CSI 345L Homework, Spring 2012

ClassA Class B Class C Class D Class E Class A Class B Class C Class A Class B Class C Inside Cover 1 127 128 191 192 223 224 239 240 255 IP Address Classes (Network 127 is reserved for loopback and internal testing) Leading bit pattern 0 09090000.00990000.00090000.00090000 Network Leading bit pattern 10 40000000.00000000.00000000.00090000 Network . Network . Host se Host Leading bit pattern 110 — 11900090.00000000.09000900.00900000 Network . Network . Network . Host (Reserved for multicast) (Reserved for experimental, used for research) Private Address Space 10.0.0.0 to 10.255.255.255 172.16.0.0 to 172.31.255.255 192.168.0.0 to 192.168.255.255 255.0.0.0 255.255.0.0 255.255.255.0 Default Subnet Masks Produced by: Robb Jones jonesr@careertech.net Frederick County Career & Technology Center Cisco Networking Academy Frederick County Public Schools Frederick, Maryland, USA Special Thanks to Melvin Baker and Jim Dorsch for taking the time to check this workbook for errors. Ye Binary To Decimal Conversion 32 16 8 4 2 41 Answers Scratch Area 0 141 0 0 4. 0 146 128 64 —————_ 16 32 119 _ 2 16 1 1 0 1 { { He g 1 1 1 1 1 1 2 / 0) 0 9) 1 0 1 119 1 1 1 0 4 1 QO Q 0) 4 1 1 00011011 10101010 01101111 41111000 00100000 01010101 00111110 00000011 11101101 11000000 Decimal To Binary Conversion Use all 8 bits for each problem 128 64 32 16 8 4 2 1 = 255 Scratch Area 7 7? ¢ OO F¢ Ft 4 @_ 238 435 34 oo 1 0 8 OO 1 OC xm HO ~“Z 7“ ee 423 m4 O 50 es 255 4 200 a 40 o 138 1 43 250 407 224 114 192 172 400 419 57 98 179 3 Address 10.250.1.4 150.10.15.0 192.14.2.0 148.17.9.1 193.42.1.1 126.8.156.0 220.200.23.1 230.230.45.58 177.100.18.4 119.18.45.0 249.240.80.78 4199.155.77.56 117.89.56.45 215.45.45.0 199.200.15.0 95.0.21.90 33.0.0.0 158.98.80.0 219.21.56.0 Address Class Identification Class Circle the network portion of these addresses: (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. 4 Network & Host Identification Circle the host portion of these addresses: 171,299.31) 198.125.87.177 223.250.200.222 17.45.222.45 126 201.54.231 191.41.35.112 55.25.169.227 92.15.155.2 23.102.45.254 48.17.9.155 00.25.1.1 195.0.21 98 25.250.135.46 171.102.77.77 55.250.5.5 218.155.230.14 10.250.1.1 Default Subnet Masks Write the correct default subnet mask for each of the following addresses: 477.100.18.4 255 .255 .0.0 119.18.45.0 191.249.234.191 223.23.223.109 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 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 itis. 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 255.0.0.0 Class B 255.255.0.0 Class C 255.255.255.0 ANDING Equations: 1AND 1=1 1 AND 0=0 OAND 1=0 OAND 0=0 Sample: What you see... IP Address: 192. 100.10. 33 What you can figure out in your head... Address Class: Cc Network Portion: 192.100.10.33 Host 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: PELE LEEPER ETP REL THE Default Subnet Mask: }11111111.1111111.11111111 9000000 Ol (255. 255. 255 0) AND: 11000000.1100100.00001010)00000000|% 100. 10 . 0) (192.100, 10.33) ANDING with the default subnet mask allows your computer to figure out the network portion of the address.