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Questions of subnet and IP addressing consisting normal conversion to simple form like decimal to binary and much more
Typology: Exercises
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100011111001010100 (^1011100101011100101100011101001) (^101111010001101000001010010110010) (^10010101011001111111010101000101) (^1101001101010011001010010101010) (^1010101000110010010101001011000) (^11010110001101011010100001011) (^0010101001101001010010)
10000110
IP Addressing
and
Subnetting
Student Name:
IP Address Classes Class A 1 – 127 (Network 127 is reserved for loopback and internal testing)Leading bit pattern 0 00000000.00000000.00000000.
Class B 128 – 191 Leading bit pattern 10 10000000.00000000.00000000. Class C 192 – 223 Leading bit pattern 110 11000000.00000000.00000000. 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. Class B 172.16.0.0 to 172.31.255. Class C 192.168.0.0 to 192.168.255.
Default Subnet Masks Class A 255.0.0. Class B 255.255.0. Class C 255.255.255.
Network. Host. Host. Host Network. Network. Host. Host Network. Network. Network. Host
Inside Cover
Produced by: Robb Jonesjonesr@careertech.net Frederick County Career & Technology CenterCisco Networking Academy Frederick County Public SchoolsFrederick, Maryland, USA for taking the time to check this workbook for errors,Special Thanks to Melvin Baker and Jim Dorsch 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
0 0 1 0 0 0 1 0
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 _____
Using the IP address and subnet mask shown write out the network address:
188.10.18.2 _____________________________ 255.255.0. 10.10.48.80 _____________________________ 255.255.255. 192.149.24.191 _____________________________ 255.255.255. 150.203.23.19 _____________________________ 255.255.0. 10.10.10.10255.0.0.0 _____________________________
186.13.23.110 _____________________________ 255.255.255. 223.69.230.250 _____________________________ 255.255.0. 200.120.135.15 _____________________________ 255.255.255. 27.125.200.151 _____________________________ 255.0.0. 199.20.150.35255.255.255.0 _____________________________
191.55.165.135 _____________________________ 255.255.255. 28.212.250.254 _____________________________ 255.255.0.
Using the IP address and subnet mask shown write out the host address:
188.10.18.2 _____________________________ 255.255.0. 10.10.48.80 _____________________________ 255.255.255. 222.49.49.11 _____________________________ 255.255.255. 128.23.230.19 _____________________________ 255.255.0. 10.10.10.10255.0.0.0 _____________________________
200.113.123.11 _____________________________ 255.255.255. 223.169.23.20 _____________________________ 255.255.0. 203.20.35.215 _____________________________ 255.255.255. 117.15.2.51 _____________________________ 255.0.0. 199.120.15.135255.255.255.0 _____________________________
191.55.165.135 _____________________________ 255.255.255. 48.21.25.54 _____________________________ 255.255.0.
Default 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 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. Class BClass C 255.255.0.0255.255.255. ANDING Equations: 1 AND 1 = 1 1 AND 0 = 00 AND 1 = 0 0 AND 0 = 0 Sample: What you see... IP Address: 192. 100. 10. 33 What you can figure out in your head... Address Class:Network Portion: (^) 192. 100. 10 C. 33 Host 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.
portion of the address.ANDING with the default subnet mask allows your computer to figure out the network
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 0 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 (255. 255. 255. 0)
(192. 100. 10. 33) (192. 100. 10. 0)
Network Host
Default Subnet Mask:^ IP Address: AND:
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 fivesmaller 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 theaddress 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 stillAND the IP address against the custom subnet mask to see what the network portion is and which subnetwork it belongs to. IP Address:Custom Subnet Mask: 192. 100. 10. 0255.255.255. Address Ranges: 192.10.10.0192.100.10.16 to 192.100.10.31 to 192.100.10. 192.100.10.32 to 192.100.10.47192.100.10.48 to 192.100.10.63 (Range in the sample below) 192.100.10.64 to 192.100.10.79192.100.10.80 to 192.100.10. 192.100.10.96 to 192.100.10.111192.100.10.112 to 192.100.10. 192.100.10.128 to 192.100.10.143192.100.10.144 to 192.100.10. 192.100.10.160 to 192.100.10.175192.100.10.176 to 192.100.10. 192.100.10.192 to 192.100.10.207192.100.10.208 to 192.100.10. 192.100.10.224 to 192.100.10.239192.100.10.240 to 192.100.10.
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 0 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 (255. 255. 255. 240)
(192. 100. 10. 33) (192. 100. 10. 32) Four bits borrowed from the hostportion of the address for the custom subnet mask. The ANDING process of the four borrowed bitsshows which range of IP addresses this particular address will fall into.
Network Network^ Sub Host Custom Subnet Mask:^ IP Address: AND:
(Invalid range) (Invalid range)
to to to to
Class C Address unsubnetted: 195.195.195.195.195. 223223223223223 ..... 5050505050 ..... 00000 195.223.50.0 to 195.223.50. Class C Address subnetted (2 bits borrowed):
Notice that the subnet andbroadcast addresses match.
Use thezero and broadcast ranges if... 2 - 2 formula and don’t use the Classful routing is used RIP version 1 is used Theconfigured on your router no ip subnet zero command is
Use thebroadcast ranges if... 2 formula and use the zero and Classless routing or VLSM is used RIP version 2, EIGRP, or OSPF is used Theconfigured on your router (default setting) ip subnet zero command is No other clues are given
When to use which formula to determine the number of subnets s 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 255 addresses in the classful C address or just the 62 usable addresses in the broadcast range? Thedetermine which formula to use, and whehter or not you can use the first and last subnets. Use CCNA and CCENT certification exams may have questions which will require you to the 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^ s formula.
Number 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
128 64 32 16 8 4 2 1 - Binary values
Number ofSubnets - 2 4 8 16 32 64 128 256 256 128 64 32 16 8 4 2 - HostsNumber of
Show your work for Problem 1 in the space below.
numbers to the left of the line to^ Add^ the binary value create the custom subnet mask.
Observe the total number ofhosts. Subtract 2 for the number ofusable 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
Show your work for Problem 3 in the space below.
/26 bits used for the network and indicates the total number of subnetworkaddress. All bits remaining belong portion of the to the host portion of the address.
128 64 32 16 8 4 2 1
Number ofSubnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
numbers to the left of the line to^ Add the binary value create the custom subnet mask. (^1024) - 1,
Observe the total number ofhosts. Subtract 2 for the number ofusable hosts.
Subtract 2 for the total number ofsubnets to get the usable number of subnets.
128 64 32 16 8 4 2 1 .....
512
Binary values -
Number ofHosts (^) -
10242048 4,0968,19216,38432,76865,
32,76816,3848,1924,0962,0481,024 512 65,
Number 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
128 64 32 16 8 4 2 1 - Binary values
Number ofSubnets - 2 4 8 16 32 64 128 256 256 128 64 32 16 8 4 2 - HostsNumber of
Show your work for Problem 4 in the space below.
Number 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
Show your work for Problem 6 in the space below.
Number ofSubnets - 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 (^) - 512 1,0242,0484,0968,19216,38432,76865,536131, 262,144524,2881,048,5762,097,1524,194, .
131,07265,53632,76816,3848,1924,0962,0481,024 512 4,194,3042,097,1521,048,576524,288262,
Number 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
Show your work for Problem 7 in the space below.
128 64 32 16 8 4 2 1
Number ofSubnets - 2 4 8 16 32 64 128 256.
. 256 128 64 32 16 8 4 2
128 64 (^32 16 8 4 2 ) .....
512 Binary values -
Number ofHosts (^) - 10242048 4,0968,19216,38432,76865,
32,76816,3848,1924,0962,0481,024 512 65,
