z 6a IPv4 Advance Subnetting, Subnet Masking & LS_IP Addressing

If you have a B Class network in which you are borrowing 10 bits from your host octets to form a subnet mask. How many subnets and how many hosts would you have? 1022 subnets, 62 hosts 62 subnets, 8190 hosts 8,190 subnets, 254 hosts 254 hosts, 126 subnets

1022 subnets, 62 hosts

How many Class A networks in the world? 126 124 128 254 256

126

(SubnetQ32): What is the maximum number of hosts you could provide to an IP address of 210.21.21.40 with a subnet mask of 255.255.255.224? 16 24 30 32 62 64

30 You calculate the hosts by the number of trailing bits available for the hosts. In this case, you have 5 trailing bits available. Hence 2^5 = 32-2 = 30 hosts.

You have a 22 bit subnet mask. How many subnets and how many hosts do you have? (This will take some thought, as you do not know if you have an A, B, or C network. All you know is that you have a 22 bit subnet mask. Hence you need to work backwards from the answers to get the most correct answer on information provided) 8190 subnets, 4096 hosts 4,194,302 subnets, 2 hosts 2,096,138 subnets, 6 hosts 16,384 subnets, 2046 hosts

4,194,302 subnets, 2 hosts

By default, a subnet mask would allow for? The maximum number of network Ids The maximum number of host IDS A balance between the number of host IDS and network IDS 254 subnets

The maximum number of host IDS

You need to come up with a TCP/IP addressing scheme for your company. Which two factors must you consider when you define the subnet mask for the network? Choose 2 answers only. The number of subnets on the network The number of host IDS on each subnet The volume of network traffic on each subnet The location of DNS servers The location of default gateways.

The number of subnets on the network The number of host IDS on each subnet

Would a computer with an IP address of 200.24.40.61, Subnet Mask 255.255.255.192 be on the same subnet as a computer with an IP address of 200.24.40.68 and a subnet mask of 255.255.255.192. Yes No

no

Would a computer with an IP address of 203.68.69.29, Subnet Mask 255.255.255.224 be on the same subnet as a computer with an IP address of 203.68.69.35 and a subnet mask of 255.255.255.224 yes no

no

Would a computer with an IP address of 198.24.36.82, Subnet Mask 255.255.255.192 be on the same subnet as a computer with an IP address of 198.24.36.126 and a subnet mask of 255.255.255.192. yes no

yes

Would a computer with an IP address of 203.68.79.98, Subnet Mask 255.255.255.224 be on the same subnet as a computer with an IP address of 203.68.79.125 and a subnet mask of 255.255.255.224 yes no

yes

(LS IPv4 Q03): Which three of the following IP addresses belong to the Class A network 140.0.0.0? (Assume the network is indicated by the default portion of the IP address) (Select three answers only. You must have all three answers correct for credit. There is no partial credit for this question). 115.88.0.55 114.58.12.0 114.0.0.15 115.0.0.66 114.122.66.12 115.77.89.4

114.58.12.0 114.0.0.15 114.122.66.12 With a Class A network, the first octet indicates the network address. All hosts on the network must have the same value in the first octet which in this case is 114.

(SubnetQ26): How many subnets can an IP address of 39.0.0.0/15 support? 64 96 126 128 254 512

126 126 Since you are using 7 bits, the number of supported subnets is 2^7 = 128 - 2 =126.

(SubnetQ09): The following information is provided:IP Address: 75.20.10.3/24 Based on this information, what is the host and network address? Host address 0.0.10.3 with a network address of 75.20.0.0 Host address of 0.20.10.0 with a network address of 75.0.0.0 Host address of 0.0.0.3 with a network address of 75.20.20.16 Host address of 0.0.0.3 with a network address of 75.20.20.8 Host address 0.0.0.3 with a network address of 75.20.10.0

Host address 0.0.0.3 with a network address of 75.20.10.0 The /24 indicates 24 bits applied to the subnet mask. In this case, it includes the first three octets or 75.20.10 as the network address. The remaining 0 bits would be representative of the host which in this case would be 3.

(SubnetQ04): What is the network number for this host: 10.34.23.5/24? 10.34.23.8 10.34.23.16 10.34.23.0 10.34.0.0 10.34.23.32

10.34.23.0 This is a class A network because of the value of 10 in the first octet. But the subnet mask is the key to this question. The 1's or the 255's in the subnet mask dictate the network address. The default mask for a Class A network is 255.0.0.0. However, in this question, the /24 means that 24 bits are being applied to the subnet mask. In this case the subnet mask would be 255.255.255.0. The 255's are telling you what part of the given IP address is network, and what part is host. Since the mask numbers are equal and fall right at the octet breaks, the network is 10.34.23.0. The host is 0.0.0.5. With the /24 you have actually created 2^16-2 = 65,534 subnets on which you can put 254 hosts on each of the subnets.

(LS IPv4 Q12): A host has the address of 100.55.177.99/16. Which of the following is the broadcast address for the subnet? 255.255.255.0 100.255.255.255 255.255.0.0 100.55.255.255 100.55.177.255

100.55.255.255 The broadcast address for the subnet (network) is the last address on the subnet. In this example, the address uses 16 bits in the subnet mask 255.255.0.0, meaning that the first two octets indicate the subnet address which is 100.55.0.0, and the last two octets are used for the host address. The last possible IP address on the subnet is 100.55.255.255, which becomes the broadcast address, and will not be valid as a host address.

(LS IPv4 Q11): What is the binary format for the following decimal IP address?IP 131.9.202.111 10000001.00001010.11000011.01010111 10000011.00001001.11001010.01101111 10000110.00001011.11000101.10101110 10000111.00001101.11001110.01011101

10000011.00001001.11001010.01101111 In order to answer this question correct you need to convert binary to decimal and in order to do that you need to remember the following numbers: 128,64,32,16,8,4,2,1.

(SubnetQ19): You have created 30 subnets with an IP 128.199.0.0/21. What is the address of the second useable subnet? 128.199.0.0 128.199.8.0 128.199.32.0 128.199.16.0 128.199.48.0

128.199.16.0 The address of the second useable subnet would be 128.199.16.0* 128.199.0.0 (invalid)* 128.199.8.0* 128.199.16.0 (2nd Useable subnet)* 128.199.24.0* 128.199.32.0* &&...............* 128.199.240.0* 128.199.248.0 (invalid)

(SubnetQ20): You have created 30 subnets with an IP 128.199.0.0/21. What is the last node on the 3rd useable subnet? 128.199.31.254 128.199.0.254 128.199.39.254 128.199.39.255 128.199.31.255

128.199.31.254 To determine the subnet ID's you must identify the incremental value (or delta) that will be used to establish your Network IDs. Since you are using 5 leading bits in the third octet, you use the value of the last trailing bit to determine your incremental value. In this case it is 8. As you can see from below, the third useable network is 128.199.24.0, and the last host on this network would then be 128.199.31.254. The broadcast address would be 128.199.31.255 Network ID ...... 1st Host .............. Last Host128.199.8.0 ....... 128.199.8.1 ~ ...... 128.199.15.254128.199.16.0...... 128.199.16.1 ~ ...... 128.199.23.254128.199.24.0 ..... 128.199.24.1 ~ ...... 128.199.31.254 (3rd useable subnet)128.199.32.0 ...... 128.199.32.1 ~ ...... 128.199.39.254&&&&&........&&&&&........128.199.240.0 .... 216.122.240.1 ~ ...... 216.122.44.254128.199.248.0 (invalid)

(LS IPv4 Q08): A host on the network has an IP address of 129.11.99.78 using the default subnet mask. How would you identify the address and mask using CIDR notation? 129.11.99.78:16 129.11.99.78/16 129.11.99.78:8 129.11.99.78/24 129.11.99.78:24 129.11.99.78/8

129.11.99.78/16 Use 129.11.99.78/16 for the address and the mask. With CIDR notation, follow the IP address with a (/) and the number of bits in the mask. The default subnet mask of this address is 255.255.0.0 which uses 16-bits in the mask. A mask value of 255.0.0.0 would use 8 bits or /8 and a mask value of 255.255.255.0 would use 24 bits or /24

(SubnetQ27): What is the maximum number of hosts that you can put on each subnet if you had an IP address of 39.0.0.0/15? 125,891 130,345 28,987 120,072 120,972 131,07

131,070 If you borrowed the 7 leading bits for the subnet mask from the second octet, then your would have 1 trailing bit remaining from the second octet and 8 trailing bits remaining from each of the third and fourth octets which you can use to configure host IDs. That would be 2^17 -2 = 131,070

(SubnetQ12): If you use a network mask of 255.255.255.240 with a given IP address of 216.122.44.0, how many supported subnets could you create? 4 8 12 14 16 18 20

14 14 subnets You must use the number of leading bits from the subnet mask to calculate the number of supported subnets. In this case it is 4 bits calculated as 2^4 = 16-2 =14 subnets What would be the Maximum Number of Hosts ID's per subnet? This would be 14 as well. You calculate the hosts by the number of remaining trailing bits. Since there are only 4 remaining trailing bits, the number of host will be 2^4 = 16-2 =14 hosts. This means you can put 14 hosts on each of the 14 subnets you have created.

(SubnetQ08): The following information is provided: IP Address: 140.150.150.32 Subnet:255.255.192.0 What is the first useable subnet for the above IP Address? 140.150.16.0 140.150.32.0 140.150.48.0 140.150.64.0 140.150.150.32 140.150.150.48 140.150.150.64

140.150.64.0 A subnet mask of 255.255.192.0, means that you have borrowed 2 bits for the subnet mask as the leading value of these 2 bits is 128 +64 = 192. The second leading bit of 64 becomes the incremental value used to calculate subnets. The first useable subnet would then be 140.150.64.0. The second useable would be 140.150.128.0

(SubnetQ24): What is the network number of a host with the address of 152.77.190.27 using the subnet mask of 255.255.240.0 152.77.32.0 152.77.64.0 152.77.96.0 152.77.128.0 152.77.160.0 152.77.192.0 152.77.224.0

152.77.160.0 152.77.160.0 In order to calculate the network number, you must first determine the incremental value of each of the subnets. You determine this by using the value of the last bit used to form you subnets. That value is 32. Hence, you increment your network and create subnets with a value of 32 as noted below. Host 190.27 sits on network 152.77.160.0 ( 152.77.160.1 ~ 152.77.191.254). The full range of networks on this network is show as follows:* 152.77.0.0 (invalid)* 152.77.32.0* 152.77.64.0* 152.77.96.0* 152.77.128.0* 152.77.160.0* 152.77.192.0* 152.77.224.0 (invalid)

(LS IPv4 Q06): Which of the following is the last IP address that can be assigned to hosts on the 166.70.0.0 network using the default subnet mask? 166.70.255.255 166.70.0.255 166.70.255.254 166.71.0.0 166.70.0.254

166.70.255.254 The last address you can assign to hosts on the 166.70.0.0 network is 166.70.255.254. The network address is a Class B address and uses a default subnet mask of 255.255.0.0. The last two octets are used for host addresses. 166.70.0.0 cannot be used as a host address because it is the network address. 166.70.255.255 cannot be used as a host address because it is the broadcast address which is used on the 166.70.0.0 network.

(LS IPv4 Q01): Which of the following is a valid IP(version 4) address? (Select two answers only. You must have both answers correct for credit. There is no partial credit for this question). 1.254.1.1024 10.384.0.3 192.168.1.512 172.16.1.26 2.2.2.2 256.0.0.1 254.7.1.417

172.16.1.26 2.2.2.2 A valid IPv4 address consists of four 8 bit (1 byte) numbers separated by periods. For example 10.0.0.65 would be a valid IP address. Because they are eight bits long, these numbers are frequently called octets.Even though we typically expressed these numbers using decimal notation, it's important to remember that these numbers are binary numbers. The lowest value one of these numbers can have is 00000000. The decimal equivalent for this number is simply 0. The highest value one of these numbers can take is 11111111. The decimal equivalent of this number is 255. Therefore, in decimal notation each octet must contain a number between 0 and 255 inclusively.

You have an IP address of 172.16.3.57 and you use 11 bits from your host octets to create a subnet mask. What are your valid hosts? 172.16.3.32 to 172.16.3.62 172.16.3.33 to 172.16.3.62 172.16.3.34 to 172.16.3.62 172.16.3.57 to 172.16.3.62

172.16.3.57 to 172.16.3.62 11 bits gives you a 255.255.255.224 mask. This equates to 2,046 subnets with 30 hosts on each subnet. The delta is 32. Hosts would fall between the range in B. 33 - 62. Can't use 63 as that would be all 1's and is a broadcast address.

You have an IP address of 172.16.4.58 and use 12 bits from your host octets to create your subnet mask. What are the valid hosts? 172.16.4.48 - 172.16.4.63 172.16.4.49 - 172.16.4.63 172.16.4.49 - 172.16.4.62 172.16.4.55 - 172.16.4.62

172.16.4.49 - 172.16.4.62 A 12 bit address of 172.16.4.58 gives a mask of 255.255.255.240. The delta is 16. The range would be 49 - 62. Cannot use 63 as it is a broadcast

(LS IPv4 Q05): Which of the following IP addresses have a default subnet mask of 255.255.0.0. (Select three answers only. You must have all three answers correct for credit. There is no partial credit for this question). 1.6.45.254 191.168.3.15 228.62.18.6 129.0.0.1 123.254.10.6 168.16.5.1

191.168.3.15 129.0.0.1 168.16.5.1 IP addresses are divided into classes. The most common of these classes are A,B and C. Each address class has a different default subnet mask. To identify the class of an IP address look at its first octet:* Class A networks use a default subnet mask of 255.0.0.0 and have a value of 0-126 as their first octet.* Class B networks use a default subnet mask of 255.255.0.0 and have a value of 128-191 as the first octet.* Class C networks use a default subnet mask of 255.255.255 and have a value of 191-223 as the first octet.

(SubnetQ03): What is the network address of a host with an IP address of: 200.200.200.41/27? 200.200.200.16 200.200.200.32 200.200.200.40 200.200.200.48 200.200.200.8 200.200.200.0

200.200.200.32 This is a class C network in which 27 bits are being used for the subnet mask. Since the default mask is 255.255.255.0 or 24 bits, you can deduce that you are borrowing 3 additional bits from the fourth octet to form the subnets. The value of the last leading bit that you borrow is 32, hence the subnets will be incremented by a value of 32. In fact, you should immediately know that you can form a total of 6 useable subnets because 2^3 = 8 -2. The first useable subnet is 200.200.200.32, while the second subnet would be 200.200.200.64. You will note that a host with an address of 200.200.200.41 will fall on the 200.200.200.32 subnet. This is because the number "41" falls between numbers "33" and "62", which are all host numbers on the 32 sub network.

(SubnetQ18): If you had an IP 128.199.0.0 and you created 30 subnets with this IP address, what would be the maximum number of hosts that would you be able to put on each subnet? 1046 2046 4046 6.086 2048

2046 Once you determine the number of leading bits needed to create the subnets, you can use the remaining trailing bits to create your hosts. In this case, you have 2 octets which forms the default Class B host IDs. You are using (or borrowing) 5 of the leading bits from the third octet to create your subnets. This leaves you with 3 trailing bits from the third octet and all 8 bits from the fourth octet that can be used to create your hosts. Hence 2^11 = 2048 -2 = 2,046 hosts on each of the 30 subnets.

(SubnetQ13): You need to calculate what the address of the first useable subnet from a given IP address and mask of 216.122.44.0/28 would be? 216.122.44.0 216.122.44.8 216.122.44.16 216.122.44.24 216.122.44.32 216.122.44.48

216.122.44.16 You can figure the delta or incremental value by the value of the last leading bit that is used to create the subnet mask. In this case, the value of that bit is 16. Hence your increment the network numbers by 16 as shown below.* 216.122.44.0 (invalid network)* 216.122.44.16 (first useable network)* 216.122.44.32* 216.122.44.48* 216.122.44.64....................................................* 216.122.44.208* 216.122.44.224* 216.122.44.240 (invalid network)

How many hosts are there on a class C network with a default subnet mask? 126 128 254 256

254

(LS IPv4 Q07): You have been told to assign the IP address 21.155.67.166 to a host on the network using the default subnet mask. Which masks should you use? 255.255.0.0 21.0.0.0 21.155.0.0 255.0.0.0 255.255.255.0 21.155.67.0

255.0.0.0 The default subnet mask for this address is 255.0.0.0. The address is a Class A address, which begins with a number between 1-126 in the first octet.21.0.0.0 is the subnet address (network address). 255.255.0.0 is the default subnet mask for a Class B network, and 255.255.255.0 is the default subnet mask for a Class C address.

You organization is assigned the network ID 114.0.0.0. by InterNIC. Your organization currently has 5 subnets with about 100,000 hosts per subnet. Your boss wants you to redesign the subnets into a total of 25 subnets to make each subnet more manageable. What should your subnet mask value be? 248.0.0.0 255.0.0.0 255.248.0.0 255.255.248.0

255.248.0.0

You have a network ID of 192.168.1.0 and you need to divide it into nine subnets. You need to provide for the largest possible number of host Ids per subnet. Which subnet mask should you assign? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248

255.255.255.240 Only mask 240 gives amount of subnet needed. 240 would give 4 bits, or 14 subnets each with 14 hosts

(SubnetQ25): You are given an IP address of 39.0.0.0 in which you must create 100 subnets. What will be your proposed subnet mask? 255.255.192.0 255.255.254.0 255.255.0.0 255.254.0.0

255.254.0.0 In this case you will need to borrow enough bits from the remaining three octets (host side octets) to create at least 100 subnets. Ask yourself what to the power of 2 will provide you with a value of 100. The answer is 7 or 7 bits. Hence 2^7 = 128 - 2 = 126. If you used 6 bits, it would not provide enough subnets to meet the requirement of 100. If 7 bits are required then you must use the value of the 7 bits to determine the subnet mask. That value is 254 (128 + 64 + 32 + 16 + 8 + 4 +2). Therefore, you subnet mask will be 255.254.0.0

You have a Class A network address with 60 subnets. You need to add 40 new subnets in the next 2 years, but still allow for the largest possible number of host Ids per subnet. Which subnet mask should you assign? 255.240.0.0 255.248.0.0 255.252.0.0 255.254.0.0

255.254.0.0 Only a mask of 254.0.0 would give total subnets of 100+. Acutally it gives 126 subnets. Actually 254.0.0 will give 7 bits or 126 subnets. Remaining bits can then be used for hosts which equals 131.070 hosts per subnet.

(LS IPv4 Q10): you are configuring the IP address for a host and have been asked to use the address 192.160.99.11/16. What subnet mask value would you use? 255.0.0.0 255.255.0.0 255.255.255.0 255.255.255.252

255.255.0.0 With CIDR notation, the number of bits in the subnet mask is indicated by the /16 following the IP address. A mask that uses 16 bits is written as 255.255.0.0 in decimal format. Each octet in the mask uses 8 bits, so a mask with 16 bits uses 2 full octets.

(SubnetQ10): You are given an address of 10.10.10.8/8. You need to create 3,000 subnets. What will be your new subnet mask? 255.255.192.0 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0

255.255.240.0 255.255.240.0 or /20 Ask yourself the question of what to the power of 2 gives you a value of 3,000. The answer is 2^12 = 4,096 - 2 = 4,096 subnets. 2^11 = 2,048 - 2 = 2,046 subnets which is quite enough. This is a Class A network, so you borrow 12 bits from the last three octets. 8 bits from the second octet and 4 bits from the third octet. Hence your mask would then be 255.255.240.0. The remaining 12 bits ( 4 from the third octet and 8 from the fourth octet) will form your host addresses.

You have a network ID of 172.16.0.0 with eight subnets. You need to allow for the largest possible number of host IDs per subnet. Which subnet mask should you assign? 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0

255.255.240.0 By using only eight subnets, you can use a 240.0 mask. This will give you 14 subnets each with 4,094 hosts.

A company is assigned the network ID 150.134.0.0 by InterNIC. The company wants to have 15 subnets and up to 1,000 hosts per subnet. What should be the value of the subnet mask? 255.255.0.0 255.255.5.0 255.255.31.0 255.255.248.0

255.255.248.0

(SubnetQ16): You have been provided with an IP address of 128.199.0.0 in which you are required to create 16 subnets. What will be your proposed subnet mask? 255.255.192.0 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0 255.255.255.224 255.255.255.240 255.255.255.248

255.255.248.0 With a requirement for 16 physical segments (subnets) you must ask yourself, what to the power of 2 equals 16? To get that answer you would say 2^5 = 32-2 = 62 subnets. Hence it would require 5 leading bits. If you were to use only 4 bits like 2^4 = 16-2 = 14 subnets, you would not be able to meet the requirement. Hence you must provide 32 subnets in order to meet the requirement for 16 subnets. You now would add the value of the 5 leading bits which would be 128 + 64 + 32 +16 + 8 = 248. This value then becomes the subnet mask.

You have a Class B network address divided into 30 subnets. You will add 25 new subnets within the next 2 years. You need 600 host IDS for each subnet. Which subnet mask should you assign? 255.192.0.0 255.254.0.0 255.255.248.0 255.255.252.0

255.255.252.0

You have a Class C network address divided into three subnets. You will need to add two more subnets in the next two years. Each subnet will have 25 hosts. Which subnet mask should you assign? 255.255.255.0 255.255.255.192 255.255.255.224 255.255.255.248

255.255.255.224 If you did have 255.255.255.0 you would have 1 network with 254 hosts. This is the default. 224 gives us 3 bits, or six subnets, each with 30 hosts.

You have a class C network address of 192.168.19.0 with four subnets. You need the largest possible number of host Ids per subnet. Which subnet mask should you assign? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248

255.255.255.224 Only answer B will give more than 4 subnets with largest amount of hosts. You have 3 bits with 224, or six subnets, each with 30 hosts.

(SubnetQ01): You have an IP address of 192.35.44.10/24. What subnet mask would you use to create 19 hosts per subnet? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.224.0 255.255.240.0

255.255.255.224 This is a Class C network. You will create subnets by borrowing bits from the fourth octet. However, this question does not ask you for subnets, but rather for a number of hosts. Hence you will have to leave a certain number of trailing bits remaining in the last octet to meet the requirement of 19 hosts. Ask yourself the question of what number to the power of 2 equals 19. The answer would be 2^5 = 32-2=30 hosts. So if you use 5 trailing bits for to create the hosts, you will have only 3 bits remaining for your subnets. The value of these 3 remaining leading bits forms the basis of the subnet mask. The value is 128+64+32 = 224, hence the subnet mask used will be 255.255.255.224.

You have a Class C network address of 192.168.88.0 and you need the largest possible number of subnets, with up to 12 hosts per subnet. Which subnet mask should you assign? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248

255.255.255.240 At 240 you need 4 bits which will give 14 subnets. The 4 bits remaining for host manipulation will equal to 16-2 hosts or 14 hosts per subnet.

(SubnetQ11): As the network administrator, you boss has required you to create 12 subnets by using an IP address of: 216.122.44.0. What will you use as your proposed subnet mask? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252

255.255.255.240 To create 12 physical segments or subnets, you must ask yourself the question of what to the power of 2 equals 12. You can see that 2^4 = 16-2=14 subnets. Since this requires 4 leading bits, you must calculate the value of these 4 bits in order to create the new subnet mask. So, 128 + 64 + 32 + 16 = 240. Hence the subnet mask is as noted above of 255.255.255.240

(SubnetQ02): You have an IP address of 192.100.100.70/24. You need to create 31 subnets. What will be your subnet mask? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252 255.255.255.253 255.255.252.0 255.255.253.0

255.255.255.252 This is a class C network. You will create subnets by borrowing bits from the fourth octet. Ask yourself the question of what number to the power of 2 equals 31. The answer would be 2^6 = 64-2=62 subnets. So if you use 6 leading bits for to create the subnets, the value is 128+64+32+16+8 +4 = 252, Hence the subnet mask used will be 255.255.255.252.

(SubnetQ28): What is ID network number of the first useable subnet using an IP address of 39.0.0.0/15? 39.0.0.2 39.2.0.0 39.0.2.0 39.4.0.0 39.8.0.0 39.32.0.0

39.2.0.0 The networks or subnet IDs are as shown below. You must figure out the incremental value to create these subnets. In order to do this you find the value of the last bit that was used to create the subnet mask. Or, another way to say that is to use the value of the last leading bit that was used to calculate the new subnet mask. That value is 2, which is then considered the incremental value of each subnet.* 39.0.0.0 (invalid)* 39.2.0.0* 39.4.0.0* 39.6.0.0* 39.8.0.0* &&&* 39.252.0.0* 39.254.0.0 (invalid)

(SubnetQ29): What is the network address for a host with an address of 39.7.200.200 when subnetting an IP address as 39.0.0.0/15? 39.2.0.0 39.4.2.0 39.4.0.0 39.6.0.0 39.6.2.0 39.6.200.0

39.6.0.0 39.6.0.0 Once you know the network numbers such as 39.2.0.0, 39.4.0.0 etc, you need to figure out where a host of 7.200.200 falls. This host falls between the 39.6.0.1 and 39.7.255.245, which as seen below is on the 39.6.0.0 network. Network ID..........1st Host..........Last Host 39.2.0.0...........39.2.0.1~................39.3.255.254 39.4.0.0...........39.4.0.1~................39.5.255.254 39.6.0.0...........39.6.0.1~................39.7.255.254 39.8.0.0...........39.8.0.1~................39.9.255.254 39.10.0.0.........39.10.0.1~...............39.11.255.254 &&&.. 39.252.0.0........39.252.0.1~............39.253.255.254

A company is assigned the network ID 150.134.0.0 by InterNIC. The company wants to have 15 subnets and up to 1,000 hosts per subnet. How many bits are needed for the custom subnet mask? 4 5 6 7

5

If you have a 19 bit subnet mask, how many subnets and how many hosts do you have? (You do not know if this is an A,B or C network, hence you have to make some assumptions. For example is 19 bits added to an already default mask, or is the entire mask comprised of 19 bits. Work backwards from each of the possible answers and consider various networks as A,B and C) 8190 subnets, 126 hosts 524,288 subnets, 32 hosts 524,286 subnets, 30 hosts 65,234 subnets, 62 hosts

524,286 subnets, 30 hosts

(SubnetQ22): How many subnets can an IP address of 152.77.0.0/19 support? 2 3 4 5 6 7 8

6 The number of subnets is calculated based on the number of borrowed bits. In this case you must borrow 3 bits. Hence 2^3 = 8-2 = 6 subnets

(SubnetQ31): If you were subnetting an IP address of 200.200.200.200 with a subnet mask of 255.255.255.224, how many supported subnets will you have created? 2 4 6 8 12 16

6 You calculate the number of subnets by knowing the number of bits being used for the subnet mask. In this case it is 3 bits. Hence 2^3 = 8-2 = 6 subnets.

If you have a Class C network in which you are using (borrowing) 6 bits to create a new subnet mask How many subnets and how many hosts do you now have? 254 subnets, 30 hosts 64 subnets, 8 hosts 62 subnets, 2 hosts, 30 subnets, 2 hosts

62 subnets, 2 hosts

(SubnetQ05): What subnet mask would you use if you need to plan for 1200 subnets with a given IP address of 75.75.0.0/16? 75.75.0.0/26 75.75.0.0/27 75.75.0.0/28 75.75.0.0/29 75.75.0.0/30

75.75.0.0/27 This is a class B network in which 16 bits are defaulted as the subnet mask. Since you must plan for 1200 subnets, you must ask yourself the question of what to the power of 2 equals 1200. The answer is 2^11 = 2,048. If you used 2^10, you would get 1,024, which would not be enough to create the required 1200 subnets. You must borrow 11 bits from the host octets to meet this requirement. In a Class B network, recall that the third and fourth octets make up the hosts. When creating subnets, you start from the leading octets to get the leading bits. So you would take all 8 bits from the third octet and 3 additional bits from the fourth octet to get your 11 bits. The value of the 8 bits in the third octet is 255, while the value of the 3 leading bits in the fourth octet is 224. Hence you create your subnet mask using all four octets with values as 255.255.255.224. In this case you will have created 2,046 subnets with 62 hosts on each subnet (25-2=62)

(SubnetQ23): What would be the maximum number of hosts that can be supported on each subnet if you are given the following IP address: 152.77.0.0/19 8290 8190 8192 4853 4800

8190 You calculate the number of hosts based on the remaining trailing bits. Since you must use 3 bits in the third octet for the subnets, 5 trailing bits remain. In addition to these 5 bits, you have all 8 bits remaining in the fourth octet. Hence you have a total of 13 trailing bits that can be used for host addresses. In this case that equates to 2^13 = 8,192 - 2 = 8,190 hosts available to each of the 6 subnets.

You have a subnet mask of 255.255.255.248. How many subnets and hosts do you have? 2,097,150 subnets with 8 hosts 8190 subnets with 6 hosts 30 subnets with 14 hosts 6 subnets with 30 hosts

8190 subnets with 6 hosts Only possible answer is that with 248 in the fourth octet, you can only have 6 hosts per subnet. You really can't tell much more about it being a Class A, B, or C, but only choice is for the 6 hosts.

(LS IPv4 Q02): Consider the following IP addresses: 124.77.8.5 131.11.0.9 190.66.250.10 196.5.89.44 Which list represents the IP address class of each listed IP address? Class A, Class B, Class C, Class C Class B, Class B, Class C, Class C Class C, Class B, Class C, Class C Class B, Class B, Class C, Class D Class A, Class B, Class B, Class C Class A, Class A, Class B, Class C

Class A, Class B, Class B, Class C You can identify the IP address class by memorizing the range of values for the first octet which are as follows: * Class A 0-126 * Class B 128 -191 * Class C 192 -223 * Class D 223 -239 * Class E 240 - 255