networks and ethernet - boise state...

552 Final Exam Questions, spring 2011 … Re-Ordered

Networks and Ethernet

What service offers secure, reliable connectivity over a shared public network infrastructure? Virtual private network The network area within which data packets originate and collide is called a Collision domain Which best describes a CSMA/CD network? A. One node transmission traverses the entire network and is received and examined by

every node. B. Signals are sent directly to the destination if the source knows both the mac and ip addresses C. One node transmission goes to the nearest router, which sends it directly to the destination D. Signals are always sent in broadcast mode

What is the name of the method in Ethernet that explains how Ethernet works? A. TCP/IP

B. CSMA/CD C. CMDA/CS D. CSMA/CA CSMA/CD – “The acronym CSMA/CD signifies carrier-sense multiple access with collision detection and describes how the Ethernet protocol regulates communication among nodes.” http://computer.howstuffworks.com/ethernet7.htm

In which circumstance are multiple copies of the same unicast frame likely to be transmitted in the school district’s Switched LAN? A. During high traffic periods

B. In an improperly implemented redundant topology C. After broken links are re-established D. When upper-layer protocols require high reliability E. When a dual ring topology is in use F. None of the above

The network administrator issues the ping 192.168.2.5 command and successfully tests connectivity to a host that has been newly connected to the network. Which protocols were used during the test? (Choose two) A. ARP

B. CDP C. DHCP D. DNS E. ICMP This is connectivity only – Data link layer checking physical connections. Yes A – ARP (Address Resolution Protocol). ARP finds the hardware address of a host from a

known IP address and we know the address. If IP doesn't find the destination host's hardware address in the ARP cache, it uses ARP to find this information.

Not B – CDP (Cisco Delivery Protocol). CDP is a proprietary Data Link layer protocol which is used to share information about other directly connected Cisco equipment. Doesn’t have to be directly connect, just needs a path.


Not C – DHCP (Dynamic Host Configuration Protocol). This protocol a host to be configured automatically and not need the intervention of a network administrator. It also provides a centralized database that keeps track of computers that are connected to the network.

Not D – DNS (Domain Name Server). A hierarchical naming system which translates domain names that are meaningful to humans into numeric identifiers associated with networking equipment.

Yes E – ICMP (Internet Control Message Protocol). One of the core protocols of the Internet Protocol Suite. It is chiefly used by the operating systems of networked computers to send error messages indicating, for example, that a requested service is not available or that a host or router could not be reached. ICMP can also be used to relay query messages.

Which protocol automates all of the following functions for hosts on a network: IP configuration, IP addresses, subnet masks, default gateways, and DNS server information? A. CDP

B. SNMP C. DHCP D. ARP E. DNS F. None of the above DHCP uses the concept of the client making a request and the server supplying the IP address to the client, plus other information such as the default gateway, subnet mask, DNS IP address, and other information. Incorrect Answers: CDP is the Cisco Discovery Protocol. It is used by Cisco devices at the data link layer to obtain

information about their directly connected neighbors. SNMP is the Simple Network Management Protocol. This is used for the maintenance,

management, and monitoring of network devices. ARP is the Address Resolution Protocol, which is used to resolve layer 2 MAC addresses to layer

3 IP addresses. DNS is the Domain Name Service. It is used to resolve domain names (for example,

www.testking.com) to IP addresses. The IP address location of primary and secondary DNS resolver servers can be obtained for hosts using DHCP.

With regard to Ethernet media access methods, which of the following are true? (Choose all that apply.) A device waits for an electronic signal before transmitting.

A device listens and waits until the media is not busy before transmitting. All devices on an Ethernet segment see data that passes on the network medium. Only the sender and the receiver devices see data that passes on the network medium. Ethernet networks allow you to configured devises with higher transmission priority. Ethernet uses the CSMA/CD algorithm:

1. A device with a frame to send listens until the Ethernet is not busy. 2. When the Ethernet is not busy, the sender begins sending the frame. 3. The sender listens to make sure that no collision occurred. 4. Once the senders hear the collision, they each send a jamming signal, to ensure that all stations recognize the collision. 5. After the jamming is complete, each sender randomizes a timer and waits that long. 6 When each timer expires, the process starts over with Step 1.


TCP/IP and OSI

What is the OSI model? A conceptual framework that specified how information travel through networks

A model that describes how data makes its way from one application program to another throughout a network A conceptual framework that specifies which network occur at each layer All of above

Which OSI layer is responsible for logical addressing? Transport (Layer 4)

Network (Layer 3) Data link (Layer 2) Physical (Layer 1) Network (Layer 3) is logical addressing and path determination, routing, reporting delivery errors.

Which OSI later defines voltage levels, maximum transmission distance, and physical connectors? Application (Layer 7)

Transport (Layer 4) Physical (Layer 1) Data transport (not the name of a layer) http://routergod.com/ccnabootcamp/osi.html http://www.webopedia.com/quick_ref/OSI_Layers.asp Physical Layer (Layer 1) - Binary transmission signals and encoding. Layout of pins, voltages, cable specifications, modulation

The TCP/IP protocol suite has specifications for which layers of the OSI model? 1 through 3

1 through 4 and 7 3 through 7 1, 3, and 4 http://www.vlsm-calc.net/models.php

Sequencing, acknowledgments, and flow control are characteristics of which OSI layer? Layer 5 (session)

Layer 4 (transport) Layer 3 (network) Layer 2 (data link) Layer 4/Transport Layer - End-to-end connections and reliability. Segmentation/desegmentation of data in proper sequence. Flow control

Which OSI layer is responsible for reliable data transport? Transport (Layer 4)

Network (Layer 3) Data Link (Layer 2) Application (Layer 7) http://routergod.com/ccnabootcamp/osi.html http://www.webopedia.com/quick_ref/OSI_Layers.asp


Binary

Which of the following represents 203.128.56.10 in binary format? 11001011.10000000.00111000.00001010 What is the decimal number 151 in binary? 10010111


Addressing

You are a technician at XYZ. You tell your newly appointed XYZ trainee about public networks IP Address classes. Your trainee now wants to know if all the IP Address Classes can be assigned for public use. What would your reply be? All except Class D and Class E addresses.

All except Class E addresses. Only Class A and Class B addresses. Only Class B and Class C addresses. Only Class A, Class C and Class E addresses. RFC 7980 defines the IP protocol, including multiple different classes of networks. IP defines three different network classes, called A, B, and C, from which individual hosts are assigned IP addresses. TCP/IP defines Class D (multicast) addresses and Class E (experimental) addresses as well. A, B, and C are public address classes, D and E are not.

Which of the following IP addresses is a private IP address? 172.20.14.36 --- within private range 172.16.0.0 - 172.31.255.255

192.168.42.34 --- within private range 192.168.0.0 - 192.168.255.255 172.33.194.30 --- out of range 12.0.0.1 --- out of range 168.172.19.39 --- out of range Private IP address ranges 10.0.0.0 - 10.255.255.255

172.16.0.0 - 172.31.255.255 192.168.0.0 - 192.168.255.255

You are a technician at XYZ. XYZ plans to implement a public network. The company's ISP suggested that XYZ register a Class IP address. The XYZ CEO wants to know how many usable IP addresses are provided in a Class C address. What would your reply be? 128 192 254 256 510

Class C address has one octet for hosts. 0 is network address, 255 is broadcast, so 254 usable addresses. Also, 2^8 – 2 = 254

Which of the following is the correct IP address configuration for Host A?

IP 192.168.100.31 255.255.255.240 default-gateway 192.168.100.18 IP 192.168.100.30 255.255.255.240 default-gateway 172.16.1.1 IP 192.168.100.20 255.255.255.240 default-gateway 192.168.100.17 IP 192.168.100.21 255.255.255.248 default-gateway 192.168.100.17 IP 192.168.100.19 255.255.255.248 default-gateway 172.16.1.1


/28 is 240 in dotted decimal, so it has to be A, B or C. The default gateway is 192.168.100.17 (host switch router) so it has to be C

You are a systems administrator and you are about to assign static IP addresses to various servers on your network. For the network 192.168.20.24/29 the router is assigned to the first usable host address, while the last usable host address goes to your Sales server. Which one of the following commands would you enter into the IP properties box of the sales server? IP address: 192.168.20.14 Subnet Mask: 255.255.255.248 Default Gateway: 192.168.20.9

IP address: 192.168.20.254 Subnet Mask: 255.255.255.0 Default Gateway: 192.168.20.1 IP address: 192.168.20.30 Subnet Mask 255.255.255.248 Default Gateway: 192.168.20.25 IP address: 192.168.20.30 Subnet Mask 255.255.255.240 Default Gateway: 192.168.20.17 IP address: 192.168.20.30 Subnet Mask 255.255.255.240 Default Gateway: 192.168.20.25 Mask /29 means there are three bits for hosts. 2^3-2 = 6. Minus 2 since first address is network and last is broadcast. 192.168.20.24 is network address so the six hosts would be .25, .26, .27, .28, .29 and .30, with 31 being broadcast address. The router is assigned to the first useable host (.25) and Sales server is last usable host (.30). Syntax for the command is destination IP (Sales server: 192.168.20.30) then mask (/29 = 255.255.255.248), then next-hop router (192.168.20.25).

XYZ is migrating to a private IP addressing scheme. Which of the following describe the use of private IP addresses? (Choose two) Addresses that cannot be routed through the public Internet.

A scheme to conserve public addresses. Addresses chosen by XYZ.com to communicate with the Internet. Addresses that can be routed through the public Internet. Addresses licensed to enterprise or ISPs by an Internet registry organization.


Broadcast and Collision Domains

Which best describes broadcasting? Sending a single frame to many stations at the same time

Sending a single frame to all routers to simultaneously update their routing tables Sending a single frame to all routers at the same time Sending a single frame to all hubs and bridges at the same time

Which of the statements mentioned below will describe the network as illustrated in the exhibit? (Select two options.)

There are two broadcast domains in the network. There are seven collision domains in the network. There are four broadcast domains in the network. There are six broadcast domains in the network. There are four collision domains in the network. There are five collision domains in the network.

Routers separate broadcast domains - there two connections to the router, so there are two

broadcast domains (A). Routers and switches separate collisions domains – there are two connections to the router and

five connections to the switch (device on the left), so there are seven collision domains. Hubs do not separate collision domains – the connections to the hub (device on the right) are all

within the collision domain created by that connection to the router. You are a technician at XYZ. You tell your newly appointed XYZ trainee that Ethernet networks are broadcast domains and collision domains. Your trainee wants to know how the hosts on an Ethernet network will know when to resume transmissions after a collision has occurred. What would your reply be? (Choose all that apply.) The destination host sends a request to the source for retransmission.

The jam signal indicates that the collision has been cleared. The hosts will attempt to resume transmission after a time delay has expired. An electrical pulse indicates that the collision has cleared. The router on the segment will signal that the collision has cleared. CSMA/CD – Collision detection means that hosts know when packets collide. Not A – Destination host doesn’t necessarily know to expect anything. Not B – The jam signal is sent by senders when they hear the collision, not when it has been cleared. Yes C – Hosts wait random amount of time, listen to see that the network is not busy, then retransmit. Not D or E – None of the devices signal that the jam has been cleared.


Study the network topology exhibit carefully, in particular the two switches XYZ 1, XYZ 2, and the router XYZ 3. Which statements are true in this scenario? Select two.

A. All the devices in both networks will receive a broadcast to 255.255.255.255 sent by host A.

B. Only the devices in network 192.168.1.0 will a broadcast to 255.255.255.255 sent by host A. C. All the devices on both networks are members of the same collision domain. D. The hosts on the 192.168.1.0 network form one collision domain, and the hosts on the 192.168.2.0 network form a second collision domain. E. Each host is in a separate collision domain.

Not A – Routers separate collision domains and the two networks are separated by a router. Therefore, devices in both networks can’t receive the same broadcast.

Yes B – Host A is in network 192.168.1.0 and all hosts within the broadcast domain receive the same broadcasts. Network 192.168.1.0 broadcast domain includes the left cable out of router XYZ3, switch XYZ1 and the five hosts in 192.168.1.0.

Not C or D, Yes E – Switches separate collision domains so devices on each network are separate collision domains.


Hubs, Bridges and Switches

Which of the following is true concerning the function of a switch? A. Switches increase the sizes of collision domains

B. Switch combine connectivity of a hub with the traffic regulation of a bridge C. Switches segment networks by separating broadcast domains D. Switches perform layer 4 path selection Not A – Switches reduce the size of collision domains Not C – Routers separate broadcast domains, not switches. Not D – Switches are a layer 2 device - Data Link Layer: Physical addressing and access to media.

Two sublayers: Logical Link Control (LLC) and Media Access Control (MAC). Layer 4 is transport – error recovery and flow control.

If host B sends a frame to host C, how will the switch act?

A. Drop the frame B. Send the frame out all ports except port 0/2 C. Return the frame to host B D. Send an ARP request for host C E. Send an ICMP Host Unreachable message to host B F. Record the destination MAC address in the switching table and send the frame directly to Host C

Host C is not in the MAC Address table so the switch will flood the ports (multicast) to see if the desired host replies. Not A – The frame isn’t dropped because the switch hasn’t yet tried to find the host. Yes B – The frame is not sent out port 0/2 because that was the receiving port and on switches, floods are

not sent out the port it was received on. Not C – Frames don’t go back to the sender. Not D – ARP (Address Resolution Protocol) is used when a sending host has an IP Address, but doesn’t

have the MAC address yet. Sending host doesn’t have the IP Address yet so it doesn’t need to resolve anything.

Not E – ICMP Host Unreachable is used when a physical connection does not exist or the indicated port or protocol is not active. From the diagram, it appears that neither of those is the case here.

Not F – The MAC address of the sender is recorded in the switching table, not the destination. Which one of the following characteristics is true regarding the use of hubs and switches? A. Hubs can have their ports be configured with VLANs

B. Using hubs is costly with regard to bandwidth availability. C. Switches cannot forward broadcasts. D. Switches are more efficient than hubs in processing frames. E. Switches increase the number of collision domains in the network. Not A – Hubs are a Layer 1 device – they don’t look at the frame, they blindly pass it along. Not a smart

device, not configured.


Not B – Hubs are cheaper than switches. I initially marked B as “yes” because I interpreted the question to mean costly in terms of bandwidth usage, not costly in terms of the price of the device. All frames are sent to all devices connected to a hub – inefficient use of bandwidth, gobbles it up needlessly.

Not C – It’s routers that can’t forward broadcasts. Switch (and dumb hubs) do forward broadcasts. Not D – Explanation from the other document says that hubs and switches can be equally efficient in

theory. In practice, switches are more efficient because they have more memory CPU and memory. I initially said “yes”, that switches are more efficient because they send out only what they need to – they look at destination and sender IP address and make logical decisions.

Yes E – Switches increase the number of collision domains because they break up broadcast domains. More collision domains mean fewer actual collisions. Each port on a switch is a separate collision domain.

When comparing and contrasting the similarities and differences between bridges and switches, which of the following are valid statements? (Choose all the valid answer choices) A. Bridges are faster than switches because they have fewer ports.

B. A switch is a multiport bridge, C. Bridges and switches learn MAC addresses by examining the source MAC address of each frame received. D. A bridge will forward a broadcast but a switch will not. E. Bridges and switches increase the size of a collision domain. F. None of the above statements are true. Not A – Hubs and switches can have the same number of ports. Not D – Only routers don’t forward broadcasts. Not E – Bridges and switches decreased the size of a collision domain by segmenting it into smaller

collision domains. Not F – two statements are true so F is not correct.

Why would a network administrator configure port security on a new switch? A. To prevent unauthorized Telnet access to a switch port.

B. To limit the number of Layer 2 broadcasts on a particular switch port. C. To prevent unauthorized hosts from accessing the LAN. D. To protect the IP and MAC address of the switch and associated ports. E. To block unauthorized access to the switch management interfaces over common TCP ports. F. None of the above You can use the port security feature to restrict input to an interface by limiting and identifying MAC addresses of the stations allowed to access the port. When you assign secure MAC addresses to a secure port, the port does not forward packets with source addresses outside the group of defined addresses. Not A – Telnet and Secure Shell (SSH) are not physical connections. Not B – Broadcasts are determined by device type, not configuration. Yes C – You can specify the IP address of hosts that are allowed to connect or have those IP

addresses added dynamically. Not D or E – protection like this comes from using encrypted passwords.


Routers and Segmenting

A router receives a packet on interface 172.16.45.66/26. The source IP of the packet is 172.16.45.127/26 and the destination is 172.16.46.191/26. How will the router handle the packet? The destination is a broadcast address, so the router will not forward the packet.

The destination is a host on another subnet, so the router will not forward the packet. The destination is a host on the same subnet, so the router will forward the packet. The destination is a network address, so the router will forward the packet.

172 is a Class C address which means 24 bits are host bits. The mask is /26 – 24 bits for hosts, 2 bits for subnet 2^2 = 4 subnets 6 bits remain for hosts --- 2^6 = 64 hosts per subnet BooleanAND of address and mask is 172.16.45.0, networks will be multiples of 64

o 172.16.45.0 --- 172.16.45.63 o 172.16.45.64 --- 172.16.45.127 o 172.16.45.128 --- 172.16.45.191 o 172.16.45.192 --- 172.16.45.255 o 172.16.46.0 --- 172.16.46.63 o 172.16.46.64 --- 172.16.46.127 o 172.16.46.128 --- 172.16.46.191 o 172.16.46.192 --- 172.16.46.255

The desination address, 172.16.46.191, is a broadcast address so the router will not forward the packet.

Row 1 172 16 45 66 Row 2 Address 10101100 00010000 00101101 01000010 Row 3 mask 11111111 11111111 11111111 11000000 Row 4 AND Row 5 Network 172 16 45 64

Row 6 Host bits 1 11111111

Row 7 Broadcast 124 16 45 255 What are the benefits of using a router to segment the network at a main office? (Select two options.) Broadcasts are not forwarded across the router.

Filtering can occur based on Layer 3 information. Broadcasts are eradicated. Adding a router to the network decreases latency. Routers are more efficient than switches. Broadcasts are forwarded across everything except routers, so A is correct. Routers are layer 3, switches are layer 2, so data will travel with greater efficiency to desired destination.

You are a network technician at XYZ. You use a Cisco router to divide the XYZ network into two separate segments, XYZ 1 and XYZ 2. The XYZ CEO is concerned about the cost, and wants to know what the purpose of your action is. What should you tell him? A. It decreases the number of broadcast domains.


B. It makes broadcasting domains more efficient between segments. C. It increases the number of collisions. D. It prevents segment 1's broadcasts from getting to segment 2. E. It connects segment 1's broadcasts to segment 2. Not A – Segmenting increases the number broadcast domains (which is better). Not B – There are no broadcast domains between segments. Not C – It decreases the number of collisions because there are smaller collision domains. Yes D – Routers segment the network, broadcasts don’t get past them. Not E – Broadcase domains don’t connect segments.

You are the network administrator at XYZ. Your newly appointed XYZ trainee wants to know what router functions are. What will your reply be? (Choose three.) A. Packet switching

B. Collision prevention on a LAN segment. C. Packet filtering D. Broadcast domain enlargement E. Broadcast forwarding F. Internetwork communication Yes A – Routers and switches pass along packets per routing tables and MAC Address tables. Not B – You can’t prevent collisions, really, you just recover from them different ways. Yes C – Routers are a Layer 3 device so they deal with packets. Routing tables and configuration

tell the router what to do with the various packets (flitering). Not D – Broadcasts can’t get past routers so they make broadcast domains smaller, not larger. Not E – Routers don’t forward broadcasts, they are the originators of the broadcast. Yes F – Routers communicate with other routers in their network to update routes and metrics.

Assume you are installing a small SOHO network. Which interface on the router will be linked to the leased line connected to the Internet? Aux

Console Fast Ethernet 0/0 Serial 0/0 Not Aux – Auxiliary Port connects to telephone line, which is connected to an analog modem, so that a remote PC can dial in to the router. Used for configuration. Not Console – Console connects to PC via rollover cable (12345678 on one end 87654321 on the other end). Used for configuration. Not FastEthernet - FastEthernet connects to devices Ethernet access devices like switches, hubs, and PCs.

Which of the following correctly describe the various functions and virtues of a router? (Select all valid answer choices) Internetwork communication

Packet filtering Packet switching Broadcast domain enlargement Broadcast forwarding Collision prevention on a LAN segment. None of the above


The school district’s LAN consists of one large flat network. You decide to segment this LAN into two separate networks with a router. What will be the affect of this change? A. The number of broadcast domains will be decreased.

B. It will make the broadcasting of traffic between domains more efficient between segments. C. It will increase the number of collisions. D. It will prevent segment 1's broadcasts from getting to segment 2. E. It will connect segment 1's broadcasts to segment 2. Router doesn’t forward broadcasts. It breaks up broadcast domain which reduces unnecessary traffic.

Not A – Number of broadcast domains increases from one to two. Not B – Link level traffic now needs to go through the router. This reduces traffic but is less

efficient transport between two segments. Not C – Since the network is basically cut in half, the number of collisions should reduce

greatly. Not E – Broadcasts from one segment are completely hidden from the other.

Router TK1 has just received a packet and needs to route it. What two actions must this router take in order to route incoming packets? (Choose two) Identify the destination network address of each packet.

Inspect the routing table to select the best path to the destination network addresses. Validate sources of routing information. Inspect the ARP table to verify a legitimate source MAC address for each packet. Verify the receipt of routed packets by the next hop router. Identify the source network address of each packet. Explanation: Figure out where it’s going and how it should get there.


Subnetting

Which of the following IP addresses for the network 124.16.32.55/28 can be assigned to hosts? (Choose three) 124.16.32.57

124.16.32.60 124.16.32.62 124.16.32.45 --- out of range 124.16.32.48 --- network address 124.16.32.63 --- broadcast address

ANDing of 124.16.32.55 and 255.255.255.240 (/28) results in network address 124.16.32.48 28 network bits, so 4 host bits. 2^h – 2 = number of hosts 2^4 – 2 = 16 – 2 =14 hosts 124.16.32.48 is network address, host range is 49-62, broadcast is 63.

Row 1 124 16 32 55 Row 2 Address 01111100 00010000 00100000 00110111 Row 3 mask 11111111 11111111 11111111 11110000 Row 4 AND 00110000 Row 5 Network 124 16 32 48

Row 6 Host bits 1 00111111

Row 7 Broadcast 124 16 32 63 Range 49 - 62

XYZ has been assigned the IP address 189.66.1.0 by its Internet Service Provider. Your newly appointed XYZ trainee wants to know how many hosts will be supported on each subnet if you divide the network by using the 255.255.255.224 subnet mask. What would your reply be? 14 16 30 32 62 64

Mask ending 224 uses 3 of the 8 bits in the last octet (/27), which leaves 5 bits for host. 2^h – 2 = number of hosts … 2^5 – 2 = 30

You are the newly appointed XYZ trainee. Your supervisor tells you that the IP address for your workstation is 172.16.209.10/22. He wants to know what the subnetwork number of your workstation is. What will your reply be? 172.16.208.0

172.16.42.0 172.16.107.0 172.16.252.0 172.16.254.0

Row 1 172 16 209 10 Row 2 Address 10101100 00010000 11010001 00001010 Row 3 mask 11111111 11111111 11111100 00000000 Row 4 AND 11010000 Row 5 Network 172 16 208 0

Row 6 Host bits 1 11010011 11111111

Row 7 Broadcast 124 16 211 255


Which of the following addresses can be assigned to network hosts when given a subnet mask of 255.255.255.224? (Choose all that apply.) 92.11.178.93

134.178.18.56 192.168.16.87 15.234.118.63 --- broadcast 201.45.116.159 --- broadcast 217.63.12.192 --- subnet address Since the subnet mask is 255.255.255.224, the number of hosts that is available is 30 (5 host bits – 3^5 – 2 = 30). Every network boundary will be a multiple of 32. This means that every subnet will be a multiple (0, 32, 64, 96, 128, 160, 192, 224) and the broadcast address for each of these subnets will be one less this number (31, 63, 95, 127, 159, 191, 223). Therefore, any IP address that does not end in one of these numbers will be a valid host IP address.

You are a network technician at XYZ. XYZ has a class C network license. The company requires 5 usable subnets. Each subnet must accommodate at least 18 hosts. Which network mask should you use? 225.225.224.0.

225.225.240.0. 225.225.255.0. 255.255.255.224 225.225.255.240 Class C network so first three octets are network – all binary 1s (=255). One octet/8 bits

remaining. Need 5 usable subnets, so 3 subnet bits (2^3 – 2 = 6) are needed. 5 bits remaining. Need 18 hosts, so 5 host bits (2^5 – 2 = 30) are needed. No bits remaining. Three S bits in last octet is 11100000, or .224.

If an Ethernet port on router TK1 was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this LAN subnet? 2046 1024 4096 8190 4094 None of the above

20 bits are used for network, which means there are 12 bits remaining for hosts. 2^12 – 2 = 4,094.

Which of the following addresses can be assigned to a host when using a subnet mask of 255.255.254.0? (Select three) 186.54.3.0 --- valid host in network 186.54.2.0

26.35.2.255 --- valid host in network 26.35.2.0 152.135.7.0 --- valid host in network 152.135.6.0 113.10.4.0 --- No, network address. 175.33.3.255 --- no, broadcast address for 175.33.2.0 17.35.36.0 --- No, network address. *Originally, I had many mistakes when working it out by hand so I used Subnet Calc app on my iPhone to check my answers. My mistake was in making host bits 1 – I made only the host bits in the last octet be 1, not all the host bits. Process: AND the bits of the subnet mask and the address to get the subnet address. Once you have the subnet address, make all the host bits “1” to get the broadcast address.


Row 1 NO 113 10 4 0 Row 2 Address A 01110001 00001010 00000100 00000000 Row 3 mask 11111111 11111111 11111110 00000000 Row 4 AND 01110001 00001010 00000100 Row 5 Network 113 10 4 0

Row 6 Host bits 1

00000101 11111111


Row 1 YES 186 54 3 0 Row 2 Address B 10111010 00110110 00000011 00000000 Row 3 mask 11111111 11111111 11111110 00000000 Row 4 AND 01110001 00001010 00000100 Row 5 Network 186 54 2 0

Row 6 Host bits 1

00000011 11111111


Row 1 NO 175 33 3 255 Row 2 Address C 10101111 00100001 00000011 11111111 Row 3 mask 11111111 11111111 11111110 00000000 Row 4 AND 00000010 Row 5 Network 175 33 2 0

Row 6 Host bits 1

00000011 11111111


Row 1 YES 26 35 2 255 Row 2 Address D 00011010 00100011 00000010 11111111 Row 3 mask 11111111 11111111 11111110 00000000 Row 4 AND 00000010 Row 5 Network 26 35 2 0

Row 6 Host bits 1

00000011 11111111


Row 1 YES 152 135 7 0 Row 2 Address E 10011000 10000111 00000111 00000000 Row 3 mask 11111111 11111111 11111110 00000000 Row 4 AND 00000110 Row 5 Network 152 135 6 0

Row 6 Host bits 1

00000111 11111111


Row 1 NO 17 35 36 0 Row 2 Address F 00010001 00100011 00100100 00000000 Row 3 mask 11111111 11111111 11111110 00000000 Row 4 AND 00100100 Row 5 Network 17 35 36 0

Row 6 Host bits 1

00100101 11111111



CLI Commands

What command must be entered on the DCE device to enable serial communication at a speed of 64 kilobits per second? Router(config-if)#clock rate 64000

Router(config)#clock rate 64000 Router(config-if)#interface-type dce Router(config-if)#bandwidth 64 Speed of communication is clock rate command. A is not correct because it’s global configuration mode. Clock rate applies to a serial interface so config-if is correct.

Which of the following applies an encrypted password of cisco to the Privileged Exec prompt? A. enable password cisco

B. enable password cisco encrypted C. enable cisco secret D. enable secret cisco Not A - enable password creates a password that is not encrypted. Not B – “encrypted” is not correct syntax Not C – Right words but in the wrong order. Use enable secet password Yes D – Syntax is correct.

What commands apply a password of “Vienna” to the first five Telnet connections on a router? A. line vty 5

login password Vienna B. line vty 0 4 login password vienna C. interface vty 0 4 login password Vienna D. line vty 0 4 login password Vienna

Virtual-type lines start numbering beginning with 0 and password is case sensitive. Not A – line vty 5 means just #5, which would be the

sixth Telnet connection. Not B – Password is case sensitive, “vienna” is not

capitalized. Not C – vty is a virtual line, not an interface Yes D – correct syntax and content

You examine the output of a show running-config command and notice two commands: enable password Cisco and enable secret CCENT. Later on, you establish a console connection to the switch and are prompted for a password to enter Privileged mode. Which password should you use? A. Cisco

B. CCENT C. Both. The enable password is first, followed by the enable secret. D. Both. The enable secret is first, followed by the enable password. E. Neither. Access to a console connection already grants Privileged mode access. The encrypted password (enable secret CCENT) overrides the password that is not encrypted (enable password cisco). Not E – Access to console is does not automatically allow access to privileged EXEC mode.


Which command can be used to verify which end of the connection is DCE? A. show interfaces

B. show dce C. show serial dce D. show controllers Not A – show interfaces will display information about the interfaces but won’t tell which is configured as DCE (Data Communications Equipment – controls clocking, etc.). Not B and C: Not valid commands. Yes D – show controllers will display the DCE end of the connection.

You wish to configure a static route to network 192.168.2.0/24. The route must leave interface Serial 0 to the next hop address of 172.16.5.3. Which of the following commands will configure the route? A. ip route 172.16.5.3 192.168.2.0 255.255.255.0 serial 0

B. ip route 192.168.2.0 255.255.255.0 serial 0 172.16.5.3 C. ip route 172.16.5.3 serial 0 192.168.2.0 255.255.255.0 D. ip route 192.168.2.0 255.255.255.0 172.16.5.3 To configure a static route, the syntax is ip route {destination IP} {mask} {next-hop-ip | interface}, where {} are required. “|” means “or”. Not A – next hop is listed first instead of destination ip Not B – lists both interface and IP. Should be one or the other, not both. Not C – out of order and too much: next-hop IP, interface, destination IP, mask. Yes D – correct syntax.

You issue a show ip route command. A sample output is provided next: R 172.16.2.0 [120/1] via 172.16.3.254, 00:00:17, Serial0/0 Which of the following statements is/are true concerning the output shown? A. This is a directly connected route.

B. This route was learned via RIP. C. The hop count metric on this route is 120. D. The hop count metric on this route is 1. E. The next hop address is 172.16.3.254. Not A – Connected routes are indicated with “C” Yes B – Routes learned by RIP are indicated with “R”. Not C, Yes D – Numbers in brackets are administrative distance first, metric second. Yes E – Next hop address comes right after “via”.

After working on a router, some problems arise and you wish to view the commands that you recently entered. Which IOS command opens the history buffer and displays the most recently entered commands? Show history

Show buffers Show typed commands Show terminal buffer Show command None of the above


You wish to configure RIPv2 on RTR1, shown in the exhibit below. You issue the router rip command, and then the version 2 command. You now need to configure the network statements for RIP. Which network commands is correct?

A. network 192.168.1.0 B. network 192.168.1.0 255.255.255.0 C. network 172.16.2.0 D. network 172.16.2.0 255.255.255.0 E. network 172.16.3.0 F. network 172.16.3.0 255.255.255.0 G. network 172.16.0.0

RIP only accepts classful networks. 192 is Class C, so A is correct. 172 is Class B so G is correct. Other addresses are subnets and not accepted. If you type in an address that is not classful, it will be silently corrected.

OSI Model

Layer # Name Mnemonic Encapsulation

Units Devices or

Components Keywords/Description

7 Application All data PC Network services for application processes, such as file, print, messaging, database services

6 Presentation People data Standard interface to data for the application layer. MIME encoding, data encryption, conversion, formatting, compression

5 Session Seem data Interhost communication. Establishes, manages and terminates connection between applications

4 Transport To segments End-to-end connections and reliability. Segmentation/desegmentation of data in proper sequence. Flow control

3 Network Need packets router Logical addressing and path determination. Routing. Reporting delivery errors

2 Data Link Data frames bridge, switch, NIC

Physical addressing and access to media. Two sublayers: Logical Link Control (LLC) and Media Access Control (MAC)

1 Physical Processing bits repeater, hub, transciever

Binary transmission signals and encoding. Layout of pins, voltages, cable specifications, modulation

OSI comparision with TCP/IP Protocol Stack

OSI # OSI Layer Name TCP/IP # TCP/IP Layer Name Encapsulation Units TCP/IP Protocols 7 Application data FTP, HTTP, POP3, IMAP, telnet, SMTP, DNS, TFTP 6 Presentation data 5 Session

4 Application

data 4 Transport 3 Transport segments TCP, UDP 3 Network 2 Internet packets IP 2 Data Link frames 1 Physical

1 Network Access bits

Source:http://www.vlsm‐calc.net/models.php


Subnetting Work Space Row 1

IP Address in Dotted Dec.

Row 2 IP Address in binary

Row 3 mask Row 4 ANDing result

Row 5 Network Address

Row 6 Make host bits 1

Row 7 Broadcast Address

Row 8 Range Minimum (network +1)

Row 9 Range Maximum (broadcast -1)

Row 1









Row 1









SubnettingWorksheet,Ferdon5/2011

Network IP address to be subnetted: ___________________________ STEP 1: Determine Network and Host bits:

Class A 1 - 126 NNNNNNNN HHHHHHHH HHHHHHHH HHHHHHHH Class B 128 - 191 NNNNNNNN NNNNNNNN HHHHHHHH HHHHHHHH Class C 192 - 223 NNNNNNNN NNNNNNNN NNNNNNNN HHHHHHHH

STEP 2: Determine how many subnets are needed - borrow bits from host. Formula: 2^s – 2 where “s” equals number of subnet bits that will be used. STEP 3: Determine how many hosts are needed. Formula: 2^h – 2 where “h” equals number of host bits that will be used. STEP 4: Any wildcard bits? (can be host or subnet). Formula: 2^h – 2 where “h” equals number of host bits that will be used. STEP 5: Fill in chart to determine subnet mask:

1. From left – “N” for each network bit 2. From left – “S” for each subnet bit 3. From right – “H” for each host bit 4. Leftovers – “X” for wildcard bit

Network, Subnet, Wildcard, Host binary dotted decimal

STEP 6: Determine “interesting octet”:

1. Enter mask and IP address 2. ANDing: For address with 255 in mask, copy IP address octet 3. ANDing: For address with 0 in mask, write 0 4. Interesting octet is the blank one

Subnet Mask IP Address to subnet Boolean AND

SubnettingWorksheet,Ferdon5/2011

STEP 7: Determine magic number: 255 – interesting octet value = magic number 255 - ______ = STEP 8: Determine subnet IP addresses, host ranges, and broadcast IP addresses:

1. Subnetwork: Add multiples of the magic number to any octet that has “S” in it until you max out at 256. 2. Broadcast: Next subnet ID minus 1, is previous subnet’s broadcast ID. 3. Host range: Subnet ID plus one, is bottom end of range. Broadcast ID minus 1, is top end of range. 4. Zero subnet is not usable (same as network address) and last subnet not usable (same as network

broadcast address).

Subnetwork

# Subnetwork ID Host Range Broadcast ID

0 ZERO SUBNET

1

2

(network broadcast)

INVALID

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