【正文】
herit the attributes and methods of all its superclasses.D) A superclass includes the attributes and methods of all its subclasses.66. Local bus can increase the data throughput. The bus that is most frequently used isA) VESA B) EISA C) MCA D) PCI67. Most frequently used algorithm for page replacement isA) LRU B) FIFO C) OPT D) PPT68. The following three concepts are essential to the OSI model:service, interface andA) networks B) applicationsC) protocols D) layers69. LANs can be connected by using bridges, which operate in theA) physical layer B) data link layerC) network layer D) transport layer70. TCP service is obtained by having both the sender and the receiver create end points, which are calledA) network address B) supernet addressC) socket D) IP address二、論述題(四個論述題可任選其一,并只選其一,多選無效,滿分30分。(4分)論述題 4在實際的網(wǎng)絡運行環(huán)境中,數(shù)據(jù)備份與恢復功能是非常重要的。(1分)內(nèi)存分配時,以頁面(塊)為單位,并按用戶程序所需頁數(shù)多少進行分配。(2分)論述題 21. 系統(tǒng)的規(guī)模。減少維護工作量的因素有:結構化技術的使用,自動化工具使用,數(shù)據(jù)庫技術應用,高性能軟件使用,新的軟件技術的應用。若只給出問題,)1. 數(shù)據(jù)存在大量冗余。如果某些課程因故刪除了,則只選修這些課程的學生信息也被刪除了,這往往也不是我們的本意,我們并不希望刪除這些學生的信息。因此,重要的網(wǎng)絡信息系統(tǒng)的軟件與重要數(shù)據(jù)保存兩個或兩個以上的網(wǎng)絡數(shù)據(jù)備份副本,并且其中一個副本需要異地存放。因為網(wǎng)絡數(shù)據(jù)備份與恢復功能是針對因自然災害、誤操作、人為破壞,以及網(wǎng)絡入侵與攻擊等各種原因,使網(wǎng)絡系統(tǒng)遭到破壞。由于主鍵屬性值不能為空值,當?shù)怯泴W生信息時,由于該學生尚未選課,C未確定,該學生的其他信息也無法登入,這不是我們的意愿。每個方案有各自的優(yōu)缺點。(符合上述要求的給10分,不足者可適當減分。當進程被選中運行時,系統(tǒng)將它從進程控制中塊中取出送入該寄存器。(2分)頁式存儲管理將內(nèi)存空間按照邏輯頁面大小劃分成等長的若干區(qū)域,每個區(qū)域為一個內(nèi)存塊。根據(jù)調(diào)查分析,確定它的屬性集合為:U={S,C,SNAME,CNAME,TEACHER,GRADE,SD}下面給出兩種確定的模式設計方案:方案一:只有一個關系模式:R(S,C,SNAME,CNAME,TEACHER,GRADE,SD)方案二:如果根據(jù)屬性之間存在的聯(lián)系或相關性,建立如下三個關系:S(S,SNAME,SD),C(C,CNAME,TEACHER),SC(S,C,GRADE)1. 試分析這兩種模式設計方案各自的優(yōu)缺點。如果一臺Internet主機有兩個或多個IP地址,則該主機屬于兩個或多個A) 邏輯網(wǎng)絡 B) 瀏覽器/服務器結構C) 通信子網(wǎng) D) 操作系統(tǒng)56. In a puter, if a logical left shift of the operand occurs, its lowest bit isA) 1 B) unchanged C) 0 D) random57. The CPU is posed of two ponents, which areA) arithmetic logic unit and controllerB) memory and controllerC) arithmetic logic unit and memoryD) controller and storageItems 58 and 59 are based on the following hash table.012345678910111226381733483525Hash function is : h(key) = key mod 13Collisions are solved by using open addressing, linear probing.58. Load factor α of the hash table is approximatelyA) B) C) D) 59. How many key parisons are needed in searching for key value 38?A) 1 B) 2 C) 3 D) 460. Which of the following operations belongs to message passing?A) read and write B) send and receiveC) wait and signal D) P and V61. I