課程名稱︰網路管理與系統管理
課程性質︰選修
課程教師：蔡欣穆
開課學院：電資學院
開課系所︰資工系
考試日期（年月日）︰2015.04.20
考試時限（分鐘）：180分鐘
試題 :
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, 機櫃一般放在不同層的同一個位置, 以方便維護, 維修(能快速找到機櫃) 與
佈線(機櫃間彼此距離短).
4. ( ) The lowest class of UTP cable to support Gigabit Ethernet is Cat.5.
Ans: Give Away, 規格上來說至 cat 5e 才全面支援 Gigabit, 但優質的 Cat 5.
網路線在短距離亦可以達到Gigabit, 因此本題送分.
5. ( ) Link aggregation can be used to increase the bandwidth, but cannot
improve the reliability of the link.
Ans: F, 若一條線斷線, 其餘的線路仍可提供通訊服務, 線路兩端並不會因此失連,
因此可增進兩端連線的可靠性.
6. ( ) A broadcast Ethernet frame has a destination address of
00-00-00-ff-ff-ff.
Ans: F, 廣播MAC 應為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, 硬碟的使用頻率與溫度和硬碟的壽命並無明顯關係.
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 約維持在 550MBps 的速度, 低於 USB3.0 之傳輸速度, 但
PCI-Express SSD 則可達到 3.2GBps 之速度, 高於 SATA II 與 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, 傳統硬碟損毀可能是由於磁頭老化, 控制器毀損或磁區受損, 前兩者可藉由
替換相應元件解決, 磁區受損則受損部份資料無法救援, 但其他則可
11. ( ) When a SSD fails, its data can no longer be retrieved, regardless of
the money you can spend.
Ans: T, SSD 正常使用中會有少量記憶體單元錯誤, 此部份非關物理受損, 在這些
未標記的損毀單元低於全部的 5% 時, 控制器仍能藉由演算法還原出正確
資料, 然此部份未標記之記憶體錯誤高於5%, 則控制器將無法分辨正確資
料, 導致讀寫失敗, 此時資料是無法救援的. 因此 SSD 須定期手動做
secure erase, 標記損毀單元並將這些單元棄之不用. (secure erase
須將 SSD 資料洗掉, 因此無法自動週期執行)
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, 本題題意不清, 有兩種解釋:
1. 對傳送端而言, 若欲傳送之 IP packet 大於 MTU, 則其會將
packet 依 MTU 大小切割後傳出.
2. 對接收端而言, 若接收到之 packet 大於 MTU, 則其會直接丟
棄.
因此本題送分.
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: 傳送端將 bit stream 轉成類比訊號透過媒介傳出, 接收端由媒介接收類比訊號並
轉回數位 bit stream
Grading Criteria: 敘述正確 1 分, 寫到轉成類比, 透過媒介 (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 與 USB3.0 等高速傳輸界面中, 為防止高速傳輸中之位元
錯誤, 加入驗證位元, 資料比為 8/10, 因此以 SATA 3 6Gbps
頻寬而言, 理論最高速為 6Gbps/8 * 8/10 = 600MBps. 本次期
中考中無論是否計算此一比例均給分 (舉例而言 SATA 3 速度
6Gbps, 4.8Gbps, 750MBps, 600MBps 均給對)
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.