分布式系统原理与范型考试-2009-12

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分布式系统练习题2009秋(1)

分布式系统练习题2009秋(1)

1、分布式系统中的体系结构样式有那几种?并简述之。

1)分层体系结构:组件组成了不同的层,其中L i层的组件可以调用下面的层L i-1。

其中的关键因素是,其控制系统是从一层到别一层的:请求是从上往下,而请求结果是从下向上。

2)基于对象的体系结构:是一种很松散的组织结构。

基本上每一个对象都对应一个组件,这些组件是通过过程调用机制来连接的。

分层和面向对象的体系结构仍然是大型软件系统最重要的样式。

3)以数据为中心的体系结构,其发展思想:进程通信需要通过一个公用(被动或主动的)仓库。

可以分成两个关键的部分:展现当前状态的中央数据结构和一个对数据进行操作的独立组件的集合4)基于事件的体系结构,进程基本上是通过事件的抟播来通信的,事件传播还可以有选择地携带数据。

基本思想是:进程发布事件,然后,中间件将确保那些订阅了这些事件的进程才接收它们,优点是进程是松散耦合的。

2、简述分布式系统设计所面临的问题及遇到的挑战。

1):难以合理设计分配策略,在集中式系统中,所有的资源都由系统系统管理和分配,但在分布式系统中,资源属于局部工作站或个人计算机,所以调度的灵活性不如集中式系统,资源的物理分布可能与服务器的分布不匹配,某些资源可能空闲,而另外一些资源可能超载。

2):部分失效问题:由于分布式系统通常是由若干部分组成的,各个部分由于各种各样的原因可能发生故障,如硬件故障。

如果一个分布式系统不对这些故障对这些问题进行有效的处理,系统某个组成部分的故障可能导致整个系统的瘫痪。

3)性能和可靠性过分依赖于网络:由于分布式系统是建立在网络之上的,而网络本身是不可靠的,可能经常发生故障,网络故障可能导致整个系统的终止;另外,网络超负荷会导致性能下降,增加系统的响应时间。

4)缺乏统一控制:一个分布式系统的控制通常是一个典型的分散式控制,没有统一的中心控制。

因此,分布式系统通常需要相应的同步机制来协调系统中各个部分的工作;5)安全保密性问题:为了获得可扩展性,分布式系统中的许多软件接口都提供给用户,这样的开放结构对于开发人员非常有价值,担同时也为破坏者打开了方便之门挑战:设计与实现一个对用户来说是透明的且具有容错能力的分布式系统。

分布式数据库系统考试

分布式数据库系统考试

分布式数据库系统考试(答案见尾页)一、选择题1. 分布式数据库系统的定义是什么?A. 一种将数据存储在多个地理位置的数据库系统中,通过分布式计算框架来管理和访问数据的一种技术。

B. 一种单一的集中式数据库系统,所有数据都存储在一个服务器上。

C. 一种将数据分割成多个部分,并分布存储在不同的服务器上的数据库系统。

D. 一种不依赖于单一服务器的数据库系统,数据可以跨多个服务器进行存储和访问。

2. 分布式数据库系统的优点包括哪些?A. 提高数据处理速度和效率。

B. 降低单点故障的风险。

C. 更好的数据冗余和容错能力。

D. 扩展性更强,可以更容易地添加新的数据和节点。

3. 以下哪个不是分布式数据库系统中的常见拓扑结构?A. 星形拓扑B. 环形拓扑C. 网状拓扑D. 树形拓扑4. 在分布式数据库系统中,什么是分片?A. 将整个数据库系统的数据分成多个部分,每个部分存放在一个单独的节点上。

B. 将数据库系统的一个或多个表按照某种规则分成多个部分。

C. 将数据库系统的数据按照某种规则分成多个部分,每个部分存放在一个单独的节点上。

D. 将数据库系统的一个或多个表按照某种规则分成多个部分,并存放在不同的节点上。

5. 在分布式数据库系统中,什么是复制?A. 将数据库系统的数据复制到多个节点上,以确保数据的可靠性和可用性。

B. 将数据库系统的数据存储在多个地理位置,以确保数据的可靠性和可用性。

C. 将数据库系统的数据按照某种规则分成多个部分,并存放在不同的节点上。

D. 将数据库系统的一个或多个表按照某种规则分成多个部分,并存储在不同的节点上。

6. 在分布式数据库系统中,什么是分布式事务?A. 一种需要在多个节点上同步更新数据的事务处理方式。

B. 一种可以在多个节点上并行处理的事务处理方式。

C. 一种需要确保数据的一致性和完整性的事务处理方式。

D. 一种可以在多个节点上同时执行的事务处理方式。

7. 分布式数据库系统中的数据一致性是指什么?A. 数据在多个节点上保持一致的状态。

分布式系统原理与范型 考试 2009 答案

分布式系统原理与范型 考试 2009 答案
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it is sent, or even at the same time it is sent, since it takes a finite, nonzero amount of time to arrive. 7. finger table (chap 5) Instead of linear approach toward key lookup, each Chord node maintains a finger table of at most m entries. If FTp denotes the finger table of node p, then FTp[i] = succ (p + 2i-1) Put in other words, the i-th entry points to the first node succeeding p by at least 2i-1. 8. out of band data (chap 3) Data is to be processed by the server before any other data from that client. 9. MapReduce (5pt)
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二、简答题(共70分) 1. Q: What is the difference between a vertical distribution and a horizontal distribution? (chap 2, 5pt) A: Vertical distribution refers to the distribution of the different layers in a multitiered architectures across multiple machines. In principle, each layer is implemented on a different machine. Horizontal distribution deals with the distribution of a single layer across multiple machines, such as distributing a single database. 2. Q: Is a server that maintains a TCP/IP connection to a client stateful or stateless? (chap 3) A: Assuming the server maintains no other information on that client, one could justifiably argue that the server is stateless. The issue is that not the server, but the transport layer at the server maintains state on the client. What the local operating systems keep track of is, in principle, of no concern to the server. 3. Q: One way to handle parameter conversion in RPC systems is to have each machine send parameters in its native representation, with the other one doing the translation, if need be. The native system could be indicated by a code in the first byte. However, since locating the first byte in the first word is precisely the problem, can this actually work? (chap 4) A: First of all, when one computer sends byte 0, it always arrives in byte 0. Thus the destination computer can simply access byte 0 (using a byte instruction) and the code will be in it. It does not matter whether this is the low-order byte or the high-order byte. An alternative scheme is to put the code in all the bytes of the first word. Then no matter which byte is examined, the code will be there. 4. Q: Routing tables in IBM WebSphere, and in many other message-queuing systems, are configured manually. Describe a simple way to do this automatically. (chap 4) A: The simplest implementation is to have a centralized component in which the topology of the queuing network is maintained. That component simply calculates all best routes between pairs of queue managers using a known routing algorithm, and subsequently generates routing tables for each queue manager. These tables can be downloaded by each manager separately. This approach works in queuing networks where there are only relatively few, but possibly widely dispersed, queue managers. 5. Q: Is an identifier allowed to contain information on the entity it refers to? (chap 5) A: Yes, but that information is not allowed to change, because that would imply changing the identifier. The old identifier should remain valid, so that changing it would imply that an entity has two identifiers, violating the second property of identifiers. 6. Q: When a node synchronizes its clock to that of another node, it is generally a good idea to take previous measurements into account as well. Why? Also, give an example of how such past readings could be taken into account. (chap 6) A: The obvious reason is that there may be an error in the current reading. Assuming that clocks need only be gradually adjusted, one possibility is to consider the last N values and compute a median or average. If the measured value falls outside a current interval, it is not taken into account (but is added to the list). Likewise, a new value can be computed by taking a weighted average, or an aging algorithm.

分布式系统原理与范型课后习题答案

分布式系统原理与范型课后习题答案

分布式系统原理与范型课后习题答案专业专心专注第一章绪论1、中间件在分布式系统中扮演什么角色,答:中间件主要是为了增强分布式系统的透明性(这正是网络操作系统所缺乏的),换言之,中间件的目标是分布式系统的单系统视图。

2、解释(分布)透明性的含义,并且给出各种类型透明性的例子。

答:分布透明性是一种现象,即一个系统的分布情况对于用户和应用来说是隐藏的。

包括:访问透明、位置透明、移植透明、重定位透明、复制透明、并发透明、故障透明和持久性透明。

3、在分布式系统中,为什么有时难以隐藏故障的发生以及故障恢复过程,答:通常,要探测一个服务器是停止服务还是该服务器的反应变慢这些情况是不可能的。

因此,一个系统可能在服务响应变慢的时候报告该服务已经停止了。

4、为什么有时候要求最大程度地实现透明性并不好,答:最大程度地实现透明性可能导致相当大的性能损失,从而导致用户无法接受。

5、什么是开放的分布式系统,开放性带来哪些好处,答:开放的分布式系统根据明确定义的规则来提供服务。

开放系统能够很容易地与其它系统协作,同时也允许应用移植到同一个系统的不同实现中。

6、请对可扩展系统的含义做出准确描述答:一个系统的可扩展包含下面几个方面:组件的数量、几何尺寸、管理域的数量与尺寸,前提是这个系统可以在上面几个方面进行增加而不会导致不可接受的性能损失。

7、可以通过应用多种技术来取得可扩展性。

请说出这些技术。

答:可扩展性可以通过分布式、复制和缓存来获得。

8、多处理器系统与多计算机系统有什么不同,答:在多处理器系统中,多个CPU访问共享的主存储器。

在多计算机系统中没有共享存储器,CPU之间只能通过消息传递来进行通信。

9、在多计算机系统中的256个CPU组成了一个16 X 16的网格方阵。

在最坏的情况下,消息的延迟时间有多长(以跳(hop)的形式给出,跳是结点之间的逻辑距离),答:假设路由是最优的,最长的路由是从网格方阵的一个角落到对角的角落。

那么这个路由的长度是30跳。

《分布式系统原理与范型--(第二版)》复习资料

《分布式系统原理与范型--(第二版)》复习资料

分布式复习资料第1章分布式系统是若干独立计算机的结合,这些计算机对于用户来说就像是单个相关系统。

硬件方面:机器本身是独立的。

软件方面:对用户来说就像与单个系统打交道。

重要特性:1、各种计算机之间的差别以及计算机之间的通信方式的差别对用户是隐藏的。

2、用户和应用程序无论在何时何地都能够以一种一致和统一的方式与分布式系统进行交互。

中间件:为了使种类各异的计算机和网络都呈现为单个的系统,分布式系统常常通过一个“软件层”组织起来。

该“软件层”在逻辑上位于由用户和应用程序组成的高层与由操作系统组成的低层之间。

如图,这样的分布式系统有时又称为中间件。

注意层次分布与组件分布式系统的最主要目标是使用户能够方便地访问远程资源,并且以一种受控的方式与其他用户共享这些资源。

透明性:如果一个分布式系统能够在用户和应用程序面前呈现为单个计算机系统,这样的分布式系统就是透明的。

透明的类型:1、访问透明性:指对不同数据表示形式以及资源访问方式的隐藏。

2、位置透明性:指用户无法判别资源在系统中的物理位置。

3、并发透明性:在资源共享时,用户不会感觉到他人也在使用自己正使用的资源。

4、故障透明性:用户不会注意到某个资源(也许他从未听说过这个资源)无法正常工作,以及系统随后从故障中恢复的过程。

开放性:一个开放式的分布式系统,是根据一系列准则来提供服务,这些准则描述了所提供服务的语法和含义。

互操作性:刻画了来自不同厂商的系统或组件的两种实现能够在何种程度上共存并且协同工作,这种共存和协同工作只能依赖于通过双方在公共标准中规定的各自所提供的服务来完成。

可移植性:刻画了这样的性能,如果为分布式系统A开发了某个应用程序,并且另一个分布式系统B与A具有相同的接口,该应用程序在不做任何修改的情况下在B上执行的可行程度。

可扩展性:当一个系统需要进行扩展时,必须解决多方面的问题。

首先考虑规模上的扩展。

在需要支持更多的用户或资源时,我们常常收到集中的服务、数据以及算法所造成的限制,如图所示。

分布式系统原理与泛型

分布式系统原理与泛型

分布式系统原理与泛型一、引言随着互联网的发展和应用的深入,分布式系统成为了当今计算机领域一项重要的技术。

分布式系统是由多个独立的处理器或计算机节点组成,这些节点通过网络进行通信和协作,共同完成特定的任务。

而泛型是一种编程范式,它通过参数化类型的方式来实现代码的复用和泛化。

本文将对分布式系统原理和泛型进行详细介绍,并探讨它们之间的关系。

二、分布式系统原理1. 定义分布式系统是由多个自治的计算机节点组成,这些节点通过网络进行通信和协作,共同完成特定的任务。

分布式系统的设计目标是提高系统的可靠性、可扩展性和性能。

2. 特点(1)分布性:分布式系统的核心特点是由多个节点组成,节点之间可以独立运行,并通过网络进行通信。

(2)并发性:分布式系统的节点可以并行地执行任务,提高系统的处理能力。

(3)故障容忍性:分布式系统可以通过冗余设计和容错机制来提高系统的可靠性,即使一个节点发生故障,系统仍然可以正常工作。

(4)透明性:分布式系统可以隐藏底层的细节,使用户感觉系统是一个整体,而不用关心具体的实现细节。

3. 架构模型(1)客户端-服务器模型:客户端向服务器发出请求,服务器处理请求并返回结果。

常见的应用有Web服务器和数据库服务器。

(2)对等模型:所有节点都是对等的,可以相互通信和协作。

常见的应用有文件共享和P2P网络。

(3)集群模型:多个节点组成一个集群,在集群中共享资源,提高系统的可靠性和性能。

常见的应用有负载均衡和高可用性系统。

三、泛型1. 定义泛型是一种编程范式,它通过参数化类型的方式来实现代码的复用和泛化。

通过使用泛型,我们可以编写出更加通用和灵活的代码,提高代码的可读性和可维护性。

2. 泛型的优势(1)代码复用:通过使用泛型,我们可以编写出适用于多种类型的通用代码,避免重复编写相似的代码。

(2)类型安全:使用泛型可以在编译期间检查类型的正确性,避免在运行时出现类型错误。

(3)可读性和可维护性:泛型代码可以更加清晰和易于理解,减少了类型转换和强制转换的代码。

分布式操作系统、分布式系统数据库试题

分布式操作系统、分布式系统数据库试题

分布式操作系统、分布式系统数据库[填空题]1什么是主从式多机操作系统?它有什么优缺点?参考答案:主从式多机操作系统的工作原理最简单,许多在单机系统上使用的软件都可以在此系统的管理下运行。

它的主要特点是监控管理程序始终由同一个主处理机执行,从机的任务分配完全由主机负责。

如果从机需要主机的服务,可向主机申请,等待主机执行相应的管理程序。

主从式操作系统对软硬件要求简单,适合于工作负荷较轻且比较明确的应用场合,特别是从机能力小于主机的非对称情况。

许多采用服务器——工作站类型的微机网络操作系统即属于主从式操作系统。

主从式系统要求系统具备一台主处理机和多台从机,缺乏灵活性,在控制和利用全部系统资源方面效率较低,而且主机故障会导致整个系统的停机。

[填空题]2什么是独立式多机操作系统?它有什么优缺点?参考答案:在这种方式下,各个处理机执行各自的监控程序和其它可执行模块,为自己的需要服务,其自治程度类似于多个单机系统。

独立式系统中各处理机控制各自的I/O设备,共享程度差,I/O设备的机构需要手工切换。

独立式系统自治程度高,不会因为个别处理机故障导致整个系统失效,但各处理机可能存在负载不平衡,而且故障的处理机重新启动并继续原来的工作往往是很困难的。

[填空题]3什么是分布式多机操作系统?它有什么优缺点?参考答案:这种方式的最初目标是最大限度地利用各个处理机,提高系统的整体处理能力。

在这种方式下,内存、I/O通道等资源都可以为系统所共享。

每个处理机都可以执行监控程序,并且可以多台同时执行,不存在固定的主从关系。

实际上,监控程序的执行者是浮动的。

当现行任务被中断或已完成时,接受新任务的调度都由各个处理机分别完成。

这样显然有利于加快系统响应,提高系统的处理能力。

采用这种方式容易实现故障状态下的降级运行,实现冗余和容错,提高系统的利用率,同时也容易做到各处理机的负载平衡,最充分地利用系统资源。

[填空题]4什么是“死锁”?出现死锁的条件有哪些?参考答案:单机系统当程序出错或某一外部条件始终不能满足时,就可能出现死循环或无休止的等待状态,即称为死锁。

分布式相关的面试题

分布式相关的面试题

分布式相关的面试题
以下是一些与分布式系统相关的面试题:
1. 什么是分布式系统?请解释其特点和优势。

2. 解释一下CAP定理是什么?为什么在分布式系统中存在CAP定理的限制?
3. 请简要解释一下一致性哈希算法的原理以及其在分布式系统中的应用。

4. 什么是分布式缓存?列举一些常见的分布式缓存系统,并比较它们的优缺点。

5. 请解释一下分布式事务是如何保证数据的一致性和可靠性的。

6. 在分布式系统中,什么是消息队列?列举一些常见的消息队列系统,并比较它们的特点和适用场景。

7. 解释一下Raft一致性算法的原理以及相对于Paxos算法的优势。

8. 请简要解释一下分布式锁是什么,并说明一下分布式锁的实现方式和应用场景。

9. 什么是微服务架构?列举一些常见的微服务框架,并说明它们的特点和适用场景。

10. 在分布式系统中,如何解决网络通信的延迟和故障问题?请举例说明。

以上问题只是分布式系统领域中的一小部分内容。

在面试中,根据具体的岗位要求和面试官的提问,可能会涉及更深入的领域和问题。

建议你在准备面试时,除了关注以上问题,还要广泛了解分布式系统的基本原理、常用技术和实际应用,并能够结合实际案例进行分析和讨论。

分布式操作系统考试题

分布式操作系统考试题

四、质量标准及验收
1、乙方应按照国家相关标准及规范进行施工,确保工程质量达到合格标准。 2、甲方有权对施工过程进行监督和检查,并对不合格的部位提出整改意见。
3、工程完工后,双方共同进行验收,验收合格后方可投入使用。
五、合同价款及支付方式
1、本工程预算为元人民币,具 体价款以实际工程量为准。
一、工程概况
1、工程名称:脚手架施工合同
3、工程范围:脚手架搭建及拆 除等施工工作。
二、合同工期
1、脚手架施工自X年月日开始, 至X年月日结束。
2、如遇不可抗力因素导致工程无法正常进行,双方应协商解决。
三、合同价款及支付方式
1、脚手架施工费用为人民币 (以下简称“元”)万元整。
2、自合同签订之日起五个工作日内,甲方向乙方支付万元作为预付款。
5、本合同未尽事宜,双方可另 行协商补充。
6、本合同一式两份,甲乙双方各执一份,具有同等法律效力。自双方签字 盖章之日起生效。建筑工程脚手架施工合同协议
甲方:
乙方:
根据《中华人民共和国合同法》及相关法律法规的规定,为明确双方在建筑 工程脚手架施工中的权利和义务,保障工程质量,现经双方协商一致,签订本协 议:
3、余款在脚手架施工结束并验 收合格后五个工作日内付清。
4、支付方式:银行转账或现金 支付。
四、质量要求及验收标准
1、乙方应按照国家相关标准和施工图纸要求进行施工,确保脚手架的牢固 性和稳定性。
2、脚手架材料应符合国家相关质量标准和施工图纸要求,并经过质量检验 合格。
3、验收标准:脚手架施工结束后,乙方应按照国家相关标准和施工图纸要 求进行自检,并通知甲方进行验收。甲方应在接到通知后五个工作日内进行验收, 并签署验收报告。

分布式数据库系统原理与应用考试

分布式数据库系统原理与应用考试

分布式数据库系统原理与应用考试(答案见尾页)一、选择题1. 分布式数据库系统的定义及特点是什么?A. 分布式数据库系统是由多个物理数据库组成的,它们可以分布在不同的地理位置。

B. 分布式数据库系统提供了一个透明的、逻辑上集中、物理上分布的数据存储,使用户感觉好像数据只存储在一个数据库中。

C. 分布式数据库系统通过数据复制和分片技术实现数据的冗余和容错。

D. 分布式数据库系统的主要目标是提高数据访问性能和数据一致性。

2. 以下哪个选项不是分布式数据库系统的一致性策略?A. 串行处理B. 两阶段提交协议C. 检索优化D. 乐观并发控制3. 在分布式数据库系统中,如何实现数据分片?A. 通过范围分区B. 通过列表分区C. 通过哈希分区D. 通过目录分区4. 分布式数据库系统中的复制策略有哪些?A. 同步复制B. 异步复制C. 混合复制D. 并发复制5. 分布式数据库系统中的数据一致性是如何保证的?A. 通过分布式事务协议B. 通过分布式锁机制C. 通过数据复制和分片D. 通过备份和恢复机制6. 什么是分布式数据库系统的CAP理论?A. 一致性、可用性和分区容错性不能同时满足B. 一致性、可用性和分区容错性可以同时满足C. 一致性、可用性和分区容错性之间存在权衡D. 以上都不是7. 在分布式数据库系统中,如何实现数据镜像?A. 通过主从复制B. 通过分片C. 通过复制集D. 通过日志备份8. 分布式数据库系统中的分片有哪几种类型?A. 范围分片B. 列表分片C. 哈希分片D. 直接分片9. 什么是分布式数据库系统中的读写分离?A. 将读操作和写操作分开在不同的节点上执行B. 将读操作和写操作集中在同一个节点上执行C. 将写操作分散到多个节点上执行,而读操作集中在一个节点上执行D. 将写操作集中在一个节点上执行,而读操作分散到多个节点上执行10. 分布式数据库系统中的故障恢复策略有哪些?A. 主从复制恢复B. 副本恢复C. 分片恢复D. 重建恢复11. 以下哪个不是分布式数据库系统的常见分区策略?A. 节点分区B. 范围分区C. 距离分区D. 列分区12. 分布式数据库系统中,分布式事务的处理方式有哪几种?A. 两阶段提交(2PC)B. 三阶段提交(3PC)C. 检查点(Checkpoint)D. 分布式事务协议(DTCP)13. 什么是分布式数据库中的复制策略?有哪些常见的复制策略?A. 主从复制B. 并发复制C. 分片复制D. 混合复制14. 在分布式数据库系统中,如何实现数据的负载均衡?A. 数据库中间件B. 分布式缓存C. 负载均衡器D. 读写分离15. 分布式数据库系统中,如何保证数据的一致性和完整性?A. 两阶段提交(2PC)B. 三阶段提交(3PC)C. 检查点(Checkpoint)D. 四阶段提交(4PC)16. 以下哪个是分布式数据库系统中的分布式锁机制?A. 乐观锁B. 悲观锁C. 行级锁D. 页级锁17. 分布式数据库系统中,如何处理跨库查询?A. 使用SQL查询B. 使用中间表C. 使用分布式查询语言(DQL)D. 使用ETL工具18. 分布式数据库系统中,如何实现数据备份和恢复?A. 定期全量备份B. 增量备份C. 差量备份D. 主从备份19. 以下哪个是分布式数据库系统的发展趋势?A. 向规模更小的分布式数据库发展B. 向更高性能的分布式数据库发展C. 向更容易扩展的分布式数据库发展D. 向更强一致性的分布式数据库发展20. 以下哪个选项是分布式数据库系统中常用的数据复制技术?A. 主从复制B. 并发复制C. 混合复制D. 非阻塞复制21. 分布式数据库系统中的分片策略有哪几种?A. 范围分片B. 列分片C. 层次分片D. 索引分片22. 在分布式数据库系统中,如何实现数据的一致性?A. 通过分布式事务协议如两阶段提交(2PC)实现B. 通过分布式锁机制实现C. 通过分布式日志和重放技术实现D. 通过数据复制和分片实现23. 分布式数据库系统面临的主要挑战包括哪些?A. 数据复制的一致性问题B. 查询优化的复杂性C. 安全性和隐私保护问题D. 系统的可靠性和容错性24. 以下哪个选项是分布式数据库系统中常用的分片算法?A. 条件分片B. 基于范围的分片C. 基于哈希的分片D. 基于权重的分片25. 分布式数据库系统中的分布式事务处理有哪些类型?A. 两阶段提交(2PC)B. 三阶段提交(3PC)C. 这些选项都不是D. 没有分布式事务处理26. 以下哪个选项是分布式数据库系统中常用的负载均衡技术?A. 轮询负载均衡B. 权重负载均衡C. 简单轮询D. 加权轮询27. 分布式数据库系统中的数据迁移有以下几种类型?A. 结构迁移B. 非结构迁移C. 逻辑迁移D. 物理迁移28. 以下哪个选项是分布式数据库系统中常用的故障恢复技术?A. 回滚操作B. 前滚操作C. 数据重同步D. 数据复制恢复29. 分布式数据库系统的定义及其与传统数据库系统的区别是什么?A. 分布式数据库系统可以在多个节点上存储数据,而传统数据库系统通常在一个节点上存储所有数据。

c++分布式面试题

c++分布式面试题

c++分布式面试题1. 什么是分布式系统?请简要说明其特点。

2. 请解释负载均衡在分布式系统中的作用。

3. 请列举几种常见的分布式系统架构。

4. 请解释CAP定理及其在分布式系统中的应用。

5. 请解释一致性哈希算法及其在分布式系统中的应用场景。

6. 请解释分布式事务的概念及其处理方式。

7. 请解释分布式锁的作用及其实现方式。

8. 请解释分布式系统的容错机制及其实现方式。

9. 请解释消息队列在分布式系统中的应用及其优点。

10. 请解释分布式数据库的概念及其与集中式数据库的区别。

11. 请解释微服务架构的概念及其优缺点。

12. 请解释服务发现和注册在分布式系统中的作用及其实现方式。

13. 请解释Zookeeper在分布式系统中的应用及其作用。

14. 请解释分布式系统的监控和诊断方法。

15. 请解释分布式系统的性能优化策略。

16. 请举例说明在实际项目中如何应用分布式技术解决问题。

17.请解释C++在分布式系统开发中的优缺点。

18. 请解释一下C++在开发分布式系统时的优点和挑战。

19. 你能详细描述一下如何使用C++来实现分布式锁吗?20. 请解释一下C++在微服务架构中的作用及其优缺点。

21. 你了解哪些主流的分布式技术框架,以及它们在实际项目中的应用?22. 请解释下你对“服务发现和注册”这个概念的理解,并说明在C++中如何实现它。

23. 请分享一下你对于C/C++在分布式系统开发中所扮演角色的观点,包括它在实践中的性能考虑。

24. 请解释一下你对CAP定理的理解,并阐述它在C++分布式系统中的实际应用。

25. 最后,请谈谈你认为在未来,C++在分布式系统开发方面会有哪些发展趋势或新的可能性?。

分布式系统及微服务架构考试

分布式系统及微服务架构考试

分布式系统及微服务架构考试(答案见尾页)一、选择题1. 分布式系统的定义是什么?A. 一种软件系统,其组件分布在多个计算机网络上B. 一种软件系统,其所有组件都运行在同一个计算机上C. 一种软件系统,其组件分布在多个计算机网络上,并且相互协作以提供分布式服务D. 一种软件系统,其所有组件都运行在同一个计算机上,并且相互协作以提供分布式服务2. 微服务架构是一种什么类型的架构?A. 面向过程的架构B. 面向对象的架构C. 面向服务的架构D. 面向功能的架构3. 分布式系统中的数据一致性是指什么?A. 所有节点在同一时间点具有相同的数据副本B. 所有节点在同一时间点具有不同的数据副本C. 所有节点在不同时间点具有相同的数据副本D. 所有节点在不同时间点具有不同的数据副本4. 微服务架构中,服务之间通信的主要协议是什么?A. HTTP/HTTPSB. RPC (远程过程调用)C. SQLD. WebSocket5. 分布式系统中,如何实现服务发现和负载均衡?A. 使用DNS进行服务发现,使用随机算法进行负载均衡B. 使用ZooKeeper进行服务发现,使用轮询算法进行负载均衡C. 使用Consul进行服务发现,使用随机算法进行负载均衡D. 使用Etcd进行服务发现,使用一致性哈希进行负载均衡6. 微服务架构中,通常使用哪种数据库?A. 关系型数据库(如MySQL、Oracle)B. 面向关系的数据库(如PostgreSQL、SQL Server)C. 面向文档的数据库(如MongoDB、Couchbase)D. 消息队列数据库(如RabbitMQ、Kafka)7. 分布式系统中的容错机制主要有哪些?A. 数据备份与恢复B. 节点冗余C. 容错算法(如Paxos、Raft)D. 分布式事务8. 微服务架构中,服务之间的依赖管理如何实现?A. 通过API网关进行服务之间的依赖管理B. 通过服务注册与发现机制进行服务之间的依赖管理C. 通过消息队列进行服务之间的依赖管理D. 通过配置中心进行服务之间的依赖管理9. 分布式系统中的安全性如何保证?A. 使用SSL/TLS加密通信B. 使用强密码策略C. 访问控制列表(ACL)或身份验证机制D. 定期进行安全审计10. 微服务架构中,如何实现服务监控和日志收集?A. 使用日志聚合工具(如ELK Stack)B. 使用监控工具(如Prometheus、Grafana)C. 使用服务注册与发现机制进行服务监控D. 使用集中式的配置中心进行服务日志收集11. 分布式系统的定义是什么?A. 一种将多个应用程序部署在一个或多个节点上的系统B. 一种将多个服务组件集成到一个应用程序中的系统C. 一种将多个计算机连接在一起的网络系统D. 一种将多个数据库连接在一起的系统12. 微服务架构是一种什么类型的架构?A. 一种单一应用程序的结构B. 一种将单个应用程序拆分成多个独立运行的服务的设计C. 一种集中式的系统架构D. 一种将多个服务器连接在一起的架构13. 分布式系统中的数据一致性是指什么?A. 数据在多个节点上保持相同值的状态B. 数据在多个节点上保持不同值的状态C. 数据在多个节点上保持同步更新的状态D. 数据在多个节点上保持临时不一致的状态14. 在分布式系统中,通常使用什么来保证数据的一致性?A. 使用分布式事务协议B. 使用分布式锁机制C. 使用分布式数据库技术D. 使用负载均衡技术15. 微服务架构中的服务注册与发现是为了实现什么功能?A. 服务之间的通信和协作B. 节点的动态管理和调度C. 服务的安全访问控制D. 服务的负载均衡和容错16. 分布式系统中的冗余设计主要是为了实现什么目标?A. 提高系统的可用性和性能B. 提高系统的可靠性和稳定性C. 提高系统的可扩展性和灵活性D. 提高系统的保密性和安全性17. 在微服务架构中,服务之间通信的主要协议是什么?A. HTTP/HTTPSB. TCP/IPC. RPC (远程过程调用)D. Message Queue18. 分布式系统中的负载均衡是一种什么技术?A. 将请求平均分配到多个节点的技术B. 将请求顺序分配到多个节点的技术C. 将请求随机分配到多个节点的技术D. 将请求根据优先级分配到多个节点的技术19. 微服务架构中的服务监控和管理是为了实现什么功能?A. 服务的实时运行状态监控B. 服务的故障自动恢复C. 服务的性能优化D. 服务的安全访问控制20. 分布式系统中的数据分片是一种什么技术?A. 将数据分散存储在不同的节点上B. 将数据集中存储在单个节点上C. 将数据分散存储在不同的磁盘或存储器上D. 将数据集中存储在多个磁盘或存储器上21. 分布式系统的定义是什么?A. 一组独立的计算机通过网络进行通信B. 一组独立的计算机通过网络进行协作C. 一组独立的计算机通过网络进行负载均衡D. 一组独立的计算机通过网络进行数据同步22. 微服务架构是一种什么类型的架构?A. 面向功能的架构B. 面向服务的架构C. 面向资源的架构D. 面向对象的架构23. 分布式系统中的数据一致性是如何保证的?A. 通过分布式事务管理B. 通过分布式锁机制C. 通过分布式状态同步D. 通过分布式文件系统24. 微服务架构中,服务之间的通信通常使用哪种协议?A. HTTP/HTTPSB. TCP/IPC. RPC(如gRPC或Thrift)D. WebSocket25. 分布式系统中的负载均衡是如何实现的?A. 负载均衡器进行流量分发B. 负载均衡器进行请求分发C. 负载均衡器进行服务分发D. 负载均衡器进行数据分发26. 微服务架构中的服务注册与发现机制是如何工作的?A. 服务注册与发现是通过静态配置实现的B. 服务注册与发现是通过动态API实现的C. 服务注册与发现是通过集中式目录实现的D. 服务注册与发现是通过DNS解析实现的27. 分布式系统中的容错机制如何实现?A. 通过冗余设计B. 通过失败转移机制C. 通过错误检测机制D. 通过数据备份机制28. 微服务架构中的熔断器模式是如何工作的?A. 通过限制请求速率来防止服务过载B. 通过隔离故障服务来防止故障扩散C. 通过动态路由来防止故障转移D. 通过熔断器来检测故障并切断依赖29. 分布式系统中的数据分片是如何实现的?A. 通过数据库分片技术B. 通过分布式数据库技术C. 通过数据复制技术D. 通过数据压缩技术30. 微服务架构中的服务监控和日志收集是如何实现的?A. 通过集中式监控工具B. 通过分布式追踪技术C. 通过日志聚合技术D. 通过自定义监控脚本31. 分布式系统具有的特性包括哪些?A. 可扩展性B. 高可用性C. 透明性D. 容错性32. 微服务架构与传统单体架构的主要区别是什么?A. 单体架构只包含一个应用程序,而微服务架构将应用程序拆分成多个独立的服务。

分布式系统原理与泛型英文版习题解答

分布式系统原理与泛型英文版习题解答

DISTRIBUTED SYSTEMS PRINCIPLES AND PARADIGMSPROBLEM SOLUTIONSANDREW S.TANENBAUMMAARTEN VAN STEENVrije UniversiteitAmsterdam,The NetherlandsPRENTICE HALLUPPER SADDLE RIVER,NJ07458SOLUTIONS TO CHAPTER1PROBLEMS1.Q:What is the role of middleware in a distributed system?A:To enhance the distribution transparency that is missing in network operat-ing systems.In other words,middleware aims at improving the single-system view that a distributed system should have.2.Q:Explain what is meant by(distribution)transparency,and give examplesof different types of transparency.A:Distribution transparency is the phenomenon by which distribution aspects in a system are hidden from users and applications.Examples include access transparency,location transparency,migration transparency,relocation tran-sparency,replication transparency,concurrency transparency,failure tran-sparency,and persistence transparency.3.Q:Why is it sometimes so hard to hide the occurrence and recovery fromfailures in a distributed system?A:It is generally impossible to detect whether a server is actually down,or that it is simply slow in responding.Consequently,a system may have to report that a service is not available,although,in fact,the server is just slow.4.Q:Why is it not always a good idea to aim at implementing the highestdegree of transparency possible?A:Aiming at the highest degree of transparency may lead to a considerable loss of performance that users are not willing to accept.5.Q:What is an open distributed system and what benefits does openness pro-vide?A:An open distributed system offers services according to clearly defined rules.An open system is capable of easily interoperating with other open sys-tems but also allows applications to be easily ported between different imple-mentations of the same system.6.Q:Describe precisely what is meant by a scalable system.A:A system is scalable with respect to either its number of components,geo-graphical size,or number and size of administrative domains,if it can grow in one or more of these dimensions without an unacceptable loss of perfor-mance.7.Q:Scalability can be achieved by applying different techniques.What arethese techniques?A:Scaling can be achieved through distribution,replication,and caching.2PROBLEM SOLUTIONS FOR CHAPTER18.Q:What is the difference between a multiprocessor and a multicomputer?A:In a multiprocessor,the CPUs have access to a shared main memory.There is no shared memory in multicomputer systems.In a multicomputer system,the CPUs can communicate only through message passing.9.Q:A multicomputer with256CPUs is organized as a16×16grid.What isthe worst-case delay(in hops)that a message might have to take?A:Assuming that routing is optimal,the longest optimal route is from one corner of the grid to the opposite corner.The length of this route is30hops.If the end processors in a single row or column are connected to each other,the length becomes15.10.Q:Now consider a256-CPU hypercube.What is the worst-case delay here,again in hops?A:With a256-CPU hypercube,each node has a binary address,from 00000000to11111111.A hop from one machine to another always involves changing a single bit in the address.Thus from00000000to00000001is one hop.From there to00000011is another hop.In all,eight hops are needed. 11.Q:What is the difference between a distributed operating system and a net-work operating system?A:A distributed operating system manages multiprocessors and homogene-ous multicomputers.A network operating system connects different indepen-dent computers that each have their own operating system so that users can easily use the services available on each computer.12.Q:Explain how microkernels can be used to organize an operating system ina client-server fashion.A:A microkernel can separate client applications from operating system ser-vices by enforcing each request to pass through the kernel.As a consequence, operating system services can be implemented by(perhaps different)user-level servers that run as ordinary processes.If the microkernel has networking capabilities,there is also no principal objection in placing those servers on remote machines(which run the same microkernel).13.Q:Explain the principal operation of a page-based distributed shared memorysystem.A:Page-based DSM makes use of the virtual memory capabilities of an operating system.Whenever an application addresses a memory location that is currently not mapped into the current physical memory,a page fault occurs, giving the operating system control.The operating system can then locate the referred page,transfer its content over the network,and map it to physical memory.At that point,the application can continue.PROBLEM SOLUTIONS FOR CHAPTER13 14.Q:What is the reason for developing distributed shared memory systems?What do you see as the main problem hindering efficient implementations?A:The main reason is that writing parallel and distributed programs based on message-passing primitives is much harder than being able to use shared memory for communication.Efficiency of DSM systems is hindered by the fact,no matter what you do,page transfers across the network need to take place.If pages are shared by different processors,it is quite easy to get into a state similar to thrashing in virtual memory systems.In the end,DSM sys-tems can never be faster than message-passing solutions,and will generally be slower due to the overhead incurred by keeping track of where pages are.15.Q:Explain what false sharing is in distributed shared memory systems.Whatpossible solutions do you see?A:False sharing happens when data belonging to two different and indepen-dent processes(possibly on different machines)are mapped onto the same logical page.The effect is that the page is swapped between the two processes,leading to an implicit and unnecessary dependency.Solutions include making pages smaller or prohibiting independent processes to share a page.16.Q:An experimentalfile server is up3/4of the time and down1/4of the time,due to bugs.How many times does thisfile server have to be replicated to give an availability of at least99%?A:With k being the number of servers,we have that(1/4)k<0.01,expressing that the worst situation,when all servers are down,should happen at most 1/100of the time.This gives us k=4.17.Q:What is a three-tiered client-server architecture?A:A three-tiered client-server architecture consists of three logical layers, where each layer is,in principle,implemented at a separate machine.The highest layer consists of a client user interface,the middle layer contains the actual application,and the lowest layer implements the data that are being used.18.Q:What is the difference between a vertical distribution and a horizontal dis-tribution?A:Vertical distribution refers to the distribution of the different layers in a multitiered architectures across multiple machines.In principle,each layer is implemented on a different machine.Horizontal distribution deals with the distribution of a single layer across multiple machines,such as distributing a single database.19.Q:Consider a chain of processes P1,P2,...,P n implementing a multitieredclient-server architecture.Process P i is client of process P i+1,and P i will return a reply to P i−1only after receiving a reply from P i+1.What are the4PROBLEM SOLUTIONS FOR CHAPTER1main problems with this organization when taking a look at the request-reply performance at process P1?A:Performance can be expected to be bad for large n.The problem is that each communication between two successive layers is,in principle,between two different machines.Consequently,the performance between P1and P2 may also be determined by n−2request-reply interactions between the other layers.Another problem is that if one machine in the chain performs badly or is even temporarily unreachable,then this will immediately degrade the per-formance at the highest level.5 SOLUTIONS TO CHAPTER2PROBLEMS1.Q:In many layered protocols,each layer has its own header.Surely it wouldbe more efficient to have a single header at the front of each message with all the control in it than all these separate headers.Why is this not done?A:Each layer must be independent of the other ones.The data passed from layer k+1down to layer k contains both header and data,but layer k cannot tell which is which.Having a single big header that all the layers could read and write would destroy this transparency and make changes in the protocol of one layer visible to other layers.This is undesirable.2.Q:Why are transport-level communication services often inappropriate forbuilding distributed applications?A:They hardly offer distribution transparency meaning that application developers are required to pay significant attention to implementing commun-ication,often leading to proprietary solutions.The effect is that distributed applications,for example,built directly on top of sockets are difficult to port and to interoperate with other applications.3.Q:A reliable multicast service allows a sender to reliably pass messages to acollection of receivers.Does such a service belong to a middleware layer,or should it be part of a lower-level layer?A:In principle,a reliable multicast service could easily be part of the trans-port layer,or even the network layer.As an example,the unreliable IP multi-casting service is implemented in the network layer.However,because such services are currently not readily available,they are generally implemented using transport-level services,which automatically places them in the middleware.However,when taking scalability into account,it turns out that reliability can be guaranteed only if application requirements are considered.This is a strong argument for implementing such services at higher,less gen-eral layers.4.Q:Consider a procedure incr with two integer parameters.The procedureadds one to each parameter.Now suppose that it is called with the same vari-able twice,for example,as incr(i,i).If i is initially0,what value will it have afterward if call-by-reference is used?How about if copy/restore is used?A:If call by reference is used,a pointer to i is passed to incr.It will be incre-mented two times,so thefinal result will be two.However,with copy/restore,i will be passed by value twice,each value initially0.Both will be incre-mented,so both will now be1.Now both will be copied back,with the second copy overwriting thefirst one.Thefinal value will be1,not2.6PROBLEM SOLUTIONS FOR CHAPTER25.Q:C has a construction called a union,in which afield of a record(called astruct in C)can hold any one of several alternatives.At run time,there is no sure-fire way to tell which one is in there.Does this feature of C have any implications for remote procedure call?Explain your answer.A:If the runtime system cannot tell what type value is in thefield,it cannot marshal it correctly.Thus unions cannot be tolerated in an RPC system unless there is a tagfield that unambiguously tells what the variantfield holds.The tagfield must not be under user control.6.Q:One way to handle parameter conversion in RPC systems is to have eachmachine send parameters in its native representation,with the other one doing the translation,if need be.The native system could be indicated by a code in thefirst byte.However,since locating thefirst byte in thefirst word is pre-cisely the problem,can this actually work?A:First of all,when one computer sends byte0,it always arrives in byte0.Thus the destination computer can simply access byte0(using a byte instruc-tion)and the code will be in it.It does not matter whether this is the low-order byte or the high-order byte.An alternative scheme is to put the code in all the bytes of thefirst word.Then no matter which byte is examined,the code will be there.7.Q:Assume a client calls an asynchronous RPC to a server,and subsequentlywaits until the server returns a result using another asynchronous RPC.Is this approach the same as letting the client execute a normal RPC?What if we replace the asynchronous RPCs with asynchronous RPCs?A:No,this is not the same.An asynchronous RPC returns an acknowledge-ment to the caller,meaning that after thefirst call by the client,an additional message is sent across the network.Likewise,the server is acknowledged that its response has been delivered to the client.Two asynchronous RPCs may be the same,provided reliable communication is guaranteed.This is generally not the case.8.Q:Instead of letting a server register itself with a daemon as is done in DCE,we could also choose to always assign it the same endpoint.That endpoint can then be used in references to objects in the server’s address space.What is the main drawback of this scheme?A:The main drawback is that it becomes much harder to dynamically allo-cate objects to servers.In addition,many endpoints need to befixed,instead of just one(i.e.,the one for the daemon).For machines possibly having a large number of servers,static assignment of endpoints is not a good idea. 9.Q:Give an example implementation of an object reference that allows aclient to bind to a transient remote object.PROBLEM SOLUTIONS FOR CHAPTER27 A:Using Java,we can express such an implementation as the following class:public class Object reference{InetAddress server address;//network address of object’s serverint server endpoint;//endpoint to which server is listeningint object identifier;//identifier for this objectURL client code;//(remote)file containing client-side stubbyte[]init data;//possible additional initialization data }The object reference should at least contain the transport-level address ofthe server where the object resides.We also need an object identifier as the server may contain several objects.In our implementation,we use a URL to refer to a(remote)file containing all the necessary client-side code.A generic array of bytes is used to contain further initialization data for that code.An alternative implementation would have been to directly put the client-code into the reference instead of a URL.This approach is followed,for example, in Java RMI where proxies are passed as reference.10.Q:Java and other languages support exceptions,which are raised when anerror occurs.How would you implement exceptions in RPCs and RMIs?A:Because exceptions are initially raised at the server side,the server stub can do nothing else but catch the exception and marshal it as a special error response back to the client.The client stub,on the other hand,will have to unmarshal the message and raise the same exception if it wants to keep access to the server transparent.Consequently,exceptions now also need to be described in an interface definition language.11.Q:Would it be useful to also make a distinction between static and dynamicRPCs?A:Yes,for the same reason it is useful with remote object invocations:it simply introduces moreflexibility.The drawback,however,is that much of the distribution transparency is lost for which RPCs were introduced in the first place.12.Q:Some implementations of distributed-object middleware systems areentirely based on dynamic method invocations.Even static invocations are compiled to dynamic ones.What is the benefit of this approach?A:Realizing that an implementation of dynamic invocations can handle all invocations,static ones become just a special case.The advantage is that onlya single mechanism needs to be implemented.A possible disadvantage is thatperformance is not always as optimal as it could be had we analyzed the static invocation.8PROBLEM SOLUTIONS FOR CHAPTER213.Q:Describe how connectionless communication between a client and aserver proceeds when using sockets.A:Both the client and the server create a socket,but only the server binds the socket to a local endpoint.The server can then subsequently do a blocking read call in which it waits for incoming data from any client.Likewise,after creating the socket,the client simply does a blocking call to write data to the server.There is no need to close a connection.14.Q:Explain the difference between the primitives mpi bsend and mpi isendin MPI.A:The primitive mpi bsend uses buffered communication by which the caller passes an entire buffer containing the messages to be sent,to the local MPI runtime system.When the call completes,the messages have either been transferred,or copied to a local buffer.In contrast,with mpi isend,the caller passes only a pointer to the message to the local MPI runtime system after which it immediately continues.The caller is responsible for not overwriting the message that is pointed to until it has been copied or transferred.15.Q:Suppose you could make use of only transient asynchronous communica-tion primitives,including only an asynchronous receive primitive.How would you implement primitives for transient synchronous communication?A:Consider a synchronous send primitive.A simple implementation is to send a message to the server using asynchronous communication,and subse-quently let the caller continuously poll for an incoming acknowledgement or response from the server.If we assume that the local operating system stores incoming messages into a local buffer,then an alternative implementation is to block the caller until it receives a signal from the operating system that a message has arrived,after which the caller does an asynchronous receive. 16.Q:Now suppose you could make use of only transient synchronous commun-ication primitives.How would you implement primitives for transient asyn-chronous communication?A:This situation is actually simpler.An asynchronous send is implemented by having a caller append its message to a buffer that is shared with a process that handles the actual message transfer.Each time a client appends a mes-sage to the buffer,it wakes up the send process,which subsequently removes the message from the buffer and sends it its destination using a blocking call to the original send primitive.The receiver is implemented similarly by offer-ing a buffer that can be checked for incoming messages by an application. 17.Q:Does it make sense to implement persistent asynchronous communicationby means of RPCs?A:Yes,but only on a hop-to-hop basis in which a process managing a queue passes a message to a next queue manager by means of an RPC.Effectively,PROBLEM SOLUTIONS FOR CHAPTER29 the service offered by a queue manager to another is the storage of a message.The calling queue manager is offered a proxy implementation of the interface to the remote queue,possibly receiving a status indicating the success or failure of each operation.In this way,even queue managers see only queues and no further communication.18.Q:In the text we stated that in order to automatically start a process to fetchmessages from an input queue,a daemon is often used that monitors the input queue.Give an alternative implementation that does not make use of a dae-mon.A:A simple scheme is to let a process on the receiver side check for any incoming messages each time that process puts a message in its own queue. 19.Q:Routing tables in IBM MQSeries,and in many other message-queuingsystems,are configured manually.Describe a simple way to do this automati-cally.A:The simplest implementation is to have a centralized component in which the topology of the queuing network is maintained.That component simply calculates all best routes between pairs of queue managers using a known routing algorithm,and subsequently generates routing tables for each queue manager.These tables can be downloaded by each manager separately.This approach works in queuing networks where there are only relatively few,but possibly widely dispersed,queue managers.A more sophisticated approach is to decentralize the routing algorithm,byhaving each queue manager discover the network topology,and calculate its own best routes to other managers.Such solutions are widely applied in com-puter networks.There is no principle objection for applying them to message-queuing networks.20.Q:How would you incorporate persistent asynchronous communication intoa model of communication based on RMIs to remote objects?A:An RMI should be asynchronous,that is,no immediate results are expected at invocation time.Moreover,an RMI should be stored at a special server that will forward it to the object as soon as the latter is up and running in an object server.21.Q:With persistent communication,a receiver generally has its own localbuffer where messages can be stored when the receiver is not executing.To create such a buffer,we may need to specify its size.Give an argument why this is preferable,as well as one against specification of the size.A:Having the user specify the size makes its implementation easier.The sys-tem creates a buffer of the specified size and is done.Buffer management becomes easy.However,if the bufferfills up,messages may be lost.The alternative is to have the communication system manage buffer size,starting10PROBLEM SOLUTIONS FOR CHAPTER2with some default size,but then growing(or shrinking)buffers as need be.This method reduces the chance of having to discard messages for lack of room,but requires much more work of the system.22.Q:Explain why transient synchronous communication has inherent scalabil-ity problems,and how these could be solved.A:The problem is the limited geographical scalability.Because synchronous communication requires that the caller is blocked until its message is received,it may take a long time before a caller can continue when the receiver is far away.The only way to solve this problem is to design the cal-ling application so that it has other useful work to do while communication takes place,effectively establishing a form of asynchronous communication.23.Q:Give an example where multicasting is also useful for discrete datastreams.A:Passing a largefile to many users as is the case,for example,when updat-ing mirror sites for Web services or software distributions.24.Q:How could you guarantee a maximum end-to-end delay when a collectionof computers is organized in a(logical or physical)ring?A:We let a token circulate the ring.Each computer is permitted to send data across the ring(in the same direction as the token)only when holding the token.Moreover,no computer is allowed to hold the token for more than T seconds.Effectively,if we assume that communication between two adjacent computers is bounded,then the token will have a maximum circulation time, which corresponds to a maximum end-to-end delay for each packet sent. 25.Q:How could you guarantee a minimum end-to-end delay when a collectionof computers is organized in a(logical or physical)ring?A:Strangely enough,this is much harder than guaranteeing a maximum delay.The problem is that the receiving computer should,in principle,not receive data before some elapsed time.The only solution is to buffer packets as long as necessary.Buffering can take place either at the sender,the receiver,or somewhere in between,for example,at intermediate stations.The best place to temporarily buffer data is at the receiver,because at that point there are no more unforeseen obstacles that may delay data delivery.The receiver need merely remove data from its buffer and pass it to the applica-tion using a simple timing mechanism.The drawback is that enough buffering capacity needs to be provided.26.Q:Imagine we have a token bucket specification where the maximum dataunit size is1000bytes,the token bucket rate is10million bytes/sec,the token bucket size is1million bytes,and the maximum transmission rate is50mil-lion bytes/sec.How long can a burst of maximum speed last?PROBLEM SOLUTIONS FOR CHAPTER211 A:Call the length of the maximum burst interval∆t.In an extreme case,the bucket is full at the start of the interval(1million bytes)and another10∆t comes in during that interval.The output during the transmission burst con-sists of50∆t million bytes,which should be equal to(1+10∆t).Consequently,∆t is equal to25msec.12SOLUTIONS TO CHAPTER3PROBLEMS1.Q:In this problem you are to compare reading afile using a single-threadedfile server and a multithreaded server.It takes15msec to get a request for work,dispatch it,and do the rest of the necessary processing,assuming that the data needed are in a cache in main memory.If a disk operation is needed, as is the case one-third of the time,an additional75msec is required,during which time the thread sleeps.How many requests/sec can the server handle if it is single threaded?If it is multithreaded?A:In the single-threaded case,the cache hits take15msec and cache misses take90msec.The weighted average is2/3×15+1/3×90.Thus the mean request takes40msec and the server can do25per second.For a mul-tithreaded server,all the waiting for the disk is overlapped,so every request takes15msec,and the server can handle662/3requests per second.2.Q:Would it make sense to limit the number of threads in a server process?A:Yes,for two reasons.First,threads require memory for setting up their own private stack.Consequently,having many threads may consume too much memory for the server to work properly.Another,more serious reason, is that,to an operating system,independent threads tend to operate in a chaotic manner.In a virtual memory system it may be difficult to build a rela-tively stable working set,resulting in many page faults and thus I/O.Having many threads may thus lead to a performance degradation resulting from page thrashing.3.Q:In the text,we described a multithreadedfile server,showing why it isbetter than a single-threaded server and afinite-state machine server.Are there any circumstances in which a single-threaded server might be better?Give an example.A:Yes.If the server is entirely CPU bound,there is no need to have multiple threads.It may just add unnecessary complexity.As an example,consider a telephone directory assistance number(like555-1212)for an area with1mil-lion people.If each(name,telephone number)record is,say,64characters, the entire database takes64megabytes,and can easily be kept in the server’s memory to provide fast lookup.4.Q:Statically associating only a single thread with a lightweight process is notsuch a good idea.Why not?A:Such an association effectively reduces to having only kernel-level threads,implying that much of the performance gain of having threads in the first place,is lost.5.Q:Having only a single lightweight process per process is also not such agood idea.Why not?PROBLEM SOLUTIONS FOR CHAPTER313 A:In this scheme,we effectively have only user-level threads,meaning that any blocking system call will block the entire process.6.Q:Describe a simple scheme in which there are as many lightweightprocesses as there are runnable threads.A:Start with only a single LWP and let it select a runnable thread.When a runnable thread has been found,the LWP creates another LWP to look for a next thread to execute.If no runnable thread is found,the LWP destroys itself.7.Q:Proxies can support replication transparency by invoking each replica,asexplained in the text.Can(the server side of)an object be subject to a repli-cated invocation?A:Yes:consider a replicated object A invoking another(nonreplicated) object B.If A consists of k replicas,an invocation of B will be done by each replica.However,B should normally be invoked only once.Special measures are needed to handle such replicated invocations.8.Q:Constructing a concurrent server by spawning a process has some advan-tages and disadvantages compared to multithreaded servers.Mention a few.A:An important advantage is that separate processes are protected against each other,which may prove to be necessary as in the case of a superserver handling completely independent services.On the other hand,process spawn-ing is a relatively costly operation that can be saved when using mul-tithreaded servers.Also,if processes do need to communicate,then using threads is much cheaper as in many cases we can avoid having the kernel implement the communication.9.Q:Sketch the design of a multithreaded server that supports multiple proto-cols using sockets as its transport-level interface to the underlying operating system.A:A relatively simple design is to have a single thread T waiting for incom-ing transport messages(TPDUs).If we assume the header of each TPDU con-tains a number identifying the higher-level protocol,the tread can take the payload and pass it to the module for that protocol.Each such module has a separate thread waiting for this payload,which it treats as an incoming request.After handling the request,a response message is passed to T,which, in turn,wraps it in a transport-level message and sends it to tthe proper desti-nation.10.Q:How can we prevent an application from circumventing a windowmanager,and thus being able to completely mess up a screen?A:Use a microkernel approach by which the windowing system including the window manager are run in such a way that all window operations are。

分布式系统试题及答案

分布式系统试题及答案

分布式系统复习题库及答案1、计算机系统的硬件异构性、软件异构性主要表现在哪几方面?参考答案:计算机系统的硬件异构性主要有三个方面的表现,即:①计算机的指令系统不同。

这意味着一种机器上的程序模块不能在另一种不兼容的机器上执行,很显然,一种机器上的可执行代码程序不能在另一种不兼容的机器上执行。

②数据表示方法不同。

例如不同类型的计算机虽然都是按字节编址的,但是高字节和低字节的规定可能恰好相反。

浮点数的表示方法也常常不一样。

③机器的配置不同。

尽管机器的类型可能相同,其硬件配置也可以互不兼容。

计算机系统的软件异构性包括操作系统异构性和程序设计语言异构性。

操作系统异构性的三个主要表现方面为:①操作系统所提供的功能可能大不相同。

例如,不同的操作系统至少提供了不同的命令集。

②操作系统所提供的系统调用在语法、语义和功能方面也不相同。

③文件系统不同。

程序设计语言的异构性表现在不同的程序设计语言用不同方法在文件中存储数据。

2、由于分布计算系统包含多个(可能是不同种类的)分散的、自治的处理资源,要想把它们组织成一个整体,最有效地完成一个共同的任务,做到这一点比起传统的集中式的单机系统要困难得多,需要解决很多新问题。

这些问题主要表现在哪些方面?参考答案:①资源的多重性带来的问题。

由于处理资源的多重性,分布计算系统可能产生的差错类型和次数都比集中式单机系统多。

最明显的一个例子是部分失效问题:系统中某一个处理资源出现故障而其他计算机尚不知道,但单机系统任何一部分出现故障时将停止整个计算。

另一个例子是多副本信息一致性问题。

可见,资源多重性使得差错处理和恢复问题变得很复杂。

资源多重性还给系统资源管理带来新的困难。

②资源的分散性带来的问题。

在分布计算系统中,系统资源在地理上是分散的。

由于进程之间的通信采用的是报文传递的方式进行的,通信将产生不可预测的、有时是巨大的延迟,特别是在远程网络所组成的分布计算系统中更是这样。

例如使用卫星通信会产生270毫秒的延迟。

PHP分布式系统面试题(3篇)

PHP分布式系统面试题(3篇)

第1篇第一部分:基础知识1. 什么是分布式系统?- 分布式系统是由多个独立节点组成,通过网络进行通信和协作,共同完成某个任务或提供某种服务的系统。

每个节点可以是一个独立的计算机或服务器,它们之间通过通信协议进行交互。

2. 分布式系统与集中式系统的区别是什么?- 集中式系统:所有资源(如数据、计算能力等)都集中在单个节点上,所有操作都由这个节点处理。

- 分布式系统:资源分布在多个节点上,每个节点负责一部分任务,节点之间通过网络通信协同工作。

3. 分布式系统的主要挑战有哪些?- 数据一致性- 服务容错- 系统扩展性- 网络通信- 安全性4. 请解释CAP定理。

- CAP定理指出,在分布式系统中,一致性(Consistency)、可用性(Availability)和分区容错性(Partition tolerance)三者中,最多只能同时满足两项。

当网络分区发生时,系统必须在一致性和可用性之间做出选择。

5. 什么是分布式锁?请列举几种实现分布式锁的方法。

- 分布式锁用于确保在分布式系统中,同一时间只有一个进程可以访问共享资源。

- 实现分布式锁的方法包括:- 基于数据库的锁- 基于Redis的锁- 基于Zookeeper的锁- 基于etcd的锁6. 请解释什么是分布式缓存?有哪些常见的分布式缓存实现?- 分布式缓存用于提高分布式系统的性能,通过在多个节点之间共享缓存来减少对后端存储系统的访问。

- 常见的分布式缓存实现包括:- Redis- Memcached- Apache Ignite- Hazelcast第二部分:PHP与分布式系统7. PHP在分布式系统中的应用场景有哪些?- PHP作为Web后端语言,在分布式系统中可以应用于以下场景:- Web服务器端脚本- API服务- 微服务- 数据处理8. 请解释PHP中的全局变量和静态变量的区别。

- 全局变量:在PHP脚本的全局作用域中定义的变量,可以在脚本中的任何地方访问。

分布式数据库原理与应用题库

分布式数据库原理与应用题库

分布式数据库原理与应用题库1. 引言随着互联网的快速发展和大数据时代的到来,数据量的爆炸式增长对数据库的存储和处理能力提出了更高的要求。

传统的单节点数据库已经难以满足这一需求,而分布式数据库应运而生。

本文将介绍分布式数据库的原理和应用,并提供一些应用题供读者练习,加深对该主题的理解。

2. 分布式数据库的原理分布式数据库是将数据存储在多个物理节点上的数据库系统,节点之间通过网络进行通信和协作。

它具有以下几个核心原理:2.1 数据分片和副本为了实现数据的分布式存储和高可用性,分布式数据库将数据进行分片处理,并将每个分片的副本存储在不同的节点上。

这样可以提高数据访问的并发性和容错性。

2.2 数据一致性和并发控制在分布式数据库中,多个节点同时操作数据可能会造成数据的不一致。

因此,分布式数据库需要实现一致性协议来保证数据的一致性,并使用并发控制技术来处理并发操作。

2.3 数据通信和数据同步分布式数据库中的节点通过网络进行数据通信和数据同步。

节点之间的通信可以通过消息传递、RPC(Remote Procedure Call)等方式实现。

数据同步可以通过数据复制和数据冗余等方式实现。

3. 分布式数据库的应用分布式数据库广泛应用于互联网、云计算、物联网等领域,它具有高可用性、可扩展性和容错性等优势。

以下是一些分布式数据库的常见应用案例:3.1 电商平台在电商平台中,用户的购物行为产生了大量的交易数据,而这些数据需要快速地进行存储和分析。

分布式数据库可以实现海量数据的存储和查询,并提供高性能的数据处理能力,从而提高用户的购物体验。

3.2 物联网物联网设备产生的数据通常具有大规模、高并发的特点。

分布式数据库可以实现对这些数据的实时监控和存储,同时提供高可靠性和高性能的数据处理能力。

3.3 金融系统金融系统需要处理大量的交易数据,并保证数据的安全和一致性。

分布式数据库可以有效地管理和存储金融数据,并提供高度可靠的事务处理能力。

分布式系统原理与应用

分布式系统原理与应用
C.可靠性与可扩展性没有关联
D.可扩展性通常以牺牲可靠性为代价
(以下为答题区域)
第二部分多选题(本题共15小题,每小题2分,共30分.在每小题给出的四个选项中,至少有一项是符合题目要求的)
1.分布式系统的设计目标包括以下哪些?()
A.提高计算速度
B.提高数据可靠性
C.降低单点故障的风险
D.提高系统复杂性
分布式系统原理与应用
考生姓名:________________答题日期:_______年__月__日得分:_________________判卷人:_________________
第一部分单选题(本题共15小题,每小题2分,共30分.在每小题给出的四个选项中,只有一项是符合题目要求的)
1.分布式系统最基本的特点是()
(以下为答题区域)
第四部分主观题(本题共2小题,每题10分,共20分)
1.请简述分布式系统中的CAP定理,并解释在分布式系统设计中如何权衡一致性、可用性和分区容错性。
2.描述微服务架构的优势和挑战。举例说明在微服务架构中如何处理服务间的通信和数据一致性问题。
标准答案
第一部分单选题
1. B
2. D
3. C
A.分布式事务
B.并行计算
C.分布式算法
D.分散式处理
9.分布式数据库的CAP定理中,C代表()
A.一致性
B.可用性
C.分区容错性
D.性能
10.以下哪种协议常用于分布式系统中的数据一致性保证?()
A. HTTP协议
B. FTP协议
C. Raft协议
D. SMTP协议
11.在分布式系统架构中,以下哪项不是微服务架构的优势?()
2.微服务架构的优势包括独立部署、容错性好、开发效率高;挑战包括服务间通信复杂、数据一致性问题。处理通信问题时可以使用API网关或消息队列;数据一致性可以通过分布式事务或最终一致性来解决。

分布式系统原理与泛型考博整理

分布式系统原理与泛型考博整理

第1章概述1.分布式系统的另一个定义,它是各自独立的计算机的集合,这些计算机看起来像是一个单的系统,就是说,它对用户是完全隐藏的,即使他有多个计算机也是如此。

请给出一个实例。

答:并行计算。

一个程序在一个分布式的系统中运行,但看起来是在单个系统中运行的。

2.中间件在分布式系统中扮演什么角色?答:中间件主要是为了增强分布式系统的透明性(这正是网络操作系统所缺乏的),换言之,中间件的目标是分布式系统的单系统视图,即使种类各异的计算机和网络都呈现为单个系统。

3.很多网络系统组织成后端办公系统和前端办公系统。

这种组织方式是如何满足分布式系统要求的?答:一个比较容易犯错的地方就是假设运行在一个组织下的分布式系统,应该运行在系统的整个组织框架下。

实际上,分布式系统被安装在一个分离的组织中。

从这层意义上讲,我们的分布式系统可以支持独立的后端处理和前端处理。

当然,这两部分可能是耦合的,并不需要要求这个耦合的部分完全透明。

4.解释(分布)透明性的含义,并且给出各种类型透明性的例子。

答:分布透明性是一种现象,即一个系统的分布情况对于用户和应用来说是隐藏的。

包括:访问透明:分布式系统中的多个计算机系统运行可能是不同的操作系统,这些操作系统的文件命名方式不同,命名方式的差异以及由此引发的文件操作方式的差异应该对用户和应用程序隐藏起来。

位置透明:从/index.htm这个url看不出parentice hall的主web服务器所在的位置,同时也看不出index.html的位置情况。

移植透明:分布式系统中的资源移动不会影响该资源的访问方式。

重定位透明:资源可以在接受访问的同时进行重新定位,而不引起用户和应用系统的注意。

移动通信用户从一个点到另一个点,可以一直使用移动设备,而无中断连接。

复制透明:对同一个资源存在多个副本这样一个事实的隐藏。

所有的副本同名。

并发透明:访问位于同一个共享数据库中的一批表。

故障透明:用户不会注意到某个资源无法正常工作,以及系统随后的恢复过程。

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Department Computer Science Distributed Systems VU University15.12.2009 MAKE SURE THAT YOUR HANDWRITING IS READABLE1a Explain what is meant by request-level and message-level interceptors in middleware.5pt Request-level interceptors are special local components to which an invocation request is passed before passing it to the underlying middleware.Such an interceptor is invocation aware in the sense that it knows with which invocation it is dealing,and for which server it is intended.Typically,such interceptors can be used to implement replicated calls.A message-level interceptor is a component that is logically placed between the middleware and the underlying operating system.It can thus handle only basic network messages,for example,by fragmenting them into smaller parts(and assembling these parts at the receiver side).1b Where does the need for adaptive middleware come from?5pt Middleware is intended to incorporate general-purpose,i.e.,application-independent,mechanisms for distributed computing.The problem is that for practical purposes,it is very difficult to separate policy from mechanism,with the effect that many middleware solutions are not right for specific applications.The result is the need to be able to tweak the middleware for the specific needs of an application.1c In the underlying feedback control loop,give an example of the analysis component in combination with the reference input.5ptAn example that is also discussed in the book,is analyzing whether measured performance is as good as it could have been when another replication scenario would have been used.In this case, the reference input is a cost function that needs to be minimized.2a What is the difference between transport-layer switching and content-aware request distribution?5pt With transport-layer switching,a front end to a server cluster accepts incoming TCP connections and hands these off to one of the back-end servers using only information that is available at the TCP-level:client address and destination port.In the case of content-aware request distribution,the switch can also inspect the content of requests(such as an HTTP URL)and use that information to decide to which back-end server the request should be forwarded.2b Explain how TCP handoff works and why it is difficult to apply to wide-area networks.5pt With TCP handoff,an incoming connection request is forwarded by a switch to a specific server, which then sends the response back directly to the client,using the network address of the switch.This last issue is problematic is a wide-area system,as it essentially involves spoofing the switch, which is often difficult to do across administrative domains.2c Explain how the content-aware request distribution can be combined with TCP handoff.5pt Your answer should explain what is happening in Figure12-9.Essential is that you mention the initial handoff to a distributor or dispatcher to decide what the best server could be based on content, after which the TCP connection is handed off to that server.The switch is subsequently informed.3a Traditional RPC mechanisms cannot handle pointers.What is the problem and how can it be ad-dressed?5pt The problem is that pointers passed as parameters refer to a memory location that is local to the caller.That location is not only often meaningless to the receipient,but more important is that the recipient will most likely not have the data structure in its memory that the caller has.There are not many things you can do about this,except copying the entire(dynamic data structure)from the caller to the callee when doing the RPC.An alternative is to replace pointers by global systemwide references,as is done with Java object references.3b Where does the need for at-least-once and at-most-once semantics come from?Why can’t we have exactly-once semantics?5pt The problem originates from having a(suspected)server crash,detected by the lack of a response in the case of an RPC.What the client-side software can do is either resend the request until itfinally gets a response(at-least-once semantics)or immediately reports the failure to the client application, thus providing at-most-once semantics.Guaranteeing exactly-once semantics is,in principle,impos-sible,because you cannot know in general whether the server crashed before or after executing the requested operation.3c Consider a client/server system based on RPC,and assume the server is replicated for performance.Sketch an RPC-based solution for hiding replication of the server from the client.5pt Simply take a client-side stub that replicates the call to the respective servers.It is essential that you mention that these calls should be done in parallel and that(for example)thefirst response is immediately passed to the client.Serializing the RPCs or waiting for all responses is OK for fault tolerance,but certainly not for performance.4a Resolve the following key lookups for the shown Chord-based P2P system:5ptsource key41542221302127201815@4:14–18;22@4:20–21–28;30@21:28–1;27@21:28;18@20:4–14–184b Adjust thefinger tables of nodes18and14when a node with ID24enters the ring.Also give the finger table of node24.5pt Node18:[20,20,24,28,4];Node14:[18,18,18,24,1];Node24:[28,28,28,1,9].4c Chord allows keys to be looked up recursively or iteratively.Explain the differences,as well as the main advantage of iterative over recursive lookup.5pt With recursive lookups,a message is forwarded from peer to peer until it reaches its destination.In contrast,with an iterative lookup,the requester is returned the next peer it should ask for the key.One can argue that in the case of Chord,iterative lookups are much better:recursive lookups do not have the advantage of proximity-awareness.Also,note that iterative lookups have the advantage of letting the client handle failures more easily.5a Explain how two-phase commit works.5pt Make sure that you explain(1)coordinator sends vote-request;(2)participants respond;(3)coordi-nator sends decision;(4)participants ack.5b Explain what happens when a participant,who is in the READY state,times out because it hasn’t received a response from the coordinator yet.5pt In that case,P can check whether any of the other particpants has made a transition to either ABORT or INIT(in which case P can abort)or COMMIT(and commit as well).The difficulty is when all others are in READY:they all need to wait until the coordinator recovers.5c If we use two-phase commit for a distributed transaction,can we allow a coordinator to issue two distributed transactions(involving the same participants)at the same time?5pt Yes:the local transaction managers at the participants will handle any concurrency issues.What is seen here is that the use of2PC is completely independent of the semantics of specific transactions.6a How can a Web hosting service help in handlingflash crowds?5pt Crucial for a correct answer is that you not only state that content is replicated,but that the origin server is assumed to be capable of redirecting requests,but perhaps no longer in also returning content-rich responses.Note that distributed request distribution is really tricky business.6b Akamai uses DNS-based redirection.Explain how resolution of the name would work.5pt The trick is that the regular DNS will resolve the name ,pointing to a DNS server that is controlled by Akamai.If we use iterative DNS name resolution,that server will know the IP address ot the requesting client,and be able to decide to which server(with logical name )it can forward the request.6c Explain the difference between content-aware and content-blind caching for Web applications by means of an example.5pt With content-aware caching,the cache has knowledge on the data model that is used by the Web application,and with that,can conduct query-containment procedures to see whether a query could possibly be addressed by the data that is already cached.For example,if an edge server had once received the query“select ALL FROM books WITH author=Irving”,it can cache that ter, when receiving a query“select ALL FROM books WITH author=Irving AND date<2008”,the edge server should be able to recognize that this is a subquery,and that it can thus look into its local cache.With content-blind caching,the cache simply attaches a unique id to an entire,specific query in order to check whether that exact query had been issued before.If so,it can possibly return the previously stored response from its cache.In our example,the two queries would each get a unqiue ID,which is then used to do a cache lookup.Grading:Thefinal grade is calculated by accumulating the scores per question(maximum:90points),and adding10bonus points.The maximum total is therefore100points.。

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