IP Addressing

ip addresses Converting binary to decimal Consider the octet 11111101. If you are familiar with counting in binary you will know that this number is 253. 2^7

2^6

2^5

2^4

2^3

2^2

2^1

2^0

128

64

32

16

8

4

2

1

1

1

1

1

1

1

0

1

128+ 64 + 32 + 16 + 8 + 4 + 0 + 1 = 253 There is an easier way to convert this number to decimal. We know that the binary number 11111111 is 255 in decimal. When we look at a number like 11111101, here subtract 2 from 255 to arrive at our decimal conversion 253. For 11101101 128+64+32+0+8+4+0+1=237 or 255- 2 -16 = 255 – 18 = 237 if you want to get good at subnetting memorize 10000000 = 128 11000000 = 192 11100000 = 224 11110000 = 240 11111000 = 248 11111100 = 252 11111110 = 254 11111111 = 255

Private Address Space The Internet Assigned Numbers Authority (IANA) has reserved the following three blocks of IP address space for private networks You can use addresses on any private LAN You CANNOT use them on the internet, internet routers will block them!

Private IP addresses range

Which of the following are true about 172.16.0.254/16?

A. IPX:MAC addresses. B. IP:classC directed broadcast. C. Private IP address:node number D. Public IP addresses:directed broadcast. E. Private IP addresses directed broadcast. Answer C Explanation The Class B network of 172.16 is a Class B Private Address Range, and the second part (0.254) is the host address, or node number/address. Incorrect Answers A. The address is an IP address format. B. The question is a Class B address, if it were Class C, the mask would be 255.255.255.0 D. 172.16 is not a public IP address. E. 0.254 is not a broadcast address Question Which of the following describe private IP addresses? (Choose two) A. Addresses chosen by a company to communicate with the Internet. B. Addresses that cannot be routed through the public Internet. C. Addresses that can be routed through the public Internet. D. A scheme to conserve public addresses. E. Addresses licensed to enterprise or ISPs by an Internet registry organization. Answer B, D Explanation Private IP address space has been allocated via RFC 1918. This means the addresses are available for any use by anyone and therefore the same private IP addresses can be reused. However they are defined as not routable on the public Internet. They are used extensively in private networks due to the shortage of publicly registered IP address space and therefore network address translation is required to connect those networks to the Internet.

IP Addressing Classes The IP address range has 5 different classes A to E,

Class A address range 0 - 126 Class B address range 128 - 191 Class C address range 192 - 223 Class D address range 224 - 239

Class A Class A has a 7-bit network number and a 24-bit local address. The highest order bit is always set to zero. This allows 128 (2^7) class A networks.

This shows the eight network bits followed by the 24 host bits.

Class B Class B network addresses have a 14-bit network number, a 16-bit local address, and they begin with "10" binary. This allows 16,384 class B networks. The first two bits of a Class B address are 1 and 0, the next fourteen bits identify the network and the last sixteen the host.

128.0.0.0 to 191.254.0.0 We can divide the host portion of a Class B address into subnet and host parts. For instance, let's split our Class B network number on the byte boundary, the eight of the host portion identifies the subnet and the remaining bits the host, as diagramed: Network Subnet Host +--------------+ +------+ +------+ | | | | | | [10xxxxxxxxxxxxxx][xxxxxxxx][xxxxxxxx] This arrangement allows 254 subnets each with 254 hosts.

Class C Class C network addresses have a 21-bit network number, an 8-bit local address, and they begin with "110" binary. This allows 2,097,152 class B networks.

192.0.1.0 to 223.255.254.0

Class D Class D network addresses are for multicasting

class D addresses 224.0.0.0 to 239.255.255.255 multicast group

Class E Class E network addresses being with 4 binary ones, and this is not allowed. The one exception to this rule is the address where all the bits are ones (255.255.255.255); this is reserved for an IP broadcast. class E addresses 240.0.0.0 to 255.255.255.254 reserved (illegal) 255.255.255.255

reserved (broadcast)

Question IP addresses use hierarchical numbering. What portion of the address that will identify the network number? A. Subnet Mask. B. Dots between octets.

C. Class of first octet. D. Assignments of DHCP. E. Address Resolution Protocol. Answer C Explanation In general, IP addresses contain two fields: one for the network and another for host. Class A addresses have a range of 1 to 126 and the network portion of the IP address is restricted to the first eight bits (octet). Class B address have a range of 128 -191. and the network portion of the IP address is contain in the first 2 octets. Class C IP addresses has a range of 192- 223 and the network portion of the IP addresses is the first three octets of the IP address. Class D addresses include the range of 224.0.0.0 to 239.255.255.255 and are used for multicast address. Class E addresses have a range of 224.0.0.0 to 247.255.255.255 and are reserved for experimental purposes. Incorrect Answers A. Although the subnet mask is used network devices to determine what part of the IP use address for the network, the subnet and the host address but it is not part of the IP address hierarchy. B. The dots are used for making the IP address readable by humans, but have no determination of the network number. D. DHCP (Dynamic Host Configuration Protocol) provides a mechanism for allocating IP addresses dynamically so that addresses can be reused when hosts no longer need them. E. Address Resolution Protocol (ARP) determines the data link layer address of the destination devices for known destination IP addresses network number.

The Subnet Mask A subnet mask is used to determine the number of bits used for the subnet and host portions of the address. The mask is a 32-bit value that uses one-bits for the network and subnet portions and zero-bits for the host portion.

Configuring Subnet Masks If you have one address and want to create six networks from it here’s where subnetting comes in benefits are Reduced network traffic- with routers, most traffic stays on the local network only packets destined for other networks pass through the router. Routers create broadcast domains the smaller the broadcast domains the less network traffic.

Understanding Classless Inter-Domain Routing (CIDR) also known as supernetting This really is a more efficient way of referring to a network. For example, if we had a network address of 192.168.0.0 with a mask of 255.255.255.128, in CIDR notation it becomes 192.168.0.0/25. The /25 means that the first 25 bits of the subnet mask are set to binary 1

192.168.10.32/8 this tells you what your subnet mask is the slash tells us how many bits are 1s. Class A default subnet mask 255.0.0.0 this means the 1st byte of the subnet mask is all 1s or 11111111 which is /8 Net bits Subnet mask

total-addresses

/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

The first address of a subnet block (all 0s) is called the network address or network ID. The last address (all 1s) is the broadcast address of the network. Typically the network address +1 or the broadcast address -1 is the gateway to the internet. Example 192.168.1.0/25 would include all address between 192.168.1.0 and 192.168.1.127 while 192.168.1.128/25 would include 192.168.1.128 and 192.168.1.255 Below is a mask table that makes it easy to look up the mask for a group of IP addresses. /25 = 255.255.255.128, Block size 256 - 128 =128 Address range for 192.168.1.0 = 0 – 127 Address range for 192.168.1.128 = 128 – 255 We stop when we reach the final address of 255 remember these addresses include the subnet and broadcast address.

Etc..

Let's take a look at two hosts on a Class B network trying to communicate with one another. Assume the IP address of the source host is 172.16.32.1/16 and the destination host address is 172.16.64.1/16. Because we are using the default subnet mask of /16 or 255.255.0.0, 172.16.32.1/16 and 172.16.64.1/16 are on the same network. As long as there is no change in

the network portion of the address, the two hosts are on the same network. The network portion of the address is determined by the subnet mask of 255.255.0.0, which tells the TCP/IP stack that the first two octets, 172.16.0.0, represent the network portion of the address.

Subnetting However, if we need more than one network, because we have remote locations or we have more hosts than we can place on a single cable? It makes sense to divide the network into smaller networks. What if we needed to create 6 subnets from our larger, single network? In a case like this, we would need to extend the default subnet mask by borrowing some of the bits from the host portion of the IP address.

The formula for determining the number of subnets = 2^n - 2 (See * ) n represents the number of bits to borrow For six useable subnets 2 ³ - 2 = 8 - 2 =

6

To create six subnets from the 172.16.0.0/16 network we need to borrow at least 3 host bits and add them to the network portion giving us 172.16.0.0/19 In binary, the subnet mask would be 11111111.11111111.11100000.00000000 So, our subnet mask is 255.255.224.0. * In the case of a custom subnet mask of 224, we create at least 6 networks. (I say at least because the actual number depends on the hardware or the software). It should be noted that in the past using subnet zero (00000—) and all-ones subnet (11111 —) was not allowed. This is not true nowadays. Since Cisco IOS Software Release 12.0 the entire address space including all possible subnets is explicitly allowed. Here are the possible network IDs in the 3rd octet for our 255.255.224.0 subnet mask Block size 256 – 224 = 32 subnet 1 00000000 0 (Normally not allowed according to RFC950) 2 00100000 32 172.16.32.0/19 3 01000000 64 172.16.64.0/19 4 01100000 96 172.16.96.0/19 5 10000000 128 172.16.128.0/19 6 10100000 160 172.16.160.0/19 7 11000000 192 172.16.192.0/19 8 11100000 224 (Normally not allowed according to RFC950) Remember, we are dealing only with the 1st 3 left-most bit positions. A subnet mask of 224 "masks" off these bits to represent the network.

Subnet Zero Traditionally we didn't count the first and last subnet (all zeroes and all ones). So a subnet mask of 255.255.255.192 would give (2^2)-2, or 2, subnets while a mask of 255.255.255.224 would give (2^3)-2, or 6, subnets. So, for 192.168.201.0/26 the first subnet would be 192.168.201.64 and the first host would be 192.168.201.65.

This is very wasteful of a scarce resource and modern software is able to utilize all definable networks. So this traditional method is no longer used (RFC1878 describes it as obsolete). With subnet zero enabled.

Now the formula for determining the number of subnets = 2^n The new CCNA and Microsoft Exams use the new method Now 255.255.255.192 gives 2^2, or 4, subnets and 255.255.255.224 gives 2^3, or 8, addresses. Note the formula to calculate the number of hosts is 2^n – 2 We – 2 as in binary the all 1’s address is reserved for ip broadcasts and all 0’s address is used to specify a network ID.

Example When the default subnet mask of 255.255.0.0 is used for hosts within the Class B network of 131.107.0.0, the ip addresses 131.107.1.11 and 131.107.2.11 are found on the same subnet and they can communicate with each other via a broadcast. However when the subnet mask is extended to 255.255.255.0 the addresses 131.107.1.11 and 131.107.2.11 are found on different subnets. To communicate with each other hosts with addresses 131.107.1.11/24 and 131.107.2.11/24 send ip packets to the default gateway, which is then responsible for routing to the destination subnet.

Class B address space not subnetted

Subnetted Class B address space

Example Accommodating Physical Topology suppose we are designing a campus network with 200 hosts spread over four buildings Voter Hall, Twilight Hall, Monroe Hall, Sunderland Hall. Each of these four buildings to include 50 hosts. The ISP has allocated us the Class C network 208.147.66.0. This means we can use 208.147.66.1 – 208.147.66.254 for hosts. # subnets = 2^n n represents the number of bits to borrow For 4 useable subnets 2^2 = 4 Therefore n =2, we have borrowed 2 host bits for our network So our Class C /24 subnet mask is now /26 Subnet mask = 255.255.255.192 How many hosts per subnet? = 6 bits left for the host address 2^6 – 2 = 62 hosts What are the valid subnets? 256 – subnet mask = block size Block Size = 256 – 192 = 64 so subnets are

0

64

128

192

What’s the broadcast address for each subnet 63 What are the valid hosts? 1-62

127 191 255 65-126 129-190 193-254

Example Given the subnet mask 255.255.255.224/27 what are the valid subnets, broadcasts and hosts Numbers of Subnets = 27- 24 = 3 bits = 2^3 = 8 Subnet mask = 11100000 = 224 Block size = 256 – 224 = 32 Hosts per subnet = 2^5 – 2 = 30 Subnet/network id Broadcast address 0 31 32 63 64 95 96 127 128 159 160 191 192 223 224 255

Host Range 1 - 30 33 - 62 65 - 94 97 - 126 129 - 158 161 - 190 193 - 222 225 - 254

Example 172.30.32.0/24 network includes the addresses 172.30.32.0 through 172.30.32.255, and the 172.30.32.0/20 network includes the addresses 172.30.32.0 through 172.30.47.255, 172.30.32.0/20,

/20= 11111111.11111111.11110000.0

Block Size =16 0, 16, 32, 48, 172.30.32.0/20 Addresses 32- 47 inc broadcast address.

Subnetting in your Head For example 192.168.10.33 = ip address, 255.255.255.224 = subnet mask Determine the subnet and broadcast address, 256 - 224 = 32 block size 0, 32, 64 The address 33 falls between two subnets 32 and 64 and must be part of the 192.168.10.32 subnet the next subnet is 64 so the broadcast address is 63, valid host range 33-62. Here’s another one 192.168.10.33, 255.255.255.240 256 - 240 = 16 subnets 0, 16, 32, 48 host address is between the 32 and 48 subnets the subnet is 192.168.10.32 and the broadcast address is 47, host range 33 – 46 One more 192.168.10.174 mask 255.255.255.240 Increment 256 – 240 = 16 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176 Host address of 174 is between 160 and 176, subnet is 160, broadcast 175, host range 161 - 174

Example Network ID 172.16.0.0 Subnet mask 255.255.240.0

CIDR 172.160.0.0/20

Subnet increment (block size) 256 - 240 = 16 The subnets are calculated by incrementing the appropriate octet of the network ID by the block size. The "appropriate octet" is the octet in the subnet mask that is not equal to 255 or 0. Net ID’s

and the broadcast address for the subnet (last IP address in the range)

172.16.0.0 172.16.16.0 172.16.32.0 172.16.48.0 172.16.64.0 172.16.80.0 172.16.96.0 172.16.112.0 172.16.128.0 172.16.144.0

172.16.15.255 172.16.31.255 172.16.47.255 172.16.63.255 172.16.79.255 172.16.95.255 172.16.111.255 172.16.127.255 172.16.143.255 172.16.159.255

172.16.160.0 172.16.176.0 172.16.192.0 172.16.208.0 172.16.224.0 172.16.240.0

172.16.175.255 172.16.191.255 172.16.207.255 172.16.223.255 172.16.239.255 172.16.255.255

To test these values, remember that a host ID ANDed with its subnet mask should equal its network ID. So if we pick a host in the 172.16.192.0 - 172.16.207.255 range, and AND it with 255.255.240.0, the result should be 172.16.192.0. An example host 172.16.200.25 mask 255.255.240.0 AND result 172.16.192.0 Trouble-Shooting Question Two routers named Atlanta and Brevard are connected by their serial interfaces as illustrated, but there is no connectivity between them. The Atlanta router is known to have a correct configuration. Given the partial configurations, identify the problem on the Brevard router that is causing the lack of connectivity.

A. transmission unit size too large B. no loopback set C. an incorrect subnet mask D. incompatible encapsulation at each end E. an incorrect IP address F. incompatible bandwidth bewteen routers Answer E Explanation Based on exhibit both Atlanta and Brevard are directly connected over serial link . Given that Atlanta is configured correctly and its S0 IP address is 192.168.10.1 /24 Whereas problem at Brevard is it is configure with incorrect IP address 192.168.11.2 /24. The IP address must be corrected to 192.168.10.2 /24 so that both routers are configured for same network and establish connectivity.

Exam Questions You are required to divide the 172.12.0.0 network into subnets. Each subnet must have the capacity of 458 IP addresses. Furthermore, according to the requirement you must provide the maximum number of subnets. Which network mask should you use? Answer - 255.255.254.0 Explanation 458 hosts required 2^8 = 256, 2^9 = 512 To obtain 458 IP addresses the number of host bits will be 9. In this maximum 512 hosts can be assigned. Keep 9 bits for host means 4th octet and last bit of the 3rd will be 0. All network bits before this bit MUST BE one, so we get 11111111.11111111.11111110.00000000 = 255.255.254.0 This gives 255.255.254.0 as the subnet mask.

SUBNET ZERO Question How many subnetworks and hosts are available per subnet if you apply a /28 mask to the 210.10.2.0 class C network? A. B. C. D. E. F.

30 networks and 6 hosts 6 networks and 30 hosts 8 networks and 32 hosts 32 networks and 18 hosts 16 networks and 14 hosts F non of the above

Answer E Explanation Class C address = 255.255.255.0 = /24 /28 subnet mask leaves 4bits for networks and 4 bits for hosts Number of hosts 2n - 2 = 2^4 – 2 = 14 Number of networks 2n = 16 ***The above example assumes that using subnet zero is legal. If subnet zero can not be used, subtract two from the number of networks*** If subnet zero is enabled, it is 2 ^ n = #of networks If subnet zero is disabled, it is 2 ^ n -2 = # of networks "no ip subnet-zero" command is used, PS: Here 'n' represents no. of bits for host portion. But whether it is disabled or enabled, formula for calculating # of hosts in each subnet remains the same. i.e, (2 ^ n) - 2. PS: Here 'n' represents no. of bits for host portion. Question You work as a network technician. You have subnetted the 213.105.72.0 network with a /28 mask. Your boss asks you how many usable subnetworks and usable host addresses per subnet this will provide. What should you tell her?

A. 62 networks and 2 hosts B. 6 networks and 30 hosts C. 8 networks and 32 hosts D. 16 networks and 16 hosts E. 14 networks and 14 hosts Answer E /28, 3 x 8 =24 with 4 bits leftover 4 bits added 2^4 – 2 = 16 - 2 = 14 networks Number of hosts =2^4 - 2 = 14 Question You work as a network technician. You have subnetted the 201.105.13.0 network with a /26 mask. Your boss asks you how many usable subnetworks and usable host addresses per subnet this will provide. What should you tell her? A. 64 networks and 4 hosts B. 4 networks and 64 hosts C. 2 networks and 62 hosts D. 62 networks and 2 hosts Answer C /26, therefore 2 bits borrowed for network Subnets= 2^2 - 2 = 4 - 2 = 2 2 bits for network portion therefore 6 bits left for number of hosts 2^6 - 2 = 64 – 2 = 62

Question You work as a network consultant. You are planning a network installation for a large organization. The design requires 100 separate subnetworks, so we have acquired a Class B network address. What subnet mask will provide the 100 subnetworks required, if 500 usable host addresses are required per subnet? A. 255.255.240.0 B. 255.255.246.0 C. 255.255.252.0 D. 255.255.254.0 E. 255.255.255.0 F. 255.255.255.192 Answer D From Network Perspective class B = /16 Over 100 subnets required 2^ 6 = 64, 2^ 7= 128 Therefore number of subnets 2^7 – 2 = 128 – 2 = 126 7 extra network bits required = 11111110 = 254 = 255.255.254.0 Note 7 extra bits for /16 = 16 + 7 = /23 also = 255.255.254.0 From Host Perspective

500 usable host addresses required 2^9 = 512 hosts therefore 9 host bits used leaving all the network bits as 1’s gives 11111111.11111111.11111110.00000000 = 255.255.254.0

Question You work as network consultant. Your customer has a class C network license. They require 5 usable subnets, each capable of accommodating at least 18 hosts. Which subnet mask should you use? Answer - 255.255.255.224 Explanation If one has to create 5 subnets, then 3 extra network bits are required for our class C address /24 2^3 -2 = 8 – 2 = 6. With 3 bits we can create 6 subnets. 11100000 = 224 Remaining 5 bits are used for Hosts. all network bits before this bit MUST BE one, so we get 11111111.11111111.11111111.11100000 One can create 30 hosts using 5 bits in host field. 2^5 – 2 = 30 This matches the requirement. Question You have a Class B network address with a subnet mask of 255.255.255.0. Which of the following statements are true regarding the resulting network? (Choose two) A. There are 254 usable hosts per subnet. B. There is one usable network. C. There are 255 usable hosts per subnet. D. There are 254 usable subnets. E. There are 30 usable subnets. F. There are 64 usable hosts per subnet. Answer A, D Class B /16 = 255.255.0.0 question says subnet mask is 255.255.255.0 8 bits borrowed therefore 2^8 -2 = 256 -2 = 254 subnets Leaving 8 host bits also = 254 Question Identify three valid host addresses in the 192.168.27.0 network with a subnet mask of 255.255.255.240. (Choose three) A. 192.168.27.33 B. 192.168.27.112 C. 192.168.27.119 D. 192.168.27.126 E. 192.168.27.175 F. 192.168.27.208 Answer A, C, D Explanation 240 = 11110000

A subnet mask of 255.255.255.240 divides the 4th octet we simply check the 4th octet to ensure that the last 4 bits (the host portion) are not 0000 or 1111 a subnet or broadcast address. A. 33 = 00100001 a valid host C. 119 = 01110111 a valid host D. 126 = 1111110 a valid host. Incorrect Answers B. 112 = 01110000. This is not a valid host address in this network. It has all host bits 0. E. 175 = 10101111. All host bits are 1’s. This is the local broadcast address and cannot be used. F. 208 = 11010000. This is not a valid host address in this network. It has all host bits 0. Another way of looking at this question 256 – 240 = 16 block size Net IDs 0, 16, 32, 48, 64, 80, 96, 112, 128,……160, 176, 192, 208 Broadcast 15, 31, 47, 63, 79, 95, 111, 127, …. 159, 175 So we can rule 112, 175 and 208

Question A Class C network address has been subnetted with a /27 mask. Which of the following addresses is a broadcast address for one of the resulting subnets? A. 201.57.78.33 B. 201.57.78.64 C. 201.57.78.97 D. 201.57.78.97 E. 201.57.78.159 F. 201.57.78.254 Answer E Explanation Which IP’s are valid Broadcast addresses for any of the 201.57.78.0/27 subnets. Broadcast address means all the host bits are 1. /27 class C = /24 3 bits borrowed for network so 5 bits are left for host Last 5 bits will be 11111 Verifying each IP 159 = 1011111. So this is broadcast address for 201.57.78.0/27 network. Incorrect answers A. 33 = 00100001. This is not broadcast address B. 64 = 01000000. This is not broadcast address C, D. 97 = 01100001.This is not broadcast address F. 254 = 11111110 .This is not broadcast address Question What is the subnetwork address for a host with IP address 201.100.5.68/28 ? A. 201.100.5.0 B. 201.100.5.32 C. 201.100.5.64 D. 201.100.5.65 E. 201.100.5.31

F. 201.100.5.1 Answer C Explanation /28 = 11110000 = 240 Block size 256 – 240 = 16 Net id - broadcast 0 - 15 16 - 31 32 - 47 48 - 63 64 - 79 80 - 95

subnet 0 1 2 3 4 5

valid hosts

65 to 78

201.100.5.68/28 is in the 201.100.5.64 subnet or /24 compared to /28 this gives us 4 network bits and 4 host bits 68 binary

01000100 11110000 subnet mask (anding) 01000000 subnet = 64

Subnet = 201.100.5.64

Question Which of the following if addresses can be assigned to host devices (Choose two) A. 205.7.8.32/27 B. 191.168.10.2/23 C. 127.0.0.1 D. 224.0.0.10 E. 203.123.45.47/28 F. 10.10.0.0/13 Answer B, F

B. /23 = 00000000 in the 4th octet 2 = 00000010 a valid host F. /13 = 11111000 in the 2nd octet 10 = 00001010 a valid host Incorrect Answers A. This is a network address. /27 = 11100000 4th octet 32 = 00100000 network address as host bits all zeros

C. That is a loop-back address. D. That is a Type D (Multicast) address. E. This is a broadcast address. /28 = 11110000 4th octet 47 = 00101111 Question A network host is configured as follows Host IP address 192.168.16.183 Subnet mask 255.255.255.224 Default gateway 192.168.16.190 Which of the following statements describe the network to which this host is attached? (Choose three) A. The default gateway is configured on a different subnet form this host. B. The host is a member of the fifth usable subnet of the 192.168.16.0 network. C. The subnetwork address for this host is 192.168.16.160/28. D. The address of the next higher subnet is 192.168.16.192. E. The address of the router interface that is attached to this subnetwork is 192.168.16.190. Answer B, D, E Explanation B. According to subnet mask, 5 bits are used for hosts. So each subnet contains 30 hosts .The fifth usable subnet range is 192.168.16.160 to 192.168.16.191. Here first address is network and later address is broad cast. Our host .183 is member of this subnetwork. D. The next subnet address after fifth usable subnet is 192.168.16.192 E. 192.168.16.190 is the router interface in the fifth usable subnet Incorrect Answers A. Default gateway is on the same subnet as the Host Default gateway 190 = 10111110 Host 183 = 10110111 C. Subnet mask is different as compared to mask defined in the question F. There are 30 usable hosts in each subnetwork. 224 = 11100000, = 3 extra network bits 256 – 224 = 32 block size

Subnet ID 0 32 64 96 128 160 192

Broadcast 31 63 95 127 159 191

subnet 0 1 2 3 4 5

Question You are configuring a subnet on the branch office in Berlin. You need to assign IP addresses to hosts in this subnet. You have been given the subnet mask of 255.255.255.224. Which IP address would be valid? (Choose three) A. 15.234.118.63 B. 92.11.178.93 C. 134.178.18.56 D. 192.168.16.87 E. 201.45.116.159 F. 217.63.12.192 Answer B, C, D Explanation B. Valid Host in subnetwork 2 ( 92.11.178.64 to 92.11.178.95) C. Valid Host in subnetwork 1(134.178.18.32 to 134.178.18.63) D. Valid host in subnetwork 2 (192.168.16.64 to 192.168.16.95) 224 = 11100000 63 = 111111 93 = 1011101 56 = 111000 87 = 1010111 159 = 10011111 192 = 11000000

= network id = broadcast

= broadcast = network id

Question What is the network address for a host with the IP address 123.200.8.68/28? A. 123.200.8.0 B. 1231.200.8.32 C. 123.200.8.64 D. 123.200.8.65 E. 123.200.8.31 F. 123.200.8.1 Answer C Explanation 3 x 8 = 24 leaving 4th octet with four network bits 68 decimal = 01000100 11110000 Anding 01000000 = 64 decimal The network is 123.200.8.64. OR /28 = 240, block size 256 – 240 = 16 NET ID 0, 16, 32, 48, 64, 80 So 68 will use the .64 subnet Incorrect Answers A. For the network to be represented as 123.200.8 then the IP address would need a /24 at the end. In this case /28 was used.

B, D, E, and F. In these cases with the IP address provided these options are impossible.

Question Using a class C address range 192.168.21.12 your network needs twenty-eight subnets. Which subnet mask should you use? A. 255.255.0.28 B. 255.255.255.0 C. 255.255.255.28 D. 255.255.255.248 E. 255.255.255.252 Answer D 28 subnets 2^5 = 32, 5 host bits borrowed 11111000 class C address + 5 bits borrowed 255.255.255.248 Question Your ISP has provided you the following class B network range 131.107.0.0/24. Which of the following statements is true regarding this network? (Choose two.) A. There are 254 usable hosts per subnet. B. There is one usable network. C. There are 255 usable hosts per subnet. D. There are 254 usable subnets. E. There are 30 usable subnets. F. There are 62 usable hosts per subnet. Answer A, D class B address 11111111.11111111.0.0 but /24 address 8 bits borrowed 2^8 -2 = 256 - 2 = 254 subnets bits left for hosts therefore 254 Question Using a class C address you need five subnets with a maximum of 17 hosts on each of these subnets. Which subnet mask would you use? A. 255.255.255.192 B. 255.255.255.224 C. 255.255.255.240 D. 255.255.255.248 Answer B Explanation class C address 255.255.255.0 5 subnets therefore 2^3 - 2 = 8 - 2 = 6 subnets (3 bits borrowed) 11100000 = 224

To check from host perspective 17 hosts 2^4 = 16 2^5 = 32 So 5 host bits leaving us with 3 network bits again 11100000 = 224 Question The network 131.107.0.0 needs to be divided into subnets where each subnet has the capacity of 458 IP addresses. What would be the correct subnet mask to accomplish this division keeping the number of subnets at a maximum? A. 255.255.0.0 B. 255.255.252.0 C. 255.255.254.0 Answer C 2^9 =512, 9 host bits leaving therefore 7 network bits 11111110.00000000 = 254 This provides 2^7 - 2 = 128 - 2 = 126 subnet In order for a Class B IP, such as 172.12.0.0, to have 458 IP available on each subnet then a subnet mask of 255.255.254.0 is required. Question What is the decimal and hexadecimal equivalent of the binary number 01010101? A. Decimal : 85 Hexadecimal : 55 B. Decimal : 125 Hexadecimal : 65 C. Decimal : 165 Hexadecimal : 75 D. Decimal : 205 Hexadecimal : 85 Answer A Explanation 01010101 binary equals 85 (64+16+4+1) decimal 0101 (leftmost 4 bits) equals 5 hexadecimal. 0101 (rightmost 4 bits) equals 5 hexadecimal. 01010101 binary equals 55 hexadecimal

Question You are configuring an IP printer that is connected to your network. You would like to use the last IP address that is on your subnet for this printer. You run ipconfig on your personal computer and the exhibit shows your IP address and its subnet mask. Based on your IP address and subnet mask what would the last possible address be on your subnet? A. 172.20.7.255 B. 172.20.7.197 C. 172.20.7.190

D. 172.20.7.129 E. 172.20.255.255

Answer C Explanation To determine the last available host draw a vertical line just after the last contiguous subnet mask 1 bit. On the next line write the address that it to the left of the line and then to the right of the line place all 1s in the remaining spaces until the last free space. Place a 0 in this place. Convert the binary to decimal and this will be the last available host. In this case the binary would be: 10101100 00010100 00000111 10 111110 This converts to 172.20.7.190 or ip 172.20.7.160 SM 255.255.255.192 256 – 192 = 64 NET ID Broadcast 0 63 64 127 128 191 160 is in this subnet last ip address is 190 192 255 255 Incorrect Answers A. This is the IP that would be used to send a broadcast to all host of the 172.20.7 subnet. B. With the facts of this question this is not a valid IP. D. This is the IP address used to send a message to all hosts on the 172.20.7 network. 10100000 11000000 (and) 10000000 = 128 network id + 1 for the broadcast =129 Question Assuming that our network is using an older version of UNIX what is the maximum number of subnets that can be assigned to networks when using the address 131.107.0.0 with a subnet mask of 255.255.240.0? A. 16 B. 32 C. 30 D. 14

E. It is an invalid subnet mask for the Network Answer D 240 = 11110000 4 bits therefore subnets = 2^4 - 2 = 16 -2 = 14 Question Using a class C address 192.168.10.X what would the subnet mask be if we needed two subnets with a maximum of 35 hosts on each subnet? A. 255.255.255.192 B. 255.255.255.224 C. 255.255.255.240 D. 255.255.255.248 Answer A 2^6 -2 = 64 hosts 6 bits for host address 2 bits for network 11000000 = 192 Question Which IP address range is allowable given an IP address of 131.107.2.56 and 28-bits of subnetting? A. 131.107.2.48 to 131.107.2.63 B. 131.107.2.48 to 131.107.2.62 C. 131.107.2.49 to 131.107.2.62 D. 131.107.2.49 to 131.107.2.63 E. 131.107.2.55 to 131.107.2.126 Answer C /28 – /24 = 4 extra network bits 11110000 = 240, block size 256 – 240 = 16 Subnets (net id) 131.107.2.0 .16 .32 .48 .64

host broadcast range 15 31 47 63 79

1-14 17-30 33-46 49-62 63-78

Explanation We have a subnet mask of 28 bits of ones followed by 4 bits of zeros, or 255.255.255.240, and gives a range 16, or 14 hosts per subnet (16-2 because we subtract out the two ranges of all zeros and all ones) This will yield subnets, 131.107.2.0 131.107.2.16 131.107.2.32 131.107.2.48

131.107.2.64 So, we can fit a network of 131.107.2.48-131.107.2.63 131.107.2.48 is the network, and the all zeros range. 131.107.2.63 is the broadcast, and is the all ones range. So, we can't user 48 or 63, and the valid host address range would be 49-62. C is correct. Incorrect Answers A, B, D, and E. These are not the proper ranges with the information provided in the question. Question Given the following IP address from the Class B address range 131.107.21.12 Your network plan requires no more than 126 hosts on a subnet that includes this address. When you configure the IP address in Cisco IOS software, which value should you use as the subnet mask? A. 255.255.0.0 B. 255.255.128.0 C. 255.255.255.128 D. 255.255.255.252 Answer C Explanation 126 hosts 2^7 - 2 = 126 7 bits in the host address therefore one bit in the network address 100000000 = 128 In the fourth octet of the subnet mask, we have 1 bit for the network, and 7 bits for the host. This high order NETWORK bit is the 128 bit, all network bits before this bit MUST BE one, so we get 255.255.255.128 Incorrect Answers A, B, and D. They do not provide a maximum of 126 hosts.

Question You have an IP of 156.233.42.56 with a subnet mask of 7 bits. How many hosts and subnets are possible? A. 126 hosts and 510 subnets B. 128 subnets and 512 hosts C. 510 hosts and 126 subnets D. 512 subnets and 128 hosts Answer C Explanation Class B network the default subnet mask is 16 bits long. There is additional 7 bits to the default subnet mask. The total number of bits in subnet are 16+7 = 23. This leaves us with 32-23 =9 bits for assigning to hosts. 7 bits of subnet mask corresponds to (2^7-2)=128-2 = 126 subnets. 9 bits belonging to host addresses correspond to (2^9-2)=512-2 = 510 hosts.

Question

Answer C The subnet mask used on this Ethernet segment is /27, which translates to 255.255.255.224 (11111111,11111111,11111111,11100000) valid hosts on the 192.168.5.33/27 subnet are 192.168.5.33 = ip address, 255.255.255.224 = subnet mask Determine the subnet and broadcast address, 256 -224 = 32 0, 32, 64 The address 33 falls between two subnets 32 and 64 and must be part of the 192.168.5.32 subnet the next subnet is 64 so the broadcast address is 63, valid host range 33-62 Therefore only choice C falls within the useable IP range. Question Given the choices below, which address represents a unicast address? A. 224.1.5.2 B. FFFF. FFFF. FFFF. C. 192.168.24.59/30 D. 255.255.255.255 E. 172.31.128.255/18 Answer E Explanation unicast is simply an address that points to one particular host, as opposed to multiple, or even all hosts like a multicast or broadcast. A. 224.1.5.2 Since the first octet of this address is 224 it falls in the multicast range so it is not a unicast address.

B. FFFF. FFFF. FFFF. This is the target MAC address of a broadcast, so again, not a unicast. C. 192.168.24.59/30 This one is here to trick you. If you work out the subnet it is 192.168.24.56 through 192.168.24.59. Since 192.168.24.59 is the last IP in the range, it is the broadcast address. In binary, all the host bits are ones. So this address actually referes to all of the hosts in the subnet and not just one. D. 255.255.255.255 This represents a broadcast address sent to anyone who is listening on a shared network segment. E. 172.31.128.255/18 Again, if we work out the subnet here the range includes 172.31.128.0 through 172.31.191.255. Since 172.31.128.255 falls in between those two numbers, it is a host IP. In other words, it refers to one particular host, making it a unicast address.

Explanation 172.16.45.14/30 /30 so /24 1st 3 octets taken up leaving 6bits for the network 172.16.45.? 14 in binary 8,4,2,1 1110 So last octet is 00001110 11111100 Anding 00001100

= 12

172.16.45.12 Or another way 11111100 = 253, block size 256 – 253 = 3 NET ID’s 0, 3, 6, 9, 12, 15 Therefore .14 is in the .12 subnet

Explanation /28 = 240 256 - 240 = 16 block size 0, 16, 32 subnets Valid host range 17 – 30, .31 = broadcast address Answer A and C

Explanation 10 subnets 2^n - 2 = number of subnets / hosts 2^3 – 2 = 6 not enough subnets 2^4 - 2 = 14 number of networks, 14 number of hosts 2^5 – 2 = 30 number of subnets ok leaves 2^3 - 2 = 6 number of hosts not ok Therefore number of bits = 4, 11110000 = 240

Explanation /22 therefore 2 x 8 =16 leaving 6 bits in the 3 octet for network portion Anding 3rd octet

=

11010010 11111100 11010000

210 in binary 208

The network consists of 5 subnets Using 3 bits for the network address 2^n – 2 = 2^3 - 2 = 6 we get 6 subnets leaving us with 5 bits for hosts 2^5 -2 = 30 11100000 = 224

2^n -2 = at least 459 2 4 8 16 32 64 128 256 512 1 2 3 4 5 6 7 8 9 host bits giving us 2^9 - 2 = 510 hosts 9 hosts bits so the last octet is all zeros (8 bits) and the last octet in the 3rd is also zero 111111110 = 254

Explanation 256 - 224 = 32 Every network boundary is multiples of 32

0, 32, 64, 96, 128, 160, 192, 224, = each subnet 31, 63, 95, 127, 159, 191, 223, broadcast address Any ip address not one of theses numbers is a valid host ip address

Answer A, C, D Explanation 256- 240 = 16 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208 15, 31, 47, 63, 79, 95, 111, 127, 143, 159, 175, 191, 207 2, 11 and 13 are valid hosts

Explanation /24 2^n =subnets 2^0 = 1,

Answer D 18 hosts 2, 4, 8, 16, 32 = 5 host bits Therefore 3 extra network bits 11100000 = 224 Default subnet mask for class C = 255.255.255.0 therefore answer is 255.255.255.224 A. B and C is wrong as the 3rd octet cannot be divided up when using class C

Explanation A class C network with 26 bit mask requires 2 bits for the network address leaving 6 bits for host addresses. 2^2 for the network = 4 and 2^6 -2 = 62 hosts Note 2^2 for calculating subnets 2^2-2 therefore ip zero is in force.

Class B = /16 From subnet perspective 300 required, so 2, 4, 8, 16, 32, 64, 128, 256, 512 An extra 9 network bits required /16 + 9 = /25 11111111.10000000 = 255.128 so answer = B 255.255.255.128 From host perspective 50 hosts 5^2 = 32, 6^2 = 64, so 6 host bits required This means 2 network bits are used 11000000 = 192 answer = E, 255.255.255.192 Alternative way of answering Explanation 300 subnets wanted 2, 4, 8, 16, 32, 64, 128, 256, 512 300 subnets have 9 network bits leaving 7 host bits = 126 hosts 11111111.11111111.11111111.10000000 = 255.255.255.128 Add another bit to give 10 network bits, this gives 1022 subnets and 6 bits for hosts = 62 hosts 11111111.11111111.11111111.11000000 = 255.255.255.192

Explanation 172.16.209.10/22 .16.209 .11010001.00001010 /22 2x8 =16 leaving 6 more 1’s in the 3rs octet .11111100.00000000 .11010000.00000000 Anding gives us 208 Or /22 = 11111111.11111111.11111100 = 255.255.253 256 – 253 = 3 block size Net ID’s 0, 3, 6, 9, 12, 15, 18, 21, 24

Explanation 115.64.4.0 = subnet mask /22

01110011.01000000.00000100.00000000 11111111.11111111.11111100.00000000 = 255.255.252.0

anding

01110011.01000000.00000100.00000000 = subnet number 115.64.4.0

the clever bit 01110011.01000000.00000111.11111111= broadcast addr 115.64.7.255 therefore valid address range 115.64.4.1 – 115.64.7.254

Explanation /20 = 11111111.11111111.11110000.0 = 240 256 – 240 = 16 the increment of the subnets 0, 16, 32, 48, 64 remember -1 for the broadcast addresses 15, 31, 47, 63 The switch ip address 172.16.80.90 is in the 172.16.80.0 subnet 172.16.95.255 is the broadcast 172.16.47.255 is the broadcast address for 172.16.32.0 subnet 172.16.79.255 is the broadcast address for 172.16.64.0 subnet

Explanation /27 = /24 leaving three 1’s in the 4th octet 11100000 = 224 256 -224 = 32 subnet increments 0, 32, 64, 96, 128, 160, 192 = subnets 31, 63, 95, 127, 159, 191 = broadcast addresses

Explanation 192.168.126.49/30 6 bits borrowed leaving 11111100 2 bits for hosts 2^2-2 =2 hosts 192.168.126.127/26 2 bits borrowed leaving 11000000 6 bits for hosts 2^6-2 = 62 hosts 192.168.126.67/29 5 bits borrowed leaving 11111000 3 bits for hosts 2^3-2 = 6 hosts 192.168.126.2/27 3 bits borrowed leaving 11100000 5 bits for hosts 5^2-2 = 30 hosts Therefore 192.168.126.35/27 = 30 hosts

Explanation 5 subnets highest number of hosts required is 16 2^n = 2 x 2 x 2 = 8 subnets this uses 3 network bits and 5 hosts bits = 2 x 2 x 2 x 2 x 2 = 30 hosts Since we only require 5 different subnets with at most 16 users this will suffice. 3 network bits = 11100000 = 224

Explanation 224 = 1110000 = 5 host bits 2^5 -2 = 30 hosts Question 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? A. IP address: 192.168.20.14 Subnet Mask: 255.255.255.248 Default Gateway: 192.168.20.9 B. IP address: 192.168.20.254 Subnet Mask: 255.255.255.0 Default Gateway: 192.168.20.1 C. IP address: 192.168.20.30 Subnet Mask 255.255.255.248 Default Gateway: 192.168.20.25 D. IP address: 192.168.20.30 Subnet Mask 255.255.255.240 Default Gateway: 192.168.20.17 E. IP address: 192.168.20.30 Subnet Mask 255.255.255.240 Default Gateway: 192.168.20.25

Answer C Explanation /29 bits. 5 bits in the 4th octet. This equates to 255.255.255.248. This network has 3 bits for hosts. 2^3 - 2 = (2*2*2 - 2 = 6) host addresses. 192.168.20.24 is the network address. Therefore the next address (192.168.20.25) would be the first host address. This address must be assigned to the router, which serves as the gateway for the network. The last available host address would be 192.168.20.30 (192.168.20.24+6). This address is assigned to the server. The broadcast address is 192.168.20.31.

Question You've been assigned a single Class C address. From this, you need 8 subnets, and your subnet mask is 255.255.255.224. Which one of the following configuration commands would you have to use before you begin? A. Router(config)# ip classless B. Router(config)# ip subnet-zero C. Router(config)# ip version 6 D. Router(config)# no ip classful E. Router(config)# ip unnumbered F. Router(config)# ip all-nets Answer B Explanation To get 8 subnets from a class C address, and a mask of 255.255.255.224 use the reserved subnet space. To do this, you need the command 'ip subnet-zero.' This will allow the router to use the very first subnet, which is normally reserved and unused as the network address. Prior to Cisco IOS® Software Release 12.0, Cisco routers, by default, did not allow an IP address belonging to subnet zero to be configured on an interface. However, if a network engineer working with a Cisco IOS software release older than 12.0 finds it safe to use subnet zero, the ip subnet-zero command in the global configuration mode can be used to overcome this restriction. As of Cisco IOS Software Release 12.0, Cisco routers now have ip subnet-zero enabled by default, but if the network engineer feels that it is unsafe to use subnet zero, the no ip subnet-zero command can be used to restrict the use of subnet zero addresses. In versions prior to Cisco IOS Software Release 8.3, the service subnet-zero command was used. It should be noted that even though it was discouraged, the entire address space including subnet zero and the all-ones subnet have always been usable. The use of the all-ones subnet was explicitly allowed and the use of subnet zero is explicitly allowed since Cisco IOS Software Release 12.0. Even prior to Cisco IOS Software Release 12.0, subnet zero could be used by entering the ip subnet-zero global configuration command. On the issue of using subnet zero and the all-ones subnet, RFC 1878 states, "This practice (of excluding all-zeros and all-ones subnets) is obsolete. Modern software will be able to utilize all definable networks." Today, the use of subnet zero and the all-ones subnet is generally accepted and most vendors support their use. However, on certain networks, particularly the ones using legacy software, the use of subnet zero and the all-ones subnet can lead to problems.

Question Three routers are connected as shown

Taking the information shown above, which command line below would correctly configure serial port0 on router2 with the LAST usable host addresses on the 192.216.32.32 subnet? A. router2(config-if)# ip address 192.216.32.63 255.255.255.248 B. router2(config-if)# ip address 192.216.32.38 255.255.255.240 C. router2(config-if)# ip address 192.216.32.39 255.255.255.248 D. router2(config-if)# ip address 192.216.32.63 255.255.255.248 no shut E. router2(config-if)# ip address 192.216.32.39 255.255.255.248 no shut F. router2(config-if)# ip address 192.216.32.38 255.255.255.248 Answer F Explanation F is correct as the last usable IP address on this subnet is 192.216.32.38. The subnet mask for a /29 is 255.255.255.248 Mask/29 11111111.11111111.11111111.11111000 255.255.255.248 Subnet 11000000.11011000.00100000.00100000 192.216.32.32 Broadcast 11000000.11011000.00100000.00100111 192.216.32.39 Address range = 192.216.32.33 192.216.32.38 Or 256 – 248 = 8 Net ID 0, 8, 16, 24, 32, 40 Broadcast 7, 15, 23, 31 39 Last useable host 6, 14, 22, 30, 38 Question

What is a valid possible IP address configuration for Host A?

A. IP 192.168.100.31 255.255.255.240 default-gateway 192.168.100.18 B. IP 192.168.100.30 255.255.255.240 default-gateway 172.16.1.1 C. IP 192.168.100.20 255.255.255.240 default-gateway 192.168.100.17 D. IP 192.168.100.21 255.255.255.248 default-gateway 192.168.100.17 E. IP 192.168.100.19 255.255.255.248 default-gateway 172.16.1.1 Answer C Explanation: The network mask for a /28 is 255.255.255.240. The default gateway is always the IP address of the router on the local subnet, and the valid IP range for this network is 192.168.100.17 - 192.168.100.30. C is the only one that meets all of these. Incorrect Answers A. The IP address 192.168.100.31 is the broadcast address. It cannot be used for the host. B. The default gateway should be the fist exit point for the network that the host is on. In this case it should be the router interface address 192.168.100.17. D. The network uses a 28 bit subnet mask (11111111.11111111.11111111.11110000). This equates to 255.255.255.240, not 255.255.255.248. E. The network uses a 28 bit subnet mask (11111111.11111111.11111111.11110000). This equates to 255.255.255.240, not 255.255.255.248. Also, the default gateway should be the fist exit point for the network that the host is on. In this case it should be the router interface address 192.168.100.17. Question

Based on the information above, which of the following would be a valid IP address of the PC? A. 192.168.5.55 B. 192.168.5.47 C. 192.168.5.40 D. 192.168.5.32 E. 192.168.5.14 Answer C Explanation 192.168.5.33/28, /28 =240 256 – 240 = 16 block size Subnets 0, 16, 32, 48

Valid host range 33 – 47 the last address 47 is the broadcast address. The router interface E0 has the IP address 192.168.5.33. Therefore it is on the 2nd network (192.169.5.32/28). The host must also be on this network. Valid IP addresses for hosts on this network are: 192.168.5.33192.168.5.46. Incorrect Answers A. 192.168.5.55 is on network 192.168.5.48. It is not on the same network as the router interface. B. This is the broadcast address. D. This is the network address. E. This is not a valid address for a 28 bit subnet mask. The first network address should be 192.168.5.16.

Question You are on the network design team and have the task of networking three locations together. Your team will be using the address range 192.168.55.0. RIP v2 will be used as the routing protocol, and "ip subnet-zero" will be configured. Your goal is to fulfill the address needs of the network while conserving unused addresses for potential future growth.

With these goals in mind, drag the host addresses on the left side to the correct router interface on the right side. Not all the addresses are going to be used, and one of the routers is already partially configured.

Answer

Explanation 192.168.55.57/27 192.168.55.29/28 192.168.55.1/30 192.168.55.132/25 192.168.55.0/30 192.168.55.127/26

3 x 8 =24 leaving 3 network bits 11100000 5 host bits 2^5 -2 = 30 hosts 4 network bits 4 host bits 2^4 -2 = 14 hosts 6 network bits 2 host bits 2^2 -2 = 2 hosts 1 network bit 7 host bits 2^7 -2 = 126 hosts 6 network bits 2 host bits 2^2 -2 = 2 hosts 2 network bits 62 hosts

Question

What command would you use to configure the correct IP address and subnet mask on R2's serial interface? A. R2(config-if)# ip address 172.16.17.1 255.255.255.0 B. R2(config-if)# ip address 172.16.18.2 255.255.252.0 C. R2(config-if)# ip address 172.16.17.2 255.255.255.252 D. R2(config-if)# ip address 172.16.16.0 255.255.255.0 Answer B Explanation /22 = 11111111.11111111.11111100.0 = 255.255.252.0 Question You work as a technician. You are configuring a Cisco router. You want to configure the IP address on an interface. Which command should you use? A. router(config-if)#ip address 142.8.2.1 subnet mask 255.255.252.0 B. router(config-if)#142.8.2.1 0.0.3.255 C. router(config-if)#ip address 142.8.2.1 255.255.252.0

D. router(config-if)#142.8.2.1 subnet mask 255.255.252.0 E. router(config-if)#ip address 142.8.2.1 0.0.3.255 F. router(config-if)#ip address 142.8.2.1 subnet mask /22

Answer C Explanation: ip address address subnet-mask - Interface configuration mode command that sets the IP address for interfaces. Only choice C uses the correct syntax.

Question Which command will assign the last usable IP address from the 192.168.32.128/28 subnetwork to a router interface? A. RA(config-if)# ip address 192.168.32.142 255.255.255.240 B. RA(config-if)# ip address 192.168.32.143 255.255.255.240 C. RA(config-if)# ip address 192.168.32.158 255.255.255.240 D. RA(config-if)# ip address 192.168.32.145 255.255.255.240 E. RA(config-if)# ip address 192.168.32.144 255.255.255.240 F. RA(config-if)# ip address 192.168.32.158 255.255.255.240 Answer A Explanation /28 = 240, block size = 256 – 240 = 16 0, 16, 32, 48, 64, 80, 96, 112, 128, 144 Broadcast = 143 Last ip = 142

Question

A network administrator is adding host 3 to the network shown in the exhibit. Which IP address can be assigned this host on this network? A. 192.1.1.14

B. 192.1.1.18 C. 192.1.1.20 D. 192.1.1.30 E. 192.1.1.31 F. 192.1.1.36 Answer B, D Explanation 256 -240 = 16 blocksize Net ID Broadcasts Hosts

0, 16, 32, 48 15, 31, 47 1-14, 16-30, 32-46

Hosts .20 and .22 are on the .16 subnet host 16 - 30 Only choices B and D are possible as C 192.1.1.20 is already used by host 1

Question

The routers in this network are running RIPv2. Which addressing scheme would satisfy the needs of this network yet waste the fewest addresses? A. Network 1: 192.168.10.0/26 Network 2: 192.168.10.64/26 Network 3: 192.168.10.128/26 Serial link 1: 192.168.20.0/24 Serial link 2: 192.168.30.0/24

B. Network 1: 192.168.10.0/26 Network 2: 192.168.10.64/28 Network 3: 192.168.10.80/29 Serial link 1: 192.168.10.88/30 Serial link 2: 192.168.10.96/30 C. Network 1: 192.168.10.0/26 Network 2: 192.168.10.64/27 Network 3: 192.168.10.96/28 Serial link 1: 192.168.10.112/30 Serial link 2: 192.168.10.116/30 D. Network 1: 192.168.10.0/27 Network 2: 192.168.10.64/28 Network 3: 192.168.10.96/29 Serial link 1: 192.168.10.112/30 Serial link 2: 192.168.10.116/30 Answer C Explanation Network 1. Required Number of hosts 50 2, 4, 8, 16, 32, 64 = 6 host bits, 6^2 -2 = 64 -2 = 62 hosts Leaves 2 network bits /24 + 2 = /26 Network 2. Required Number of Hosts 20 2, 4, 8, 16, 32 = 5 host bits = 30 hosts this leaves 3 network bits 24 + 3 = /27 Network 3. Required Number of Hosts 10 2, 4, 8, 16 = 4 host bits = 14 hosts this leaves 4 network bits 24 + 4 = /28

Question

Based on the information above, which IP address should be assigned to the host? A. 192.168.5.5 B. 192.168.5.32 C. 192.168.5.40 D. 192.168.5.63 E. 192.168.5.75 Answer C Explanation: Host address should be in same subnet of Connected Router's Interface. In exhibit Router's ethernet address is in 192.168.5.33/27 subnet then host address should be also in same subnet. /27, 11100000 = 224, 256 – 224 = 32 block size Net id 0, 32, 64, 96 Broadcast 1, 31, 63, 95 Hosts 1-30, 32-62, 64-94 .33 is in the 32 bit subnet, so only answer C is correct as .40 is also in the same subnet.

Question DRAG DROP We are redesigning the network that connects three locations. The administrator gave the networking team 192.168.3.0 to use for addressing the entire network. After subnetting the address, the team is ready to assign the addresses. The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time conserve unused addresses for future growth. With those goals in mind, drag the host addresses on the left to the correct router interface. One of the routers is partially configured. Not all of the host address addresses on the left are necessary.

Answer

Explanation 2, 4, 8, 16, 32, 64, 128 13 hosts so 4 host bits are required, 4 network bits used = 11110000 = /28 42 hosts so 6 host bits are required, 2 network bits used = 11000000 = /26 111 hosts so 7 host bits are required, 1 network bit used = 10000000 = /25 For the WAN connection between routers, S0/1 if you use the /30 bits for Network address you will get 2 usable host address.

Question ABC is redesigning the network that connects its three locations. The administrator gave the network team 192.168.15.0 to use for addressing the entire network. After subnetting the address, the team is ready to assign the addresses. The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time conserve unused addresses for future growth. With those goals in mind, drag the host addresses on the left to the correct router interface. One of the routers is partially configured. Not all of the host addresses on the left are necessary.

Answer

Explanation 2, 4, 8, 16, 32, 64, 128 118 hosts so 7 host bits are required, 1 network bit used = 10000000 = /25 59 hosts so 6 host bits are required, 2 network bits used = 11000000 = /26 9 hosts so 4 host bits are required, 4 network bits used = 11110000 = /28 For the WAN connection between routers, S0/0 if you use the /30 bits for Network address you will get 2 usable host address.

Question A Class C network address has been subnetted with a /27 mask. Which of the following addresses is a broadcast address for one of the resulting subnets? A. 201.57.78.64 B. 201.57.78.87 C. 201.57.78.33 D. 201.57.78.254 E. 201.57.78.159 F. 201.57.78.97 Answer E Explanation: A subnet mask of /27 (255.255.255.224) will have 3 bits used for the network portion and 5 bits for the host portion. This will create 2^3 = 8 networks with 2^5 = 32 hosts per network. From this we know that the number of subnets will be a multiple of 32, making the subnets: 201.57.78.32 201.57.78.64 201.57.78.96 201.57.78.128 201.57.78.160 201.57.78.192 201.57.78.224

201.57.78.256 Since the broadcast address is always the last IP address in the subnet we need to only subtract 1 from each of the IP addresses above to find the broadcast. From the list above we see that 201.57.78.159 is the only available option, as this is the broadcast address for the network previous to the 201.57.78.160 network. Question

What IP address should be assigned to Workstation A? A. 192.168.1.159/28 B. 192.168.1.160/28 C. 192.168.1.145/28 D. 192.168.1.144/28 E. 192.168.1.143/28 Answer C Explanation. The available subnets and IP ranges that are available using a /28 (255.255.255.240) subnet mask is shown below Based on this information, we need to choose an IP address within the 145-158 range, since the IP address of the Fa0/0 on the router is 192.168.1.158, leaving only answer choice C as feasible.

Question Refer to the exhibit. What are the broadcast addresses for each subnet?

A. Admin - 172.16.31.0 QA - 172.16.1.127 Development - 172.16.2.255 Sales - 172.16.32.255

B. Admin - 172.16.31.255 QA - 172.16.1.255 Development - 172.16.3.255 Sales - 172.16.63.255 C. Admin - 172.16.31.255 QA - 172.16.1.127 Development - 172.16.3.255 Sales - 172.16.63.255 D. Admin - 172.16.31.0 QA - 172.16.1.255 Development - 172.16.2.255 Sales - 172.16.32.255 Answer C Admin - 172.16.31.255 QA - 172.16.1.127 Development - 172.16.3.255 Sales - 172.16.63.255 Explanation Admin 172.16.16.0/20 (/20 mask = 255.255.11110000.0) =255.255.240.0 256 -240 = 16 blocksize so subnets are a multiple of 16 Net ID’s

0, 16, 32,

15 Broadcasts 31 47

SubNet ID’s 172.16.0.0 – 172.16.16.0 – 172.16.32.0 -

Broadcasts 172.16.15.255 172.16.31.255 172.16.47.255

QA

172.16.1.64/26

/26 = 255.255.255.192 256 -192 = 64 blocksize Net ID’s

0 64

Broadcasts

Net ID’s 172.16.1.0 172.16.1.64

Broadcasts 172.16.1.63 172.16.1.127

63 127

Question Refer to the exhibit. What is the first usable IP address that can be assigned to the WGROUP3 switch?

A. 172.16.50.96/27 B. 172.16.50.97/27 C. 172.16.50.98/27 D. 172.16.50.99/27 Answer B 172.16.50.97/27 Explanation /27 = 255.255.255.224 Blocksize = 32 Net ID’s 172.16.50.0 172.16.50.32 172.16.50.64 172.16.50.96

Broadcasts 172.16.50.31 172.16.50.63 172.16.50.95 172.16.50.128

Hosts 1- 30 33-62 65-94 97-127