Appendix A
Answers to Written Labs
Chapter 1: Understanding Basic Networking
Answers to Written Lab 1
1. Bus, ring, and star
2. Ring
3. Server
4. Client/server
5. Point-to-point
6. Switch
7. Star and extended star
8. Virtual LAN
9. A segment
10. Bus
Chapter 2: Internetworking
Answers to Written Lab 2.1
1. The Application layer is responsible for finding the network resources broadcast from a server and adding flow control and error control (if the application developer chooses).
2. The Physical layer takes frames from the Data Link layer and encodes the 1s and 0s into a digital signal for transmission on the network medium.
3. The Network layer provides routing through an internetwork and logical addressing.
4. The Presentation layer makes sure that data is in a readable format for the Application layer.
5. The Session layer sets up, maintains, and terminates sessions between applications.
6. PDUs at the Data Link layer are called frames and provide physical addressing plus other options to place packets on the network medium.
7. The Transport layer uses virtual circuits to create a reliable connection between two hosts.
8. The Network layer provides logical addressing, typically IP addressing and routing.
9. The Physical layer is responsible for the electrical and mechanical connections between devices.
10. The Data Link layer is responsible for the framing of data packets.
11. The Session layer creates sessions between different hosts’ applications.
12. The Data Link layer frames packets received from the Network layer.
13. The Transport layer segments user data.
14. The Network layer creates packets out of segments handed down from the Transport layer.
15. The Physical layer is responsible for transporting 1s and 0s (bits) in a digital signal.
16. Transport
17. Transport
18. Data Link
19. Network
20. 48 bits (6 bytes) expressed as a hexadecimal number
Answers to Written Lab 2.2
| Description | Device or OSI Layer |
| This device sends and receives information about the Network layer. | Router |
| This layer creates a virtual circuit before transmitting between two end stations. | Transport |
| This device uses hardware addresses to filter a network. | Bridge or switch |
| Ethernet is defined at these layers. | Data Link and Physical |
| This layer supports flow control, sequencing and acknowledgments. | Transport |
| This device can measure the distance to a remote network. | Router |
| Logical addressing is used at this layer. | Network |
| Hardware addresses are defined at this layer. | Data Link (MAC sublayer) |
| This device creates one big collision domain and one large broadcast domain. | Hub |
| This device creates many smaller collision domains, but the network is still one large broadcast domain. | Switch or bridge |
| This device can never run full duplex. | Hub |
| This device breaks up collision domains and broadcast domains. | Router |
Answers to Written Lab 2.3
1. Hub: One collision domain, one broadcast domain
2. Bridge: Two collision domains, one broadcast domain
3. Switch: Four collision domains, one broadcast domain
4. Router: Three collision domains, three broadcast domains
Chapter 3: Ethernet Technologies
Answers to Written Lab 3.1
1. Convert from decimal IP address to binary format.
Complete the following table to express 192.168.10.15 in binary format.
| Decimal | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Binary |
| 192 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 11000000 |
| 168 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 10101000 |
| 10 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 00001010 |
| 15 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 00001111 |
Complete the following table to express 172.16.20.55 in binary format.
| Decimal | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Binary |
| 172 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 10101100 |
| 16 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 00010000 |
| 20 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 00010100 |
| 55 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 00110111 |
Complete the following table to express 10.11.12.99 in binary format.
| Decimal | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Binary |
| 10 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 00001010 |
| 11 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 00001011 |
| 12 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 00001100 |
| 99 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 01100011 |
2. Convert the following from binary format to decimal IP address.
Complete the following table to express 11001100.00110011.10101010.01010101 in decimal IP address format.
| Binary | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Decimal |
| 11001100 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 204 |
| 00110011 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 51 |
| 10101010 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 170 |
| 01010101 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 85 |
Complete the following table to express 11000110.11010011.00111001.11010001 in decimal IP address format.
| Binary | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Decimal |
| 11000110 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 198 |
| 11010011 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 211 |
| 00111001 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 57 |
| 11010001 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 209 |
Complete the following table to express 10000100.11010010.10111000.10100110 in decimal IP address format.
| Binary | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Decimal |
| 10000100 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 132 |
| 11010010 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 210 |
| 10111000 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 184 |
| 10100110 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 166 |
3. Convert the following from binary format to hexadecimal.
Complete the following table to express 11011000.00011011.00111101.01110110 in hexadecimal.
| Binary | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Hexadecimal |
| 11011000 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | D8 |
| 00011011 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1B |
| 00111101 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 3D |
| 01110110 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 76 |
Complete the following table to express 11001010.11110101.10000011.11101011 in hexadecimal.
| Binary | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Hexadecimal |
| 11001010 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | CA |
| 11110101 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | F5 |
| 10000011 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 83 |
| 11101011 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | EB |
Complete the following table to express 10000100.11010010.01000011.10110011 in hexadecimal.
| Binary | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Hexadecimal |
| 10000100 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 84 |
| 11010010 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | D2 |
| 01000011 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 43 |
| 10110011 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | B3 |
Answers to Written Lab 3.2
When a collision occurs on an Ethernet LAN, the following happens:
1. A jam signal informs all devices that a collision occurred.
2. The collision invokes a random backoff algorithm.
3. Each device on the Ethernet segment stops transmitting for a short time until the timers expire.
4. All hosts have equal priority to transmit after the timers have expired.
Answers to Written Lab 3.3
1. Crossover
2. Straight-through
3. Crossover
4. Crossover
5. Straight-through
6. Crossover
7. Crossover
8. Rolled
Answers to Written Lab 3.4
At a transmitting device, the data encapsulation method works like this:
1. User information is converted to data for transmission on the network.
2. Data is converted to segments, and a reliable connection is set up between the transmitting and receiving hosts.
3. Segments are converted to packets or datagrams, and a logical address is placed in the header so each packet can be routed through an internetwork.
4. Packets or datagrams are converted to frames for transmission on the local network. Hardware (Ethernet) addresses are used to uniquely identify hosts on a local network segment.
5. Frames are converted to bits, and a digital encoding and clocking scheme is used.
Chapter 4: TCP/IP DoD Model
1. TCP
2. Host-to-Host
3. UDP
4. TCP
5. ICMP
6. Frames
7. Segment
8. Port numbers
9. IP addresses
10. 1, 2
Chapter 5: IP Adressing
1. 4 billion
2. NAT
3. 1 through 126
4. Loopback or diagnostics
5. Turn all host bits off.
6. Turn all host bits on.
7. 10.0.0.0 through 10.255.255.255
8. 172.16.0.0 through 172.31.255.255
9. 192.168.0.0 through 192.168.255.255
10. 128-bit colon-delimited hexadecimal
Chapter 6: Easy Subnetting
Answers to Written Lab 6.1
1. 192.168.100.25/30. A /30 is 255.255.255.252. The valid subnet is 192.168.100.24, broadcast is 192.168.100.27, and valid hosts are 192.168.100.25 and 26.
2. 192.168.100.37/28. A /28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts 33–46.
3. 192.168.100.66/27. A /27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64. The host is in the 32 subnet, broadcast address of 63. Valid host range of 33–62.
4. 192.168.100.17/29. A /29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid hosts 17–22.
5. 192.168.100.99/26. A /26 is 255.255.255.192. The fourth octet has a block size of 64. 0, 64, 128. The host is in the 64 subnet, broadcast of 127. Valid hosts 65–126.
6. 192.168.100.99/25. A /25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid hosts 1–126.
7. A default Class B is 255.255.0.0. A Class B 255.255.255.0 mask is 256 subnets, each with 254 hosts. We need fewer subnets. If we used 255.255.240.0, this provides 16 subnets. Let’s add one more subnet bit. 255.255.248.0. This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21.
8. A /29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0, 8, 16. The host is in the 8 subnet, broadcast is 15.
9. A /29 is 255.255.255.248, which is 5 subnet bits and 3 host bits. This is only 6 hosts per subnet.
10. A /23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet; the broadcast address is 16.3.255.
Answers to Written Lab 6.2
| Classful Address | Subnet Mask | Number of Hosts per Subnet (2n – 2) |
| /16 | 255.255.0.0 | 65,534 |
| /17 | 255.255.128.0 | 32,766 |
| /18 | 255.255.192.0 | 16,382 |
| /19 | 255.255.224.0 | 8,190 |
| /20 | 255.255.240.0 | 4,094 |
| /21 | 255.255.248.0 | 2,046 |
| /22 | 255.255.252.0 | 1,022 |
| /23 | 255.255.254.0 | 510 |
| /24 | 255.255.255.0 | 254 |
| /25 | 255.255.255.128 | 126 |
| /26 | 255.255.255.192 | 62 |
| /27 | 255.255.255.224 | 30 |
| /28 | 255.255.255.240 | 14 |
| /29 | 255.255.255.248 | 6 |
| /30 | 255.255.255.252 | 2 |
Answers to Written Lab 6.3
| Decimal IP Address | Address Class | Number of Subnet and Host Bits | Number of Subnets (2x) | Number of Hosts (2x – 2) |
| 10.25.66.154/23 | A | 15/9 | 32,768 | 510 |
| 172.31.254.12/24 | B | 8/8 | 256 | 254 |
| 192.168.20.123/28 | C | 4/4 | 16 | 14 |
| 63.24.89.21/18 | A | 10/14 | 1,024 | 16,384 |
| 128.1.1.254/20 | B | 4/12 | 16 | 4,094 |
| 208.100.54.209/30 | C | 6/2 | 64 | 2 |
Chapter 7: Introduction to Nexus
Answers to Written Lab 7.1
1. The console port is a serial port used for out-of-band configuration.
2. The management port is a dedicated Ethernet port that allows for remote out-of-band configuration.
3. L1/L2 ports are not used.
Answers to Written Lab 7.2
Each of these are virtualized with a different technology.
1. VLAN
2. Trunking
3. Switch virtual interfaces (SVIs)
4. Virtual Routing and Forwarding (VRF)
5. Virtual Device Context (VDC)
Answers to Written Lab 7.3
At a transmitting device, the data encapsulation method works like this:
1. HSRP is a layer 3 process.
2. STP is a layer 2 process.
3. Pim is a layer 2 process.
4. Cisco Discovery Protocol is a layer 2 process.
5. OSPF is a layer 3 process.
6. UDLD is a layer 2 process.
Answers to Written Lab 7.4
Remember, these are purely virtual devices and only software. There is no hardware.
1. Zero
2. None
3. None
4. None
Answers to Written Lab 7.5
VRF allows for multiple routing tables on a single device. This is useful because you may wish for a different interface to treat layer 3 traffic differently. VDC effectively creates another switch with its own administration and configuration. This is very useful in a multi-tenant environment or anywhere you want to have administration separated.
Chapter 8: Configuring Nexus
1. no shutdown
2. write erase boot
3. username todd role network-admin password todd
4. username todd role network-operator password cisco
5. reload
6. switchname Chicago, or hostname Chicago
Chapter 9: IP Routing
Answers to Written Lab 9
1. False. The MAC address would be the router interface, not the remote host.
2. It will use the gateway interface MAC at layer 2 (L2) and the actual destination IP at layer 3 (L3).
3. True.
4. None; not on this planet.
5. IP will discard the packet, and ICMP will send a destination unreachable packet out the interface on which the packet was received.
Chapter 10: Routing Protocols
1. EIGRP
2. 15
3. 30 seconds
4. RIP
5. Address-family
Chapter 11: Layer 2 Switching Technologies
1. Client
2. show vtp status
3. Broadcast
4. Collision
5. Transparent
6. Trunking allows you to make a single port part of multiple VLANs at the same time.
7. Frame identification (frame tagging) uniquely assigns a user-defined ID to each frame. This is sometimes referred to as a VLAN ID or color.
8. True
9. Access link
10. interface-vlan
Chapter 12: Redundant Switched Technologies
1. show mac address-table
2. show spanning-tree or show spanning-tree vlan vlan#
3. 802.1w
4. STP, or RSTP in NX-OS
5. Rapid-PVST+
6. spanning-tree port type edge
7. show interface port-channel number
8. spanning-tree port type switch
9. show spanning-tree and show spanning-tree summary
10. RSTP and MSTP
Chapter 13: Security
1. ip access-list 101, deny ip 172.16.0.0 0.0.255.255 any, permit ip any any
2. ip access-group 101 out
3. ip access-list 101, deny ip host 192.168.15.5 any, permit ip any any
4. show access-lists
5. ip access-list 110
Deny tcp host 172.16.10.1 host 172.16.30.5 eq 23
permit ip any any
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