Question 3.1.1
a. 123 - 01111011
b. 202 - 11001010
c. 67 - 01000011
d. 7 - 00000111
e. 252 - 11111100
f. 91 - 01011011
g. 116.127.71.3 - 01110100.01111111.01000111.00000011
h. 255.255.255.0 – 11111111.11111111.11111111.00000000
i. 192.143.255.255 – 11000000.10001111.11111111.11111111
j. 12.101.9.16 – 00001100.01100101.00001001.00010001
Question 3.2.1
a. 1110 - 14
b. 100110 - 38
c. 11111111 - 255
d. 11010011 - 211
e. 01000001 - 65
f. 11001110 - 204
g. 01110101 - 117
h. 10001111 - 143
i. 11101001.00011011.10000000.10100100 – 233.27.128.164
j. 10101010.00110100.11100110.00010111 – 170.52.230.23
Question 3.3.1
1.
What is the decimal and binary range of the first octet of all possible Class B IP
addresses?
Decimal: From: 128
Binary:
From: 10000000
To: 191
To: 10111111
2. Which octet(s) represent the network portion of a Class C IP address? The first three
(NNNH)
3.
Which octet(s) represent the host portion of a Class A IP address? The last three
(NHHH)
4.
What is the maximum number of useable hosts with a Class C network address? 254
5.
How many Class B networks are there? 16,382
6.
How many hosts can each Class B network have? 65,534
7.
How many octets are there in an IP address? 4 How many bits per octet? 8
Question 3.3.2
Fill in the following table:
Notes
A=NNNH
B=NNHH
C=NNNH
Broadcast=255
Network=0
Host IP Address
Address
Class
Network
Address
Host
Address
Network Broadcast
Address
Default Subnet
Mask
216.14.55.137
C
216.14.55.0
137
216.14.55.255
255.255.255.0
123.1.1.15
A
123.0.0.0
1.1.15
123.255.255.255
255.0.0.0
150.127.221.244
B
150.127.0.0
221.244
150.127.255.255
255.255.0.0
194.125.35.199
C
194.125.35.0
199
194.125.35.255
255.255.255.0
175.12.239.244
B
175.12.0.0
239.244
175.12.255.255
255.255.0.0
Question 3.3.3
IP Host Address
Valid Address?
(Yes/No)
Why or Why Not
150.100.255.255
No
Class B. It is a broadcast address not a host address. (.255.255)
175.100.255.18
Yes
Class B - IP address.
195.234.253.0
No
Class C. network address (ends in .0)
100.0.0.23
Yes
Class A network address
188.258.221.176
No
The highest a decimal range can be is 255
127.34.25.189
No
Reserved for loopback and diagnostic functions
224.156.217.73
No
Reserved for multicasting.
Question 3.4.1:
Subnet No.
Subnet Bits
Borrowed
Binary Value
Subnet Bits
Decimal and
Subnet No.
Host Bits Possible
Binary Values
(Range) (5 Bits)
Subnet/Host
Decimal
Range
Use?
000
0
00000-11111
0-31
No
001
32
00000-11111
32-63
Yes
010
64
00000-11111
64-95
Yes
rd
011
96
00000-11111
96-127
Yes
th
100
128
00000-11111
128-159
Yes
th
101
160
00000-11111
160-191
Yes
th
110
192
00000-11111
192-223
Yes
th
111
224
00000-11111
224-255
No
0 Subnet
st
1 Subnet
2
nd
Subnet
3 Subnet
4 Subnet
5 Subnet
6 Subnet
7 Subnet
1. Which octet(s) represent the network portion of a Class C IP address? First 3
2. Which octet(s) represent the host portion of a Class C IP address? Last 1
3. What is the binary equivalent of the Class C network address in the scenario? 197.15.22.0
Decimal network address:
197.15.22.0
Binary network address:
11000101000011110001011000000000
4. How many high-order bits were borrowed from the host bits in the fourth octet? 3
5. What subnet mask must be used? Show the subnet mask in decimal and binary.
Decimal subnet mask:
Binary subnet mask:
255.255.255.224
11111111111111111111111111100000
6. What is the maximum number of subnets that can be created with this subnet mask? 8
7. What is the maximum number of useable subnets that can be created with this mask? 6
8. How many bits were left in the fourth octet for host IDs? 5
9. How many hosts per subnet can be defined with this subnet mask? 30
10. What is the maximum number of hosts that can be defined for all subnets with this
scenario? Assume the lowest and highest subnet numbers and the lowest and highest
host ID on each subnet cannot be used. 30x6=180
11. Is 197.15.22.63 a valid host IP address with this scenario? Why or why not? No, it is
the broadcast number.
12. Is 197.15.22.160 a valid host IP address with this scenario? Why or why not? No, it is
the subnet ID number.
13. Host A has an IP address of 197.15.22.126. Host B has an IP address of
197.15.22.129. Are these hosts on the same subnet? Why? No, 197.15.22.126 =
subnet 3 and 197.15.22.129 = subnet 4.
Question 3.5.1:
1. How many bits were borrowed from the host portion of this address? 16
2. What is the subnet mask for this network?
Dotted decimal? 255.255.255.0
Binary? 11111111111111111111111100000000
3. How many usable subnetworks are there? 2^16 -2 = 65,534
4. How many usable hosts are there per subnet? 2^8-2=254
5. What is the host range for usable subnet sixteen? 10.0.16.1-10.0.16.254
6. What is the network address for usable subnet sixteen? 10.0.16.0
7. What is the broadcast address for usable subnet sixteen? 10.0.16.255
8. What is the broadcast address for the last usable subnet? 10.255.254.255
9. What is the broadcast address for the major network? 10.255.255.255
Question 3.6.1:
1. How many subnets are needed for this network? 36+24+10=70
2. What is the minimum number of bits that can be borrowed? 7
3. What is the subnet mask for this network?
Dotted decimal? 255.255.254.0
Binary? 11111111111111111111111000000000
4. Slash format /23
5. How many usable subnetworks are there? 2^7-2=126
6. How many usable hosts are there per subnet? 2^9-2=510
Question 3.6.2:
Subnetwork #
Subnetwork ID
Host Range
Broadcast ID
172.16.2.0
1
172.16.2.1-172.16.3.254
172.16.3.255
172.16.4.0
2
172.16.4.1-172.16.5.254
172.16.5.255
172.16.6.0
3
172.16.6.1-172.16.7.254
172.16.7.255
172.16.246.0
123
172.16.246.1-172.16.247.254
172.16.247.255
172.16.248.0
124
172.16.248.1-172.16.249.254
172.16.249.255
172.16.250.0
125
172.16.250.1-172.16.251.254
172.16.251.255
172.16.252.0
126
172.16.252.1-172.16.253.254
172.16.253.255
1. What is the host range for subnet two? 172.16.4.1-172.16.5.254
2. What is the broadcast address for the 126th subnet? 176.16.253.255
3. What is the broadcast address for the major network? 176.16.255.255
Question 3.7.1:
1. How many subnets are needed for this network? 5
2. What is the subnet mask for this network?
Dotted decimal? 255.255.255.224
Binary? 11111111111111111111111111100000
3. Slash format /27
4. How many usable hosts are there per subnet? 2^5-2=30
Question 3.7.2:
Subnetwork #
Subnetwork ID
Host Range
Broadcast ID
0
192.168.1.0
192.168.1.1-192.168.1.30
192.168.1.31
1
192.168.1.32
192.168.1.32-192.168.1.62
192.168.1.63
2
192.168.1.64
192.168.1.65-192.168.1.94
192.168.1.95
3
192.168.1.96
192.168.1.97-192.168.1.126
192.168.1.127
4
192.168.1.128
192.168.1.129-192.168.1.158
192.168.1.159
5
192.168.1.160
192.168.1.161-192.168.1.190
192.168.1.191
6
192.168.1.192
192.168.1.193-192.168.1.222
192.168.1.223
7
192.168.1.224
192.168.1.225-192.168.1.254
192.168.1.255
1. What is the host range for subnet six? 192.168.1.193-192.168.1.222
2. What is the broadcast address for the 3rd subnet? 192.168.1.127
3. What is the broadcast address for the major network? 192.168.1.255