3第三章习题
3 习题
第三章 分子物理学习 题一、单选题１、两种理想气体，温度相等，则（ ）。
A ．内能必然相等B ．分子的平均平动动能必然相等C ．分子的平均总动能必然相等D ．分子的平均能量必然相等２、若用p 、ρ和T 分别表示理想气体的压强、质量密度和热力学温度，k 、R 、N 0分别为玻尔兹曼常数、普适气体常数和阿伏伽德罗常数，则理想气体的摩尔质量M 可表示为（ ）。
A ．pρM 0N = B ．T pρM R N 0= C . T k p ρM =D ．T pρM k N 0=３、在温度t ℃时，单原子分子理想气体的内能为气体分子的（ ）。
A ．部分势能与部分动能之和 B ．全部平动动能 C ．全部势能 D ．全部振动能量４、两容器中分别贮有两种双原子理想气体，已知它们的压强相同，体积也相同，则（ ）。
A ．它们的内能一定相等 B ．它们中温度较高的内能较多 C ．它们中分子数较多的内能较多 D ．它们中质量较大的内能较多５、一摩尔氮气，在平衡态时，设其温度为T ，则氮气分子（视为刚性分子）的平均总动能k e 和氮气的内能E 分别为（ ）。
A ．k e =23k T , E =23R T ；B ．k e =25k T , E =23R T ；C ．k e =25k T , E =25R T ； D ．k e =3 kT, E =21R T ； ６、氢气和氧气的温度相同时，二者分子的方均根速率之比为（ ）。
A ．2﹕1 B ．4﹕1 C ．8﹕1 D ．1﹕2７、理想气体的温度为T 时，气体分子的平均平动动能为（ ）。
A 、21k T B . 23k T C ．2i k T D ．2i R T ８、两瓶不同种类的理想气体，若分子的平均平动能相同，则（ ）。
A ．温度一定相同B ．温度一定不相同C ．压强一定相同D ．分子数密度一定相同９、若p 、ρ、m 和T 分别表示理想气体的压强、密度、分子质量和热力学温度，则下列式子中表示气体分子方均根速率的是（ ）。
3静力学第三章习题答案
第三章 部分习题解答3-10 AB ,AC 和DE 三杆连接如图所示。
杆DE 上有一插销H 套在杆AC 的导槽内。
试求在水平杆DE 的一端有一铅垂力F 作用时，杆AB 所受的力。
设DE BC HE DH DB AD ===,,，杆重不计。
解：假设杆AB ，DE 长为2a 。
取整体为研究对象，受力如右图所示，列平衡方程：∑=0C M02=⋅a F By0=By F取杆DE 为研究对象，受力如图所示，列平衡方程：∑=0HM0=⋅-⋅a F a F DyF F Dy =∑=0B M 02=⋅-⋅a F a F DxF F Dx 2=取杆AB 为研究对象，受力如图所示，列平衡方程：∑=0y F0=++By Dy Ay F F FF F Ay -=(与假设方向相反)∑=0A M02=⋅+⋅a F a F Bx DxF F Bx -=(与假设方向相反) ∑=0B M02=⋅-⋅-a F a F Dx AxF F Ax -=(与假设方向相反)3-12AD AC AB ,,和BC 四杆连接如图所示。
在水平杆AB 上作用有铅垂向下的力F 。
接触面和各铰链均为光滑的，杆重不计，试求证不论力F 的位置如何，杆AC 总是受到大小等于F 的压力。
解：取整体为研究对象，受力如图所示，列平衡方程：∑=0C M0=⋅-⋅x F b F DF bx F D =F CF C yF DF CxF CyF BxF ByF DxF DyF HyF BxF ByF DyF DxF Ax F Ay取杆AB 为研究对象，受力如图所示，列平衡方程：∑=0A M0=⋅-⋅x F b F BF bx F B =杆AB 为二力杆，假设其受压。
取杆AB 和AD 构成的组合体为研究对象，受力如图所示，列平衡方程：∑=0E M02)2(2)(=⋅--⋅+⋅+bF x b F b F F AC D B解得F F AC =，命题得证。
注意：销钉A 和C 联接三个物体。
第三章习题解答
第3章 力学基本定律与守恒律 习题及答案1．作用在质量为10 kg 的物体上的力为i t F)210(+=N ，式中t 的单位是s ，(1)求4s 后，这物体的动量和速度的变化．(2)为了使这力的冲量为200 N ·s ，该力应在这物体上作用多久，试就一原来静止的物体和一个具有初速度j 6-m ·s -1的物体,回答这两个问题．解: (1)若物体原来静止，则i t i t t F p t 1401s m kg 56d )210(d -⋅⋅=+==∆⎰⎰,沿x 轴正向，ip I imp v111111s m kg 56s m 6.5--⋅⋅=∆=⋅=∆=∆ 若物体原来具有6-1s m -⋅初速，则⎰⎰+-=+-=-=t tt F v m t m F v m p v m p 000000d )d (,于是⎰∆==-=∆t p t F p p p 0102d,同理， 12v v ∆=∆,12I I=这说明，只要力函数不变，作用时间相同，则不管物体有无初动量，也不管初动量有多大，那么物体获得的动量的增量(亦即冲量)就一定相同，这就是动量定理． (2)同上理，两种情况中的作用时间相同，即⎰+=+=tt t t t I 0210d )210(亦即 0200102=-+t t 解得s 10=t ，(s 20='t 舍去)2．一颗子弹由枪口射出时速率为10s m -⋅v ，当子弹在枪筒内被加速时，它所受的合力为 F =(bt a -)N(b a ,为常数)，其中t 以秒为单位：(1)假设子弹运行到枪口处合力刚好为零，试计算子弹走完枪筒全长所需时间；(2)求子弹所受的冲量．(3)求子弹的质量． 解: (1)由题意，子弹到枪口时，有0)(=-=bt a F ,得ba t =(2)子弹所受的冲量⎰-=-=tbt at t bt a I 0221d )(将bat =代入，得 ba I 22=(3)由动量定理可求得子弹的质量202bv a v I m == 3．如图所示，一质量为m 的球，在质量为M 半径为R 的1/4圆弧形滑槽中从静止滑下。
3第三章 不饱和烃习题
第三章 不饱和烃习题(P112)(一)用系统命名法命名下列各化合物：(2) 对称甲基异丙基乙烯12345CH 3CH=CHCH(CH 3)24-甲基-2-戊烯2,2,5-三甲基-3-己炔3-异丁基-4-己烯-1-炔(二)用Z,E-标记法命名下列各化合物：(Z)-1-氟-1-氯-2-溴-2-碘乙烯(三)写出下列化合物的构造式，检查其命名是否正确，如有错误予以改正，并写出正确的系统名称。
(1) 顺-2-甲基-3-戊烯(2) 反-1-丁烯CH 2=CHCH 2CH 3顺-4-甲基-3-戊烯1-丁烯 (无顺反异构)(3) 1-溴异丁烯(4) (E)-3-乙基-3-戊烯2-甲基-1-溴丙烯3-乙基-2-戊烯(无顺反异构)(四)完成下列反应式：解：红色括号中为各小题所要求填充的内容。
(硼氢化反应的特点：顺加、反马、不重排)(五) 用简便的化学方法鉴别下列各组化合物：(六) 在下列各组化合物中，哪一个比较稳定？为什么？解：(B)中甲基与异丙基的空间拥挤程度较小，更加稳定。
解：(A)中甲基与碳-碳双键有较好的σ-π超共轭，故(A)比较稳定。
解：(C)的环张力较小，较稳定。
解：(A)的环张力较小，较稳定。
(七) 将下列各组活性中间体按稳定性由大到小排列成序：解：(1)C＞A＞B (2)B＞C＞A(八) 下列第一个碳正离子均倾向于重排成更稳定的碳正离子，试写出其重排后碳正离子的结构。
(九) 在聚丙烯生产中，常用己烷或庚烷作溶剂，但要求溶剂中不能有不饱和烃。
如何检验溶剂中有无不饱和烃杂质？若有，如何除去？解：可用Br2/CCl4或者KMnO4/H2O检验溶剂中有无不饱和烃杂质。
若有，可用浓硫酸洗去不饱和烃。
(十) 写出下列各反应的机理：解：解：该反应为自由基加成反应：… …终止：略。
(箭头所指方向为电子云的流动方向!)(十一) 预测下列反应的主要产物，并说明理由。
双键中的碳原子采取sp 2杂化，其电子云的s 成分小于采取sp杂化的叁键碳，离核更远，流动性更大，解释：在进行催化加氢时，首先是H 2及不饱和键被吸附在催化剂的活性中心上，而且，叁键的吸附速度大于双键。
第三章习题
第三章婴儿心理的发展第一部分习题一、选择题1. 胎儿的大脑形成时间是( )A.20周左右B.22周左右C.25周左右D.18周左右2. 新生儿出生后就会显示痛苦表情，但是新生儿愤怒的表情出现于( )A.5个月B.8个月C.4个月D.6个月3.认为不愉快反应通常是自然动作的简单增加，由所有不利于机体安全的刺激所引起的是心理学家( )A.林传鼎B.陈鹤琴C.斯皮兹D.霍姆林斯基4. 儿童的基本情绪主要包括微笑、哭泣、兴趣、惊奇、厌恶等，其中最基本的积极情绪是( )A.微笑、惊奇B.微笑、兴趣 C、微笑、厌恶 D.微笑、哭泣5. 乳儿期指的胎儿出生至( )A.2 岁B.3岁C.1岁D.1.5岁6. 提出情绪分化理论认为婴儿出生时就具有五大情绪是心理学家( )A.孟昭兰B.伊扎德C.斯皮兹D.林传鼎7.乳儿能辨别彩色与非彩色在( )A.4至5个月B.3至4个月C.2至3个月D.5至6个月8. 哪个不属于幼儿前期幼儿言语发展阶段( )A.单词句阶段B.多词句阶段C.复杂句阶段D.简单句阶段9. 以聊天，给胎儿讲故事，与胎儿一起欣赏文学作品，画册等方式进行的胎教是( )A.言语胎教B.音乐胎教C.抚摸胎教D.文学胎教10.婴儿期的年龄范围是( )A.0至2岁B.0至4岁C.0至1岁D.0至3岁二、判断题( )1.乳儿期是儿童动作发展最迅速的阶段，其发展是按照一定的顺序和规律进行的。
( )2.乳儿的动作起初是无意的，当他做出各种动作时，既无目的也不知道自己在干什么。
6个月以后逐渐出现有目的的动作。
( )3.教育进度应按每个儿童的特点来决定，不应强求一致或操之过急。
( )4.恐惧是婴儿自出生就有的情绪反应，是一种本能的、反射性的反应。
( )5.刚生下来的宝宝对光线是没有反应的。
( )6.微笑是婴儿的第一个社会性行为，是情绪愉快的表现。
( )7.新生儿的哭很普通，不需要担心或紧张。
( )8.新生儿是指自脐带结扎至出生后满36天的婴儿。
第三章习题——精选推荐
第三章习题3.1 确定T＝300K时GaAs中Ec和Ec＋kT之间的总量子态数量。
3.2 定T＝300K时GaAs中Ev和Ev＋kT之间的总量子态数量。
3.3求出Ec＋kT处导带有效状态密度与Ev－kT处价带有效状态密度的比值。
3.4（a）如果EF＝Ec，试求E＝Ec＋kT处的状态概率；（b）如果EF＝Ev试求E＝Ev－kT处的状态概率。
3.5试确定比费米能级高（a）1kT，（b）5kT和（c）10kT的能带被电子占据的概率。
3.6试确定比费米能级低（a）1kT，（b）5kT和（c）10kT的能带被电子占据的概率。
3.7证明高于费米能级ΔE的量子态被占据的概率与低于费米能级ΔE的量子态为空的概率相等。
3.8某种材料T＝300K时的费米能级为6.25eV。
该材料中的电子符合费米－狄拉克分布函数。
（a）求6.50eV处能级被电子占据的概率。
（b）如果温度上升为T＝950K，重复前面的计算（假设E F不变）。
（c）如果比费米能级低0.30eV处能级为空的概率时1％，此时温度为多少？3.9 铜在T＝300K时的费米能级为7.0eV。
铜中的电子符合费米－狄拉克分布函数。
（a）求7.15eV处能级被电子占据的概率。
（b）如果温度上升为T＝1000K，重复前面的计算（假设EF不变）。
（c）当E＝6.85eV，T＝300K时，重复前面计算。
（d）求T＝300K和T＝1000K时E＝EF的概率。
3.10考虑图3.40所示的能级。
令T＝300K。
（a）如果E1－EF＝0.30eV，确定E＝E1被电子占据的概率以及E＝E2为空的概率。
（b）如果EF－E2＝0.40eV，重复前面的计算。
3.11假设T＝300K时费米能级恰好处于禁带中央。
（a）分别计算Si，Ge和GaAs中导带底被占据的概率。
（b）分别计算Si，Ge和GaAs中价带顶为空的概率。
3.12计算低于费米能级0.55eV的能级被电子占据的概率为106 时的温度。
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:。
新版高中数学人教A版必修3习题：第三章概率 3.1.2(1)
3.1.2概率的意义课时过关·能力提升一、基础巩固1.概率是指()A.事件发生的可能性大小B.事件发生的频率C.事件发生的次数D.无任何意义2.若某篮球运动员的投篮命中率为98%,则估计该运动员投篮1 000次命中的次数为()A.20B.98C.980D.9981000次命中的次数约为1000×98%=980.3.天气预报中预报某地明天降雨的概率为90%,则()A.降雨的可能性是90%B.90%太大,一定降雨C.该地有90%的区域降雨D.降雨概率为90%没有什么意义90%说明明天降雨的可能性是90%.4.已知某学校有教职工400名,从中选举40名教职工组成教职工代表大会,每名教职工当选的概率是110,则下列说法正确的是()A.10名教职工中,必有1人当选B.每名教职工当选的可能性是1 10C.数学教研组共有50人,该组当选教工代表的人数一定是5D.以上说法都不正确5.从一批准备出厂的电视机中随机抽取10台进行质量检查,其中有1台是次品.若用C表示抽到次品这一事件,则下列说法正确的是()A.事件C发生的概率为1 10B.事件C发生的频率为1 10C.事件C发生的概率接近1 10D.每抽10台电视机,必有1台次品6.某医院治疗一种疾病的治愈率为15,若前4位病人都未治愈,则第5位病人的治愈率为()A.1B.4 5C.15D.015,表明每位病人被治愈的可能性均为15,并不是5人中必有1人治愈.故选C.7.在乒乓球、足球等比赛中,裁判员经常用掷硬币或抽签法决定谁先发球,这种方法.(填“公平”或“不公平”),这两种方法都是公平的.因为采用掷硬币得正面、反面的概率相等;采用抽签法,抽到某一签的概率相等.8.某市运动会前夕,质检部门对这次运动会所用的某种产品进行抽检,得知其合格率为99%.若该运动会所需该产品共20 000件,则其中的不合格产品约有件.1-99%=1%,则不合格产品约有20000×1%=200(件).9.某射击教练评价一名运动员时说:“你射中的概率是90%.”则下面两个解释中能代表教练的观点的为.①该射击运动员射击了100次,恰有90次击中目标②该射击运动员射击一次,中靶的机会是90%90%说明中靶的可能性是90%,所以①不正确,②正确.10.为了估计水库中鱼的尾数,使用以下的方法:先从水库中捕出2 000尾鱼,给每尾鱼做上记号,不影响其存活,然后放回水库.经过适当的时间,让其和水库中的其他鱼充分混合,再从水库中捕出500尾,查看其中有记号的鱼,有40尾.试根据上述数据,估计水库中鱼的尾数.n(n∈N*),每尾鱼被捕到的可能性相等,给2000尾鱼做上记号后,从水库中任捕一尾鱼,带记号的概率为2000n.又从水库中捕500尾鱼,有40尾带记号,于是带记号的频率为40500.则有2000n≈40500,解得n≈25000.所以估计水库中有25000尾鱼.二、能力提升1.在给病人动手术之前,外科医生会告知病人或家属一些情况,其中有一项是说这种手术的成功率大约是99%.下列解释正确的是()A.100个手术有99个手术成功,有1个手术失败B.这个手术一定成功C.99%的医生能做这个手术,另外1%的医生不能做这个手术D.这个手术成功的可能性是99%99%,说明手术成功的可能性是99%.2.根据山东省教育研究机构的统计资料,今在校学生近视率约为37.4%.某眼镜商要到一中学给学生配眼镜,若已知该校学生总数为600人,则该眼镜商应带眼镜的数目为()A.374副B.224.4副C.不少于225副D.不多于225副,该校近视生人数约为37.4%×600=224.4,结合实际情况,眼镜商应带眼镜数不少于225副.3.某套数学试题中,有12道选择题,每道选择题有4个选项,其中只有1个选项是正确的,则随机选择其中一个选项正确的概率是14.某家长说:“要是都不会做,每题都随机选择其中一个选项,则一定有3道题答对.”这句话() A.正确 B.错误C.不一定D.无法解释,答对的概率是14说明了对的可能性大小是14.做12道选择题,即进行了12次试验,每个结果都是随机的,那么答对3道题的可能性较大,但是并不一定答对3道题.也可能都选错,或有1,2,4,…,甚至12个题都选择正确.4.玲玲和倩倩下象棋,为了确定谁先走第一步,玲玲对倩倩说:“拿一个飞镖射向如图所示的靶中,若射中区域所标的数字大于3,则我先走第一步,否则你先走第一步”.你认为这个游戏规则公平吗?.(填“公平”或“不公平”),所标的数字大于3的区域有5个,而小于或等于3的区域只有3个,所以玲玲先走的概率是58,倩倩先走的概率是38.所以不公平.★5.某地区牛患某种病的概率为0.25,且每头牛患病与否是互不影响的,今研制一种新的预防药,任选12头牛做试验,结果这12头牛服用这种药后均未患病,则此药.(填“有效”或“无效”)头牛都在服药后未患病,由极大似然法,可得此药有效.6.试解释下列情况的概率的意义:(1)某商场为促进销售,实行有奖销售活动,凡购买其商品的顾客中奖率是0.20;(2)一生产厂家称:我们厂生产的产品合格率是0.98.解::(1)“中奖率是0.20”是指购买其商品的顾客中奖的可能性是20%.(2)“产品的合格率是0.98”是指该厂生产的产品合格的可能性是98%.★7.某种彩票的抽奖是从写在36个球上的36个号码中随机摇出7个.有人统计了过去中特等奖的号码,声称某一号码在历次特等奖中出现的次数最多,它是一个幸运号码,人们应该买这一号码;也有人说,若一个号码在历次特等奖中出现的次数最少,由于每个号码出现的机会相等,则应该买这一号码.你认为他们的说法对吗?36个号码的36个球大小、质量是一致的,严格地说,为了保证公平,每次用的36个球, ,除非能保证用过一次后,球没有磨损、变形.因此,当把这36个球看成每次抽奖中只用了一次时,不难看出,以前抽奖的结果对今后抽奖的结果没有任何影响,他们的说法都是错误的.。
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第三章练习题
一、选择题
1．计算机使用总线结构便于增减外设，同时 C 。
A．减少了信息传输量
B．提高了信息的传输速度
C．减少了信息传输线的条数
2．在三种集中式总线控制中， C 方式响应时间最快。
A．链式查询
B．计数器定时查询
C．独立请求
3．三种集中式总线控制中， A 方式对电路故障最敏感。
A．链式查询
B．计数器定时查询
C．独立请求
4．所谓三总线结构的计算机是指 B 。
A．地址线、数据线和控制线三组传输线
B．I/O总线、主存总线和DMA总线三组传输线
C．I/O总线、主存总线和系统总线三组传输线
5．连接计算机与计算机之间的总线属于 C 总线。
A．片内B．系统C．通信
6．在计数器定时查询方式下，若每次计数从上一次计数的终止点开始，则 B 。
A．设备号小的优先级高
B．每个设备使用总线的机会相等
C．设备号大的优先级高
7．在计数器定时查询方式下，若计数从0开始，则 A 。
A．设备号小的优先级高
B．每个设备使用总线的机会相等
C．设备号大的优先级高
8．在独立请求方式下，若有N个设备，则 B 。
A．有一个总线请求信号和一个总线响应信号
B．有N个总线请求信号和N个总线响应信号
C．有一个总线请求信号和N个总线响应信号
9．在链式查询方式下，若有N个设备，则 C 。
A．有N条总线请求线
B．无法确定有几条总线请求线
C．只有一条总线请求线
10．总线通信中的同步控制是 B 。
A．只适合于CPU控制的方式
B．由统一时序控制的方式
C．只适合于外围设备控制的方式
11.在各种异步通信方式中， C 速度最快。
A．全互锁B．半互锁C．不互锁
12.总线的异步通信方式 A 。
A．不采用时钟信号，只采用握手信号
B．既采用时钟信号，又采用握手信号
C．既不采用时钟信号，又不采用握手信号
13.总线的半同步通信方式 B 。
A．不采用时钟信号，只采用握手信号
B．既采用时钟信号，又采用握手信号
C．既不采用时钟信号，又不采用握手信号
14.“总线忙”信号的建立者是 A 。
A．获得总线控制权的设备
B．发出“总线请求”信号的设备
C．总线控制器
D．CPU
15．下列选项中英文缩写为总线标准的是 D 。
（2010年全国考研题）
A．PCI、CRT、USB、EISA
B．ISA、CPI、VESA、EISA
C．ISA、SCSI、RAM、MIPS
D．ISA、EISA、PCI、PCI-Express
16．系统总线分为数据总线、地址总线、控制总线，划分的根据是 A 。
A．系统总线传输信息的不同
B．系统总线所处的位置
C．系统总线的传送方向
D．系统总线的传送方式
17．CPU芯片中的总线属于 B 。
A．系统总线
B．片内总线
C．板级总线
D．通信总线
二、填空题
1．只能将信息从总线的一端传送到另一端，不能反向传输。
2．衡量总线性能的一个重要指标是，可以理解为总线的数据传输速率，即单位时间内总线传输数据的位数，通常用每秒传输信息的字节数来衡量。
3．在计数器定时查询的总线控制方式中，的设备可以取得总线的使用权。