1
0011001100001
101010010001111100
10111001010111001011000111010011011110100011010000010100101100101001010101100111111101010100010111010011010100110010100101010101010101000110010010101001011000110101100011010110101000010110010101001101001010010
IP Addressingand
Subnetting Workbook
Version 1.5
11111110
10010101
00011011
11010011
Instructor’s Edition
10000110
1 01
IP Address Classes
Class A 1 – 127 (Network 127 is reserved for loopback and internal testing) Leading bit pattern 0 00000000.00000000.00000000.00000000
Network . Host . Host . Host
Class B 128 – 191 Leading bit pattern 10 10000000.00000000.00000000.00000000Network . Network . Host . Host
Class C 192 – 223 Leading bit pattern 110 11000000.00000000.00000000.00000000Network . Network . Network . Host
Class D 224 – 239 (Reserved for multicast)
Class E 240 – 255 (Reserved for experimental, used for research)
Private Address Space
Class A 10.0.0.0 to 10.255.255.255
Class B 172.16.0.0 to 172.31.255.255
Class C 192.168.0.0 to 192.168.255.255
Default Subnet Masks
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0
Produced by: Robb Jones
[email protected] County Career & Technology
Center Cisco Networking AcademyFrederick County Public Schools
Frederick, Maryland, USA
Special Thanks to Melvin Baker and Jim Dorsch for taking the time to check this workbook for errors,
and to everyone who has sent in suggestions to improve the series.
Workbooks included in the series:
IP Addressing and Subnetting Workbooks ACLs - Access Lists
WorkbooksVLSM Variable-Length Subnet Mask IWorkbooks
Instructors (and anyone else for that matter) please do not post the Instructors version on public websites.
When you do this you are giving everyone else worldwide the answers. Yes, students look for answers this
way.It also discourages others; myself included, from posting high quality
materials.
Inside Cover
1
00011011 27
10101010 170
01101111 111
11111000 248
00100000 32
01010101 85
00111110 62
00000011 3
11101101 237
11000000 192
Binary To Decimal Conversion
128 64 32 16 8 4 2 1 Answers Scratch Area
1 0 0 1 0 0 1 0 146 128 6416 32
0 1 1 1 0 1 1 1 119 2 16146 4
1 1 1 1 1 1 1 1 2 1
1 1 0 0 0 1 0 1 197 119
1 1 1 1 0 1 1 0 2 4 6
0 0 0 1 0 0 1 1 1 9
1 0 0 0 0 0 0 1 129
0 0 1 1 0 0 0 1 49
0 1 1 1 1 0 0 0 120
1 1 1 1 0 0 0 0 240
0 0 1 1 1 0 1 1 59
0 0 0 0 0 1 1 1 7
2
Decimal To Binary ConversionUse all 8 bits for each
problem
128 64 32 16 8 4 2 1 = 255 Scratch Area
1 1 1 0 1 1 1 0
0 0 1 0 0 0 1 0
0 1 1 1 1 0 1 1
0 0 1 1 0 0 1 0
1 1 1 1 1 1 1 1
1 1 0 0 1 0 0 0
0 0 0 0 1 0 1 0
1 0 0 0 1 0 1 0
0 0 0 0 0 0 0 1
0 0 0 0 1 1 0 1
1 1 1 1 1 0 1 0
0 1 1 0 1 0 1 1
1 1 1 0 0 0 0 0
0 1 1 1 0 0 1 0
1 1 0 0 0 0 0 0
1 0 1 0 1 1 0 0
0 1 1 0 0 1 0 0
0 1 1 1 0 1 1 1
0 0 1 1 1 0 0 1
0 1 1 0 0 0 1 0
1 0 1 1 0 0 1 1
0 0 0 0 0 0 1 0
238
34
123
50
255
200
10
138
1
13
250
107
224
114
192
172
100
119
57
98
179
2
238-128110-6446
-32 14 -8
6-42
-2 0
34-32
2-20
3
Address Class Identification
Address Class
10.250.1.1 __A___
150.10.15.0 __B___
192.14.2.0 __C___
148.17.9.1 __B___
193.42.1.1 __C___
126.8.156.0 __A___
220.200.23.1 __C___
230.230.45.58 __D___
177.100.18.4 __B___
119.18.45.0 __A___
249.240.80.78 __E___
199.155.77.56 __C___
117.89.56.45 __A___
215.45.45.0 __C___
199.200.15.0 __C___
95.0.21.90
__A___ 33.0.0.0 __A___
158.98.80.0 __B___
219.21.56.0
4
__C___
5
Network & Host Identification
Circle the network portion of these addresses:
177.100.18.4
119.18.45.0
209.240.80.78
199.155.77.56
117.89.56.45
215.45.45.0
192.200.15.0
95.0.21.90
33.0.0.0
158.98.80.0
217.21.56.0
10.250.1.1
150.10.15.0
192.14.2.0
148.17.9.1
193.42.1.1
126.8.156.0
220.200.23.1
Circle the host portion of these addresses:
10.15.123.50
171.2.199.31
198.125.87.177
223.250.200.222
17.45.222.45
126.201.54.231
191.41.35.112
155.25.169.227
192.15.155.2
123.102.45.254
148.17.9.155
100.25.1.1
195.0.21.98
25.250.135.46
171.102.77.77
55.250.5.5
218.155.230.14
10.250.1.1
6
Network Addresses
Using the IP address and subnet mask shown write out the network address:
188.10.18.2 ______1_8_8__.__1_0__._0__.__0_________255.255.0.0
10.10.48.80 ______1_0__.__1_0__._4__8__. _0_________255.255.255.0
192.149.24.191 ______1_9_2___. _1_4__9__. _2_4___.
_0______255.255.255.0
150.203.23.19 ______1_5__0__._2__0_3__.__0__. _0_______255.255.0.0
10.10.10.10 ______1_0__.__0__. _0__.__0___________255.0.0.0
186.13.23.110
______1_8_6__.__1_3__._2__3__. _0________ 255.255.255.0
223.69.230.250 ______2_2__3__._6__9__. _0__._0_________255.255.0.0
200.120.135.15 ______2_0__0__._1__2_0__.__1_3_5___.
_0____255.255.255.0
27.125.200.151 ______2_7__.__0__._0__.__0___________255.0.0.0
199.20.150.35 ______1_9_9__.__2_0___. _1_5__0__.
_0______255.255.255.0
7
191.55.165.135 ______1_9_1__.__5_5___. _1_6_5___.
_0______255.255.255.0
28.212.250.254 ______2_8__.__2_1_2___. _0__._0_________255.255.0.0
8
Host Addresses
Using the IP address and subnet mask shown write out the host address:
188.10.18.2 ______0__._0___. _1_8__._2____________255.255.0.0
10.10.48.80 ______0__._0___. _0__._8__0___________255.255.255.0
222.49.49.11 ______0__._0___. _0__._1_1____________255.255.255.0
128.23.230.19 ______0__._0___.
_2_3__0__._1_9_________255.255.0.0
10.10.10.10 ______0__._1__0__. _1_0__.__1_0_________255.0.0.0
200.113.123.11 ______0__._0___.
_0__._1_1____________ 255.255.255.0
223.169.23.20 ______0__._0___. _2_3___. _2_0__________255.255.0.0
203.20.35.215 ______0__._0___. _0__._2__1_5__________255.255.255.0
117.15.2.51 ______0__._1__5__._2__.__5_1__________255.0.0.0
199.120.15.135 ______0__._0___. _0__._1_3__5__________255.255.255.0
191.55.165.135 ______0__._0___. _0__._1_3__5__________255.255.255.0
48.21.25.54 ______0__._0___. _2_5___.
9
_5__4_________255.255.0.0
10
Default Subnet Masks
Write the correct default subnet mask for each of the following addresses:
177.100.18.4 ______2_5__5__.__2_5__5__._0___.
_0______
119.18.45.0 _______2__5_5___. _0__.__0__. _0________
191.249.234.191
______2_5__5__._2__5__5__._0__.__0______
223.23.223.109 ____2__5_5___.
_2_5__5__.__2_5__5__._0_____
10.10.250.1 _______2__5_5___. _0__.__0__. _0________
126.123.23.1 _______2__5_5___. _0__.__0__. _0________
223.69.230.250 ____2__5_5___.
_2_5__5__.__2_5__5__._0_____
192.12.35.105 ____2__5_5___.
_2_5__5__.__2_5__5__._0_____
77.251.200.51 _______2__5_5___. _0__.__0__. _0________
189.210.50.1
______2_5__5__._2__5__5__._0__.__0______
88.45.65.35 _______2__5_5___. _0__.__0__. _0________
128.212.250.254
______2_5__5__._2__5__5__._0__.__0______
193.100.77.83 ____2__5_5___.
11
_2_5__5__.__2_5__5__._0_____
125.125.250.1 _______2__5_5___. _0__.__0__. _0________
1.1.10.50 _______2__5_5___. _0__.__0__. _0________
220.90.130.45 ____2__5_5___.
_2_5__5__.__2_5__5__._0_____
134.125.34.9
______2_5__5__._2__5__5__._0__.__0______
95.250.91.99 _______2__5_5___. _0__.__0__. _0________
12
ANDING With Default subnet masks
Every IP address must be accompanied by a subnet mask. By now you should be able to look at an IP address and tell what class it is. Unfortunately your computer doesn’t think that way. For your computer to determine the network and subnet portion of an IP address it must“AND” the IP address with the subnet mask.
Default Subnet Masks:Class A 255.0.0.0Class B 255.255.0.0Class C 255.255.255.0
ANDING Equations:1 AND 1 = 11 AND 0 = 00 AND 1 = 00 AND 0 = 0
Sample:
What you see...
IP Address: 192 . 100 . 10 . 33
What you can figure out in your head...
Address Class: CNetwork Portion: 192 . 100 . 10 . 33Host Portion: 192 . 100 . 10 . 33
In order for you computer to get the same information it must AND the IP address with the subnet mask in binary.
Network Host
IP Address: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1
(192 . 100 . 10 . 33)
Default Subnet Mask: 1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 0 0 0 0 0 0 0 0
(255 . 255 . 255 . 0)
AND: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 0 0 0 0 0 0
(192 . 100 . 10 . 0)
ANDING with the default subnet mask allows your computer to figure out the network portion of the address.
13
ANDING With Custom subnet masks
When you take a single network such as 192.100.10.0 and divide it into five smaller networks (192.100.10.16, 192.100.10.32, 192.100.10.48, 192.100.10.64, 192.100.10.80) the outside world still sees the network as 192.100.10.0, but the internal computers and routers see five smaller subnetworks. Each independent of the other. This can only be accomplished by using a custom subnet mask. A custom subnet mask borrows bits from the host portion of the address to create a subnetwork address between the network and host portions of an IP address. In this example each range has 14 usable addresses in it. The computer must still AND the IP address against the custom subnet mask to see what the network portion is and which subnetwork it belongs to.
IP Address: 192 . 100 . 10 . 0Custom Subnet Mask: 255.255.255.240
Address Ranges: 192.10.10.0 to 192.100.10.15 192.100.10.16 to 192.100.10.31192.100.10.32 to 192.100.10.47 (Range in the sample below)192.100.10.48 to 192.100.10.63192.100.10.64 to 192.100.10.79192.100.10.80 to 192.100.10.95192.100.10.96 to 192.100.10.111192.100.10.112 to 192.100.10.127192.100.10.128 to 192.100.10.143192.100.10.144 to 192.100.10.159192.100.10.160 to 192.100.10.175192.100.10.176 to 192.100.10.191192.100.10.192 to 192.100.10.207192.100.10.208 to 192.100.10.223192.100.10.224 to 192.100.10.239192.100.10.240 to 192.100.10.255
NetworkSub
Network Host
IP Address: Custom Subnet Mask:
AND:
1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1 (192 . 100 . 10 .
33)
1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 0 (255 . 255 . 255 . 240)
1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 0 (192 . 100 . 10 .
32)
Four bits borrowed from the host portion of the address for the custom subnet mask.
The ANDING process of the four borrowed bits shows which range of IP addresses this particular address will fall into.
14
In the next set of problems you will determine the necessary information to determine the correct subnet mask for a variety of IP addresses.
15
How to determine the number of subnets and the number of hosts per subnet
Two formulas can provide this basic information:
Number of subnets = 2 s (Second subnet formula: Number of subnets = 2s - 2)
Number of hosts per subnet = 2 h - 2
Both formulas calculate the number of hosts or subnets based on the number of binary bits used. For example if you borrow three bits from the host portion of the address use the number of subnets formula to determine the total number of subnets gained by borrowing the three bits. This would be 23 or 2 x 2 x 2 = 8 subnets
To determine the number of hosts per subnet you would take the number of binary bits used in the host portion and apply this to the number of hosts per subnet formula If five bits are in the host portion of the address this would be 25 or 2 x 2 x 2 x 2 x 2 = 32 hosts.
When dealing with the number of hosts per subnet you have to subtract two addresses from the range. The first address in every range is the subnet number. The last address in every range is the broadcast address. These two addresses cannot be assigned to any device in the network which is why you have to subtract two addresses to find the number of usable addresses in each range.
For example if two bits are borrowed for the network portion of the address you can easily determine the number of subnets and hosts per subnets using the two formulas.
195. 223 . 50 . 0 0 0 0 0 0 0 0
The number of subnets created by borrowing 2 bits is 22 or 2 x 2 = 4 subnets.
The number of hosts created by leaving 6 bits is 26 - 2 or2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62usable hosts per subnet.
What about that second subnet formula:
Number of subnets = 2 s
- 2
In some instances the first and last subnet range of addresses are reserved. This is similar to the first and last host addresses in each range of addreses.
The first range of addresses is the zero subnet. The subnet number for the zero subnet is also the subnet number for the classful subnet address.
The last range of addresses is the broadcast subnet. The broadcast address for the last subnet in the broadcast subnet is the same as the classful broadcast address.
16
Class C Address unsubnetted:
195. 223 . 50 . 0
195.223.50.0 to 195.223.50.255
Class C Address subnetted (2 bits borrowed):
Notice that the subnet and broadcast addresses match.
195. 223 . 50 . 0 0 0 0 0 0 0 0
(Invalid range)
(Invalid range)
(0)(1)(2)(3)
195.223.50.0 to195.223.50.64 to195.223.50.128 to195.223.50.192 to
195.223.50.63195.223.50.127195.223.50.191195.223.50.255
The primary reason the the zero and broadcast subnets were not used had to do pirmarily with the broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255 addresses in the classful C address or just the 62 usable addresses in the broadcast range?
The CCNA and CCENT certification exams may have questions which will require you to determine which formula to use, and whehter or not you can use the first and last subnets. Use the chart below to help decide.
When to use which formula to determine the number of subnets
Use the 2s
- 2 formula and don’t use the zero and broadcast ranges if...
Use the 2s
formula and use the zero and broadcast ranges if...
Classful routing is used Classless routing or VLSM is used
RIP version 1 is used RIP version 2, EIGRP, or OSPF is used
The no ip subnet zero command is configured on your router
The ip subnet zero command is configured on your router (default setting)No other clues are given
Bottom line for the CCNA exams; if a question does not give you any clues as to whether or not to allow these two subnets, assume you can use them.
This workbook has you use the number of subnets = 2s
formula.
17
Problem 1
Custom Subnet Masks
Number of needed subnets 14Number of needed usable hosts 14
Network Address 192.10.10.0
Address class ____C______
Default subnet mask ____2__5__5__._2__5__5__._2__5_5___. _0______
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._2__4_0___
Total number of subnets _________1_6_________
Total number of host addresses _________1_6_________
Number of usable addresses _________1_4_________
Number of bits borrowed _________4__________
Show your work for Problem 1 in the space below.
Number of256 128 64 32 16 8 4 2 - Hosts
Number ofSubnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
192 . 10 . 10 . 0 0 0 0 0 0 0 0
Add the binary value numbers to the left of the line to create the custom subnet mask.
1286432
+16
240
16
18
Observe the total number of hosts.
-2 Subtract 2 for the
number of
14 usable hosts.
19
Problem 2
Custom Subnet Masks
Number of needed subnets 1000Number of needed usable hosts 60
Network Address 165.100.0.0
Address class ____B______
Default subnet mask ____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._1_9__2___
Total number of subnets _______1_,0__2_4________
Total number of host addresses _________6_4_________
Number of usable addresses _________6_2_________
Number of bits borrowed _________1_0_________
Show your work for Problem 2 in the space below.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
Add the binary value numbers to the left of the line to create the custom subnet mask.
128643216
842
+1 128+64 192
64
20
Observe thetotal
number
of hosts.
-2 Subtract 2 for the number of
62 usable hosts.
255
1286432
Add the binary value 16numbers to the left of the line to 8create the custom subnet mask. 4
2+1
64
21
Problem 3
Custom Subnet Masks
Network Address 148.75.0.0 /26
Address class _____B_____
/26 indicates the total number of bits used for the network and subnetwork portion of the address. All bits remaining belong to the host portion of the address.
Default subnet mask _____2_5__5__.__2_5__5__._0___.
_0_________
Custom subnet mask _____2_5__5__.__2_5__5__.__2_5__5__._1__9_2___
Total number of subnets _______1_,0__2_4________
Total number of host addresses _________6_4_________
Number of usable addresses _________6_2_________
Number of bits borrowed _________1_0_________
Show your work for Problem 3 in the space below.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128+64
192 Observe the total number ofhosts.
-2 Subtract 2 for the number of
62 usable hosts.
22
255 1024-2
1,022
Subtract 2 for the total number of subnets to get the usable number of subnets.
23
Problem 4
Custom Subnet Masks
Number of needed subnets 6Number of needed usable hosts 30
Network Address 210.100.56.0
Address class ____C___
Default subnet mask ____2__5_5___. _2__5_5___.
_2_5__5___. _0______ Custom subnet mask
____2__5_5___. _2__5_5___. _2_5__5___.
_2_2__4___
Total number of subnets
________8___________ Total number of host
addresses ________3__2_________ Number of
usable addresses ________3__0_________
Number of bits borrowed ________3___________
Show your work for Problem 4 in the space below.
Number of256 128 64 32 16 8 4 2 - Hosts
Number ofSubnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
210 . 100 . 56 . 0 0 0 0 0 0 0 0
12864 8
+3 2 -2224 6
32 -2 30
24
Problem 5
Custom Subnet Masks
Number of needed subnets 6Number of needed usable hosts 30
Network Address 195.85.8.0
Address class ____C___
Default subnet mask ____2__5_5___. _2__5_5___.
_2_5__5___. _0______ Custom subnet mask
____2__5_5___. _2__5_5___. _2_5__5___.
_2_2__4___
Total number of subnets
________8___________ Total number of host
addresses ________3__2_________ Number of
usable addresses ________3__0_________
Number of bits borrowed ________3___________
Show your work for Problem 5 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
195 . 85 . 8 . 0 0 0 0 0 0 0 0
25
12864
+32224
32 -2 30
8 -2
6
26
Problem 6
Custom Subnet Masks
Number of needed subnets 126Number of needed usable hosts 131,070
Network Address 118.0.0.0
Address class ___A____
Default subnet mask ____2_5__5__._0___.
_0__.__0_____________ Custom subnet mask
____2_5__5__._2__5__4_._0___. _0___________
Total number of subnets
________1_2__8________ Total number of host
addresses ________1_3_1__,0__7_2____ Number of
usable addresses ________1_3_1__,0__7_0____
Number of bits borrowed ________7___________
Show your work for Problem 6 in the space below.
Number of
Hosts -. 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256 . .
Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1286432
27
1684
+2 254
128 -2 126
131,072 - 2 131,070
18 255
Problem 7
Custom Subnet Masks
Number of needed subnets 2000Number of needed usable hosts 15
Network Address 178.100.0.0
Address class _____B_____
Default subnet mask
_____2_5__5__.__2_5__5__._0___. _0_________
Custom subnet mask
___
__2_5__5__.__2_5__5__._2__5__5__._2__2_4___
Total number of subnets
_________2_,_0_4__8_____ Total number of host
addresses _________3_2_________ Number of
usable addresses _________3_0_________ Number
of bits borrowed _________1_1_________
Show your work for Problem 7 in the space below.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
29
128643216
842
+1
2,048 - 2 2,046
32 -2 30
18 255
Problem 8
Custom Subnet Masks
Number of needed subnets 3Number of needed usable hosts 45
Network Address 200.175.14.0
Address class ___C____
Default subnet mask
____2_5__5__._2__5__5__._2__5__5__._0_______
Custom subnet mask
__
__2_5__5__._2__5__5__._2__5__5__._1__9_2___
_
Total number of subnets
________4___________ Total number of host
addresses ________6_4__________ Number of
usable addresses ________6_2__________ Number
of bits borrowed ________2___________
Show your work for Problem 8 in the space below.
Number of256 128 64 32 16 8 4 2 - Hosts
Number ofSubnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
200 . 175 . 14 . 0 0 0 0 0 0 0 0
31
128 4 +6 4 -2240 2
64 -2 62
20
Problem 9
Custom Subnet Masks
Number of needed subnets 60Number of needed usable hosts 1,000
Network Address 128.77.0.0
Address class ____B___
Default subnet mask
____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask
____2__5__5__._2__5__5__._2__5_2___. _0______
Total number of subnets
_________6_4_________ Total number of host
addresses _________1_,0__2_4______ Number of
usable addresses _________1_,0__2_2______
Number of bits borrowed _________6__________
Show your work for Problem 9 in the space below.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
192 62 221
128643216
8 +4 252
64 1,024- 2 - 2 62 1,022
22
Problem 10
Custom Subnet Masks
Number of needed usable hosts 60Network Address 198.100.10.0
Address class ___C____
Default subnet mask
____2_5__5__._2__5__5__._2__5__5__._0_______
Custom subnet mask
__
__2_5__5__._2__5__5__._2__5__5__._1__9_2____
Total number of subnets
________4___________ Total number of host
addresses ________6_4__________ Number of
usable addresses ________6_2__________ Number
of bits borrowed ________2___________
Show your work for Problem 1 0 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
198 . 100 . 10 . 0 0 0 0 0 0 0 0
192 62 223
128 64 4+6 4 - 2 -2
24
Problem 1 1
Custom Subnet Masks
Number of needed subnets 250Network Address 101.0.0.0
Address class ___A____
Default subnet mask
___2__5_5___._0__.__0__._0______________ Custom
subnet mask ___2__5_5___._2__5_5___.
_0__.__0__________
Total number of subnets
________2_5__6________ Total number of host
addresses ________6_5__,5__3_6_____ Number of
usable addresses ________6_5__,5__3_4_____
Number of bits borrowed ________8___________
Show your work for Problem 1 1 in the space below.
Number of
Hosts -. 256 128 64 32 16 8 4 2
Number of
Subnets - 2 4 8 16 32 64 128 256 . .Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1286432
1684
2 +1 255
192 62 225
256 -2 254
65,536 - 65,534
224 62 223
Problem 12
Custom Subnet Masks
Number of needed subnets 5Network Address 218.35.50.0
Address class
___C____
Default subnet mask
____2_5__5__._2__5__5__._2__5__5__._0_______
Custom subnet mask
__
__2_5__5__._2__5__5__._2__5__5__._2__2_4___
_
Total number of subnets
________8___________ Total number of host
addresses ________3_2__________ Number of
usable addresses ________3_0__________ Number
of bits borrowed ________3___________
Show your work for Problem 1 2 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
218 . 35 . 50 . 0 0 0 0 0 0 0 0
24
12864
+3264 4 - 2 -2
224 62 225
Problem 13
Custom Subnet Masks
Number of needed usable hosts 25Network Address 218.35.50.0
Address class
___C____
Default subnet mask ___2__5_5___._2__5_5___.
_2__5_5___. _0_______ Custom subnet mask
___2__5_5___._2__5_5___. _2__5_5___.
_2__2_4____
Total number of subnets
________8___________ Total number of host
addresses ________3_2__________ Number of
usable addresses ________3_0__________ Number
of bits borrowed ________3___________
Show your work for Problem 1 3 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
218 . 35 . 50 . 0 0 0 0 0 0 0 0
26
12864 8
+3 2 -2224 6
32 -2 30
25
Problem 14
Custom Subnet Masks
Number of needed subnets 10Network Address 172.59.0.0
Address class
___B____
Default subnet mask
____2_5__5__._2__5__5__._0___. _0__________
Custom subnet mask
____2_5__5__._2__5__5__._2__4_0___. _0_______
Total number of subnets
________1_6__________ Total number of host
addresses ________4_,_0_9__6______ Number of
usable addresses ________4_,_0_9__4______
Number of bits borrowed ________4___________
Show your work for Problem 1 4 in the space below.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
26
1286432
+16 240
16 -2 14
4,096 -2 4,094
27
Problem 15
Custom Subnet Masks
Number of needed usable hosts 50Network Address 172.59.0.0
Address class
___B____
Default subnet mask ___2__5_5___._2__5_5___.
_0__.__0__________ Custom subnet mask
___2__5_5___._2__5_5___. _2__5_5___.
__1_9_2____
Total number of subnets
________1_,0__2_4_______ Total number of host
addresses ________6_4__________ Number of
usable addresses ________6_2__________ Number
of bits borrowed ________1_0__________
Show your work for Problem 1 5 in the space below.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 6432
168
28
42
+1 255
128 64
+6 4 -2192 62
1,024 - 2 1,022
29
Problem 16
Custom Subnet Masks
Number of needed usable hosts 29Network Address 23.0.0.0
Address class
___A____
Default subnet mask ____2_5__5__._0___.
_0__.__0_____________ Custom subnet mask
__
__2_5__5__._2__5__5__._2__5__5__._2__2_4___
_
Total number of subnets
________5_2__4__,2__8_8___ Total number of host
addresses ________3_2__________ Number of
usable addresses ________3_0__________ Number
of bits borrowed ________1_9__________
Show your work for Problem 1 6 in the space below.
Number of
Hosts -. 256 128 64 32 16 8 4 2
Number of
Subnets - 2 4 8 16 32 64 128 256 . .
Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
23 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
30
12864
+32224
32 524,288- 2 - 2 30 524,286
31
Problem 1
Subnetting
Number of needed subnets 14Number of needed usable hosts 14
Network Address 192.10.10.0
Address class ____C______
Default subnet mask ____2__5__5__._2__5__5__._2__5_5___. _0______
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._2__4_0___
Total number of subnets _________1_6_________
Total number of host addresses _________1_6_________
Number of usable addresses _________1_4_________
Number of bits borrowed _________4__________
What is the 4thsubnet range? 192.10.10.48 to 192.10.10.63
What is the subnet number for the 8th subnet?
What is the subnet broadcast address for
the 13th subnet?
___1_9_2___. _1_0__.__1_0__._1_1__2___
___1_9_2___. _1_0__.__1_0__._2__0_7___
32
What are the assignable addresses for the 9th
subnet? _1_9__2_._1_0_._1_0__.1_2__9___t_o___1_9_2__.1_0__.1_0__.1__4_2___
33
Show your work for Problem 1 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
192. 10 . 10 . 0 0 0 0 0 0 0 0
(0) 0 0 0 0 192.10.10.0 to 192.10.10.15(1) 0 0 0 1 192.10.10.16 to 192.10.10.31(2) 0 0 1 0 192.10.10.32 to 192.10.10.47(3) 0 0 1 1 192.10.10.48 to 192.10.10.63(4) 0 1 0 0 192.10.10.64 to 192.10.10.79(5) 0 1 0 1 192.10.10.80 to 192.10.10.95(6) 0 1 1 0 192.10.10.96 to 192.10.10.111(7) 0 1 1 1 192.10.10.112 to 192.10.10.127(8) 1 0 0 0 192.10.10.128 to 192.10.10.143(9) 1 0 0 1 192.10.10.144 to 192.10.10.159(10)
1 0 1 0 192.10.10.160 to 192.10.10.175(11) 1 0 1 1 192.10.10.176 to 192.10.10.191(12)
1 1 0 0 192.10.10.192 to 192.10.10.207(13)
1 1 0 1 192.10.10.208 to 192.10.10.223(14)
1 1 1 0 192.10.10.224 to 192.10.10.239(15)
1 1 1 1 192.10.10.240 to 192.10.10.255
Custom subnet mask
1286432
+16240
16
Usable subnets -2
14
16
Usable hosts -2
14
The binary value of the last bit borrowed is the range. In this problem the range is 16.
The first address in each subnet range is the subnet number.
The last address in each subnet range is the subnet broadcast address.
34
Problem 2
Subnetting
Number of needed subnets 1000Number of needed usable hosts 60
Network Address 165.100.0.0
Address class ____B______
Default subnet mask ____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask ____2__5__5__._2__5__5__._2__5_5___._1_9__2___
Total number of subnets _______1_,0__2_4________
Total number of host addresses _________6_4_________
Number of usable addresses _________6_2_________
Number of bits borrowed _________1_0_________
What is the 15thsubnet range? 165.100.3.128 to 165.100.3.191
What is the subnet number for the 6th subnet?
What is the subnet broadcast address for
the 6th subnet?
___1_6_5___. _1_0__0__._1__._6__4____
___1_6_5___. _1_0__0__._1__._1__2_7___
What are the assignable addresses for the 9th
35
subnet? _1_6__5_._1_0__0_._2_._1__t_o___1_6__5_._1_0_0__.0__.6__2______
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) . 0 165.100.0.0 to 165.100.0.63
64 128-2 6462 32
16
1 01
01
1
1 . 01 . 01 . 1 01 . 1 1
1 1 1 1 1 1 1 . 1 0 165.100.255.128 to 165.100.255.1911 1 1 1 1 1 1 . 1 1 165.100.255.192 to 165.100.255.255
Sh
ow
you
rw
ork
for
Pro
blem
2in
the
space
belo
w.
31
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Usable hosts
Custom subnet mask
128 8
+644
192+1
(1) 1
(2)(3)(4)(5)(6)(7)(8) 1 0 . 0 0
165.100.0.64 to 165.100.0.127165.100.0.128 to 165.100.0.191165.100.0.192 to 165.100.0.255
165.100.1.0 to 165.100.1.63165.100.1.64 to 165.100.1.127165.100.1.128 to 165.100.1.191165.100.1.192 to 165.100.1.255
165.100.2.0 to 165.100.0.63The binary value of the last bit borrowed is the range. In this problem the range is 64.
The first address in each subnet range is the subnet number.
The last address in each subnet range is the subnet broadcast address.
(1022) 1(1023) 1
255(9) 1 0 . 0 1(10) 1 0 . 1 0(11) 1 0 . 1 1(12) 1 1 . 0 0(13) 1 1 . 0 1(14) 1 1 . 1 0(15) 1 1 . 1 1
165.100.2.64 to 165.100.0.127165.100.2.128 to 165.100.0.191165.100.2.192 to 165.100.0.255
165.100.3.0 to 165.100.3.63165.100.3.64 to 165.100.3.127165.100.3.128 to 165.100.3.191165.100.3.192 to 165.100.3.255
Down to
32
Problem 3
Subnetting
Number of needed subnets 2Network Address 195.223.50.0
Address class ____C______
Default subnet mask
____2__5__5__._2__5__5__._2__5_5___.
_0______ Custom subnet mask
__
__2__5__5__._2__5__5__._2__5_5___._1_9__2__
_
Total number of subnets
_________4__________ Total number of host
addresses _________6_4_________ Number of
usable addresses _________6_2_________ Number
of bits borrowed _________2__________
What is the 3rdsubnet range? 195.223.50.128 - 195.223.50.191
What is the subnet number for the 2nd subnet?
What is the subnet broadcast address for
the 1st subnet?
_____1_9_5__._2_2__3_._5_0__.6__4____
___ __1_9_5__._2_2__3_._5_0__.6__3____
What are the assignable addresses for the 3rd
subnet? _1_9__5_._2_2__3_._5_0__.1_2__9__-___1_9_5__.2__2_3__.5__0_._1_9_0_
34
Show your work for Problem 3 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
195. 223 . 50 . 0 0 0 0 0 0 0 0
(0) 0 195.223.50.0 to 195.223.50.63(1) 1 195.223.50.64 to 195.223.50.127(2) 1
0 195.223.50.128 to 195.223.50.191(3) 1 1 195.223.50.192 to 195.223.50.255
128 +64 192 64
-2 62
33
34
Problem 4
Subnetting
Number of needed subnets 750Network Address 190.35.0.0
Address class ____B______
Default subnet mask
____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask
__
__2__5__5__._2__5__5__._2__5_5___._1_9__2__
_
Total number of subnets
_________1_,0__2_4______ Total number of host
addresses _________6_4_________ Number of
usable addresses _________6_2_________ Number
of bits borrowed _________1_0_________
What is the 15thsubnet range? 190.35.3.128 to 190.35.3.191
What is the subnet number for the 13th subnet?
What is the subnet broadcast address for
the 10th subnet?
What are the assignable addresses for the 6th subnet?
______1_9__0_._3_5__.3__.0________
______1_9__0_._3_5__.2__.1_2__7_____
__1__9_0_._3_5__.1_._6
_5____t_o___1_9__0_.3__5_._1_._1_2_6_____
(0) . 0 190.35.0.0 to 190.35.0.63(1) 1 190.35.0.64 to 190.35.0.127(2) 1 0 190.35.0.128 to 190.35.0.191(3) 1 1 190.35.0.192 to 190.35.0.255(4) 1 . 0 0 190.35.1.0 to 190.35.1.63(5) 1 . 0 1 190.35.1.64 to 190.35.1.127(6) 1 . 1 0 190.35.1.128 to 190.35.1.191(7) 1 . 1 1 190.35.1.192 to 190.35.1.255(8) 1 0 . 0 0 190.35.2.0 to 190.35.2.63(9) 1 0 . 0 1 190.35.2.64 to 190.35.2.127(10)
1 0 . 1 0 190.35.2.128 to 190.35.2.191(11) 1 0 . 1 1 190.35.2.192 to 190.35.2.255(12)
1 1 . 0 0 190.35.3.0 to 190.35.3.63(13)
1 1 . 0 1 190.35.3.64 to 190.35.3.127(14)
1 1 . 1 0 190.35.3.128 to 190.35.3.191(15)
1 1 . 1 1 190.35.3.192 to 190.35.3.255
Sh
ow
you
rw
ork
for
Pro
blem
4in
the
space
belo
w.
35
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
190 . 35 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128643216
842
+1252
64 128
-2 +6462 252
36
Problem 5
Subnetting
Number of needed usable hosts 6Network Address 126.0.0.0
Address class ____A______
Default subnet mask ____2__5__5__._0___.
_0__._0_____________ Custom subnet mask
____2__5__5__._2__5__5__._2__5_5___.
_2__4_8___
Total number of subnets
_________2_,_0_9__7_,1_5__2_ Total number of
host addresses _________8_______
Number of usable addresses
_________6__________ Number of bits
borrowed _________2_1_________
What is the 2ndsubnet range? 126.0.0.8 to 126.0.0.15
What is the subnet number for the 5th subnet?
What is the subnet broadcast address for
the 7th subnet?
_____1_2__6_._0_._0_._3_2_________
_____1_2__6_._0_._0_._5_5_________
What are the assignable addresses for the 10th
subnet? ______1_2_6__.0__.0__.7_3____t_o___1_2__6_.0__.0__.7__8_____
(0) 0 126.0.0.0 to 126.0.0.7(1) 1 126.0.0.8 to 126.0.0.15(2) 1 0 126.0.0.16 to 126.0.0.23(3) 1 1 126.0.0.24 to 126.0.0.31(4) 1 0 0 126.0.0.32 to 126.0.0.39(5) 1 0 1 126.0.0.40 to 126.0.0.47(6) 1 1 0 126.0.0.48 to 126.0.0.55(7) 1 1 1 126.0.0.56 to 126.0.0.63(8) 1 0 0 0 126.0.0.64 to 126.0.0.71(9) 1 0 0 1 126.0.0.72 to 126.0.0.79(10)
1 0 1 0 126.0.0.80 to 126.0.0.87(11) 1 0 1 1 126.0.0.88 to 126.0.0.95(12)
1 1 0 0 126.0.0.96 to 126.0.0.103(13)
1 1 0 1 126.0.0.104 to 126.0.0.111(14)
1 1 1 0 126.0.0.112 to 126.0.0.119(15)
1 1 1 1 126.0.0.120 to 126.0.0.127
Sh
ow
you
rw
ork
for
Pro
blem
5in
the
space
belo
w.
37
Number of
Hosts -. 256 128 64 32 16 8 4 2
Number of
Subnets - 2 4 8 16 32 64 128 256 . .
Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
126. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128643216
842
+1 255
8-26
12864
3216+8
248
38
Problem 6
Subnetting
Number of needed subnets 10Network Address 192.70.10.0
Address class
____C______
Default subnet mask
____2__5__5__._2__5__5__._2__5_5___.
_0______ Custom subnet mask
__
__2__5__5__._2__5__5__._2__5_5___._2__4_0__
_
Total number of subnets
_________1_6_________ Total number of host
addresses _________1_6_________ Number of usable
addresses _________1_4_________
Number of bits borrowed _________4__________
What is the 9th subnet range?
What is the subnet number for the 4th subnet?
What is the subnet broadcast address for
the 12th subnet?
192.70.10.128 to 192.70.10.143
____1_9_2__.7__0_._1_0_._4__8_______
____1_9_2__.7__0_._1_0__.1_9__1______
What are the assignable addresses for the 10th
subnet? _1_9__2_._7_0_._1_0_._1_4_5_____t_o__1__9_2_._7_0__.1_0__.1_5__8__
40
Show your work for Problem 6 in the space below.Number of
256 128 64 32 16 8 4 2 - HostsNumber of
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
192 . 70 . 10 . 0 0 0 0 0 0 0 0
(0) 0 192.70.10.0 to 192.70.10.15(1) 1 192.70.10.16 to 192.70.10.31(2) 1 0 192.70.10.32 to 192.70.10.47(3) 1 1 192.70.10.48 to 192.70.10.63(4) 1 0 0 192.70.10.64 to 192.70.10.79(5) 1 0 1 192.70.10.80 to 192.70.10.95(6) 1 1 0 192.70.10.96 to 192.70.10.111(7) 1 1 1 192.70.10.112 to 192.70.10.127(8) 1 0 0 0 192.70.10.128 to 192.70.10.143(9) 1 0 0 1 192.70.10.144 to 192.70.10.159(10)
1 0 1 0 192.70.10.160 to 192.70.10.175(11) 1 0 1 1 192.70.10.176 to 192.70.10.191(12)
1 1 0 0 192.70.10.192 to 192.70.10.0207(13)
1 1 0 1 192.70.10.208 to 192.70.10.223(14)
1 1 1 0 192.70.10.224 to 192.70.10.239(15)
1 1 1 1 192.70.10.240 to 192.70.10.255
128 +64 240
16 -2 14
39
40
Problem 7
Subnetting
Network Address 10.0.0.0 /16
Address class ____A______
Default subnet mask ____2__5__5__._0___. _0__._0_____________
Custom subnet mask ____2__5__5__._2__5__5__._0__.__0_________
Total number of subnets _________2_5__6_______
Total number of host addresses _________6_5__,5__3_6____
Number of usable addresses _________6_5__,5__3_4____
Number of bits borrowed _________8__________
What is the 11thsubnet range? 10.10.0.0 to 10.10.255.255
What is the subnet number for the 6th subnet?
What is the subnet broadcast address for
the 2nd subnet?
_______1_0_._5__.0__.0__________
_______1_0_._1_._2_5__5_._2_5__5____
What are the assignable addresses for the 9th
subnet? ___1_0__.8_._0_._1___t_o___1_0__.8_._2_5__5__.2__5_4________
Subnets - 2 4 8 16 32 64 128 256 . .
Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
10. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) 0 10.0.0.0 to 10.0.255.255(1) 1 10.1.0.0 to 10.1.255.255(2) 1 0 10.2.0.0 to 10.2.255.255(3) 1 1 10.3.0.0 to 10.3.255.255(4) 1 0 0 10.4.0.0 to 10.4.255.255
128 (5) 1 0 1 10.5.0.0 to 10.5.255.255
64 (6) 1 1 0 10.6.0.0 to 10.6.255.255
32 (7) 1 1 1 10.7.0.0 to 10.7.255.255
16 (8) 1 0 0 0 10.8.0.0 to 10.8.255.255
8 (9) 1 0 0 1 10.9.0.0 to 10.9.255.255
4 1 0 1 0 10.10.0.0 to 10.10.255.255
2 (11) 1 0 1 1 10.11.0.0 to 10.11.255.255
+1 1 1 0 0 10.12.0.0 to 10.12.255.255
255 1 1 0 1 10.13.0.0 to 10.13.255.255(14) 1 1 1 0 10.14.0.0 to 10.14.255.255
36 (15) 1 1 1 1 10.15.0.0 to
Sh
ow
you
rw
ork
for
Pro
blem
7in
the
space
belo
w.
41
Number of
Hosts - . 256 128 64 32 16 8 4 2
Number of
65,5 - 2 65,534
42
Problem 8
Subnetting
Number of needed subnets 5Network Address 172.50.0.0
Address class ____B______
Default subnet mask
____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask
____2__5__5__._2__5_5___._2__2_4___. _0______
Total number of subnets
_________8__________ Total number of host
addresses _________8_,1__9_2______ Number of
usable addresses _________8_,1__9_0______
Number of bits borrowed _________3__________
What is the 4thsubnet range? 172.50.96.0 to 172.50.127.255
What is the subnet number for the 5th subnet?
What is the subnet broadcast address for
the 6th subnet?
____1_7__2_._5_0__.1__2_8_._0_______
____1_7__2_._5_0__.1__9_1_._2_5__5____
What are the assignable addresses for the 3rd
subnet? _1_7__2_._5_0__.6__4_._1___t_o___1_7_2__.5__0_._9_5__.2__5_4____
Sh
ow
you
rw
ork
for
Pro
blem
8in
the
space
belo
w.
43
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172 . 50 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) 0 172.50.0.0 to 172.50.31.255(1) 1 172.50.32.0 to 172.50.63.255(2) 1 0 172.50.64.0 to 172.50.95.255(3) 1 1 172.50.96.0 to 172.50.127.255(4) 1 0 0 172.50.128.0 to 172.50.159.255(5) 1 0 1 172.50.160.0 to 172.50.191.255(6) 1 1 0 172.50.192.0 to 172.50.223.255(7) 1 1 1 172.50.224.0 to 172.50.255.255
8,192 - 2 8,190
12864
+32224
44
Problem 9
Subnetting
Number of needed usable hosts 28Network Address 172.50.0.0
Address class ____B______
Default subnet mask
____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask
____2__5__5__._2__5_5___._2__5_5___.
_2__2_4___
Total number of subnets
_________2_,_0_4__8_____ Total number of host
addresses _________3_2_________ Number of
usable addresses _________3_0_________ Number
of bits borrowed _________1_1_________
What is the 2ndsubnet range? 172.50.0.32 to 172.50.0.63
What is the subnet number for the 10th subnet?
What is the subnet broadcast address for
the 4th subnet?
____1_7__2_._5_0__.1_._3_2_________
____1_7__2_._5_0__.0__.1__2_7_______
What are the assignable addresses for the 6th
subnet? _1__7_2_._5_0__.0__.1_6__1___t_o__1__7_2_._5__0_.0__.1__9_0_____
(0) . 0 172.50.0.0 to 172.50.0.31(1) 1 172.50.0.32 to 172.50.0.63(2) 1 0 172.50.0.64 to 172.50.0.95(3) 1 1 172.50.0.96 to 172.50.0.127(4) 1 0 0 172.50.0.128 to 172.50.0.159(5) 1 0 1 172.50.0.160 to 172.50.0.191(6) 1 1 0 172.50.0.192 to 172.50.0.223(7) 1 1 1 172.50.0.224 to 172.50.0.255(8) 1 . 0 0 0 172.50.1.0 to 172.50.1.31(9) 1 . 0 0 1 172.50.1.32 to 172.50.1.63(10)
1 . 0 1 0 172.50.1.64 to 172.50.1.95(11) 1 . 0 1 1 172.50.1.96 to 172.50.1.127(12) 1 .
0 0 172.50.1.128 to 172.50.1.159(13) 1 . 1 0 1 172.50.1.160 to 172.50.1.191(14) 1 .
11 0 172.50.1.192 to 172.50.1.223
(15) 1 . 1
1 1 172.50.1.224 to 172.50.1.255
Sh
ow
you
rw
ork
for
Pro
blem
9in
the
space
belo
w.
45
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172 . 50 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128643216
842
+1252
32-230
12864+32224
46
Problem 10
Subnetting
Number of needed subnets 45Network Address 220.100.100.0
Address class ____C______
Default subnet mask
____2__5__5__._2__5__5__._2__5_5___.
_0______ Custom subnet mask
__
__2__5__5__._2__5__5__._2__5_5___._2__5_2__
_
Total number of subnets
_________6_4_________ Total number of host
addresses _________4__________ Number of usable
addresses _________2__________ Number of bits
borrowed _________6__________
What is the 5thsubnet range? 220.100.100.16 to 220.100.100.19
What is the subnet number for the 4th subnet?
What is the subnet broadcast address for
the 13th subnet?
_____2_2_0__.1_0__0_.1__0_0_._1_2_____
_____2_2_0__.1_0__0_.1__0_0_._5_1_____
What are the assignable addresses for the 12th
subnet? __2_2__0_.1_0__0_.1_0__0_._4_5___t_o__2__2_0_._1_0_0_._1_0_0_._4_6__
(0) 0 220.100.100.0 to 220.100.100.3(1) 1 220.100.100.4 to 220.100.100.7(2) 1 0 220.100.100.8 to 220.100.100.11(3) 1 1 220.100.100.12 to 220.100.100.15(4) 1 0 0 220.100.100.16 to 220.100.100.19(5) 1 0 1 220.100.100.20 to 220.100.100.23(6) 1 1 0 220.100.100.24 to 220.100.100.27(7) 1 1 1 220.100.100.28 to 220.100.100.31(8) 1 0 0 0 220.100.100.32 to 220.100.100.35(9) 1 0 0 1 220.100.100.36 to 220.100.100.39(10)
1 0 1 0 220.100.100.40 to 220.100.100.43(11) 1 0 1 1 220.100.100.44 to 220.100.100.47(12)
1 1 0 0 220.100.100.48 to 220.100.100.51(13)
1 1 0 1 220.100.100.52 to 220.100.100.55(14)
1 1 1 0 220.100.100.56 to 220.100.100.59(15)
1 1 1 1 220.100.100.60 to 220.100.100.63
Sh
ow
you
rw
ork
for
Pro
blem
10in
the
space
belo
w.
47
Number of256 128 64 32 16 8 4 2 - Hosts
Number ofSubnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
220 . 100 . 100 . 0 0 0 0 0 0 0 0
128643216
8+4
252
4-22
48
Problem 1 1
Subnetting
Number of needed usable hosts 8,000Network Address 135.70.0.0
Address class ____B______
Default subnet mask
____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask
____2__5__5__._2__5_5___._2__2_4___. _0______
Total number of subnets
_________8__________ Total number of host
addresses _________8_,1__9_2______ Number of
usable addresses _________8_,1__9_0______
Number of bits borrowed _________3__________
What is the 6thsubnet range? 135.70.160.0 to 135.70.191.255
What is the subnet number for the 7th subnet?
What is the subnet broadcast address for
the 3rd subnet?
_____1_3_5__.7__0_._1_9_2__.0_______
_____1_3_5__.7__0_._9_5__.2__5_5_____
What are the assignable addresses for the 5th
subnet? __1_3_5__._7_0_._1_2_8__.1___t_o___1_3_5__.7__0_.1__5_9__.2__5_4__
Sh
ow
you
rw
ork
for
Pro
blem
11in
the
space
belo
w.
49
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
135 . 70 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) 0 135.70.0.0 to 135.70.31.255(1) 1 135.70.32.0 to 135.70.63.255(2) 1 0 135.70.64.0 to 135.70.95.255(3) 1 1 135.70.96.0 to 135.70.127.255(4) 1 0 0 135.70.128.0 to 135.70.159.255(5) 1 0 1 135.70.160.0 to 135.70.191.255(6) 1 1 0 135.70.192.0 to 135.70.223.255(7) 1 1 1 135.70.224.0 to 135.70.255.255
8,192 -2 8,190
12864
+32224
50
Problem 12
Subnetting
Number of needed usable hosts 45Network Address 198.125.50.0
Address class ____C______
Default subnet mask
____2__5__5__._2__5__5__._2__5_5___.
_0______ Custom subnet mask
__
__2__5__5__._2__5__5__._2__5_5___._1_9__2__
_
Total number of subnets
_________4__________ Total number of host
addresses _________6_4_________ Number of
usable addresses _________6_2_________
Number of bits borrowed _________2__________
What is the 2nd subnet range?
What is the subnet number for the 2nd subnet?
What is the subnet broadcast address for
the 4th subnet?
198.125.50.64 to 98.125.50.127
__
__1_9_8__.1_2__5__.5__0_._6_4______
____1_9_8__.1_2__5__.5__0_._2_5__5____
What are the assignable addresses for the 3rd
subnet? _1_9__8_.1_2__5__.5__0_._1_2_9___t_o___1_9__8_.1__2_5__.5__0_._1_9_0
52
Show your work for Problem 1 2 in the space below.
Number of
Number of256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
198 . 125 . 50 . 0 0 0 0 0 0 0 0
(0) 0 198.125.50.0 to 198.125.50.63(1) 1 198.125.50.64 to 198.125.50.127(2) 1 0 198.125.50.128 to 198.125.50.191(3) 1 1 198.125.50.192 to 198.125.50.255
128+64192
64 -2 62
51
52
Problem 13
Subnetting
Network Address 165.200.0.0 /26
Address class
____B______
Default subnet mask
____2__5__5__._2__5__5__._0__.__0_________
Custom subnet mask
__
__2__5__5__._2__5__5__._2__5_5___._1_9__2__
_
Total number of subnets
_________1_,0__2_4______ Total number of host
addresses _________6_4_________ Number of
usable addresses _________6_2_________ Number
of bits borrowed _________1_0_________
What is the 10thsubnet range? 165.200.2.64 to 165.200.2.127
What is the subnet number for the 11th subnet?
What is the subnet broadcast address for the 1023rd subnet?
_____1_6_5__.2_0__0_.2__.1_2__8______
_____1_6_5_._2_0__0_.2__5_5__.1_9_1____
What are the assignable addresses for the 1022nd
subnet? _1_6_5__.2__0_0_._2_5_5__.6__5___to___1_6_5__.2_0__0_._2_5_5__.1_2__6
(0) . 0 165.200.0.0 to 165.200.0.63(1) 1 165.200.0.64 to 165.200.0.127(2) 1 0 165.200.0.128 to 165.200.0.191(3) 1 1 165.200.0.192 to 165.200.0.255(4) 1 0 0 165.200.1.0 to 165.200.1.63(5) 1 0 1 165.200.1.64 to 165.200.1.127(6) 1 1 0 165.200.1.128 to 165.200.1.191(7) 1 1 1 165.200.1.192 to 165.200.1.255(8) 1 0 . 0 0 165.200.2.0 to 165.200.2.63(9) 1 0 . 0 1 165.200.2.64 to 165.200.2.127(10)
1 0 . 1 0 165.200.2.128 to 165.200.2.191(11)1 0 . 1 1 165.200.2.192 to 165.200.2.255
0 0 165.200.3.0 to 165.200.3.630 1 165.200.3.64 to 165.200.3.1271 0 165.200.3.128 to 165.200.3.1911 1 165.200.3.192 to 165.200.3.255
(12)
1 1(13)
1 1(14)
1 1(15)
1 1
165.200.255.64 to 165.200.255.127165.200.155.128 to 165.200.255.191165.200.255.192 to 165.200.255.255
Sh
ow
you
rw
ork
for
Pro
blem
13in
the
space
belo
w.
53
.
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
165 . 200 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
12864
3216
842
+1 252
64 128
-2 +64 62 252
.
..
(1021) 1 1 1 1 1 1 1 1 . 0 1(1022) 1 1 1 1 1 1 1 1 . 1 0(1023) 1 1 1 1 1 1 1 1 . 1 1
54
Problem 14
Subnetting
Number of needed usable hosts 16Network Address 200.10.10.0
Address class ____C______
Default subnet mask
____2__5__5__._2__5__5__._2__5_5___.
_0______ Custom subnet mask
____2__5__5__._2__5_5___._2__5_5___.
_2__2_4___
Total number of subnets
_________8__________ Total number of host
addresses _________3_2_________ Number of
usable addresses _________3_0_________ Number
of bits borrowed _________3__________
What is the 7thsubnet range? 200.10.10.192 to 200.10.10.223
What is the subnet number for the 5th subnet?
What is the subnet broadcast address for
the 4th subnet?
____2__0__0_._1_0_._1_0__.1_2__8_____
__
__2__0__0_._1_0_._1_ 0__.1_2__7_____
What are the assignable addresses for the 6th
subnet? __2_0__0_._1_0_._1_0_._1_6_1___t_o___2_0__0_.1__0_._1_0_._1_9_0___
56
Show your work for Problem 1 4 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
200 . 10 . 10 . 0 0 0 0 0 0 0 0
(0) 0 200.10.10.0 to 200.10.10.31(1) 1 200.10.10.32 to 200.10.10.63(2) 1 0 200.10.10.64 to 200.10.10.95(3) 1 1 200.10.10.96 to 200.10.10.127(4) 1 0 0 200.10.10.128 to 200.10.10.159(5) 1 0 1 200.10.10.160 to 200.10.10.191(6) 1 1 0 200.10.10.192 to 200.10.10.223(7) 1 1 1 200.10.10.224 to 200.10.10.255
12864 32
+3 2 -2224 30
55
56
Problem 15
Subnetting
Network Address 93.0.0.0 \19
Address class ____A______
Default subnet mask ____2__5__5__._0___.
_0__._0_____________ Custom subnet mask
____2__5__5__._2__5_5___._2__2_4___. _0______
Total number of subnets
_________2_,_0_4__8_____ Total number of host
addresses _________8_,1__9_2______ Number of
usable addresses _________8_,1__9_0______
Number of bits borrowed _________1_1_________
What is the 15thsubnet range? 93.1.192.0 to 93.1.223.255
What is the subnet number for the 9th subnet?
What is the subnet broadcast address for
the 7th subnet?
______9_3__.1_._0_._0___________
______9_3__.0__.2__2_3__.2__5_5_____
What are the assignable addresses for the 12th
subnet? ___9_3__.1_._9_6_._1___t_o__9__3_.1__.1_2__7_.2__5__4________
(0) . 0 93.0.0.0 to 93.0.31.255(1) 1 93.0.32.0 to 93.0.63.255(2) 1 0 93.0.64.0 to 93.0.95.255(3) 1 1 93.0.96.0 to 93.0.127.255(4) 1 0 0 93.0.128.0 to 93.0.159.255(5) 1 0 1 93.0.160.0 to 93.0.191.255(6) 1 1 0 93.0.192.0 to 93.0.223.255(7) 1 1 1 93.0.224.0 to 93.0.255.255(8) 1 . 0 0 0 93.1.0.0 to 93.1.31.255
1 93.1.32.0 to 93.1.63.2550 93.1.64.0 to 93.1.95.2551 93.1.96.0 to 93.1.127.2550 93.1.128.0 to 93.1.159.2551 93.1.160.0 to 93.1.191.2550 93.1.192.0 to 93.1.223.255
(9) 1 . 0 0(10) 1 . 0 1(11) 1 . 0 1(12) 1 .
1 0(13) 1 . 1 0(14) 1 . 1 1
Sh
ow
you
rw
ork
for
Pro
blem
15in
the
space
belo
w.
57
Number of
Hosts -. 256 128 64 32 16 8 4 2
Number of
Subnets - 2 4 8 16 32 64 128 256 . .
Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
93. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
12864
3216
842
+1255
12864
+32224
8,192 - 2 8,190
58
Practical Subnetting 1Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnet s , and allow enough extra subnets and hosts for 100% growth in both areas. Circle each subnet on the graphic and answer the questions below.
Marketing 24 Hosts
F0/0Router A
IP Address 172.16.0.0
S0/0/0 S0/0/1
F0/0
Reasearch 60 Hosts
Router B
F0/1
Management 15 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 100% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 100% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
____________
__B____________________
__2__5_5_._2_5__5_._2_2_4__.
0________
____4_____
_+___4_____
_=___8_____
____6_0____
_+___6_0____
_=__1_2_0____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Research
IP address range for Marketing
IP address range for Management
IP address range for Router A to Router B serial connection _
__1_7_2__.1_6_._0_._0__t_o_1_7_2__.3_1__.2_5__5___
__1__7_2_.1_6__.3_2__.0__t_o__1_7_2_._6_3_._2_5__5__
__1_7__2_.1_6__.6_4__.0__t_o__1_7_2_._9_5_._2_5__5__
__1_7_2__.1_6_._9_6_._0__t_o_1_7_2__.1_2__7_.2_5__5__
(0) 0 172.16.0.0 to 172.16.31.255(1) 1 172.16.32.0 to 172.16.63.255(2) 1 0 172.16.64.0 to 172.16.95.255(3) 1 1 172.16.96.0 to 172.16.127.255(4) 1 0 0 172.16.128.0 to 172.16.159.255(5) 1 0 1 172.16.160.0 to 172.16.191.255(6) 1 1 0 172.16.192.0 to 172.16.223.255(7) 1 1 1 172.16.224.0 to 172.16.255.255
Sh
ow
you
rw
ork
for
Practical
Su
bn
etting
1in
the
space
belo
w.
59
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172 . 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
4x1.0
4
60 x1.0 60
60
Practical Subnetting 2Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets andhosts for 30% growth in all areas. Circle each subnet on the graphic and answer the questions below.
F0/0Router A
IP Address 135.126.0.0S0/0/0
S0/0/1
S0/0/1 F0/0 Router B F0/1
Science Lab 10 Hosts
S0/0/0
F0/1Router C
Tech Ed Lab 20 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 30% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 30% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
English Department 15 Hosts
______________B_______________
______2_5__5_._2_5_5__.2__5_5__.2_2__4______
_____5____
_+____2____
_=____7____
____2_0____
_+___6_____
_=___2_6____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Tech Ed
IP address range for English
IP address range for Science
IP address range for Router A to Router B serial connection
IP address range for Router A
_1_3__5_._1_2_6_._0_.0___to__1_3_5__.1_2__6_.0__.3_1__
_1_3_5_._1_2_6_._0_._3_2__t_o__1_3_5_._1_2_6__.0_._6_3_
_1_3_5_._1_2_6_._0_._6_4__t_o__1_3_5_._1_2_6_._0_._9_5_
1_3__5_.1_2__6_.0__.9_6__t_o__1_3_5__.1_2__6_.0__.1_2_7_
to Router B serial connection1_3_5_._1_2_6_._0_._1_2_8__t_o__1_3_5_._1_2_6_._0_.1__5_9
0 135.126.0.0 to 135.126.0.311 135.126.0.32 to 135.126.0.63
1 0 135.126.0.64 to 135.126.0.951 1 135.126.0.96 to 135.126.0.127
1 0 0 135.126.0.128 to 135.126.0.1591 0 1 135.126.0.160 to 135.126.0.1911 1 0 135.126.0.192 to 135.126.0.2231 1 1 135.126.0.224 to 135.126.0.2550 0 0 135.126.1.0 to 135.126.1.310 0 1 135.126.1.32 to 135.126.1.630 1 0 135.126.1.64 to 135.126.1.950 1 1 135.126.1.96 to 135.126.1.1271 0 0 135.126.1.128 to 135.126.1.1591 0 1 135.126.1.160 to 135.126.1.1911 1 0 135.126.1.192 to 135.126.1.2231 1 1 135.1261.224 to 135.126.1.255
Sh
ow
you
rw
ork
for
Pro
blem
2in
the
space
belo
w.
61
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
5 x.3 1.5
(Round up to 2)
20 x.3
6
(0) .(1)(2)
(3)(4)(5)(6)(7)(8) 1 .(9) 1 .(10) 1 .(11) 1 .
(12) 1 .(13) 1 .(14) 1 .(15) 1 .
62
Practical Subnetting 3Based on the information in the graphic shown, design a classfull network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets and hosts for 25% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 172.16.0.0
Administrative
F0/0
Router A
S0/0/1F0/0 Sales
185 Hosts30 Hosts F0/1 S0/0/0
Router B
Marketing 50 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 25% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 25% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
______________B_______________
_______2__5_5__.2_5__5_._2_5__5_.
_0_______
____4_____
_+___1_____
_=___5_____
___1_8_5____
_+___4_7____
_=__2_3__2___
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Sales
IP address range for Marketing
IP address range for Administrative
IP address range for Router A to Router B serial connection
__1_7_2__.1_6__.0_._0__t_o_1_7__2_.1_6__.0_._2_5__5__
__1_7_2__.1_6_._1_.0___t_o__1_7
_2_._1_6_._1_.2_5__5__
__1_7_2_._1_6_.2__.0___t_o__1_7
_2_._1_6_._2_.2__5_5__
__1_7_2_._1_6_.3__.0___t_o__1_7
_2__.1_6_._3_.2__5_5__
(0)(1)(2)(3)(4)(5)(6)(7)(8) 1(9) 1(10)
1(11) 1(12)
1(13)
1(14)
1(15)
1
0 172.16.0.0 to 1172.16.0.2551 172.16.1.0 to 1172.16.1.255
1 0 172.16.2.0 to 1172.16.2.2551 1 172.16.3.0 to 1172.16.3.255
1 0 0 172.16.4.0 to 1172.16.4.2551 0 1 172.16.5.0 to 1172.16.5.2551 1 0 172.16.6.0 to 1172.16.6.2551 1 1 172.16.7.0 to 1172.16.7.2550 0 0 172.16.8.0 to 1172.16.8.2550 0 1 172.16.9.0 to 1172.16.9.2550 1 0 172.16.10.0 to 1172.16.10.2550 1 1 172.16.11.0 to 1172.16.11.2551 0 0 172.16.12.0 to 1172.16.12.2551 0 1 172.16.13.0 to 1172.16.13.2551 1 0 172.16.14.0 to 1172.16.14.2551 1 1 172.16.15.0 to 1172.16.15.255
Sh
ow
you
rw
ork
for
Pro
blem
3in
the
space
belo
w.
63
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172. 16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
.
4x..25
1
225 .x.2 5 .
56.25 .(Round up to 57) .
.
.
.
.
64
Practical Subnetting 4Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnet s , and allow enough extra subnets and hosts for 70% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 135.126.0.0
F0/0 S0/0/0Router A
S0/0/1
S0/0/1
F0/0 Router B
S0/0/0
DallasF0/1
Router C F0/0
150 Hosts New York
Washington D.C.220 Hosts
325 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 70% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 70% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
______________B_______________
_______2__5_5__.2__5_5_._2_4__0_.0________
____5_____
_+___4_____
_=___9_____
___3_2__5___
_+__2_2__8___
_=__5_5__3___
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for New York 1
_3__5_._1_2_6_._0_.0__t_o__1_3_5__.1_2__6_.1_5__.2__5_5
IP address range for Washington D. C. 1_3_5__.1_2__6_.1_6__.0__t_o__1_3_5_._1_2_6_._3_1_._2_5_5
IP address range for Dall
as1_3_5__.1_2_6__.3_2__.0__t_o__1_3_5__.1_2_6__.4_7_.
_2_5_ 5
IP address range for Router Ato Router B serial connection
1_3_5__.1_2_6__.4_8_._0__t_o__1_3_5_._1_2_6_._6_3_._2_5
_ 5
IP address range for Router Ato Router C serial connection 1_3_5__.1_2_6__.6_4__.0__t_o__1_3_5__.1_2_6__.7_9_._2_5_5
1 0 0 135.126.192.0 to 135.126.207.2551 0 1 135.126.208.0 to 135.126.223.2551 1 0 135.126.224.0 to 135.126.239.2551 1 1 135.126.240.0 to 135.126.1255.255
(12)
1(13)
1(14)
1(15)
1
.
.
.
0 0 0 135.126.128.0 to 135.126.143.2550 0 1 135.126.144.0 to 135.126.159.2550 1 0 135.126.160.0 to 135.126.175.255
(11) 1 . 0 1 1 135.126.176.0 to 135.126.191.255
(8) 1(9) 1(10)
1
Sh
ow
you
rw
ork
for
Pro
blem
4in
the
space
belo
w.
65
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
135. 126 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) . 0 135.126.0.0 to 135.126.15.255(1) 1 135.126.16.0 to 135.126.31.255(2) 1 0 135.126.32.0 to 135.126.47.255(3) 1 1 135.126.48.0 to 135.126.63.255(4) 1 0 0 135.126.64.0 to 135.126.79.255(5) 1 0 1 135.126.80.0 to 135.126.95.255(6) 1 1 0 135.126.96.0 to 135.126.111.255(7) 1 1 1 135.126.112.0 to 135.126.127.255
.
...
66
Practical Subnetting 5Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subne t , and allow enough extra subnets and hosts for 100% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 210.15.10.0
F0/0F0/1
Science Room 10 Hosts
Tech Ed Lab 18 Hosts
English classroom 15 Hosts Art Classroom
12 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 100% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 100% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
______________C_______________
______2__5_5_._2_5__5_._2_5_5__.1_9__2______
____2_____
_+___2_____
_=___4_____
___3__0____
_+__3__0____
_=__6__0____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router F0/0 Port
IP address range for Router F0/1 Port
__2_1_0_._1_5_._1_0_._0__t_o_2__1_0_.1_5__.1_0__.6_3__
_2_1_0_._1_5_._1_0_._6_4__t_o_2
__1_0_._1_5_.1_0__.1_2__7
68
Show your work for Problem 5 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
210. 15 . 10 . 0 0 0 0 0 0 0 0
(0) 0 210.15.10.0 to 210.15.10.63(1) 1 210.15.10.64 to 210.15.10.127(2) 1
0 210.15.10.128 to 210.15.10.191(3) 1 1 210.15.10.192 to 210.15.10.255
67
68
Practical Subnetting 6Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnet s , and allow enough extra subnets and hosts for 20% growth in all areas. Circle each subnet on the graphic and answer the questions below.
Router A
F0/0
S0/0/0
S0/0/1
IP Address 10.0.0.0
S0/0/1
S0/0/0 Router B
F0/1
Technology Building 320 Hosts
S0/0/0Art & Drama
S0/0/1
Router CAdministration
75 Hosts F0/0 F0/135 Hosts
Science Building 225 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 20% growth(Round up to the next whole number)
Total number of subnets needed
______________A_______________
_________2_5__5_._2_4_0__.0_._0_________
_____7____
_+____2____
_=____9____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Technology
IP address range for Science
IP address range for Arts & Drama
IP Address range Administration
IP address range for Router A to Router B serial connection
IP address range for Router A
to Router C serial connection
IP address range for Router B to Router C serial connection
__1_0_._0_._0_.0__t_o__1_0_._1_5
_._2_5__5_._2_5_5___
__1_0_._1_6_.0__.0__t_o__1_0_.3__1_.2__5_5_._2_5__5__
_1_0__.3_2__.0__.0__t_o__1_0_._4_7_.2__5_5__.2__5_5__
__1_0_.4__8_.0_._0__t_o__1_0_._6
_3_.2__5_5__.2_5__5__
__1_0_.6__4_.0__.0__t_o__1_0_._7
_9_.2__5_5__.2_5__5__
_1_0__.8_0__.0_._0__t_o__1_0_.9__5_.2__5_5__.2__5_5__
_1_0__.9_6_._0_._0__t_o_1_0__.1_1_1_._2_5_5__.2__5_5__
Sh
ow
you
rw
ork
for
Pro
blem
6in
the
space
belo
w.
69
Number of
Hosts -. 256 128 64 32 16 8 4 2
Number of
Subnets - 2 4 8 16 32 64 128 256 . .
Binary values -128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
10. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) 0 10.0.0.0 to 10.15.255.255(1) 1 10.16.0.0 to 10.32.255.255(2) 1 0 10.32.0.0 to 10.47.255.255(3) 1 1 10.48.0.0 to 10.63.255.255(4) 1 0 0 10.64.0.0 to 10.79.255.255(5) 1 0 1 10.80.0.0 to 10.95.255.255(6) 1 1 0 10.96.0.0 to 10.111.255.255(7) 1 1 1 10.112.0.0 to 10.127.255.255(8) 1 0 0 0 10.128.0.0 to 10.143.255.255(9) 1 0 0 1 10.144.0.0 to 10.159.255.255(10)
1 0 1 0 10.160.0.0 to 10.175.255.255(11) 1 0 1 1 10.176.0.0 to 10.191.255.255(12)
1 1 0 0 10.192.0.0 to 10.207.255.255(13)
1 1 0 1 10.208.0.0 to 10.223.255.255(14)
1 1 1 0 10.224.0.0 to 10.239.255.255(15)
1 1 1 1 10.240.0.0 to 10.255.255.255
70
Practical Subnetting 7Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets and hosts for 125% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 177.135.0.0S0/0/0
Router A
F0/0S0/0/0 F0/0
Router B
F0/1
Marketing 75 Hosts
Administration33 Hosts Sales
255 Hosts
Research 135 Hosts
Deployment 63 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 125% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 125% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
______________B_______________
_______2__5_5__.2_5__5_._2_5__2_.0________
____4_____
_+___5_____
_=___9_____
___3_6__3___
_+__4_5__4___
_=__8_1_7____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router A Port F0/0
IP address range for Research
IP address range for Deployment
IP address range for Router A
_1_7_7_._1_3_5__.0_._0__t_o_1_7__7_.1_3_5__.3_._2_5__5_
_1_7_7__.1_3_5__.4_._0__t_o_1__7_
7_.1_3_5__.7_._2_5__5_
_
1_7_7_._1_3_5_._8_.0__t_o__1_7_7
_.1__3_5_._1_1_.2_5__5_to Router B serial connection 1_7_7_._1_3_5__.1_2_._0__t_o_1_7_7__.1_3_5__.1_5__.2_5__5
(0)(1)(2)(3)(4)(5)(6)(7)(8) 1(9) 1
(10) 1(11) 1(12) 1(13) 1(14) 1(15) 1
0 177.135.0.0 to 177.135.3.2551 177.135.4.0 to 177.135.7.255
1 0 177.135.8.0 to 177.135.11.2551 1 177.135.12.0 to 177.135.15.255
1 0 0 177.135.16.0 to 177.135.19.2551 0 1 177.135.20.0 to 177.135.23.2551 1 0 177.135.24.0 to 177.135.27.2551 1 1 177.135.28.0 to 177.135.31.2550 0 0 177.135.32.0 to 177.135.35.2550 0 1 177.135.36.0 to 177.135.39.2550 1 0 177.135.40.0 to 177.135.43.2550 1 1 177.135.44.0 to 177.135.47.2551 0 0 177.135.48.0 to 177.135.51.2551 0 1 177.135.52.0 to 177.135.55.2551 1 0 177.135.56.0 to 177.135.59.2551 1 1 177.135.60.0 to 177.135.63.255
Sh
ow
you
rw
ork
for
Pro
blem
7in
the
space
belo
w.
71
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
177.135 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
.
.
.
.
.
.
.
.
.
72
Practical Subnetting 8Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number subnet s , and allow enough extra subnets and hosts for 85% growth in all areas. Circle each subnet on the graphic and answer the questions below.
IP Address 192.168.1.0F0/0 S0/0/0
Router A S0/0/1 F0/1
F0/0 Router B
New York 8 Hosts
Boston 5 Hosts
Research & Development 8 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 85% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 85% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
______________C_______________
______2_5__5_._2_5_5__.2__5_5__.2_2__4______
____3_____
_+___3_____
_=___6_____
____1_3____
_+___1_2____
_=___2_5____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Router A F0/0 IP address range for New York
IP address range for Router A to Router B serial connection
__1_9_2__.1_6__8_.1_._0__to__1_9_2__.1_6_8_._1_.3__1__
__1_9_2_.1__6_8_.1_._3_2__t_o__1_9_2_._1_6_8_.1_._6_3__
_1_9__2_.1_6__8_.1_._6_4__t_o__1_9_2_._1_6_8_.1_._9_5__
74
Show your work for Problem 8 in the space below.
Number of
Number of 256 128 64 32 16 8 4 2 - Hosts
Subnets - 2 4 8 16 32 64 128 256
128 64 32 16 8 4 2 1 - Binary values
192. 168 . 1 . 0 0 0 0 0 0 0 0
(0) 0 192.168.1.0 to 192.168.1.31(1) 1 192.168.1.32 to 192.168.1.63(2) 1 0 192.168.1.64 to 192.168.1.95(3) 1 1 192.168.1.96 to 192.168.1.127(4) 1 0 0 192.168.1.128 to 192.168.1.159(5) 1 0 1 192.168.1.160 to 192.168.1.1191(6) 1 1 0 192.168.1.192 to 192.168.1.223(7) 1 1 1 192.168.1.224 to 192.168.1.255
73
Practical Subnetting 9Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of hosts per subn e t , and allow enough extra subnets andhosts for 15% growth in all areas. Circle each subnet on the graphic and answer the questions below.
Router A
IP Address 148.55.0.0S0/0/0
S0/0/1 F0/1
S0/0/1 F0/0 Router B
S0/0/0
Router C
Dallas 1500 Hosts
F0/0
Router D S0/0/0S0/0/1
Ft. Worth 2300 Hosts Address class
Custom subnet mask
______________B_______________
_______2__5_5__.2__5_5_._2_4__0_.0
________
Minimum number of subnets needed
Extra subnets required for 15% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 15% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
____5_____
_+___1_____
_=___6_____
__2__3_0_0___
_+__3_4__5___
_=_2__6_4_5___
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Ft. Worth
IP address range for Dallas
IP address range for Router A to Router B serial connection
IP address range for Router A to Router C serial connection
IP address range for Router C74 to Router D serial connection
_1_4__8_.5__5_.0__.0_.__t_o_1_4__8_.5__5_._1_5_.2__5_5_
_1_4_8_._5_5__.1_6_._0_._t_o__1
_4_8_._5_5__.3_1_._2_5__5
_1_4_8_.5__5_._3_2_._0_._t_o__1
_4_8_._5_5__.4__7_.2_5__5
_1_4_8_._5_5_._4_8_._0_._t_o__1
_4_8_._5_5__.6_3__.2_5__5
_1_4_8_._5_5_._6_4_._0_._t_o__1
_4_8__.5_5__.7_9__.2_5__5
(11) 1 . 0 1 1 148.55.176.0 to 148.55.191.255(12)
1 .1 0 0 148.55.192.0 to 148.55.207.2551 0 1 148.55.208.0 to 148.55.223.2551 1 0 148.55.224.0 to 148.55.239.2551 1 1 148.55.240.0 to 148.55.255.255
(13)
1 .(14)
1 .(15)
1 .
Sh
ow
you
rw
ork
for
Pro
blem
9in
the
space
belo
w.
75
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
148. 55 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
(0) . 0 148.55.0.0 to 148.55.15.255(1) 1 148.55.16.0 to 148.55.31.255(2) 1 0 148.55.32.0 to 148.55.47.255(3) 1 1 148.55.48.0 to 148.55.63.255(4) 1 0 0 148.55.64.0 to 148.55.79.255(5) 1 0 1 148.55.80.0 to 148.55.95.255(6) 1 1 0 148.55.96.0 to 148.55.111.255(7) 1 1 1 148.55.112.0 to 148.55.127.255(8) 1 . 0 0 0 148.55.128.0 to 148.55.143.255(9) 1 . 0 0 1 148.55.144.0 to 148.55.159.255(10)
1 . 0 1 0 148.55.160.0 to 148.55.175.255
76
Practical Subnetting 10Based on the information in the graphic shown, design a network addressing scheme that will supply the minimum number of subnet s , and allow enough extra subnets and hosts for 110% growth in all areas. Circle each subnet on the graphic and answer the questions below.
Sales115 Hosts
IP Address 172.16.0.0
S0/0/0
Marketing 56 Hosts
F0/0F0/0 Router A S0/0/1
Router B
F0/1
Management 25 Hosts
Research 35 Hosts
Address class
Custom subnet mask
Minimum number of subnets needed
Extra subnets required for 110% growth(Round up to the next whole number)
Total number of subnets needed
Number of host addressesin the largest subnet group
Number of addresses needed for 110% growth in the largest subnet
(Round up to the next whole number)
Total number of address needed for the largest subnet
______________B_______________
______2_5__5_._2_5__5_.2__5_5__.2__4_0______
_____4____
_+____5____
_=____9____
___1_4_0____
_+__1_5_4____
_=__2_9_4____
Start with the first subnet and arrange your sub-networks from the largest group to the smallest.
IP address range for Sales/Managemnt
IP address range for Marketing
IP address range for
Research
IP address range for Router A to Router B serial connection _
_1_7_2__.1_6_._0_.0__t_o__1_7_2__.1_6_._1_5_._2_5_5__
__1_7_2_._1_6_.1_6__.0__t_o__1_7_2_._1_6_.3_1__.2_5__5_
_1_7__2_.1_6__.3_2__.0__t_o__1_7_2_._1_6_.4__7_.2__5_5_
__1_7_2_.1_6__.4_8__.0__t_o__1_7_2_._1_6_.6__3_.2__5_5_
(0)(1)(2)(3)(4)(5)(6)(7)(8) 1(9) 1(10)
1(11) 1(12)
1(13)
1(14)
1(15)
1
0 172.16.0.0 to 172.16.15.2551 172.16.16.0 to 172.16.31.255
1 0 172.16.32.0 to 172.16.47.2551 1 172.16.48.0 to 172.16.63.255
1 0 0 172.16.64.0 to 172.16.79.2551 0 1 172.16.80.0 to 172.16.95.2551 1 0 172.16.96.0 to 172.16.111.2551 1 1 172.16.112.0 to 172.16.127.2550 0 0 172.16.128.0 to 172.16.143.2550 0 1 172.16.144.0 to 172.16.159.2550 1 0 172.16.160.0 to 172.16.175.2550 1 1 172.16.176.0 to 172.16.191.2551 0 0 172.16.192.0 to 172.16.207.2551 0 1 172.16.208.0 to 172.16.223.2551 1 0 172.16.224.0 to 172.16.239.2551 1 1 172.16.240.0 to 172.16.255.255
Sh
ow
you
rw
ork
for
Pro
blem
10in
the
space
belo
w.
77
Number of Hosts - . 256 128 64 32 16 8 4 2
Number ofSubnets - 2 4 8 16 32 64 128 256.
Binary values - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172.16 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
.
.
.
.
.
.
.
.
.
78
Valid and Non-Valid IP Addresses
Using the material in this workbook identify which of the addresses below are correct and usable. If they are not usable addresses explain why.
IP Address: 0.230.190.192 _T__h_e_n_e_t_w__o_r_k__I_D___c_a_n_n_o_t__b_e__0__. __ Subnet Mask: 255.0.0.0 ________________________________ Reference Page Inside Front Cover
IP Address: 192.10.10.1 _O__K_____________________________ Subnet Mask: 255.255.255.0 ________________________________ Reference Pages 28-29
IP Address: 245.150.190.10 _2__4_5__i_s__r_e_s_e_r_v_e_d___f_o_r___________ Subnet Mask: 255.255.255.0 _e_x_p_e_r_i_m__e_n_t_a_l_u_s_e_._______________ Reference Page Inside Front Cover
IP Address: 135.70.191.255 _T__h_is__i_s__t_h_e__b_r_o_a_d__c_a_s_t__a_d_d_r__e_s_s_ Subnet Mask: 255.255.254.0 _f_o_r__t_h_i_s__r_a_n_g_e_._________________ Reference Pages 48-49
IP Address: 127.100.100.10 _1_2__7__i_s_r__e_s_e_r_v_e_d__f_o_r__l_o_o_p_b_a__c_k__ Subnet Mask: 255.0.0.0 _t_e_s_t_i_n_g_.________________________ Reference Pages Inside Front Cover
IP Address: 93.0.128.1 _O__K_____________________________ Subnet Mask: 255.255.224.0 ________________________________ Reference Pages 56-57
IP Address: 200.10.10.128 _T_h_i_s_i_s_t_h_e__s_u_b_n_et__a_d_d_r_e_s_s__fo_r__t_h_e__ Subnet Mask: 255.255.255.224 _3_r_d__u_s_a_b_le__r_a_n_ge__o_f__2_0_0_._1_0_._1_0_.0____Reference Pages 54-55
IP Address: 165.100.255.189 _O__K_____________________________ Subnet Mask: 255.255.255.192 ________________________________ Reference Pages 30-31
IP Address: 190.35.0.10 _T_h_is__a_d_d_r_e_s_s_i_s_t_a_k_e_n_f_r_o_m__th_e__f_ir_s_t__ Subnet Mask: 255.255.255.192
79
_r_a_ng_e__f_or__t_h_is__s_ub_n_e_t__w_h_ic_h__is__in_v_a_l_id_._ Reference Pages 34-35
IP Address: 218.35.50.195 _T_h_i_s__h_a_s__a__c_l_a_s_s__B__s_u_b_n__e_t______ Subnet Mask: 255.255.0.0 _m_a__s_k_.__________________________ Reference Page Inside Front Cover
IP Address: 200.10.10.175 /22 _A__c_l_a_s_s__C__a_d__d_r_e_s_s__m__u_s_t__u_s_e__a__Reference Pages 54-55 and/or Inside Front Cover
_m_i_n_i_m__u_m__o_f__2__4__b_i_t_s_.___________
IP Address: 135.70.255.255 _T_h_i_s__i_s__a__b_r_o_a_d_c_a__s_t_a__d_d_r_e_s_s__. __ Subnet Mask: 255.255.224.0 ________________________________ Reference Pages 48-49
80
IP Address Breakdown/24 /25 /26 /27 /28 /29 /30
8+8+8255.255.255.0
256 Hosts
8+8+8+1255.255.255.128
128 Hosts
8+8+8+2255.255.255.192
64 Hosts
8+8+8+3255.255.255.224
32 Hosts
8+8+8+4255.255.255.240
16 Hosts
8+8+8+5255.255.255.248
8 Hosts
8+8+8+6255.255.255.252
4 Hosts
0-255
0-127
0-63
0-150-7
0-34-7
8-158-11
12-15
16-3116-23
16-1920-23
24-3124-2728-31
32-4732-39
32-3536-39
40-4740-4344-47
48-6348-55
48-5152-55
56-6356-5960-63
64-127
64-7964-71
64-6768-71
72-7972-7576-79
80-9580-87
80-8384-87
88-9588-9192-95
96-11196-103
96-99100-103
104-111104-107108-111
112-127112-119
112-115116-119
120-127120-123124-127
128-255
128-191
128-143128-135
128-131132-135
136-143136-139140-143
144-159144-151
144-147148-151
152-159152-155156-159
160-17516-167
160-163164-167
168-175168-171172-175
176-191176-183
176-179180-183
184-191184-187188-191
192-255
192-207192-199
192-195196-199
200-207200-203204-207
208-223208-215
208-211212-215
216-223216-219220-223
224-239224-231
224-227228-231
232-239232-235236-239
240-255240-247
240-243244-247
248-255248-251252-255
81
Visualizing Subnets Using The Box Method
The box method is the simplest way to visualize the breakdown of subnets and addresses into smaller sizes.
Start with a square. The whole square is a single subnet comprised of 256addresses.
/24255.255.255.0
256 Hosts 1 Subnet
Split the box in half and you get two subnets with 128 addresses,
/25255.255.255.128
128 Hosts 2 Subnets
Divide the box into quarters and you get four subnets with 64 addresses,
/26255.255.255.192
64 Hosts4 Subnets
80
Split each individual square and you get eight subnets with 32 addresses,
/27255.255.255.224
32 Hosts8 Subnets
Split the boxes in half again and you get sixteen subnets with sixteenaddresses,
/28255.255.255.240
16 Hosts16 Subnets
The next split gives you thirty two subnets with eight addresses,
/29255.255.255.248
8 Hosts32 Subnets
The last split gives sixty four subnets with four addresses each,
/30255.255.255.252
4 Hosts64 Subnets
CIDR# of Bits
BorrowedSubnet Mask
Total # of Subnets
Total # of Hosts
Usable # of Hosts
/24 0 255.255.255.0 1 256 254/25 1 255.255.255.128 2 128 126/26 2 255.255.255.192 4 64 62/27 3 255.255.255.224 8 32 30/28 4 255.255.255.240 16 16 14/29 5 255.255.255.248 32 8 6/30 6 255.255.255.252 64 4 2
83
Class A Addressing Guide
CIDR# of Bits
BorrowedSubnetMask
Total # ofSubnets
Total # ofHosts
Usable # ofHosts
/8 0 255.0.0.0 1 16,777,216 16,777,214/9 1 255.128.0.0 2 8,388,608 8,388,606
/10 2 255.192.0.0 4 4,194,304 4,194,302/11 3 255.224.0.0 8 2,097,152 2,097,150/12 4 255.240.0.0 16 1,048,576 1,048,574/13 5 255.248.0.0 32 524,288 524,286/14 6 255.252.0.0 64 262,144 262,142/15 7 255.254.0.0 128 131,072 131,070/16 8 255.255.0.0 256 65,536 65,534/17 9 255.255.128.0 512 32,768 32,766/18 10 255.255.192.0 1,024 16,384 16,382/19 11 255.255.224.0 2,048 8,192 8,190/20 12 255.255.240.0 4,096 4,096 4,094/21 13 255.255.248.0 8,192 2,048 2,046/22 14 255.255.252.0 16,384 1,024 1,022/23 15 255.255.254.0 32,768 512 510/24 16 255.255.255.0 65,536 256 254/25 17 255.255.255.128 131,072 128 126/26 18 255.255.255.192 262,144 64 62/27 19 255.255.255.224 524,288 32 30/28 20 255.255.255.240 1,048,576 16 14/29 21 255.255.255.248 2,097,152 8 6/30 22 255.255.255.252 4,194,304 4 2
Class B Addressing Guide
CIDR# of Bits
BorrowedSubnetMask
Total # ofSubnets
Total # ofHosts
Usable # ofHosts
/16 0 255.255.0.0 1 65,536 65,534/17 1 255.255.128.0 2 32,768 32,766/18 2 255.255.192.0 4 16,384 16,382/19 3 255.255.224.0 8 8,192 8,190/20 4 255.255.240.0 16 4,096 4,094/21 5 255.255.248.0 32 2,048 2,046/22 6 255.255.252.0 64 1,024 1,022/23 7 255.255.254.0 128 512 510/24 8 255.255.255.0 256 256 254/25 9 255.255.255.128 512 128 126/26 10 255.255.255.192 1,024 64 62/27 11 255.255.255.224 2,048 32 30/28 12 255.255.255.240 4,096 16 14/29 13 255.255.255.248 8,192 8 6/30 14 255.255.255.252 16,384 4 2
Class C Addressing Guide
82
Inside Cover
Top Related