Ditka - ICDN1 - Chapter 21

Question 1

An IP subnetting design effort is under way at a company. So far, the senior engineer has decided to use Class B network 172.23.0.0. The design calls for 100 subnets, with the largest subnet needing 500 hosts. Management reauires that the design accomedate 50 percent growth in the number of subnets and the size of the largest subnet. The requirements also state that a single mask must be used throughout the Class B network. How many masks meet the requirements.

a) 0
b) 1
c) 2
d) 3+

Answer 1

A. With 50 percent growth, the mask needs to define enough subnet bits to create 150 subnets. As a result, the mask needs at least 8 subnet bits (7 subnet bits supply
27, or 128, subnets, and 8 subnet bits supply 28, or 256, subnets). Similarly, the need for 50 percent growth in the size for the largest subnet means that the host part needs enough bits to number 750 hosts/subnet. Nine host bits are not enough (29 – 2 = 510), but 10 host bits supply 1022 hosts/subnet (210 – 2 = 1022). With 16 network bits existing because of the choice to use a Class B network, the design needs a total of 34 bits (at least) in the mask (16 network, 8 subnet, 10 host), but only 32 bits exist—so no single mask meets the requirements.

Question 2

An IP subnetting design requires 200 subnets and 120 hosts/subnet for he largest subnets, and requires that a single mask be used throughout the the one private IP network that will be used. The design also requires planning for 20 percent growth in the number of subnets and number of hosts/subnet in the largest subnet. Which of the following answers lists a private IP network and mask that that if chosen, would meet the requirements?

a) 10.0.0.0/25
b) 10.0.0.0/22
c) 172.16.0.0/23
d) 192.168.7.0/24

Answer 2

B. With a growth of 20 percent, the design needs to support 240 subnets. To meet that need, 7 subnet bits do not meet the need (27 = 128), but 8 subnet bits do meet
the need (28 = 256). Similarly, the minimum number of host bits is also 8, because the need, after the 20 percent growth, would be 144 hosts/subnet. That number requires 8 host bits (28 – 2 = 254). These numbers are minimum numbers of subnet and host bits.
The right answer, 10.0.0.0/22, has 8 network bits because the network class is Class A, 14 subnet bits (/22 – 8 = 14), and 10 host bits (32 – 22 = 10). This mask supplies at
least 8 subnet bits and at least 8 host bits. The masks in the other answers either do not supply at least 8 host bits or do not supply at least 8 subnet bits.

Question 3

An engineer has planned to use Class B network 172.19.0.0 and a single subnet mask throughout the network. The answers list the masks considered by the engineer. Choose the mask that, among the answers, supplies the largest number of hosts per subnet, while also supplying enough subnet bits to support 1000 subnets.

a) 255.255.255.0
b) /26
c) 255.255.252.0
d) /28

Answer 3

B. To support 1000 subnets, 10 subnet bits (210 = 1024) are needed. The design uses a Class B network, which means that 16 network bits exist as well. So, the shortest mask that meets the requirements is 255.255.255.192, or /26, comprised of 16 network plus 10 subnet bits. The /28 answer also supplies enough subnets to meet the need, but compared to /26, /28 supplies fewer host bits and so fewer hosts/subnet.

Question 4

An engineer has calculated the list of subnet IDs, in consecutive order, for network 172.30.0.0, assuming that the /22 mask is used throughout the network. Which of the following are true? (Choose two answers.)

a) Any two consecutive subnet IDs differ by a value of 22 in the third octet.
b) Any two consecutive subnet IDs differ by a value of 16 in the fourth octet.
c) The list contains 64 subnet IDs.
d) The last subnet ID is 172.30.252.0.

Answer 4

C and D. The mask converts to 255.255.252.0, so the difference from subnet ID to subnet ID (called the magic number in this chapter) is 256 – 252 = 4. So, the subnet
IDs start with 172.30.0.0, then 172.30.4.0, then 172.30.8.0, and so on, adding 4 to the third octet. The mask, used with a Class B network, implies 6 subnet bits, for 64
total subnet IDs. The last of these, 172.30.252.0, can be recognized in part because the third octet, where the subnet bits sit, has the same value as the mask in that third octet.

Question 5

Which of the following are valid subnet IDs for the network 192.168.9.0 using mask /29, assuming that mask /29 is used throughout the network?

a) 192.168.9.144
b) 192.168.9.58
c) 192.168.9.242
d) 192.168.9.9

Answer 5

A. The first (numerically lowest) subnet ID is the same number as the classful network number, or 192.168.9.0. The remaining subnet IDs are each 8 larger than the previous subnet ID, in sequence, or 192.168.9.8, 192.168.9.16, 192.168.9.24, 192.168.9.32, and so on, through 192.168.9.248.

Question 6

Which of the following is not a valid subnet ID for network 172.19.0.0 using mask /24, assuming that mask /24 is used throughout the network?

a) 172.19.0.0
b) 172.19.1.0
c) 172.19.255.0
d) 172.19.0.16

Answer 6

D. Using mask /24 (255.255.255.0), the subnet IDs increment by 1 in the third octet. The reasoning is that with a Class B network, 16 network bits exist, and with mask /24, the next 8 bits are subnet bits, so the entire third octet contains subnet bits. All the subnet IDs will have a 0 as the last octet, because the entire fourth octet consists of host bits. Note that 172.19.0.0 (the zero subnet) and 172.19.255.0 (the broadcast subnet) might look odd but are valid subnet IDs.