ICND1 – Subnetting 2

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Note: If you are not sure about Subnetting, please read my Subnetting Made Easy tutorial.

Question 1

[am4show have=’p2;’]Refer to the exhibit. The junior network support staff provided the diagram as a recommended configuration for the first phase of a four-phase network expansion project. The entire network expansion will have over 1000 users on 14 network segments and has been allocated this IP address space:

192.168.1.1 through 192.168.5.255
192.168.100.1 through 198.168.100.255

What are three problems with this design? (Choose three)

A – The AREA 1 IP address space is inadequate for the number of users.
B – The AREA 3 IP address space is inadequate for the number of users.
C – AREA 2 could use a mask of /25 to conserve IP address space.
D – The network address space that is provided requires a single network-wide mask.
E – The router-to-router connection is wasting address space.
F – The broadcast domain in AREA 1 is too large for IP to function.

 

Answer: A C E[/am4show]

Explanation

AREA 1 has 500 users but it uses class C which only supports 254 users (from 192.168.1.1 to 192.168.1.254)-> A is correct.

AREA 3 also uses class C and as mentioned above it supports 254 users so it is enough for 200 users -> B is incorrect.

In AREA 2 there are only 60 users < 64 = 2⁶ so we can use a subnet mask which has 6 bits 0 -> /26. Of course we can use larger subnets (like /25) for future expansion -> C is correct.

A large network should never use a single network-wide mask. It should be some different subnet masks to make the network flexible and easy to be summarized -> D is incorrect.

For router-to-router connection we should use a subnet mask of /30 which supports 2 hosts per subnet. This subnet mask is ideal for router-to-router connection -> E is correct.

There is no limit for IP to function if we know how to organize our network -> F is incorrect.

Question 2

[am4show have=’p2;’]Refer to the exhibit. The enterprise has decided to use the network address 172.16.0.0. The network administrator needs to design a classful addressing scheme to accommodate the three subnets, with 30, 40, and 50 hosts, as shown. What subnet mask would accommodate this network?

Net bits	Subnet mask	total-addresses per subnet
/20	255.255.240.0	4096
/21	255.255.248.0	2048
/22	255.255.252.0	1024
/23	255.255.254.0	512
/24	255.255.255.0	256
/25	255.255.255.128	128
/26	255.255.255.192	64
/27	255.255.255.224	32
/28	255.255.255.240	16
/29	255.255.255.248	8
/30	255.255.255.252	4

 

A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248

 

Answer: A[/am4show]

Explanation

The maximum number of hosts in this question is 50 hosts so we have to use /26 subnet mask or above.

Question 3

[am4show have=’p2;’]The network manager has requested a 300-workstation expansion of the network. The workstations are to be installed in a single broadcast domain, but each workstation must have its own collision domain. The expansion is to be as cost-effective as possible while still meeting the requirements. Which three items will adequately fulfill the request? (Choose three)

A. one IP subnet with a mask of 255.255.254.0
B. two IP subnets with a mask of 255.255.255.0
C. seven 48-port hubs
D. seven 48-port switches
E. one router interface
F. seven router interfaces

 

Answer: A D E[/am4show]

Explanation

To support 300 workstations in a single broadcast domain, we need to use a subnet mask which supports 512 hosts = 2⁹ -> /23 or 255.255.254.0 in decimal form -> A is correct.

If we use 48-port switches we need 300/48 = 6.25 -> seven 48-port switches are enough because we also need trunking between them -> D is correct.

We only need one router interface and it is connected with one of seven switches -> E is correct.

Question 4

[am4show have=’p2;’]Which router command will configure an interface with the IP address 10.10.80.1/19?

A. router(config-if)# ip address 10.10.80.1/19
B. router(config-if)# ip address 10.10.80.1 255.255.0.0
C. router(config-if)# ip address 10.10.80.1 255.255.255.0
D. router(config-if)# ip address 10.10.80.1 255.255.224.0
E. router(config-if)# ip address 10.10.80.1 255.255.240.0
F. router(config-if)# ip address 10.10.80.1 255.255.255.240

 

Answer: D[/am4show]

Explanation

/19 = 255.255.224.0. The fast way to find out this subnet mask is to remember /16 = 255.255.0.0 and we need 3 more bits 1 for 3rd octet: 1110 0000 which is 224.

Question 5

[am4show have=’p2;’]Refer to the exhibit. A TFTP server has recently been installed in the Atlanta office. The network administrator is located in the NY office and has made a console connection to the NY router. After establishing the connection they are unable to backup the configuration file and IOS of the NY router to the TFTP server. What is the cause of this problem?

A. The NY router has an incorrect subnet mask.
B. The TFTP server has an incorrect IP address.
C. The TFTP server has an incorrect subnet mask.
D. The network administrator computer has an incorrect IP address.

 

Answer: C[/am4show]

Question 6

[am4show have=’p2;’]Which two statements describe the IP address 10.16.3.65/23? (Choose two)

A. The subnet address is 10.16.3.0 255.255.254.0.
B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
E. The network is not subnetted.

 

Answer: B D[/am4show]

Explanation

Increment: 2 (/23 = 11111111.11111111.11111110.00000000 = 255.255.254.0)
Network address: 10.16.2.0 (because 2 = 2 * 1 and 2 < 3)
Broadcast address: 10.16.3.255 (because 2 + 2 – 1 = 3 for the 3rd octet)

-> The lowest (first assignable) host address is 10.16.2.1 and the broadcast address of the subnet is 10.16.3.255 255.255.254.0

Question 7

[am4show have=’p2;’]Refer to the exhibit. The internetwork is using subnets of the address 192.168.1.0 with a subset mask of 255.255.255.224. The routing protocol in use is RIP version 1. Which address could be assigned to the FastEthernet interface on RouterA?

A. 192.168.1.31
B. 192.168.1.64
C. 192.168.1.127
D. 192.168.1.190
E. 192.168.1.192

 

Answer: D[/am4show]

Explanation

255.255.255.224 = 1111 1111.1111 1111.1111 1111.1110 0000 (binary form)

Increment: 32

First subnetwork: 192.168.1.0 -> 192.168.1.31 (A is incorrect because 192.168.1.31 is a broadcast address)
Second subnetwork: 192.168.1.32 -> 192.168.1.63
Third subnetwork: 192.168.1.64 -> 192.168.1.95 (B is incorrect because 192.168.1.64 is a network address)
Fourth subnetwork: 192.168.1.96 -> 192.168.1.127 (C is incorrect because 192.168.1.127 is a broadcast address)
Fifth subnetwork: 192.168.1.128 -> 192.168.1.159
Sixth subnetwork: 192.168.1.160 -> 192.168.1.191 (D is correct because 192.168.1.190 is the last assignable host address of this subnetwork)
Seventh subnetwork: 192.168.1.192 -> 192.168.1.224 (E is incorrect because 192.168.1.192 is a network address)

Question 8

[am4show have=’p2;’]A network administrator is connecting PC hosts A and B directly through their Ethernet interfaces as shown in the graphic. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts? (Choose two)

A. A crossover cable should be used in place of the straight-through cable.
B. A rollover cable should be used in place of the straight-through cable.
C. The subnet masks should be set to 255.255.255.192
D. A default gateway needs to be set on each host.
E. The hosts must be reconfigured to use private IP addresses for direct connections of this type.
F. The subnet masks should be set to 255.255.255.0

 

Answer: A F[/am4show]

Question 9

[am4show have=’p2;’]If an Ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?

A. 1024
B. 2046
C. 4094
D. 4096
E. 8190

 

Answer: C[/am4show]

Question 10

[am4show have=’p2;’]Refer to the exhibit. After configuring two interfaces on the HQ router, the network administrator notices an error message. What must be done to fix this error?

HQ#configure terminal HQ(config)# interface fastethernet 0/0 HQ(config-if)# ip address 192.168.1.17 255.255.255.0 HQ(config-if)# no shutdown HQ(config-if)# interface serial 0/0 HQ(config-if)# ip address 192.168.1.65 255.255.255.240 HQ(config-if)# no shutdown % 192.168.1.0 overlaps with FastEthernet0/0

A. The serial interface must be configured first.
B. The serial interface must use the address 192.168.1.2
C. The subnet mask of the serial interface should be changed to 255.255.255.0
D. The subnet mask of the FastEthernet interface should be changed to 255.255.255.240
E. The address of the FastEthernet interface should be changed to 192.168.1.66

 

Answer: D[/am4show]

Question 11

[am4show have=’p2;’]Refer to the exhibit. Host B has just been added to the network and must acquire an IP address. Which two addresses are possible addresses that will allow host B to communicate with other devices in the network? (Choose two)

A. 192.168.10.32
B. 192.168.10.38
C. 192.168.10.46
D. 192.168.10.47
E. 192.168.10.49
F. 192.168.10.51

 

Answer: B C[/am4show]

Explanation

The IP address of host B must be in the range of 192.168.10.32/28 subnet, which ranges from 192.168.10.32 to 192.168.10.47 (Increment: 16), except the IP addresses of 192.168.10.32, 192.168.10.46 (which are the network and broadcast addresses of the subnet), 192.168.10.33, 192.168.10.34 (which have been assigned to the interface’s router and the switch). Therefore there are only two IP addresses of 192.168.10.38 & 192.168.10.46.

Question 12

[am4show have=’p2;’]A network administrator has subnetted the 172.16.0.0 network using a subnet mask of 255.255.255.192. A duplicate IP address of 172.16.2.120 has accidentally been configured on a workstation in the network. The technician must assign this workstation a new IP address within that same subnetwork. Which address should be assigned to the workstation?

A. 172.16.1.80
B. 172.16.2.80
C. 172.16.1.64
D. 172.16.2.64
E. 172.16.2.127
F. 172.16.2.128

 

Answer: B[/am4show]

Question 13

[am4show have=’p2;’]What is the maximum number of bits that can be borrowed to create subnets if a Class B network address is being used?

A. 2
B. 6
C. 8
D. 14
E. 16

 

Answer: D[/am4show]

Question 14

[am4show have=’p2;’]The internetwork shown in the diagram is experiencing network connectivity problems. What is the cause of the problem?

A. The cabling connecting host A to Switch3 is too long.
B. The address of host B is a broadcast address.
C. The IP address of interface Fa0/0 of Router1 is not a usable address.
D. The cable connecting Switch2 and Switch3 should be a crossover.
E. The IP address of Server 1 is in the wrong subnet.

 

Answer: D[/am4show]

Question 15

[am4show have=’p2;’]What is the subnet address for the IP address 172.19.20.23/28?

A. 172.19.20.0
B. 172.19.20.15
C. 172.19.20.16
D. 172.19.20.20
E. 172.19.20.32

 

Answer: C[/am4show]

Explanation

From the /28 we can find all information we need:

Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 172.19.20.16 (because 16 < 23)
Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)

In fact we don’t need to find out the broadcast address because the question only asks about subnet address (network address).