！！！！！！！！ 来源：张学利的日志 第一章绪论 一、填空题 1、建筑物的耐火等级分为（）级。 2、砖混结构是承重墙为（），楼板和屋顶为（）的建筑。 3、建筑按使用功能分为（）、（）、（）三大类。 4、模数分为（）和（），基本模数的数值为（），1M=（）。 5、一般民用建筑由（）、（）、( )、( )、( )、（）和门窗组成。 6、耐火等级标准主要根据房屋主要构件的( )和它的( )来确定。 7、新的建筑方针：（）、（）、（）、（）（简称八字方针）。 8、地震的强弱程度用（）和（）。国家规定（）地区必须抗震设防。 9、横向定位轴线之间的距离称为（），一般是按（）的模数数列选定的；纵向定位轴线之间的距离称为（），一般是按（）的模数数列选定的。 10、（）是实现建筑工业化的前提。 11、楼房的层高是指该层楼面上表面至（）的垂直距离。 12、7～9层为（）建筑。 二、判断题 1、内骨架结构、外墙为承重墙，不需设构造和圈梁。（） 2、建筑物的二级耐久年限为100年以上。（） 3、标志尺寸应符合模数、数列的规定，用以标注建筑物定位轴线之间的距离。（） 4、地面竖向定位轴线应与楼地面面层上表面重合。（）
5、建筑物的模数系列中“3M”数列常用于确定民用建筑中开间、进深、门窗洞口的尺寸（） 6、标志尺寸等于构造尺寸加减允许偏差。（） 7、构造尺寸是指建筑构配件的设计尺寸，它符合模数。（） 8、震级越大，烈度越大；距震中越远，烈度越小。（） 三、选择题 1、建筑物最下面的部分是（） Ａ首层地面Ｂ首层墙或柱Ｃ基础Ｄ地基 2、符合模数数列规定的尺寸为（） Ａ构造尺寸Ｂ标志尺寸Ｃ实际尺寸Ｄ允许偏差值 3、按建筑物主体结构的耐久年限，二级建筑物为（） Ａ25～50年Ｂ40～80年Ｃ50～100年Ｄ100年以上 4、多层住宅一般选用的结构形式为（） Ａ砖木结构Ｂ钢筋混凝土结构Ｃ砖混结构Ｄ钢结构 5、下列（）组数字符合建筑模数统一制的要求。 Ⅰ3000mm Ⅱ3330mm Ⅲ50mm Ⅳ1560mm ＡⅠⅡＢⅠⅢＣⅡⅢＤⅠⅣ 6、民用建筑中的开间，进深等其他模数尺寸是选用（） Ａ1／2ＭＢ1ＭＣ3ＭＤ6Ｍ 7、民用建筑按其用途分为（） Ａ居住建筑及公共建筑Ｂ居住建筑Ｃ大型建筑Ｄ大量民用建筑8、下列说法正确的是（）