A. ip route 209.165.201.0 255.255.255.224 209.165.202.130
B. ip route 0.0.0.0 0.0.0.0 209.165.200.224
C. ip route 209.165.200.224 255.255.255.224 209.165.202.129 254
D. ip route 0.0.0.0 0.0.0.0 209.165.202.131
Correct Answer: C
ccna 200-301 network simulator ns21tv the legend.to all routing entries (there is a special section on routing later in the book, the concern here is only with IP address calculation). o R3 has 12+3 (interconnected segments between routers) = 15 entries in its routing table, too many routing table entries will take up more memory, consume more CPU, and bring network instability. Route summarization techniques can be used to reduce the size of the R3 routing table.The route summarization is achieved by converting all the detailed route entries into binary form and taking out the common parts. Here is an example of 4 entries on R1, as shown in Figure 2-5.6.Step 1: 4 steps to convert the 4 detailed route entries into binary form.Step 2: Draw a vertical line after the common part of all the detailed entries. A vertical line is drawn.Step 3:Take out the common part and fill in the bit after it with "0", here it is 191.1.0.0c and count the number of bits to the left of the vertical line is 22, the summary network address is 191.1.0.0/22. Similarly, the summary route entry to R2 is 191.1.4.0/22, The route summary entry to R4 is 191.1.8.0/22oStep 4: Eliminate the detail entries on R3 and keep only the aggregated entries (after learning the routing section, the reader is ready to do the configuration). The routing table entries go from 15 before aggregation to 3 (aggregated routes) + 3 (interconnected segments in the router) = 6 after aggregation, and the routing table is greatly reduced.191.1.0.0/24- 191.1.1.0/24 -191.1.2.0/24 \ 191.1.0.0/22 Part < . )0 10100000 0X-19118.824 191.1.9.0/24 * 10111111 00000001000000 )1 11000000 . 191.1.10.0/24 191.1 K 10111111 00000001000000 0 10000000 . JgJ'J5;0/24 191.1 6.0/24 191.1.4.0/22 191.1.3.0 1011111100000001 000000 . 1 00000000 . Figure 2-5-5 Route Aggregation Chart 256 Route Aggregation Algorithm The calculation of IP addresses is a significant portion of the CCNA exam, so understanding these few examples is not enough; you need to complete the problems later in the chapter and read the solutions and ideas carefully. In addition, VLSM, introduced in Section 6.3 of this book, is also part of the calculation of IP addresses problem. >2.6 Encapsulation and Decapsulation** Encapsulation and decapsulation of data is on the CCNA exam and is required almost every time. Understanding the encapsulation and decapsulation of data is also quite important for you to understand the transmission of packets in the network, and this section combines an example to explain the flow of packets in the network. In fact, the flow of packets in the network is a re The process of encapsulation and decapsulation is repeated. For the explanation of this section, please refer to the video file Video \2.2.wrf" in the CD. The specific implementation steps are listed below. Figure 2-6.1 Flow of packets through the networkFa0/0: Fa0/0:
A. ip route 209.165.201.0 255.255.255.224 209.165.202.130
B. ip route 0.0.0.0 0.0.0.0 209.165.200.224
C. ip route 209.165.200.224 255.255.255.224 209.165.202.129 254
D. ip route 0.0.0.0 0.0.0.0 209.165.202.131
Correct Answer: C
A. to analyze traffic and drop unauthorized traffic from the Internet
B. to transmit wireless traffic between hosts
C. to pass traffic between different networks
D. forward traffic within the same broadcast domain
Correct Answer: C
A. switchport mode trunk
B. switchport mode dynamic desirable
C. switchport mode dynamic auto
D. switchport nonegotiate
Correct Answer: B
A. transfers a backup configuration file from a server to a switch using a username and password
B. transfers files between file systems on a router
C. transfers a configuration files from a server to a router on a congested link
D. transfers IOS images from a server to a router for firmware upgrades
Correct Answer: D
A. different nonoverlapping channels
B. different overlapping channels
C. one overlapping channel
D. one nonoverlapping channel
Correct Answer: D
Exam Code: 200-301
Exam Duration: 120 minutes
Exam Topics:
Latest Update: 11.21,2024
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