Final Exam - Practice Examples Flashcards
How many host bits needed for 99 hosts?
7 bits
how many host bits needed for 120 hosts?
7 bits
How many host bits needed for 45 hosts?
6 bits
How many host bits needed for 8hosts?
4 bits
How many host bits needed for 5 hosts?
3 bits
How many host bits needed for 32 hosts?
6 bits
For Class C 1 to 6 bits may be borrowed leaving 7 down to 2 bits(Minimum 2 bits—Calculate the # of bits, then subtract 2 bits for the network ID and Broadcast. This value is # valid hosts. 2 2 - 2 = 2 valid hosts 2 3 - 2 = 6 valid hosts (8-2) 2 4 - 2 = 14 valid hosts (16-2) 2 5 - 2 = 30 valid hosts (32-2) 2 6 - 2 = 62 valid hosts (64-2) 2 7 - 2 = 126 valid hosts (128-2)
For Class C 1 to 6 bits may be borrowed leaving 7 down to 2 bits(Minimum 2 bits—Calculate the # of bits, then subtract 2 bits for the network ID and Broadcast. This value is # valid hosts. 2 2 - 2 = 2 valid hosts 2 3 - 2 = 6 valid hosts (8-2) 2 4 - 2 = 14 valid hosts (16-2) 2 5 - 2 = 30 valid hosts (32-2) 2 6 - 2 = 62 valid hosts (64-2) 2 7 - 2 = 126 valid hosts (128-2)
How many bits borrowed for 3 subnets?
2 bits
How many bits borrowed for 15 subnets
4
How many bits borrowed for 33 subnets
6
How many bits borrowed for 60 subnets
6
How many bits borrowed for 70 subnets
n/a
How many bits borrowed for 6 subnets
3 bits
(Class C can borrow 1 to 6 bits—must leave a minimum of 2 bits) 2 1 = 2 subnets 2 2 = 4 subnets 2 3 = 8 subnets 2 4 = 16 subnets 2 5 = 32 subnets 2 6 = 64 subnets
(Class C can borrow 1 to 6 bits—must leave a minimum of 2 bits) 2 1 = 2 subnets 2 2 = 4 subnets 2 3 = 8 subnets 2 4 = 16 subnets 2 5 = 32 subnets 2 6 = 64 subnets
For Class A and Class B, many more bits are available to borrow or to be remaining. The chart can
be extended to accommodate larger values.
If a company needs 1000 hosts on a network, how many bits are required? Continuing the chart above… 2 7 - 2 = 126 valid hosts (128-2) 2 8 – 2 = 254 valid hosts (256-2) 2 9 – 2 = 510 valid hosts (512-2( 2 10 – 2 = 1022 valid hosts (1024-2)
Class C– Organize your calculations for each step since a calculation may help answer more than one
question.
Using Binary
The first 3 octets will remain the same since this is a class C network to start with and the last octet
was used for creating subnets and hosts.
The /27 indicates 3 bits borrowed: 205.25.25. _ _ _| _ _ _ _ _
Showing the last octet may help: 205.25.25. 0 1 0 0 1 0 1 1
64 | 8 2 1 = 75 Network is 64—bits before the line
Using increment values
/27 means 3 bits borrowed and last bit to be borrowed was 32, so subnets are
205.25.25.0
205.25.25.32
205.25.25.64 broadcast 205.25.25.95 host range 205.25.25.65—205.25.25.94 (75 between 64 & 96)
205.25.25.96
205.25.25.128
205.25.25.160
205.25.25.192
205.25.25.224
What is the default subnet mask for 205.25.25.75/27?
First octet value is 205 which means a class C so default mask is 255.255.255.0 (default prefix is /24)
hat is the custom subnet mask for 205.25.25.75/27?
255.255.255.224
For the IP address of 205.25.25.75/27, how many bits were borrowed to create the subnet?
Default prefix is /24 and custom prefix is /27 so 3 bits were borrowed.
What is the broadcast address for the subnet (network) that 205.25.25.75/27 is a part of?
Using increment value: (Previously calculated)
- 25.25.0
- 25.25.32
- 25.25.64 broadcast 205.25.25.95 (one less than next subnet)
- 25.25.96
- 25.25.128
- 25.25.160
- 25.25.192
- 25.25.224
What is the total # of valid hosts for the subnet (network) that 205.25.25.75/27 is a part of?
3 bits borrowed, leaves remaining 5 bits, 25
-2 = 32 – 2 = 30 valid hosts
What is the subnet # for the subnet that 205.25.25.75/27 resides on?
(Previously calculated) Subnet #0--205.25.25.0 Subnet #1--205.25.25.32 Subnet #2--205.25.25.64 (205.25.25.75—75 between 64 and 96) Subnet #3--205.25.25.96 Subnet #4--205.25.25.128 Subnet #5--205.25.25.160 Subnet #6--205.25.25.192 Subnet #7--205.25.25.224
What is the network ID of 199.198.197.51/28? 199.198.197.48/28
*Using binary:
The first 3 octets will remain the same since this is a class C network to start with and the last octet
was used for creating subnets and hosts.
The /28 indicates 3 bits borrowed: 199.198.197. _ _ _ _| _ _ _ _
Showing the last octet may help: 199.198.197. 0 0 1 1| 0 0 1 1
32+16 +2+1This tells us the network is 48
*Using Increments:
/28 means borrowing 4 bits and last bit borrowed is 16 so 16 is the Increment.
Subnet #0 199.198.197.0 /28
Subnet #1 199.198.197.16/28
Subnet #2 199.198.197.32/28
Subnet #3 199.198.197.48/28 Broadcast 199.198.197.63 Host Range 199.198.197.49-199.198.197.62
Subnet #4 199.198.197.64/28
…
Subnet #15 199.198.197.240 /28
What is the default subnet mask for 199.198.197.51/28?
First octet value is 199 which means a class C so default mask is 255.255.255.0 (default prefix is /24)
What is the custom subnet mask for 199.198.197.51/28
255.255.255.240
For the IP address of 199.198.197.51/28, how many bits were borrowed to create the subnet?
Default prefix is /24 and custom prefix is /28 so 4 bits were borrowed.
What is the broadcast address for the subnet (network) that 199.198.197.51/28 is a part of?
Using increment value: /28 means borrowing 4 bits and last bit borrowed is 16 so 16 is the Increment. Subnet #0 199.198.197.0 /28 Subnet #1 199.198.197.16/28 Subnet #2 199.198.197.32/28 Subnet #3 199.198.197.48/28 ---If next network is .64 then broadcast is 199.198.197.63/28 Subnet #4 199.198.197.64/28 … Subnet #15 199.198.197.240 /28
What is the total # of valid hosts for the subnet (network) that 199.198.197.51/28 is a part of?
4 bits borrowed, leaves remaining 4 bits, 24
-2 = 16 – 2 = 14 valid hosts
What is the subnet # for the subnet that 205.25.25.75/27 resides on? Subnet#3
Previous work shows last octet with the subnet values: Subnet#0 199.198.197.0 /28 Subnet#1 199.198.197.16/28 Subnet#2 199.198.197.32/28 Subnet#3 199.198.197.48/28 Subnet#4 199.198.197.64/28 … Subnet#15 199.198.197.240 /28
What is the network ID of 166.166.5.61/24? 166.166.5.0/24
The address 166.166.5.61 with /24 is on the 5th subnet (3rd octet) all hosts are in 4th octet.
*Using binary:
The first 2 octets will remain the same since this is originally a Class B address. 8 bits were borrowed,
so the value of the 3rd octet identifies the subnet#. The entire 4th octet is zeroes for the Network ID.
The /24 indicates a total of 8 bits were borrowed: 166.166.0 0 0 0 0 1 0 1 . 0 0 0 0 0 0 0 0
Showing the last octet may help: 166.166. 4 + 1. 0 0 0 0 0 0 0 0
(3rd octet value 4 +1 tells us the subnet is 5
*Using Increments:
/24 means borrowing 8 bits and last bit borrowed is 1 so 1 is the Increment.
Subnet #0 166.166.0.0/24
Subnet #1 166.166.1.0/24
Subnet #2 166.166.2.0/24
Subnet #3 166.166.3.0/24
Subnet #4 166.166.4.0/24
Subnet #5 166.166.5.0/24 Broadcast is 166.166.5.255 Host Range 166.166.5.1-166.166.5.254
Subnet #6 166.166.6.0/24
Subnet #7 166.166.7.0/24
…
Subnet#255 166.166.255.0/24
What is the default subnet mask for 166.166.5.0/24
255.255.0.0
What is the custom subnet mask for 166.166.5.0/24
255.255.255.0
For the IP address of 166.166.5.61/24, how many bits were borrowed to create the subnet?
Default prefix is /16 and custom prefix is /24 so 8 bits were borrowed.
What is the broadcast address for the subnet (network) that 166.166.5.61/24 is a part of?
*Using Increments (Previously calculated) Subnet #0 166.166.0.0/24 Subnet #1 166.166.1.0/24 Subnet #2 166.166.2.0/24 Subnet #3 166.166.3.0/24 Subnet #4 166.166.4.0/24 Subnet #5 166.166.5.0/24 Broadcast is 166.166.5.255 Subnet #6 166.166.6.0/24 … Subnet #255 166.166.255.0/24
What is the total # of valid hosts for the subnet (network) that 166.166.5.61/24 is a part of?
8 bits borrowed, leaves remaining 8 bits, 28
-2 = 256 – 2 = 254 valid hosts
What is the subnet # for the subnet that 166.166.5.61/24 resides on? Subnet#5
Using Increments (Previously calculated) Subnet #0 166.166.0.0/24 Subnet #1 166.166.1.0/24 Subnet #2 166.166.2.0/24 Subnet #3 166.166.3.0/24 Subnet #4 166.166.4.0/24 *Subnet #5 166.166.5.0/24 Subnet #6 166.166.6.0/24 Subnet #7 166.166.7.0/24
