3.第三章

合集下载

第三章自由基共聚合-3

第三章自由基共聚合-3

f1
1 f2

[M1 ] [M1] [M2 ]
(3—12)
F1
1 F2

d[M1 ] d[M1] d[M2 ]
(3—13)
将式(3—11)、(3—12)、(3—13)合并并整理,可得到以摩尔 分数表示的共聚物组成方程。
F1

r1f12 f1f2
r1f12

2f1f 2

r2f
2 2
反过来, r1 < 1, r2 > 1的情况是类似的,只是曲线处于 对角线下方。
该类例子很多: 如丁二烯—苯乙烯体系( r1=1.35, r2=0.58, 50℃);氯乙烯—醋酸乙烯酯体系( r1=1.68, r2=0.23 );甲基 丙烯酸甲酯—丙烯酸甲酯体系( r1=1.91, r2=0.5 )。
即: 体系中自由基总浓度及两种自由基浓度都保持不变。 除引发速率和终止速率相等外,还要求M1·和M2 ·两自由基
相互转变的速率相等。
13
第三章 自由基共聚合
链引发 链增长
链终止
R + M1 ki1 R + M2 ki2
R M1 R M2
RiM1 RiM2
M1 + M1 k11 M1 + M2 k12 M2 + M1 k21 M2 + M2 k22

0
(3—4) (3—5)
根据稳态假定,M1·和M2·的引发速率分别等于各自的 终止速率:
RiM1 2R t11 R t12
(3—6)
RiM2 2R t22 R t12
(3—7)
16
第三章 自由基共聚合
根据稳态假定,M1·转变成M2·和M2·转变成M1·的速率相 等,即:

第三章 多相流流型及判别方法

第三章 多相流流型及判别方法

体积与质量含气率:
x
多相混输技术的研究及其应用 2015-7-3 5
第三章 多相流流型及判别方法
对于截面含气率有:

1 1 1 w 1 Al wl g 1 s 1 A w w g g l
一、两相混合物密度 气液两相混合物密度有两种表示方法: (一)流动密度
(二)体积流量 单位时间内流过管路横截面的流体体积称为体积流量。对于气液两相混 输管路有:
Q Qg Ql
二、流速 (一)气相和液相速度 气相速度: 液相速度:
wg
Qg Ag
wl
Ql Al
(二)气相和液相的折算速度 气相折算速度:
多相混输技术的研究及其应用 2015-7-3 2
第三章 多相流流型及判别方法
y -0.6746608 x 4.2203391
x -1.028449 y 6.319154 y -0.2228661 x 3.361187
3)流型判别程序流程图
多相混输技术的研究及其应用 2015-7-3 24
第三章 多相流流型及判别方法
开始 输入已知数据 计算Bx,By,x,y P(x,y)在L1之下 吗? 否 P(x,y)在C4及L3 之右吗? 否 P(x,y)在L2之下 吗? 否 P(x,y)在C2之下 吗? 否 P(x,y)在C3之下 吗? 否 雾状流 按流型计算相关参数 输出流型 结束 是 环状流 是 冲击流 是 气团流 是 气泡流 是 波状流 否 P(x,y)在C1之下 吗? 是
散布流 不对称散 布流 移动床流
固定床流
图3-12 水平管液固两相流流型示意图
多相混输技术的研究及其应用 2015-7-3 18

[3] 第三章 复变函数的积分

[3] 第三章 复变函数的积分
3
第一节 解析函数的概念
➢ 一、积分的定义
有向曲线:设C为平面给定的一条光滑(或按段光滑)的曲线,如果选
定C的两个可能方向的一个作为正方向(或正向),则我们就把C称为有 向曲线.与曲线C反方向的曲线记为 C 1
简单闭曲线的正方向:当曲线上的点P顺此方向前进时,邻近P点的曲线
内部始终位于P点的左方,
如果把z0除去,虽然在除去z0的C的内部,函数处处解析,
但是这个区域已经不是单连通区域.
由此可猜想:积分的值与路径无关或沿闭曲线积分值为零的条件 与被积函数的解析性及区域的单连通性有关.究竟关系如何,下面我 们讨论此问题.
16
➢一、柯西积分定理
定理3.2(柯西—古萨基本积分定理) 设函数f (z)在单连通区域D内 处处解析,那么函数f (z)在D内沿任何一条封闭曲线C的积分为零,
G
(
Q x
P y
)dxdy.
证明: 因为函数f (z)在区域D内解析,故f (z)存在,(下面在f (z)连续的假设下证明)
因为u与v的一阶偏导数存在且连续,故应用格林公式得:
C f (z)dz C (u iv)(dx idy) C u(x, y)dx v(x, y)dy iC v(x, y)dx u(x.y)dy
k 1
k 1
k 1
其中sk是小弧段z¼ k1zk的长, | zk | xk2 yk2 sk
n
n
n
lim |
0
k 1
f
( k )zk
|
lim
0
|
k 1
f
( k ) || zk
|
lim
0
|
k 1
f
( k ) ||

第三章 生产过程组织3

第三章 生产过程组织3

.
企业生产过程的核心是基本生产过程,其他部分根据企业的生产规模、 管理模式、专业化程度等具体情况,包括在企业生产过程之中,或由专 门的单位来完成。
第三章 生产过程组织
2.纵向展开 ①工艺阶段 工艺过程是指直接改变劳动对象的性质、形状、大小等的过程。 ②工序 工艺工序 非工艺工序 工序:是指一个(或一组)工人,在一台机床(或一个工作地点), 对同一个(或同时对几个)工件所连续完成的那一部分工艺过程。 ③工步 工步:是在加工表面不变、加工工具不变、切削用量不变的条件下所 连续完成的那一部分工序
备货性生产和订货性生产的的主要区别
项目 产品特点 生产流程 库存 计划 设备 人员 备货性生产 量大、标准、好预测 稳定、标准、均衡 连接生产和市场的纽带 优化的标准计划 专用高效设备 专业化 定货性生产 量小、多变、难预测 不稳定、无标准、难均衡 不设成品库存 不便详细,近细远粗 通用设备 多种操作技能
第三章 生产过程组织
2.成批生产 特点:品种较多,单产品产量较小;成批轮番生产。 工作地或设备的有效工作时间短。机械化、自动化水平不高; 对工人的技术要求较高。 3.单件小批成产 特点: (1)品种繁多,每一品种的产量极少,生产重复率低。 (2)设计方面:一品一设计,设计质量不高。 (3)工艺方面:一品一工艺,工艺质量不高。 (4)生产组织方面:粗略分工,专业化水平低。设备集群式布 置,产品路线长。 (5)生产管理方面:粗略工时定额;协作关系不稳定,质量和 交货期不易保证。例行管理少,例外管理多。人员多。
小于0.025 小于
大量系数法
★计算工序的大量系数,参考表3-1,确定工
作地类型。工序的大量系数计算公式如下:
Kb=ti/r
式中: Kb—工序的大量系数; ti—工序的单件时间(分/件); r—零件平均生产节拍(分/件)。 — ★零件平均节拍可用下式计算:

生物苏教版必修3课件:第三章第三节 生物群落的演替(共35张PPT)

生物苏教版必修3课件:第三章第三节 生物群落的演替(共35张PPT)

解析:选A。在原生演替过程中,生物总量是
逐渐增加的。先驱物种一般是较为低等的植物,
但某个阶段的生物种类数量在演替到这个阶段
时逐渐增加,由此阶段向另一个阶段演替时逐
渐减少。必须指出,演替能否达到顶极群落
(树林或森林)又决定于气候条件。
影响群落演替的因素
1.内部因素 (1)种内关系 ①种内互助:同一物种内部存在互相帮助,会 提高本物种的生存能力,在与其他物种的竞争 中会提高其优势地位,如蚂蚁、蜜蜂。 ②种内斗争:同一物种内部也会因食物、生存 空间等限制因素而发生斗争,种群数量增长的S 型曲线就能说明这一问题。
(1)火烧后的森林原有植被虽已被破坏,但由于原
有土壤条件、原始物种等的存在,因而该种群的
恢复过程应为次生演替。
(2)演替能否恢复到原始群落由多种因素制约,如
环境条件、变化程度,原始物种是否存在等,若 改变,则难以恢复。
(3)群落演替的总趋势是物种多样性增加。 (4)弃耕农田保留有原有土壤条件和部分原有物 种,属于次生演替;在保留有土壤条件的情况下 首先生长的植物应是草本植物;如果条件允许演
例2
【尝试解答】 【解析】
D
过度放牧、毁林开荒和围湖造田都可
导致生物群落结构简单化、生产力下降等,这些 属于逆行演替。而退耕还湖则会导致进展演替。
跟踪训练 (2011 年新疆伊犁高二检测)下列有关人类 活动对群落演替影响的表述,不正确的是( ) . A.人类的许多活动影响群落的演替 B.人类可建立人工群落,将演替的方向和速度置于 人为控制下 C.人类活动对生物群落演替的影响远远超过其他某 些自然因素 D.人类的活动对群落的演替均是具有破坏性的
思考感悟 2.是否所有的群落最终都可以演替为森林阶段? 试举例说明。

3. 第三章课后习题及答案

3. 第三章课后习题及答案

第三章1. (Q1) Suppose the network layer provides the following service. The network layer in the source host accepts a segment of maximum size 1,200 bytes and a destination host address from the transport layer. The network layer then guarantees to deliver the segment to the transport layer at the destination host. Suppose many network application processes can be running at the destination host.a. Design the simplest possible transport-layer protocol that will get application data to thedesired process at the destination host. Assume the operating system in the destination host has assigned a 4-byte port number to each running application process.b. Modify this protocol so that it provides a “return address” to the destination process.c. In your protocols, does the transport layer “have to do anything” in the core of the computernetwork.Answer:a. Call this protocol Simple Transport Protocol (STP). At the sender side, STP accepts from thesending process a chunk of data not exceeding 1196 bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number.b. The segment now has two header fields: a source port field and destination port field. At thesender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address,a source port number, and a destination port number. STP creates a segment which contains theapplication data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number.c. No, the transport layer does not have to do anything in the core; the transport layer “lives” inthe end systems.2. (Q2) Consider a planet where everyone belongs to a family of six, every family lives in its own house, each house has a unique address, and each person in a given house has a unique name. Suppose this planet has a mail service that delivers letters form source house to destination house. The mail service requires that (i) the letter be in an envelope and that (ii) the address of the destination house (and nothing more ) be clearly written on the envelope. Suppose each family has a delegate family member who collects and distributes letters for the other family members. The letters do not necessarily provide any indication of the recipients of the letters.a. Using the solution to Problem Q1 above as inspiration, describe a protocol that thedelegates can use to deliver letters from a sending family member to a receiving family member.b. In your protocol, does the mail service ever have to open the envelope and examine theletter in order to provide its service.Answer:a.For sending a letter, the family member is required to give the delegate the letter itself, theaddress of the destination house, and the name of the recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an e nvelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate than gives the letter to the family member with this name.b.No, the mail service does not have to open the envelope; it only examines the address on theenvelope.3. (Q3) Describe why an application developer might choose to run an application over UDP rather than TCP.Answer:An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP.4. (P1) Suppose Client A initiates a Telnet session with Server S. At about the same time, Client B also initiates a Telnet session with Server S. Provide possible source and destination port numbers fora. The segment sent from A to B.b. The segment sent from B to S.c. The segment sent from S to A.d. The segment sent from S to B.e. If A and B are different hosts, is it possible that the source port number in the segment fromA to S is the same as that fromB to S?f. How about if they are the same host?Yes.f No.5. (P2) Consider Figure 3.5 What are the source and destination port values in the segmentsflowing form the server back to the clients’ processes? What are the IP addresses in the network-layer datagrams carrying the transport-layer segments?Answer:Suppose the IP addresses of the hosts A, B, and C are a, b, c, respectively. (Note that a,b,c aredistinct.)To host A: Source port =80, source IP address = b, dest port = 26145, dest IP address = a To host C, left process: Source port =80, source IP address = b, dest port = 7532, dest IP address = cTo host C, right process: Source port =80, source IP address = b, dest port = 26145, dest IP address = c6. (P3) UDP and TCP use 1s complement for their checksums. Suppose you have the followingthree 8-bit bytes: 01101010, 01001111, 01110011. What is the 1s complement of the sum of these 8-bit bytes? (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.) Show all work. Why is it that UDP takes the 1s complement of the sum; that is , why not just sue the sum? With the 1s complement scheme, how does the receiver detect errors? Is it possible that a 1-bit error will go undetected? How about a 2-bit error?Answer:One's complement = 1 1 1 0 1 1 1 0.To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected (e.g., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1).7. (P4) Suppose that the UDP receiver computes the Internet checksum for the received UDPsegment and finds that it matches the value carried in the checksum field. Can the receiver be absolutely certain that no bit errors have occurred? Explain.Answer:No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated. If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error.8. (P5) a. Suppose you have the following 2 bytes: 01001010 and 01111001. What is the 1scomplement of sum of these 2 bytes?b. Suppose you have the following 2 bytes: 11110101 and 01101110. What is the 1s complement of sum of these 2 bytes?c. For the bytes in part (a), give an example where one bit is flipped in each of the 2 bytesand yet the 1s complement doesn’t change.0 1 0 1 0 1 0 1 + 0 1 1 1 0 0 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 + 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 1Answer:a. Adding the two bytes gives 10011101. Taking the one’s complement gives 01100010b. Adding the two bytes gives 00011110; the one’s complement gives 11100001.c. first byte = 00110101 ; second byte = 01101000.9. (P6) Consider our motivation for correcting protocol rdt2.1. Show that the receiver, shown inthe figure on the following page, when operating with the sender show in Figure 3.11, can lead the sender and receiver to enter into a deadlock state, where each is waiting for an event that will never occur.Answer:Suppose the sender is in state “Wait for call 1 from above” and the receiver (the receiver shown in the homework problem) is in state “Wait for 1 from below.” The sender sends a packet with sequence number 1, and transitions to “Wait for ACK or NAK 1,” waiting for an ACK or NAK. Suppose now the receiver receives the packet with sequence number 1 correctly, sends an ACK, and transitions to state “Wait for 0 from below,” waiting for a data packet with sequence number 0. However, the ACK is corrupted. When the rdt2.1 sender gets the corrupted ACK, it resends the packet with sequence number 1. However, the receiver is waiting for a packet with sequence number 0 and (as shown in the home work problem) always sends a NAK when it doesn't get a packet with sequence number 0. Hence the sender will always be sending a packet with sequence number 1, and the receiver will always be NAKing that packet. Neither will progress forward from that state.10. (P7) Draw the FSM for the receiver side of protocol rdt3.0Answer:The sender side of protocol rdt3.0 differs from the sender side of protocol 2.2 in that timeouts have been added. We have seen that the introduction of timeouts adds the possibility of duplicate packets into the sender-to-receiver data stream. However, the receiver in protocol rdt.2.2 can already handle duplicate packets. (Receiver-side duplicates in rdt 2.2 would arise if the receiver sent an ACK that was lost, and the sender then retransmitted the old data). Hence the receiver in protocol rdt2.2 will also work as the receiver in protocol rdt 3.0.11. (P8) In protocol rdt3.0, the ACK packets flowing from the receiver to the sender do not havesequence numbers (although they do have an ACK field that contains the sequence number of the packet they are acknowledging). Why is it that our ACK packets do not require sequence numbers?Answer:To best Answer this question, consider why we needed sequence numbers in the first place. We saw that the sender needs sequence numbers so that the receiver can tell if a data packet is a duplicate of an already received data packet. In the case of ACKs, the sender does not need this info (i.e., a sequence number on an ACK) to tell detect a duplicate ACK. A duplicate ACK is obvious to the rdt3.0 receiver, since when it has received the original ACK it transitioned to the next state. The duplicate ACK is not the ACK that the sender needs and hence is ignored by the rdt3.0 sender.12. (P9) Give a trace of the operation of protocol rdt3.0 when data packets and acknowledgmentpackets are garbled. Your trace should be similar to that used in Figure 3.16Answer:Suppose the protocol has been in operation for some time. The sender is in state “Wait for call fro m above” (top left hand corner) and the receiver is in state “Wait for 0 from below”. The scenarios for corrupted data and corrupted ACK are shown in Figure 1.13. (P10) Consider a channel that can lose packets but has a maximum delay that is known.Modify protocol rdt2.1 to include sender timeout and retransmit. Informally argue whyyour protocol can communicate correctly over this channel.Answer:Here, we add a timer, whose value is greater than the known round-trip propagation delay. We add a timeout event to the “Wait for ACK or NAK0” and “Wait for ACK or NAK1” states. If the timeout event occurs, the most recently transmitted packet is retransmitted. Let us see why this protocol will still work with the rdt2.1 receiver.• Suppose the timeout is caused by a lost data packet, i.e., a packet on the senderto- receiver channel. In this case, the receiver never received the previous transmission and, from the receiver's viewpoint, if the timeout retransmission is received, it look exactly the same as if the original transmission is being received.• Suppose now that an ACK is lost. The receiver will eventually retransmit the packet on atimeout. But a retransmission is exactly the same action that is take if an ACK is garbled. Thus the sender's reaction is the same with a loss, as with a garbled ACK. The rdt 2.1 receiver can already handle the case of a garbled ACK.14. (P11) Consider the rdt3.0 protocol. Draw a diagram showing that if the network connectionbetween the sender and receiver can reorder messages (that is, that two messagespropagating in the medium between the sender and receiver can be reordered), thenthe alternating-bit protocol will not work correctly (make sure you clearly identify thesense in which it will not work correctly). Your diagram should have the sender on theleft and the receiver on the right, with the time axis running down the page, showingdata (D) and acknowledgement (A) message exchange. Make sure you indicate thesequence number associated with any data or acknowledgement segment.Answer:15. (P12) The sender side of rdt3.0 simply ignores (that is, takes no action on) all received packetsthat are either in error or have the wrong value in the ack-num field of anacknowledgement packet. Suppose that in such circumstances, rdt3.0 were simply toretransmit the current data packet . Would the protocol still work? (hint: Consider whatwould happen if there were only bit errors; there are no packet losses but prematuretimeout can occur. Consider how many times the nth packet is sent, in the limit as napproaches infinity.)Answer:The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost (and from the receiver standpoint, it never knows which of these events, if either, will occur). To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as n approaches infinity.16. (P13) Consider a reliable data transfer protocol that uses only negative acknowledgements.Suppose the sender sends data only infrequently. Would a NAK-only protocol bepreferable to a protocol that uses ACKs? Why? Now suppose the sender has a lot ofdata to send and the end to end connection experiences few losses. In this second case ,would a NAK-only protocol be preferable to a protocol that uses ACKs? Why?Answer:In a NAK only protocol, the loss of packet x is only detected by the receiver when packetx+1 is received. That is, the receivers receives x-1 and then x+1, only when x+1 is received does the receiver realize that x was missed. If there is a long delay between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol.On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are never sent – a significant reduction in feedback in the NAK-only case over the ACK-only case.17. (P14) Consider the cross-country example shown in Figure 3.17. How big would the windowsize have to be for the channel utilization to be greater than 80 percent?Answer:It takes 8 microseconds (or 0.008 milliseconds) to send a packet. in order for the sender to be busy 90 percent of the time, we must have util = 0.9 = (0.008n) / 30.016 or n approximately 3377 packets.18. (P15) Consider a scenario in which Host A wants to simultaneously send packets to Host Band C. A is connected to B and C via a broadcast channel—a packet sent by A is carriedby the channel to both B and C. Suppose that the broadcast channel connecting A, B,and C can independently lose and corrupt packets (and so, for example, a packet sentfrom A might be correctly received by B, but not by C). Design a stop-and-wait-likeerror-control protocol for reliable transferring packets from A to B and C, such that Awill not get new data from the upper layer until it knows that B and C have correctlyreceived the current packet. Give FSM descriptions of A and C. (Hint: The FSM for Bshould be essentially be same as for C.) Also, give a description of the packet format(s)used.Answer:In our solution, the sender will wait until it receives an ACK for a pair of messages (seqnum and seqnum+1) before moving on to the next pair of messages. Data packets have a data field and carry a two-bit sequence number. That is, the valid sequence numbers are 0, 1, 2, and 3. (Note: you should think about why a 1-bit sequence number space of 0, 1 only would not work in the solution below.) ACK messages carry the sequence number of the data packet they are acknowledging.The FSM for the sender and receiver are shown in Figure 2. Note that the sender state records whether (i) no ACKs have been received for the current pair, (ii) an ACK for seqnum (only) has been received, or an ACK for seqnum+1 (only) has been received. In this figure, we assume that theseqnum is initially 0, and that the sender has sent the first two data messages (to get things going).A timeline trace for the sender and receiver recovering from a lost packet is shown below:Sender Receivermake pair (0,1)send packet 0Packet 0 dropssend packet 1receive packet 1buffer packet 1send ACK 1receive ACK 1(timeout)resend packet 0receive packet 0deliver pair (0,1)send ACK 0receive ACK 019. (P16) Consider a scenario in which Host A and Host B want to send messages to Host C. HostsA and C are connected by a channel that can lose and corrupt (but not reorder)message.Hosts B and C are connected by another channel (independent of the channelconnecting A and C) with the same properties. The transport layer at Host C shouldalternate in delivering messages from A and B to the layer above (that is, it should firstdeliver the data from a packet from A, then the data from a packet from B, and so on).Design a stop-and-wait-like error-control protocol for reliable transferring packets fromA toB and C, with alternating delivery atC as described above. Give FSM descriptionsof A and C. (Hint: The FSM for B should be essentially be same as for A.) Also, give adescription of the packet format(s) used.Answer:This problem is a variation on the simple stop and wait protocol (rdt3.0). Because the channel may lose messages and because the sender may resend a message that one of the receivers has already received (either because of a premature timeout or because the other receiver has yet to receive the data correctly), sequence numbers are needed. As in rdt3.0, a 0-bit sequence number will suffice here.The sender and receiver FSM are shown in Figure 3. In this problem, the sender state indicates whether the sender has received an ACK from B (only), from C (only) or from neither C nor B. The receiver state indicates which sequence number the receiver is waiting for.20. (P17) In the generic SR protocol that we studied in Section 3.4.4, the sender transmits amessage as soon as it is available (if it is in the window) without waiting for anacknowledgment. Suppose now that we want an SR protocol that sends messages twoat a time. That is , the sender will send a pair of messages and will send the next pairof messages only when it knows that both messages in the first pair have been receivercorrectly.Suppose that the channel may lose messages but will not corrupt or reorder messages.Design an error-control protocol for the unidirectional reliable transfer of messages.Give an FSM description of the sender and receiver. Describe the format of the packetssent between sender and receiver, and vice versa. If you use any procedure calls otherthan those in Section 3.4(for example, udt_send(), start_timer(), rdt_rcv(), and soon) ,clearly state their actions. Give an example (a timeline trace of sender and receiver)showing how your protocol recovers from a lost packet.Answer:21. (P18) Consider the GBN protocol with a sender window size of 3 and a sequence numberrange of 1024. Suppose that at time t, the next in-order packet that the receiver isexpecting has a sequence number of k. Assume that the medium does not reordermessages. Answer the following questions:a. What are the possible sets of sequence number inside the sender’s window at timet? Justify your Answer.b .What are all possible values of the ACK field in all possible messages currentlypropagating back to the sender at time t? Justify your Answer.Answer:a.Here we have a window size of N=3. Suppose the receiver has received packet k-1, and hasACKed that and all other preceeding packets. If all of these ACK's have been received by sender, then sender's window is [k, k+N-1]. Suppose next that none of the ACKs have been received at the sender. In this second case, the sender's window contains k-1 and the N packets up to and including k-1. The sender's window is thus [k- N,k-1]. By these arguments, the senders window is of size 3 and begins somewhere in the range [k-N,k].b.If the receiver is waiting for packet k, then it has received (and ACKed) packet k-1 and the N-1packets before that. If none of those N ACKs have been yet received by the sender, then ACKmessages with values of [k-N,k-1] may still be propagating back. Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK for k-N-1.Once the receiver has sent an ACK for k-N-1 it will never send an ACK that is less that k-N-1.Thus the range of in-flight ACK values can range from k-N-1 to k-1.22. (P19) Answer true or false to the following questions and briefly justify your Answer.a. With the SR protocol, it is possible for the sender to receive an ACK for a packet thatfalls outside of its current window.b. With CBN, it is possible for the sender to receiver an ACK for a packet that fallsoutside of its current window.c. The alternating-bit protocol is the same as the SR protocol with a sender and receiverwindow size of 1.d. The alternating-bit protocol is the same as the GBN protocol with a sender andreceiver window size of 1.Answer:a.True. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at t0 . At t1 (t1 > t0)the receiver ACKS 1, 2, 3. At t2 (t2 > t1) the sender times out and resends 1, 2, 3. At t3 the receiver receives the duplicates and re-acknowledges 1, 2, 3. At t4 the sender receives the ACKs that the receiver sent at t1 and advances its window to 4, 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the receiver sent at t2 . These ACKs are outside its window.b.True. By essentially the same scenario as in (a).c.True.d.True. Note that with a window size of 1, SR, GBN, and the alternating bit protocol arefunctionally equivalent. The window size of 1 precludes the possibility of out-of-order packets (within the window). A cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.23. (Q4) Why is it that voice and video traffic is often sent over TCP rather than UDP in today’sInternet. (Hint: The Answer we are looking for has nothing to do with TCP’s congestion-control mechanism. )Answer:Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls24. (Q5) Is it possible for an application to enjoy reliable data transfer even when the applicationruns over UDP? If so, how?Answer:Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however.25. (Q6) Consider a TCP connection between Host A and Host B. Suppose that the TCP segmentstraveling form Host A to Host B have source port number x and destination portnumber y. What are the source and destination port number for the segments travelingform Host B to Host A?Answer:Source port number y and destination port number x.26. (P20) Suppose we have two network entities, A and B. B has a supply of data messages thatwill be sent to A according to the following conventions. When A gets a request fromthe layer above to get the next data (D) message from B, A must send a request (R)message to B on the A-to-B channel. Only when B receives an R message can it send adata (D) message back to A on the B-to-A channel. A should deliver exactly one copy ofeach D message to the layer above. R message can be lost (but not corrupted) in the A-to-B channel; D messages, once sent, are always delivered correctly. The delay alongboth channels is unknown and variable.Design(give an FSM description of) a protocol that incorporates the appropriatemechanisms to compensate for the loss-prone A-to-B channel and implementsmessage passing to the layer above at entity A, as discussed above. Use only thosemechanisms that are absolutely necessary.Answer:Because the A-to-B channel can lose request messages, A will need to timeout and retransmit its request messages (to be able to recover from loss). Because the channel delays are variable and unknown, it is possible that A will send duplicate requests (i.e., resend a request message that has already been received by B). To be able to detect duplicate request messages, the protocol will use sequence numbers. A 1-bit sequence number will suffice for a stop-and-wait type of request/response protocol.A (the requestor) has 4 states:• “Wait for Request 0 from above.” Here the requestor is waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R0, to B, starts a timer and make s a transition to the “Wait for D0” state. When in the “Wait for Request 0 from above” state, A ign ores anything it receives from B.• “Wait for D0”. Here the requestor is waiting for a D0 data message from B. A timer is always running in this state. If the timer expires, A sends another R0 message, restarts the timer and remains in this state. If a D0 message is received from B, A stops the time and transits to the “Wait for Request 1 from above” state. If A receives a D1 data message while in this state, it is ignored.• “Wait for Request 1 from above.” Here the requestor is again waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R1, to B, starts a timer and makes a transition to the “Wait for D1” state. When in the “Wait for Request 1 from above” state, A ignores anything it receives from B.• “Wait for D1”. Here the requestor is waiting for a D1 data message from B. A timer is always running in this state. If the timer expires, A sends another R1 message, restarts the timer and remains in this state. If a D1 message is received from B, A stops the timer and transits to the “Wait for Request 0 from above” state. If A receives a D0 data message while in this state, it is ignored.The data supplier (B) has only two states:。

第三章人力规划3

第三章人力规划3
T:指由于技术进步或设备改进后节省的人力资源。
例:某公司目前员工是200人,在三年后由于业务发展 需要增加100人,但由于技术提高后可节省25人,试问三年 后公司需要的人力资源数是多少? 根据公式,即得:
NHR = 200+100–25 = 275(人)
特点:根据经验来预测,要求预测人员素 质高,比较适用于技术稳定的企业的中短期 人力资源的预测。
O:指目前每人的平均其他支出;
a%:指未来企业计划每年人类资源成本增加的平均百分数; T:指未来一段时间的年限。

例:某公司计划三年后人力成本总额控制在300万元/ 月。目前员工的平均工资是1000元/月;平均奖金是 200元/月;平均福利是720元/月;平均其他收入是80 元/月。公司计划对人力资源成本投入按5%的比率增 长。请预测该公司三年后的人力资源数量。
人力资源需求预测程序图
3、需求预测的类型 (1)短期预测 : 一年以内
(2)中期预测:
(3)长期预测:
一年到三年
五年以上
4、人力资源需求预测的方法
经验预测法 描述法 工作研究法 德尔菲法
自上而下 自下而上
定性预测方法 人 力 资 源 需 求 预 则 方 法 定量预测方法
工作定额法趋 势分析法 生产率分析法 技能组合法 时间序列法 回归分析法
100
190 150
70
100 80
80
110 110
120
200 160
70
60 80
80
80 110
120
120 150
该公司对人力的需求量究竟是多少?方法有以下
三种: ①计算最低需求、最可能需求和最高需求的算术 平均值,得到人力需求量。 NHR=(80+110+150)÷3=114(人) ②计算加权平均值,得到人力需求量。一般地最 低需求、最可能需求和最高需求所赋的权重分别 是0.2、0.5、0.3。 NHR=80×0.2+110×0.5+150×0.3=116(人) ③用中位数计算人力资源需求量。 将第三次判断的结果按预测值的大小由低到高排 列:

第3章 第三章随机向量

第3章  第三章随机向量
且对给定的 成立,故X和Y相互独立. 例4 设 (X, Y)的联合分布密度函数为
3 x, 0 x 1, x y x, p ( x, y ) 2 0, 其他 .
问X, Y是否独立? 解
x 3 2 x x d y 3 x , 0 x 1, p X ( x ) p ( x, y ) d y 2 0, 其他 .
例3 设 (X, Y) 的联合分布列如下, 问X, Y是否独立?
X Y
0 1 2
1 2 20 2 20 4 20
0 1 20 1 20 2 20
2 2 20 2 20 4 20

X p
易得X和Y的边缘分布律分别为:
0 1 4 1 1 4 2 2 4 Y p 1 2 5 0 1 5 2 2 5
3.4 条件分布与随机变量的独立性



e
dt
1 e 2
( x ).
pY ( y )
1 e 2
y2 2
( y ).
本节
上页
下页
3.3 连续型随机向量及分布
本章
上页
下页
3.4 条件分布与随机变量的独立性
1.离散型条件分布
2.连续型条件分布
3.随机变量的独立性
本章
上页
下页
3.4 条件分布与随机变量的独立性
( xi , yi )(i, j 1,2,), 且 P( X xi ,Y y j ) pij ,
则我们把它称为(X,Y)的联合分布列.
本节
上页
下页
3.2 离散型随机向量及分布
联合分布列:
X
Y
x1 xi
  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

由于弯曲带板定义式计算十分复杂,我国《海船 规范》规定,安装在平板上的主要构件带板的有效 面积为: 2
A 10 fbtp
(cm )
f 0.3(l / b)2 / 3,但不大于1;b—主要构件支承面积
t 平均宽度,m;l —主要构件长度,m; p —带板的 平均厚度,mm。 中国船舶检验局颁布的《内河钢船建造规范》 (1991)规定:强骨材带板宽度取其跨度的1/6,即 be=l/6,但不大于负荷平均宽度亦不小于普通骨材 间距。
§3 典型船体结构 的局部强度计算
1. 船底结构的强度计算
船底是船体梁的下翼板,受到很大的总纵弯曲 应力,此外还承受机器重量、货物重量、压载水及 舷外水压力等横向荷重作用。在波浪中高速航行的 船舶底部,特别是首部附近的船底还受到很大的冲 击力。 在总纵强度校核时,船底纵桁应力要与总纵弯 曲应力合成,此时船底板架的计算载荷应取相应的 总纵弯曲计算时的载荷状态和波浪位置的水头高度。 在局部强度计算时,船底板架计算水头为舷外水压 与货物反压力之差值。
h d hB / 2
d为载重吃水(m),hB为半波高。
§2
船体骨架 的带板
船体结构中的骨架都是焊接在钢板上的,当骨 架受力发生变形时,与它连接的板也一起参加骨架 抵抗变形。因此,为估算骨架的承载能力,也应当 把一定宽度的板作为它的组成部分来计算骨架梁的 剖面积、惯性矩和剖面模数等几何要素,这部分板 称为带板或附连翼板。 因骨架的受力情况不同,带板宽度有两种不同 的定义和数值,即 (1)压杆的(稳定性)带板宽度,We; (2)梁的(弯曲)带板宽度,be。
M 3Q1B /8
Q=qbL,Q1=qaB,b为纵桁间距,q为载荷强度; γ1、γ2、γ3影响系数,根据板架长宽比和中桁材 与旁桁材惯性比值查表确定。
2.甲板结构的强度计算
最上层连续甲板是船体梁的上甲板,它对保证船 体总纵强度起重要作用,称为强力甲板。 下甲板主 要承受货物重量,因此首先应保证其局部强度。无 论哪层甲板都承受均布荷重。 民船中不载货的上层露天甲板,承受甲板上浪的 水压力,其水头高度可按规范规定计算。《海船规 范》规定,露天强力甲板计算水头高度为1.2~1.5m 之间,且不小于下式计算值: 1 100 3L h0 1.2 ( 150 ) (m) 500 D d L-船长(m),D -型深(m),d -吃水(m)。
2) 船底纵骨弯曲应力计算
船底纵骨由肋板支持,纵骨结构、载荷对称于肋板,因此 可以把纵骨当作两端固定受均布荷重q的在肋板上的单跨梁 计算,其弯矩为: qba 2
M 0 12 (N M) 2 M qba (N M) 1 24
(支座处)
(跨中处)
a-纵骨跨距;b -纵骨间距;q -分布载荷强度,分别取
总之,正确分析结构变形特点才能作到力学上 等价,这是模型化的关键。同时还应注意结构对称 化的应用。
3.载荷模型化
载荷模型化的目的是,选择船舶在营运中可能遇 到的较危险的和经常性的荷重情况,并能用有限参 数来描述实际载荷。具体应考虑如下问题: (1) 作用于结构上的载荷工况; (2) 计算载荷的性质(不变荷重、静变荷重、动变荷 重和冲击荷重)与载荷类型(经常性荷重、偶然性 荷重); (3)载荷大小,并决定施加在哪些构件上; (4)载荷的组合与搭配。
第3章 船体结构局部 强度计算
§1 局部强度计 算的力学模型
船体在外力作用下除发生总纵弯曲变形外,各 局部结构,如船底、甲板、船侧和舱壁板架以及横 向肋骨框架也会因局部载荷作用而发生变形、失稳 或破坏。研究它们的强度问题你入局部强度。局部 强度的内容很多,除上述板梁和框架外,各种骨材 以及壳板的强度计算也是局部强度讨论的对象。 在进行局部强度计算时,首先,应根据结构受力 与变形特点,把实际复杂的结构抽象为可以用力学 方法计算的简化模型(称为力学模型或计算模型); 然后,对共力学模型进行内力和应力分析并进行强 度校核。力学模型的建立是与计算方法相联系的, 用船舶结构力学方法进行局部强度计算时,只能将 船体各部分结构简化为板架、刚架、连续梁和板等 结构进行计算,而且载荷也只能取比较简单情况。
1)受压骨架带板宽度
b
a
s
We /2
cr s
由船舶结构力学知,长为b宽为a,筒形弯曲刚度为 D的矩形板格的临界压力为F cr =kπ 2 D/b 2 。若令有效 宽度内的压应力达到板格的临界应力 cr 。和板的屈 2 2 2 s,则 服极限 4 D 4 Et
2 于是可得带板宽度为: e
中拱和中垂时的水压力。于是纵骨弯曲应力为:
3 M / W (MPa)
W为纵骨带带板的剖面模数,mm3。
3) 船底板架计算
船底板架由多根交叉构件和很多主向梁组成的板 架。其结构强度比强力甲板靠近船体剖面中和轴线。 因此在船体中拱变形时船底板架不易失稳,其主要 矛盾是强度问题。 对于横骨架式板架,主向梁(实肋板)承受肋板 间距范围内荷重,交叉构件只承受节点反力,如图 所示。 实肋板 组合肋板
1) 船底外板的强度计算
受均布水压力作用的船底板,可作为四周刚性固 底纵桁 y 定的刚性板来计算。 实 实 (1)横骨架式船底板 肋 肋 c板 1 2 板 检查三点。 x 若c/s>2,则长边中点(2点)的 最大应力(沿船长方向)用下式计算: 底纵桁 s ( 4 )x 0.5q(s / t )2 (N/mm2 ,MPa) 板中点(1点)沿船长方向的应力为:
3.骨架的支座简化
将局部构件或结构从整体结构中分离出来进行局 部强度计算,需考虑相邻构件对计算结构的影响, 即支座。在船体结构计算中,通常有三种支座情况: (1)自由支持在刚性支座上;(2)刚性固定;(3)弹性 支座和弹性固定。 肋板 纵骨 如船底纵骨:
因实肋板刚性远大于纵骨, 且变形以肋板为支点左右 对称,因此计算船底纵骨 强度时可按两端刚性固定 的单跨梁来进行。
s cr
W (1 t ) 12(1 )W
2

2 e
4 2 E We t 2 12 (1 ) S
2)骨架弯曲带板宽度
骨架弯曲时与腹板连接的面板也跟着伸长或缩短, 板变形的主要原因是腹板边缘给它的剪切,其次才 是弯曲影响。在腹板上面的面板部分弯曲应力最大, 沿面板宽度离开腹板逐渐减小,这种现象称为“剪 切滞后”效应。带板宽度be 就是将面板宽度b中的弯 曲应力化成腹板上面的面板中的应力时所需要的面 板宽度。计算 be 时所用的应力 x是骨架弯曲时其带 板x方向(骨架方向)的正应力。将 x沿y方向(横向) 从 零积分到b/2就得到轴向力x的一半。因此,弯曲带 b/2 b/2 板宽度定义为: 2 x dy t 2 x dy X be 0 0 max t max t max
计算板架时,其长度、宽度取相应的支持构件间距离,如 船底板架和甲板板架的长度取横舱壁之间的距离,宽度取组 成肋骨框架梁中和轴的跨距,或取为船宽。 对于如图所示的肋骨刚架,其长度、宽度取组成肋骨框架 梁的中和轴线交点间距离,用中和轴线代替实际构件,不计梁 拱及舭部的弯曲。肘板和开孔 (人孔、减轻孔等) 而引起的构 件剖面变化也不予考虑,即在内力(弯矩、切力)计算时把每 一构件作为等直梁处理。
a a Q=γhaB=qaB b b b b B b b L Rj1 Rj2 Rj3 Rj4 L 交叉构件 b b 主向 主向梁
对于纵骨架式板架,载荷通过纵骨传给实肋板, 交叉构件也只承受节点反力,如图所示。
a a
实肋板 纵骨
Q=γhaB=qaB
主向梁
b b b B b b L Rj1 Rj2 Rj3 b
(4)计算工具 使用的计算工具愈先进,计算简图 则可以更精确些,电子计算机的使用使许多复杂的 计算图形可采用。 此外,必须注意,从实际结构得出合理的计算简 图是一方面;另一方面,在选定计算简图之后,还 应采用适当的结构措施,使所设计出的结构体现计 算简图的要求。
2.构件几何尺寸的简化
在进行局部强度计算时,不可能也没有必要对 实际结构的各种因素加以考虑。在确定几何要素 (如 跨距、宽度、带板尺寸、剖面模数等)时,将结构 作 一些“理想化”处理。
l
而甲板纵骨,在船舶中垂弯曲时受轴向压力作用。纵骨 稳定性计算时,根据其变形特点可作为两端自由支持的单跨 梁来计算。 肋骨框架由于肋板刚度远大于肋骨,故肋骨下端可作刚性 固定;当甲板上无荷重,又可进一步按船舶结构力学方法, 可算出其弹性固定端的转角和柔性系数而简化为弹性固定的 单跨梁。 板架的交叉构件(龙骨、纵桁)在横舱壁处的固定条件取 决于相邻板架的刚度、跨度和载荷之比。精确计算相邻板架 的相互影响,必须对它们进行连续板架计算;但实用上,通 常引入横舱壁的支座固定系数χ考虑相邻板架的影响。在多 数情况下,交叉构件在横舱壁处可以认为是刚性固定的。船 底板架在舷侧处的固定情况可通过肋骨刚架计算确定,通常 计算中可近似认为自由支持在舷侧,因为肋骨的刚度比肋板 小得多。
b
b
l
R之比为L/B<0.8时,中桁材舱壁处 与跨中的弯矩十分接近,因此可将中桁材当单跨梁 处理进行强度校核计算。其弯矩为:
M 0 QB /12 (支座处) M1 QB / 24 (跨中处)
对于L/B>0.8的板架,可按近似公式计算: 中桁材弯矩: M 0 1QB /12 M 1 2QB / 24 央肋板在中伤材处弯矩:
船体结构是在线弹性范围内进行强度校核,因此 在复杂载荷作用时可以应用迭加原理计算。 局部强度计算载荷主要有货物重量和水压力,一 般不计结构自重影响。 货物重量通常用水头高度h表示,即h=H/1.35(m), H为货舱载货高度(m)。 对于水压力,一般以船舶静置于波浪上的静水压 力作为计算载荷,因此水头高度为:
1.建立计算力学模型的原则
船体结构的强度计算, 首先应根据结构的实际受 力情况,将具体结构抽象化为计算简图—力学模型, 然后对计算简图采用力学分析力法进行结构分析。 所谓结构的计算简图,就是将实际结构经过简 化的计算模型。由于实际结构的繁杂性,完全按结 构的实际情况进行力学分析是不可能的,也是不必 要的。因此,对实际结构进行力学计算之前,必须 对结构进行简化,略去不重要的细节,表现其基本 特点,用一个简化的图形代替实际结构。但其力学 模型必须:(1)反映实际结构的工作性能;(2)便于计 算。
相关文档
最新文档