课程名称︰网路管理与系统管理 课程性质︰选修 课程教师：蔡欣穆 开课学院：电资学院 开课系所︰资工系 考试日期（年月日）︰2015.04.20 考试时限（分钟）：180分钟 试题 : Network Administration and System Administration, Spring 2015 Midterm Examination (Handwritten Part) Solution 30 points Time: 6:30pm - 9:30pm, Monday, April 20, 2015 ˙ Remember to put your name, student ID, and page number on both pages of the exam paper. ˙ You are free to use the Internet and your laptops when taking this exam, but NOT allowed to communicate with others (online or in person). Problem 1. True or False problems. You only need to specify T or F for each problem. (15 points) 1. ( ) A computer does not need special driver for a network interface card to receive untagged VLAN packets. Ans: T 2. ( ) Always deploy the best class of network cable when constructing a building network. Ans: T 3. ( ) For better reliability and robustness, it is better to put IDF in different floors at different vertical locations in a building. Ans: F, 机柜一般放在不同层的同一个位置, 以方便维护, 维修(能快速找到机柜) 与 布线(机柜间彼此距离短). 4. ( ) The lowest class of UTP cable to support Gigabit Ethernet is Cat.5. Ans: Give Away, 规格上来说至 cat 5e 才全面支援 Gigabit, 但优质的 Cat 5. 网路线在短距离亦可以达到Gigabit, 因此本题送分. 5. ( ) Link aggregation can be used to increase the bandwidth, but cannot improve the reliability of the link. Ans: F, 若一条线断线, 其余的线路仍可提供通讯服务, 线路两端并不会因此失连, 因此可增进两端连线的可靠性. 6. ( ) A broadcast Ethernet frame has a destination address of 00-00-00-ff-ff-ff. Ans: F, 广播MAC 应为ff-ff-ff-ff-ff-ff. 7. ( ) A hard drive labeled with 2 TB capacity can store 2^31 bytes of data. Ans: F, 2 TB = 2 * 10^3, 2 TiB = 2^31. 8. ( ) If a hard drive is frequently used by a bittorrent application, it has a higher probability to fail in the long term. Ans: F, 硬碟的使用频率与温度和硬碟的寿命并无明显关系. 9. ( ) The sequential read speed of the fastest SSD on the market today can already saturate both SATA II's and USB 3.0's bandwidth. Ans: T, SATA SSD 约维持在 550MBps 的速度, 低於 USB3.0 之传输速度, 但 PCI-Express SSD 则可达到 3.2GBps 之速度, 高於 SATA II 与 USB 3.0 10. ( ) When a hard drive fails, its data can no longer be retrieved, regardless of the money you can spend. Ans: F, 传统硬碟损毁可能是由於磁头老化, 控制器毁损或磁区受损, 前两者可藉由 替换相应元件解决, 磁区受损则受损部份资料无法救援, 但其他则可 11. ( ) When a SSD fails, its data can no longer be retrieved, regardless of the money you can spend. Ans: T, SSD 正常使用中会有少量记忆体单元错误, 此部份非关物理受损, 在这些 未标记的损毁单元低於全部的 5% 时, 控制器仍能藉由演算法还原出正确 资料, 然此部份未标记之记忆体错误高於5%, 则控制器将无法分辨正确资 料, 导致读写失败, 此时资料是无法救援的. 因此 SSD 须定期手动做 secure erase, 标记损毁单元并将这些单元弃之不用. (secure erase 须将 SSD 资料洗掉, 因此无法自动周期执行) 12. ( ) In a RAID 5 system, more than 1 drive failure would result in total data loss. Ans: T, More than 2 drive failure would result in data loss. 1 drive failure only result in few data bits error. 13. ( ) 168.95.1.1 is in a class C network. Ans: F, Class B. 14. ( ) 10.4.217.0 is a private IP address. Ans: T 15. ( ) When the size of an IP packet is larger than the MTU at the link layer, the packet will simply be discarded. Ans: Given Away, 本题题意不清, 有两种解释: 1. 对传送端而言, 若欲传送之 IP packet 大於 MTU, 则其会将 packet 依 MTU 大小切割後传出. 2. 对接收端而言, 若接收到之 packet 大於 MTU, 则其会直接丢 弃. 因此本题送分. Problem 2. What is the manufacturer of the device with Ethernet address 78:2b:cb:78:97:ab? (1 point) Ans: DELL Problem 3. In a 172.30.0.0/255.254.0.0 network, (1) what is the maximum number of hosts that can be put in this network, given that each one needs to have a unique IP address? (1 point) (2) Write down the network address and the broadcast address of this network. (2 points) Ans: 1. (256 - 254) * (256 - 0) * (256 - 0) - 2 = 131070. 2. Network address: 172.30.0.0 3. Broadcast address: 172.31.255.255 Problem 4. Use a sentence to describe the main task of physical layer in the 5-layer Internet protocol stack. (2 points) Ans: 传送端将 bit stream 转成类比讯号透过媒介传出, 接收端由媒介接收类比讯号并 转回数位 bit stream Grading Criteria: 叙述正确 1 分, 写到转成类比, 透过媒介 (media) 各 0.5 分 Problem 5. Please write down the theoretical speed of the following interfaces and order them from the fastest to the slowest: (A) USB 3.0 (B) USB 2.0 (C) USB 1.0 (D) Thunderbolt 2.0 (E) SATA 1 (F) SATA 2 (G) SATA 3 (H) FireWire (IEEE 1394) 400 (5 points) Ans: 1. (D) Thunderbolt 2.0 - 20Gbps / 2.5GBps 2. (G) SATA 3 - 6Gbps / 600MBps 3. (A) USB 3.0 - 5Gbps / 500MBps 4. (F) SATA 2 - 3Gbps / 300MBps 5. (E) SATA 1 - 1.5Gbps / 150MBps 6. (B) USB 2.0 - 480Mbps / 60MBps 7. (H) FireWire (IEEE 1394) 400 - 400Mbps / 50MBps 8. (C) USB 1.0 - 1.5Mbps / 187.5KBps Grading Criteria: SATA 与 USB3.0 等高速传输界面中, 为防止高速传输中之位元 错误, 加入验证位元, 资料比为 8/10, 因此以 SATA 3 6Gbps 频宽而言, 理论最高速为 6Gbps/8 * 8/10 = 600MBps. 本次期 中考中无论是否计算此一比例均给分 (举例而言 SATA 3 速度 6Gbps, 4.8Gbps, 750MBps, 600MBps 均给对) Problem 6. Explain (1) what is a journaling filesystem (2) what is its main benefit. (4 points) Ans: A journaling file system is a file system that keeps track of the changes that will be made in a journal (which is usually a circular log in a dedicated area of the file system) before committing them to the main file system. In the event of a system crash or power failure, such file systems are quicker to bring back online and less likely to become corrupted - from wikipedia. --

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