June 29, 2016 | Author: Paul | Category: Types, Research, Internet & Technology

#### Description

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!

Which of the following are true about 172.16.0.254/16?

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

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

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

/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

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

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

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

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

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

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

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

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.

D. 172.20.7.129 E. 172.20.255.255

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

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

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

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

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 Three routers are connected as shown

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

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

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.

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

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.

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.

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,

SubNet ID’s 172.16.0.0 – 172.16.16.0 – 172.16.32.0 -

QA

172.16.1.64/26

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

0 64

Net ID’s 172.16.1.0 172.16.1.64

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