第60題： 下面的光纖以太網(wǎng)標準中，支持1000m以上傳輸距離的是（ ）。
A.1000BASE-FX
B.1000BASE-CX
C.1000BASE-SX
D.1000BASE-LX
答案解析與討論：www.xomuzic.com/st/1307828647.html

第61題： IEEE802.11采用了CSMA/CA協(xié)議，采用這個協(xié)議的原因是（ ）。
A.這個協(xié)議比CSMA/CD更安全
B.這種協(xié)議可以開放更多業(yè)務(wù)
C.這種協(xié)議可能解決隱蔽站的問題
D.這個協(xié)議比其它協(xié)議更有效率
答案解析與討論：www.xomuzic.com/st/1307912861.html

第62題： 配置路由器默認路由的命令是（ ）。
A.iproute220.117.15.0255.255.255.00.0.0.0
B.iproute220.117.15.0255.255.255.0220.117.15.1
C.iproute0.0.0.0255.255.255.0220.117.15.1
D.iproute0.0.0.00.0.0.0220.117.15.
答案解析與討論：www.xomuzic.com/st/130804178.html

第64題： 一家連鎖店需要設(shè)計一種編址方案來支持全國各個門店銷售網(wǎng)絡(luò)，門店有300家左右，每個門店一個子網(wǎng)，每個子網(wǎng)終端最多50臺電腦，該連鎖店從ISP處得到一個B類地址，應(yīng)該采用的子網(wǎng)掩碼是（ ）。
A.255.255.255.128
B.255.255.252.0
C.255.255.248.0
D.255.255.255.224
答案解析與討論：www.xomuzic.com/st/1308225547.html

第65題： 網(wǎng)絡(luò)系統(tǒng)設(shè)計過程中，物理網(wǎng)絡(luò)設(shè)計階段的任務(wù)是（ ）。
A.依據(jù)邏輯網(wǎng)絡(luò)設(shè)計的要求，確定設(shè)備的具體物理分布和運行環(huán)境
B.分析現(xiàn)有網(wǎng)絡(luò)和新網(wǎng)絡(luò)的各類資源分布，掌握網(wǎng)絡(luò)所處的狀態(tài)
C.根據(jù)需求規(guī)范和通信規(guī)范，實施資源分配和安全規(guī)劃
D.理解網(wǎng)絡(luò)應(yīng)該具有的功能和性能，最終設(shè)計出符合用戶需求的網(wǎng)絡(luò)
答案解析與討論：www.xomuzic.com/st/1308323231.html

第66題： Traditional IP packet forwarding analyzes the （ 1）IP address contained in then network layee header of each packet as the 　packet rravels from its sourse to its final destination. A router analyzes the destination IP address independently at each hop in the network. Dynamic （ 2） protocols or staitic configuration builds the database needed to analyze the destination IP address(the routing table). The process of implementing traditional IP routing also is called hop-by-hop destination-based （ 3）routing. Although successful,and obviously widely deployed,certain restrictions,which have been realized for some time,exist for this method of packet forwarding that diminish its （ 4）.New techniques are therefore required to address and expand the functionality of an IP-based network infrastructure. This first chapter concertrates on idenfifying these restrictions.and presents a new architecture,known as multiprotocol（ 5）switching,that provides solutions to some of this restrictions.

( 1).A.datagram
B.destination
C.connection
D.service
( 2).A.routing
B.forwarding
C.transmission
D.managerment
( 3).A.anycast
B.multicast
C.broadcast
D.unicast
( 4).A.reliability
B.flexibility
C.stability
D.capability
( 5).A.cost
B.cast
C.mark
D.label

答案解析與討論：www.xomuzic.com/st/1308411314.html