49445184 IP Addressing and Subnetting Workbook Instructors Version 1 5
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11111110 1010100 10001111100 1011100101011100 101100011101001 1011110100011010 00001010010110010 1001010101100111 1111010101000101 1101001101010011 001010010101010 1010101000110010 010101001011000 110101100011010 11010100001011 001010100110 1001010010 10011000 0101 11010011 10000110 10010101 00011011 001 IP Addressing and Subnetting Workbook Instructor’s Edition Version 1.5
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Transcript of 49445184 IP Addressing and Subnetting Workbook Instructors Version 1 5
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11111110
101010010001111100
10111001010111001011000111010011011110100011010000010100101100101001010101100111111101010100010111010011010100110010100101010101010101000110010010101001011000110101100011010110101000010110010101001101001010010
100110000101
1101001110000110
1001010100011011
001IP Addressing
andSubnetting
Workbook
Instructors Edition
Version 1.5
-
IP Address Classes
Class A 1 127 (Network 127 is reserved for loopback and internal testing)Leading bit pattern 0 00000000.00000000.00000000.00000000
Class B 128 191 Leading bit pattern 10 10000000.00000000.00000000.00000000
Class C 192 223 Leading bit pattern 110 11000000.00000000.00000000.00000000
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
Network . Host . Host . Host
Network . Network . Host . Host
Network . Network . Network . Host
Inside Cover
Produced by: Robb [email protected]
Frederick County Career & Technology CenterCisco Networking Academy
Frederick County Public SchoolsFrederick, Maryland, USA
Special Thanks to Melvin Baker and Jim Dorschfor 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 WorkbooksACLs - Access Lists Workbooks
VLSM 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.
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Binary To Decimal Conversion
128 64 32 16 8 4 2 1 Answers Scratch Area1 0 0 1 0 0 1 0 146
0 1 1 1 0 1 1 1 119
1 1 1 1 1 1 1 1 255
1 1 0 0 0 1 0 1 197
1 1 1 1 0 1 1 0 246
0 0 0 1 0 0 1 1 19
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
00011011 27
10101010 170
01101111 111
11111000 248
00100000 32
01010101 85
00111110 62
00000011 3
11101101 237
11000000 192
128162
146
643216421
119
1
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Decimal To Binary Conversion
128 64 32 16 8 4 2 1 = 255 Scratch Area_________________________________________ 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
-3214-86
-42
-20
1 1 1 0 1 1 1 0 34-32
2-20
0 0 1 0 0 0 1 0
Use all 8 bits for each problem
2
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
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Address Class Identification
Address Class
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 _____
230.230.45.58 _____
177.100.18.4 _____
119.18.45.0 _____
249.240.80.78 _____
199.155.77.56 _____
117.89.56.45 _____
215.45.45.0 _____
199.200.15.0 _____
95.0.21.90 _____
33.0.0.0 _____
158.98.80.0 _____
219.21.56.0 _____
A
B
C
B
C
A
C
D
B
A
E
C
A
C
C
A
A
B
C3
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Network & Host Identification
Circle the network portionof 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 ofthese 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
4
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5Network Addresses
Using the IP address and subnet mask shown write out the network address:
188.10.18.2 _____________________________255.255.0.0
10.10.48.80 _____________________________255.255.255.0
192.149.24.191 _____________________________255.255.255.0
150.203.23.19 _____________________________255.255.0.0
10.10.10.10 _____________________________255.0.0.0
186.13.23.110 _____________________________255.255.255.0
223.69.230.250 _____________________________255.255.0.0
200.120.135.15 _____________________________255.255.255.0
27.125.200.151 _____________________________255.0.0.0
199.20.150.35 _____________________________255.255.255.0
191.55.165.135 _____________________________255.255.255.0
28.212.250.254 _____________________________255.255.0.0
188 . 10 . 0 . 0
10 . 10 . 48 . 0
192 . 149 . 24 . 0
150 . 203 . 0 . 0
10 . 0 . 0 . 0
186 . 13 . 23 . 0
223 . 69 . 0 . 0
200 . 120 . 135 . 0
27 . 0 . 0 . 0
199 . 20 . 150 . 0
191 . 55 . 165 . 0
28 . 212 . 0 . 0
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Host Addresses
Using the IP address and subnet mask shown write out the host address:
188.10.18.2 _____________________________255.255.0.0
10.10.48.80 _____________________________255.255.255.0
222.49.49.11 _____________________________255.255.255.0
128.23.230.19 _____________________________255.255.0.0
10.10.10.10 _____________________________255.0.0.0
200.113.123.11 _____________________________255.255.255.0
223.169.23.20 _____________________________255.255.0.0
203.20.35.215 _____________________________255.255.255.0
117.15.2.51 _____________________________255.0.0.0
199.120.15.135 _____________________________255.255.255.0
191.55.165.135 _____________________________255.255.255.0
48.21.25.54 _____________________________255.255.0.0
0 . 0 . 18 . 2
0 . 0 . 0 . 80
0 . 0 . 0 . 11
0 . 0 . 230 . 19
0 . 10 . 10 . 10
0 . 0 . 0 . 11
0 . 0 . 23 . 20
0 . 0 . 0 . 215
0 . 15 . 2 . 51
0 . 0 . 0 . 135
0 . 0 . 0 . 135
0 . 0 . 25 . 54
6
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Default Subnet Masks
Write the correct default subnet mask for each of the following addresses:
177.100.18.4 _____________________________
119.18.45.0 _____________________________
191.249.234.191 _____________________________
223.23.223.109 _____________________________
10.10.250.1 _____________________________
126.123.23.1 _____________________________
223.69.230.250 _____________________________
192.12.35.105 _____________________________
77.251.200.51 _____________________________
189.210.50.1 _____________________________
88.45.65.35 _____________________________
128.212.250.254 _____________________________
193.100.77.83 _____________________________
125.125.250.1 _____________________________
1.1.10.50 _____________________________
220.90.130.45 _____________________________
134.125.34.9 _____________________________
95.250.91.99 _____________________________
255 . 255 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 0 . 0
255 . 255 . 255 . 0
255 . 0 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 255 . 0
255 . 255 . 255 . 0
255 . 0 . 0 . 0
255 . 255 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 0 . 0
255 . 255 . 255 . 0
255 . 0 . 0 . 0
255 . 0 . 0 . 0
255 . 255 . 255 . 0
255 . 255 . 0 . 0
255 . 0 . 0 . 0
7
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ANDING WithDefault subnet masks
Every IP address must be accompanied by a subnet mask. By now you should be able to lookat an IP address and tell what class it is. Unfortunately your computer doesnt think that way.For your computer to determine the network and subnet portion of an IP address it mustAND 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 withthe subnet mask in binary.
ANDING with the default subnet mask allows your computer to figure out the networkportion of the 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 11 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 01 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
(255 . 255 . 255 . 0)
(192 . 100 . 10 . 33)
(192 . 100 . 10 . 0)
Network Host
IP Address:Default Subnet Mask:
AND:
8
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ANDING WithCustom 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 outsideworld still sees the network as 192.100.10.0, but the internal computers and routers see fivesmaller subnetworks. Each independent of the other. This can only be accomplished by usinga custom subnet mask. A custom subnet mask borrows bits from the host portion of theaddress to create a subnetwork address between the network and host portions of an IPaddress. In this example each range has 14 usable addresses in it. The computer must stillAND the IP address against the custom subnet mask to see what the network portion is andwhich subnetwork it belongs to.
IP Address: 192 . 100 . 10 . 0Custom Subnet Mask: 255.255.255.240
Address Ranges: 192.10.10.0 to 192.100.10.15192.100.10.16 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
In the next set of problems you will determine the necessary information to determine thecorrect subnet mask for a variety of IP addresses.
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 11 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 01 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
(255 . 255 . 255 . 240)
(192 . 100 . 10 . 33)
(192 . 100 . 10 . 32)
Four bits borrowed from the hostportion of the address for thecustom subnet mask.
The ANDING process of the four borrowed bitsshows which range of IP addresses thisparticular address will fall into.
SubNetwork HostNetwork
IP Address:Custom Subnet Mask:
AND:
9
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How to determine the number of subnets and thenumber of hosts per subnet
Two formulas can provide this basic information:
Number of subnets = 2 (Second subnet formula: Number of subnets = 2 - 2)
Number of hosts per subnet = 2 - 2
Both formulas calculate the number of hosts or subnets based on the number of binary bitsused. For example if you borrow three bits from the host portion of the address use thenumber of subnets formula to determine the total number of subnets gained by borrowing thethree bits. This would be 2 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 inthe host portion and apply this to the number of hosts per subnet formula If five bits are in thehost portion of the address this would be 2 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 fromthe range. The first address in every range is the subnet number. The last address in everyrange is the broadcast address. These two addresses cannot be assigned to any device inthe network which is why you have to subtract two addresses to find the number of usableaddresses in each range.
For example if two bits are borrowed for the network portion of the address you can easilydetermine the number of subnets and hosts per subnets using the two formulas.
h
3
ss
5
10
195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0The number of subnetscreated by borrowing 2bits is 2 or 2 x 2 = 4subnets.
2
The number of hosts created byleaving 6 bits is 2 - 2 or2 x 2 x 2 x 2 x 2 x 2 = 64 - 2 = 62usable hosts per subnet.
6
What about that second subnet formula:
Number of subnets = 2 - 2
In some instances the first and last subnet range of addresses are reserved. This is similar tothe 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 isalso the subnet number for the classful subnet address.
The last range of addresses is the broadcast subnet. The broadcast address for the lastsubnet in the broadcast subnet is the same as the classful broadcast address.
s
-
11
195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0195.223.50.0195.223.50.64195.223.50.128195.223.50.192
195.223.50.63195.223.50.127195.223.50.191195.223.50.255
(Invalid range)
(Invalid range)
totototo
(0)(1)(2)(3)
Class C Address unsubnetted:
195. 223 . 50 . 0195. 223 . 50 . 0195. 223 . 50 . 0195. 223 . 50 . 0195. 223 . 50 . 0
195.223.50.0 to 195.223.50.255
Class C Address subnetted (2 bits borrowed):Notice that the subnet andbroadcast addresses match.
Use the 2 - 2 formula and dont use thezero and broadcast ranges if...
Classful routing is used
RIP version 1 is used
The no ip subnet zero command isconfigured on your router
Use the 2 formula and use the zero andbroadcast ranges if...
Classless routing or VLSM is used
RIP version 2, EIGRP, or OSPF is used
The ip subnet zero command isconfigured on your router (default setting)
No other clues are given
When to use which formula to determine the number of subnetss s
The primary reason the the zero and broadcast subnets were not used had to do pirmarily withthe broadcast addresses. If you send a broadcast to 195.223.255 are you sending it to all 255addresses 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 todetermine which formula to use, and whehter or not you can use the first and last subnets. Usethe chart below to help decide.
Bottom line for the CCNA exams; if a question does not give you any clues as to whether or notto allow these two subnets, assume you can use them.
This workbook has you use the number of subnets = 2 formula.s
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Custom Subnet Masks
Problem 1Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
1414192.10.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0192 . 10 . 10 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
Show your work for Problem 1 in the space below.
Add the binary value numbers to the left of the line tocreate the custom subnet mask.
C
255 . 255 . 255 . 0
255 . 255 . 255 . 240
16
16
14
4
Observe the total number ofhosts.
Subtract 2 for the number ofusable hosts.
16-214
12
1286432
+16240
-
13
Custom Subnet Masks
Problem 2Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
100060165.100.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
Show your work for Problem 2 in the space below.
Add the binary value numbers to the left of the line tocreate the custom subnet mask.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
Observe the total number ofhosts.
Subtract 2 for the number ofusable hosts.
64-262
128643216842
+1255
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
128+64192
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
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Custom Subnet Masks
Problem 3Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
148.75.0.0 /26
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 3 in the space below.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
/26 indicates the total number ofbits used for the network andsubnetwork portion of theaddress. All bits remaining belongto the host portion of the address.
148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0148 . 75 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
Add the binary value numbers to the left of the line tocreate the custom subnet mask.
1024-2
1,022
Observe the total number ofhosts.
Subtract 2 for the number ofusable hosts.
64-262
Subtract 2 for the total number ofsubnets to get the usable number ofsubnets.
128643216842
+1255
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
128+64192
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
14
-
Custom Subnet Masks
Problem 4Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
630210.100.56.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0210 . 100 . 56 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
Show your work for Problem 4 in the space below.
C
255 . 255 . 255 . 0
255 . 255 . 255 . 224
8
32
30
3
12864
+32224
8-26
32-230
15
-
16
Custom Subnet Masks
Problem 5Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
630195.85.8.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0 195 . 85 . 8 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
Show your work for Problem 5 in the space below.
C
255 . 255 . 255 . 0
255 . 255 . 255 . 224
8
32
30
3
12864
+32224
32-230
8-26
-
17
Custom Subnet Masks
Problem 6Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
126131,070118.0.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 6 in the space below.
118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0118. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
Number of Subnets - 2 4 8 16 32 64 128 256 .
. 256 128 64 32 16 8 4 2
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
Number ofHosts -
A
255 . 0 . 0 . 0
255 . 254. 0 . 0
128
131,072
131,070
7
12864321684
+2254
128-2
126
131,072 -2131,070
512
1,0242,0484,0968,192
16,38432,76865,536
131,072262,144524,2881,048,5762,097,1524,194,304.
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
-
18
Custom Subnet Masks
Problem 7Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
200015178.100.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 7 in the space below.
B
255 . 255 . 0 . 0
255 . 255 . 255 . 224
2,048
32
30
11
128643216842
+1255
2,048 -22,046
32-230
178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
-
19
Custom Subnet Masks
Problem 8Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
345200.175.14.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 8 in the space below.
C
255 . 255 . 255 . 0
255 . 255 . 255 . 192
4
64
62
2
200 . 175 . 14 . 0 0 0 0 0 0 0 0200 . 175 . 14 . 0 0 0 0 0 0 0 0200 . 175 . 14 . 0 0 0 0 0 0 0 0200 . 175 . 14 . 0 0 0 0 0 0 0 0200 . 175 . 14 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 162 4 8 162 4 8 162 4 8 162 4 8 16 32 64 128 256
Number of256 128 64 32 1616161616 8 4 2 - Hosts
4-22
64-262
128+64240
-
Custom Subnet Masks
Problem 9Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
601,000128.77.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 9 in the space below.
20
B
255 . 255 . 0 . 0
255 . 255 . 252 . 0
64
1,024
1,022
6
1286432168
+4252
1,024 -21,022
64-262
128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 . 77 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
-
Custom Subnet Masks
Problem 10 Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
60198.100.10.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 10 in the space below.
21
C
255 . 255 . 255 . 0
255 . 255 . 255 . 192
4
64
62
2
198 . 100 . 10 . 0 0 0 0 0 0 0 0198 . 100 . 10 . 0 0 0 0 0 0 0 0198 . 100 . 10 . 0 0 0 0 0 0 0 0198 . 100 . 10 . 0 0 0 0 0 0 0 0198 . 100 . 10 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 162 4 8 162 4 8 162 4 8 162 4 8 16 32 64 128 256
Number of256 128 64 32 1616161616 8 4 2 - Hosts
64-262
4-22
128+64192
-
Number of Subnets - 2 4 8 16 32 64 128 256 .
Custom Subnet Masks
Problem 11Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
250101.0.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 11 in the space below.
22
A
255 . 0 . 0 . 0
255 . 255 . 0 . 0
256
65,536
65,534
8
101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0101. 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
. 256 128 64 32 16 8 4 2
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
Number ofHosts -
128643216842
+1255
256-2
254
65,536 -265,534
512
1,0242,0484,0968,192
16,38432,76865,536
131,072262,144524,2881,048,5762,097,1524,194,304.
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
-
Custom Subnet Masks
Problem 12Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
5218.35.50.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 12 in the space below.
23
C
255 . 255 . 255 . 0
255 . 255 . 255 . 224
8
32
30
3
218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 162 4 8 162 4 8 162 4 8 162 4 8 16 32 64 128 256
Number of256 128 64 32 1616161616 8 4 2 - Hosts
64-262
4-22
12864
+32224
-
Custom Subnet Masks
Problem 13Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
25218.35.50.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 13 in the space below.
C
255 . 255 . 255 . 0
255 . 255 . 255 . 224
8
32
30
3
218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0218 . 35 . 50 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 162 4 8 162 4 8 162 4 8 162 4 8 16 32 64 128 256
Number of256 128 64 32 1616161616 8 4 2 - Hosts
8-26
32-230
12864
+32224
24
-
Custom Subnet Masks
Problem 14Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
10172.59.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 14 in the space below.
B
255 . 255 . 0 . 0
255 . 255 . 240 . 0
16
4,096
4,094
4
1286432
+16240
16-214
4,096-2
4,094
172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
25
-
Custom Subnet Masks
Problem 15Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
50172.59.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 15 in the space below.
26
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
128643216842
+1255
64-262
1,024 -21,022
172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0172 . 59 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1
Number of Subnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts -
102420484,0968,192
16,384
32,76865
,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,5
36
128+64192
-
Custom Subnet Masks
Problem 16Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
2923.0.0.0
_______
_______________________________
_______________________________
___________________
___________________
___________________
___________________
Show your work for Problem 16 in the space below.
27
A
255 . 0 . 0 . 0
255 . 255 . 255 . 224
524,288
32
30
19
Number of Subnets - 2 4 8 16 32 64 128 256 .
23 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 023 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 023 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 023 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 023 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
. 256 128 64 32 16 8 4 2
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
Number ofHosts -
12864
+32224
32-230
524,288 -2524,286
512
1,0242,0484,0968,192
16,38432,76865,536
131,072262,144524,2881,048,5762,097,1524,194,304.
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
-
_______________________________________________
________________________
________________________
______________________________________
192.10.10.48 to 192.10.10.63
192 . 10 . 10 . 112
192 . 10 . 10 . 207
192.10.10.129 to 192.10.10.142
28
Subnetting
Problem 1Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
1414192.10.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
C
255 . 255 . 255 . 0
255 . 255 . 255 . 240
16
16
14
4
What is the 4thsubnet range?
What is the subnet numberfor the 8th subnet?
What is the subnetbroadcast address for
the 13th subnet?
What are the assignableaddresses for the 9th
subnet?
-
Show your work for Problem 1 in the space below.
192.10.10.0 to 192.10.10.15192.10.10.16 to 192.10.10.31192.10.10.32 to 192.10.10.47192.10.10.48 to 192.10.10.63192.10.10.64 to 192.10.10.79192.10.10.80 to 192.10.10.95192.10.10.96 to 192.10.10.111192.10.10.112 to 192.10.10.127192.10.10.128 to 192.10.10.143192.10.10.144 to 192.10.10.159192.10.10.160 to 192.10.10.175192.10.10.176 to 192.10.10.191192.10.10.192 to 192.10.10.207192.10.10.208 to 192.10.10.223192.10.10.224 to 192.10.10.239192.10.10.240 to 192.10.10.255
192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0 192. 10 . 10 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
16-214
16-214
1286432
+16240
The binary value of the last bit borrowed is the range. In thisproblem 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 broadcastaddress.
Custom subnetmask
Usable hostsUsable subnets
29
0101010101010101
0011001100110011
0000111100001111
0000000011111111
(0)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)
-
_______________________________________________
________________________
________________________
______________________________________
165.100.3.128 to 165.100.3.191
165 . 100 . 1 . 64
165 . 100 . 1 . 127
165.100.2.1 to 165.100.0.62
30
Subnetting
Problem 2Number of needed subnets
Number of needed usable hostsNetwork Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
100060165.100.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
What is the 15thsubnet range?
What is the subnet numberfor the 6th subnet?
What is the subnetbroadcast address for
the 6th subnet?
What are the assignableaddresses for the 9th
subnet?
-
165
. 10
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
016
5 .
100
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
165
. 10
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
016
5 .
100
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
165
. 10
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
016
5.1
00.0
.0to
1
65.1
00.0
.63
165
.100
.0.6
4to
1
65.1
00.0
.127
165
.100.0
.12
8to
1
65.1
00.0
.191
165
.100.0
.192
to
165
.100
.0.2
55
165
.100.1
.0to
1
65.1
00.1
.63
165
.100
.1.6
4to
1
65.1
00.1
.127
165
.100.1
.12
8to
1
65.1
00.1
.191
165
.100.1
.192
to
165
.100
.1.2
55
165
.100.2
.0to
1
65.1
00.0
.63
165
.100
.2.6
4to
1
65.1
00.0
.127
165
.100.2
.12
8to
1
65.1
00.0
.191
165
.100.2
.192
to
165
.100
.0.2
55
165
.100.3
.0to
1
65.1
00.3
.63
165
.100
.3.6
4to
1
65.1
00.3
.127
165
.100.3
.12
8to
1
65.1
00.3
.191
165
.100.3
.192
to
165
.100
.3.2
55
Dow
n to
165
.100
.25
5.1
28
to
16
5.1
00.2
55
.191
165
.100
.25
5.1
92 t
o 1
65.1
00.2
55
.25
5
The
bina
ry v
alue
of t
he la
st b
it bo
rrow
ed is
the
rang
e. In
this
pro
blem
the
rang
e is
64.
The
first
add
ress
in e
ach
subn
et r
ange
is th
esu
bnet
num
ber.
The
last
add
ress
in e
ach
subn
et r
ange
is th
esu
bnet
bro
adca
st a
ddre
ss.
31
64 -2 62U
sabl
eho
sts
128
64 32 16 8 4 2 +1
25
5
128
+64
192
Cus
tom
subn
et m
ask
Show your work for Problem 2 in the space below.
128
64
32 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,768
65,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1
1 1 1 1 0 0 0 0 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
. . . . . . . . . . . . . . .
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
1
1 1
1 1
1 1
1 1
1 1
1(1
022)
(102
3)
-
_______________________________________________
________________________
________________________
______________________________________
32
Subnetting
Problem 3Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
2195.223.50.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 3rdsubnet range?
What is the subnet numberfor the 2nd subnet?
What is the subnetbroadcast address for
the 1st subnet?
What are the assignableaddresses for the 3rd
subnet?
C
255 . 255 . 255 . 0
255 . 255 . 255 . 192
4
64
62
2
195.223.50.128 - 195.223.50.191
195.223.50.64
195.223.50.63
195.223.50.129 - 195.223.50.190
-
Show your work for Problem 3 in the space below.
33
128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 2 4 8 16 32 64 128 256
Number of256 128 64 32 16 8 4 2 - Hosts
195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 0 195. 223 . 50 . 0 0 0 0 0 0 0 001
1 01 1
195.223.50.0195.223.50.64195.223.50.128195.223.50.192
195.223.50.63195.223.50.127195.223.50.191195.223.50.255
128+64192 64-2
62
totototo
(0)(1)(2)(3)
-
_______________________________________________
________________________
________________________
______________________________________
34
Subnetting
Problem 4Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
750190.35.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 15thsubnet range?
What is the subnet numberfor the 13th subnet?
What is the subnetbroadcast address for
the 10th subnet?
What are the assignableaddresses for the 6th
subnet?
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
190.35.3.128 to 190.35.3.191
190.35.3.0
190.35.2.127
190.35.1.65 to 190.35.1.126
-
Show your work for Problem 4 in the space below.
35
128
64 32 16 8 4 2
+
12
52
64 -2 62
190
. 35
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
019
0 . 3
5 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
190
. 35
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
019
0 . 3
5 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
190
. 35
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,768
65,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
128
+64
25
2
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
190.3
5.0
.0to
190.
35.0
.63
190.
35.0
.64
to19
0.3
5.0
.12
719
0.3
5.0
.12
8to
190.3
5.0
.191
190.3
5.0
.192
to19
0.3
5.0
.25
519
0.3
5.1
.0to
190.
35.1
.63
190.
35.1
.64
to19
0.3
5.1
.12
719
0.3
5.1
.12
8to
190.3
5.1
.191
190.3
5.1
.192
to19
0.3
5.1
.25
519
0.3
5.2
.0to
190.
35.2
.63
190.
35.2
.64
to19
0.3
5.2
.12
719
0.3
5.2
.12
8to
190.3
5.2
.191
190.3
5.2
.192
to19
0.3
5.2
.25
519
0.35
.3.0
to19
0.35
.3.6
319
0.35
.3.6
4to
190.
35.3
.12
719
0.3
5.3
.12
8to
190.3
5.3
.191
190.
35.3
.192
to19
0.35
.3.2
55
. . . . . . . . . . . . .
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
-
_______________________________________________
________________________
________________________
______________________________________
36
Subnetting
Problem 5Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
6126.0.0.0
__________
_______________________________
_______________________________
___________________
________________
___________________
___________________
What is the 2ndsubnet range?
What is the subnet numberfor the 5th subnet?
What is the subnetbroadcast address for
the 7th subnet?
What are the assignableaddresses for the 10th
subnet?
A
255 . 0 . 0 . 0
255 . 255 . 255 . 248
2,097,152
8
6
21
126.0.0.8 to 126.0.0.15
126.0.0.32
126.0.0.55
126.0.0.73 to 126.0.0.78
-
Show your work for Problem 5 in the space below.
37
Num
ber
of S
ubne
ts
-
2
4
8
16
32
64
12
8 2
56
.12
6. 0
0 0
0 0
0 0
0 .
0 0
0 0
0 0
0 0
. 0 0
0 0
0 0
0 0
126.
0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
012
6. 0
0 0
0 0
0 0
0 .
0 0
0 0
0 0
0 0
. 0 0
0 0
0 0
0 0
126.
0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
012
6. 0
0 0
0 0
0 0
0 .
0 0
0 0
0 0
0 0
. 0 0
0 0
0 0
0 0
. 25
6 12
8 6
4 3
2
16
8
4
2
Bin
ary
valu
es -
128
64
32
16
8
4
2
1
. . . . . 1
28
64
32
16
8
4
2
1
. . . . . 1
28
64
32
16
8
4
2
1
-
5121,0242,0484,0968,192
16,38432,76
865,53
6
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
.
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
128
64 32 16 8 4 2
+
12
55 8
-2 6
128
64 32 16
+8
248
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
Num
ber
ofH
osts
126.
0.0.
012
6.0.
0.8
126.
0.0.
1612
6.0.
0.24
126.
0.0.
3212
6.0.
0.40
126.
0.0.
4812
6.0.
0.56
126.
0.0.
6412
6.0.
0.72
126.
0.0.
8012
6.0.
0.88
126.
0.0.
9612
6.0.
0.10
412
6.0.
0.11
212
6.0.
0.12
0
to to to to to to to to to to to to to to to to
126.
0.0.
712
6.0.
0.15
126.
0.0.
2312
6.0.
0.31
126.
0.0.
3912
6.0.
0.47
126.
0.0.
5512
6.0.
0.63
126.
0.0.
7112
6.0.
0.79
126.
0.0.
8712
6.0.
0.95
126.
0.0.
103
126.
0.0.
111
126.
0.0.
119
126.
0.0.
127
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
-
_______________________________________________
________________________
________________________
______________________________________
38
Subnetting
Problem 6Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
10192.70.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 9thsubnet range?
What is the subnet numberfor the 4th subnet?
What is the subnetbroadcast address for
the 12th subnet?
What are the assignableaddresses for the 10th
subnet?
C
255 . 255 . 255 . 0
255 . 255 . 255 . 240
16
16
14
4
192.70.10.128 to 192.70.10.143
192.70.10.48
192.70.10.191
192.70.10.145 to 192.70.10.158
-
Show your work for Problem 6 in the space below.
39
. 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 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 162 4 8 162 4 8 162 4 8 162 4 8 16 32 64 128 256
Number of256 128 64 32 1616161616 8 4 2 - Hosts
16-214
128+64240
0101010101010101
11001100110011
111100001111
11111111
192.70.10.0192.70.10.16192.70.10.32192.70.10.48192.70.10.64192.70.10.80192.70.10.96192.70.10.112192.70.10.128192.70.10.144192.70.10.160192.70.10.176192.70.10.192192.70.10.208192.70.10.224192.70.10.240
192.70.10.15192.70.10.31192.70.10.47192.70.10.63192.70.10.79192.70.10.95192.70.10.111192.70.10.127192.70.10.143192.70.10.159192.70.10.175192.70.10.191192.70.10.0207192.70.10.223192.70.10.239192.70.10.255
totototototototototototototototo
192 . 70 . 10 .192 . 70 . 10 .192 . 70 . 10 .192 . 70 . 10 .192 . 70 . 10 .(0)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)
-
40
Subnetting
Problem 7Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
10.0.0.0 /16
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 11thsubnet range?
What is the subnet numberfor the 6th subnet?
What is the subnetbroadcast address for
the 2nd subnet?
What are the assignableaddresses for the 9th
subnet?
A
255 . 0 . 0 . 0
255 . 255 . 0 . 0
256
65,536
65,534
8
10.10.0.0 to 10.10.255.255
10.5.0.0
10.1.255.255
10.8.0.1 to 10.8.255.254
_______________________________________________
________________________
________________________
______________________________________
-
Show your work for Problem 7 in the space below.
41
Num
ber
of S
ubne
ts
-
2
4
8
16
32
64
12
8 2
56
.10
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
010
. 0 0
0 0
0 0
0 0
. 0
0 0
0 0
0 0
0 . 0
0 0
0 0
0 0
0
. 25
6 12
8 6
4 3
2
16
8
4
2
Bin
ary
valu
es -
128
64
32
16
8
4
2
1
. . . . . 1
28
64
32
16
8
4
2
1
. . . . . 1
28
64
32
16
8
4
2
1
-
5121,0242,0484,0968,192
16,38432,76
865,53
6
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
.
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
131,072
262,144
524,288
1,048,576
2,097,152
4,194,304
128
64 32 16 8 4 2
+
12
55
65,5
36
-
265
,534
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
Num
ber
ofH
osts
10.0
.0.0
10.1
.0.0
10.2
.0.0
10.3
.0.0
10.4
.0.0
10.5
.0.0
10.6
.0.0
10.7
.0.0
10.8
.0.0
10.9
.0.0
10.1
0.0
.010
.11.
0.0
10.1
2.0
.010
.13.
0.0
10.1
4.0.0
10.1
5.0
.0
10.0
.25
5.2
55
10.1
.25
5.2
55
10.2
.25
5.2
55
10.3
.25
5.2
55
10.4
.25
5.2
55
10.5
.25
5.2
55
10.6
.25
5.2
55
10.7
.25
5.2
55
10.8
.25
5.2
55
10.9
.25
5.2
55
10.1
0.25
5.2
55
10.1
1.2
55
.25
510
.12.2
55
.25
510
.13.
25
5.2
55
10.1
4.25
5.2
55
10.1
5.2
55
.25
5
to to to to to to to to to to to to to to to to
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
-
_______________________________________________
________________________
________________________
______________________________________
42
Subnetting
Problem 8Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
5172.50.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 4thsubnet range?
What is the subnet numberfor the 5th subnet?
What is the subnetbroadcast address for
the 6th subnet?
What are the assignableaddresses for the 3rd
subnet?
B
255 . 255 . 0 . 0
255 . 255 . 224 . 0
8
8,192
8,190
3
172.50.96.0 to 172.50.127.255
172.50.128.0
172.50.191.255
172.50.64.1 to 172.50.95.254
-
Show your work for Problem 8 in the space below.
43
128
64 +32
22
48,
192
-
28,
190
172
. 5
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
017
2 .
50
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 5
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
017
2 .
50
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 5
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,768
65,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
0 1 0 1 0 1 0 1
1 1 0 0 1 1
1 1 1 1
172
.50.0
.017
2.5
0.3
2.0
172
.50.
64.0
172
.50.9
6.0
172
.50.1
28.
017
2.5
0.1
60.0
172
.50.1
92.0
172
.50.2
24.0
to to to to to to to to
172
.50.3
1.2
55
172
.50.
63.2
55
172
.50.9
5.2
55
172
.50.1
27.
25
517
2.5
0.1
59.
25
517
2.5
0.1
91.2
55
172
.50.2
23.
25
517
2.5
0.2
55
.25
5
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
-
_______________________________________________
________________________
________________________
______________________________________
44
Subnetting
Problem 9Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
28172.50.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 2ndsubnet range?
What is the subnet numberfor the 10th subnet?
What is the subnet broadcastaddress for
the 4th subnet?
What are the assignableaddresses for the 6th
subnet?
B
255 . 255 . 0 . 0
255 . 255 . 255 . 224
2,048
32
30
11
172.50.0.32 to 172.50.0.63
172.50.1.32
172.50.0.127
172.50.0.161 to 172.50.0.190
-
Show your work for Problem 9 in the space below.
45
128
64 32 16 8 4 2
+
12
52 32 -2 30
172
. 5
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
017
2 .
50
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 5
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
017
2 .
50
. 0
0 0
0
0
0 0
0
. 0
0
0 0
0
0
0 0
172
. 5
0 . 0
0
0
0 0
0
0
0 .
0 0
0
0
0 0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,768
65,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
128
64 +32
22
4
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
172.
50.0
.017
2.50
.0.32
172.
50.0
.6417
2.50
.0.96
172.
50.0
.128
172.
50.0
.160
172.
50.0
.192
172.
50.0
.224
172.
50.1.
017
2.50
.1.32
172.
50.1.
6417
2.50
.1.96
172.
50.1.
128
172.
50.1.
160
172.
50.1.
192
172.
50.1.
224
to to to to to to to to to to to to to to to to
172.
50.0
.3117
2.50
.0.63
172.
50.0
.9517
2.50
.0.12
717
2.50
.0.15
917
2.50
.0.19
117
2.50
.0.2
2317
2.50
.0.2
5517
2.50
.1.31
172.
50.1.
6317
2.50
.1.95
172.
50.1.
127
172.
50.1.
159
172.
50.1.
191
172.
50.1.
223
172.
50.1.
255
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
-
_______________________________________________
________________________
________________________
______________________________________
46
Subnetting
Problem 10Number of needed subnets
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
45220.100.100.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 5thsubnet range?
What is the subnet numberfor the 4th subnet?
What is the subnetbroadcast address for
the 13th subnet?
What are the assignableaddresses for the 12th
subnet?
C
255 . 255 . 255 . 0
255 . 255 . 255 . 252
64
4
2
6
220.100.100.16 to 220.100.100.19
220.100.100.12
220.100.100.51
220.100.100.45 to 220.100.100.46
-
Show your work for Problem 10 in the space below.
47
22
0 . 1
00 .
100
. 0
0
0 0
0
0
0 0
22
0 . 1
00 .
100
. 0
0
0 0
0
0
0 0
22
0 . 1
00 .
100
. 0
0
0 0
0
0
0 0
22
0 . 1
00 .
100
. 0
0
0 0
0
0
0 0
22
0 . 1
00 .
100
. 0
0
0 0
0
0
0 0
128
64
32
16
8
4
2
1
-
Bin
ary
valu
es
Num
ber
of
Sub
nets
-
2
4
8
16
2
4
8
16
2
4
8
16
2
4
8
16
2
4
8
16
32
6
4 12
8 2
56
N
umbe
r of
25
6 12
8 64
32
1
6161616 16 8
4
2 -
H
osts
4 -2 212
864 32 16 8 +4
25
2
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
220.
100.
100.
022
0.10
0.10
0.4
220.
100.
100.
822
0.10
0.10
0.12
220.
100.
100.
1622
0.10
0.10
0.20
220.
100.
100.
2422
0.10
0.10
0.28
220.
100.
100.
3222
0.10
0.10
0.36
220.
100.
100.
4022
0.10
0.10
0.44
220.
100.
100.
4822
0.10
0.10
0.52
220.
100.
100.
5622
0.10
0.10
0.60
to to to to to to to to to to to to to to to to
220.
100.
100.
322
0.10
0.10
0.7
220.
100.
100.
1122
0.10
0.10
0.15
220.
100.
100.
1922
0.10
0.10
0.23
220.
100.
100.
2722
0.10
0.10
0.31
220.
100.
100.
3522
0.10
0.10
0.39
220.
100.
100.
4322
0.10
0.10
0.47
220.
100.
100.
5122
0.10
0.10
0.55
220.
100.
100.
5922
0.10
0.10
0.63
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
-
_______________________________________________
________________________
________________________
______________________________________
48
Subnetting
Problem 11Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
8,000135.70.0.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 6thsubnet range?
What is the subnet numberfor the 7th subnet?
What is the subnetbroadcast address for
the 3rd subnet?
What are the assignableaddresses for the 5th
subnet?
B
255 . 255 . 0 . 0
255 . 255 . 224 . 0
8
8,192
8,190
3
135.70.160.0 to 135.70.191.255
135.70.192.0
135.70.95.255
135.70.128.1 to 135.70.159.254
-
Show your work for Problem 11 in the space below.
49
8,19
2 -28,
190
135
. 70
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
013
5 .
70 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
135
. 70
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
013
5 .
70 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
135
. 70
. 0
0
0 0
0
0
0 0
. 0
0
0
0 0
0
0
012
8 6
4 3
2 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,768
65,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
128
64 +32
22
4
0 1 0 1 0 1 0 1
1 1 0 0 1 1
1 1 1 1
135
.70.0
.013
5.7
0.32
.013
5.7
0.64
.013
5.7
0.9
6.0
135
.70.1
28.
013
5.7
0.1
60.0
135
.70.1
92.0
135
.70.2
24.
0
to to to to to to to to
135
.70.
31.2
55
135
.70.
63.2
55
135
.70.9
5.2
55
135
.70.1
27.
25
513
5.7
0.1
59.
25
513
5.7
0.1
91.2
55
135
.70.
22
3.2
55
135
.70.
25
5.2
55
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
-
_______________________________________________
________________________
________________________
______________________________________
50
Subnetting
Problem 12Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
45198.125.50.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 2ndsubnet range?
What is the subnet numberfor the 2nd subnet?
What is the subnetbroadcast address for
the 4th subnet?
What are the assignableaddresses for the 3rd
subnet?
C
255 . 255 . 255 . 0
255 . 255 . 255 . 192
4
64
62
2
198.125.50.64 to 98.125.50.127
198.125.50.64
198.125.50.255
198.125.50.129 to 198.125.50.190
-
Show your work for Problem 12 in the space below.
51
198 . 125 . 50 . 0 0 0 0 0 0 0 0198 . 125 . 50 . 0 0 0 0 0 0 0 0198 . 125 . 50 . 0 0 0 0 0 0 0 0198 . 125 . 50 . 0 0 0 0 0 0 0 0198 . 125 . 50 . 0 0 0 0 0 0 0 0128 64 32 16 8 4 2 1 - Binary values
Number of Subnets - 2 4 8 162 4 8 162 4 8 162 4 8 162 4 8 16 32 64 128 256
Number of256 128 64 32 1616161616 8 4 2 - Hosts
64-262
128+64192
0101
11
198.125.50.0198.125.50.64198.125.50.128198.125.50.192
198.125.50.63198.125.50.127198.125.50.191198.125.50.255
totototo
(0)(1)(2)(3)
-
_______________________________________________
________________________
________________________
______________________________________
52
Subnetting
Problem 13Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
165.200.0.0 /26
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 10thsubnet range?
What is the subnet numberfor the 11th subnet?
What is the subnetbroadcast address for
the 1023rd subnet?
What are the assignableaddresses for the 1022nd
subnet?
B
255 . 255 . 0 . 0
255 . 255 . 255 . 192
1,024
64
62
10
165.200.2.64 to 165.200.2.127
165.200.2.128
165.200.255.191
165.200.255.65 to 165.200.255.126
-
Show your work for Problem 13 in the space below.
53
128
64 32 16 8 4 2
+
12
52
64 -2 62
165
. 2
00 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
165
. 2
00 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
165
. 2
00 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
165
. 2
00 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
165
. 2
00 .
0 0
0
0
0 0
0
0 .
0
0 0
0
0
0 0
0
128
64
32 16
8
4
2
1
Num
ber
of S
ubne
ts
-
2
4
8
16 3
2 6
4 1
28
25
6 .
. 25
6 12
8 6
4 32
16
8
4
2
128
64
32
16
8
4
2
1
.... .
512
Bin
ary
valu
es -
Num
ber
ofH
osts
-
102420484,0968,192
16,384
32,768
65,536
512
1,024
2,048
4,096
8,192
16,384
32,768
65,536
128
+64
25
2
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 1 1 0 0 0 0 1 1 1 1
1 1 1 1 1 1 1 1
. . . . . . . . .
165.
200.
0.0
165.
200.
0.64
165.
200.
0.12
816
5.20
0.0.
192
165.
200.
1.0
165.
200.
1.64
165.
200.
1.128
165.
200.
1.192
165.
200.
2.0
165.
200.
2.64
165.
200.
2.12
816
5.20
0.2.
192
165.
200.
3.0
165.
200.
3.64
165.
200.
3.12
816
5.20
0.3.
192
165.
200.
255.
6416
5.20
0.15
5.12
816
5.20
0.25
5.19
2
to to to to to to to to to to to to to to to to to to to
165.
200.
0.63
165.
200.
0.12
716
5.20
0.0.
191
165.
200.
0.25
516
5.20
0.1.
6316
5.20
0.1.1
2716
5.20
0.1.1
9116
5.20
0.1.
255
165.
200.
2.63
165.
200.
2.12
716
5.20
0.2.
191
165.
200.
2.25
516
5.20
0.3.
6316
5.20
0.3.
127
165.
200.
3.19
116
5.20
0.3.
255
165.
200.
255.
127
165.
200.
255.
191
165.
200.
255.
255
1 1
1 1
1 1
1 1
. 0
11
1 1
1 1
1 1
1 .
1 0
1 1
1 1
1 1
1 1
. 1
1
(10
21)
(10
22
)(1
02
3)
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
-
_______________________________________________
________________________
________________________
______________________________________
54
Subnetting
Problem 14Number of needed usable hosts
Network Address
Address class
Default subnet mask
Custom subnet mask
Total number of subnets
Total number of host addresses
Number of usable addresses
Number of bits borrowed
16200.10.10.0
__________
_______________________________
_______________________________
___________________
___________________
___________________
___________________
What is the 7thsubnet range?
What is the subnet numberfor the 5th subnet?
What is the subnetbroadcast address for
the 4th subnet?
What are the assignableaddresses for the 6th
subnet?
C
255 . 255 . 255 . 0
255 . 255 . 255 . 224
8
32
30
3
200.10.10.192 to 200.10.
