DNS and IP tending to.


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Educator: Sam Nanavaty. How does a switch know where to course the data when you just sort in a URL (e.g., www.yahoo.com) in the Internet program?. Teacher: Sam Nanavaty. Teacher: Sam Nanavaty. What is an IP address?. Educator: Sam Nanavaty. IP tending to. 32 bits in length4 octetsFive classes A,B,C,D and EOnly A,B and C utilized for naming devices.D utilized for multicasting groupsE is
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DNS and IP tending to Instructor: Sam Nanavaty

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How does a switch know where to course the data when you essentially sort in a URL (e.g., www.yahoo.com) in the Internet program? Teacher: Sam Nanavaty

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Instructor: Sam Nanavaty

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What is an IP address? Teacher: Sam Nanavaty

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IP tending to 32 bits long 4 octets Five classes A,B,C,D and E Only A,B and C utilized for naming gadgets. D utilized for multicasting bunches E is held for test purposes Instructor: Sam Nanavaty

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# of hosts = 2 n - 2 (where n = # of bits in host Id) Why? Have ID = all 1\'s is not allowed as this alludes to communicate address Host Id = all 0\'s is not allowed as this alludes to "this system" Instructor: Sam Nanavaty

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Examples Identify Network address, class, and figure out whether it is legitimate host address also. 222.10.1.1 127.12.1.98 97.1.255.255 197.17.0.255 0.12.252.1 10.0.1.0 220.0.0.254 Instructor: Sam Nanavaty

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Subnetting Steal the bits from the host part of the IP deliver and add it to the system divide (gives more systems and less has per system) Can you think about the advantages of subnetting? Teacher: Sam Nanavaty

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Subnet cover assurance Convert IP addr and default veil into parallel Identify your base system addr Determine what number of bits are expected to accomplish fancied number of subnets and broaden the 1\'s in the subnet veil by this sum Now you know subnet partition and the host segment of the IP address. Ensure that the said veil still gives enough has per subnet (counting some space for development) Instructor: Sam Nanavaty

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Masking principle Mask dependably is relegated from left to ideal in the bit arrange Every cover contains adjoining 1\'s i.e., 255.255.255.192 is right, notwithstanding, 255.240.255.192 is mistaken Hosts on a similar system, must utilize the same subnet cover Subnets that are all 0\'s or all 1\'s are NOT permitted naturally (in private environment you may utilize these nonetheless) Instructor: Sam Nanavaty

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Subnet cover assurance Instructor: Sam Nanavaty

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Answers Instructor: Sam Nanavaty

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Calculating the scope of locations given a subnet veil IP Network address : 192.168.1.0 Subnet Mask: 255.255.255.248 ( 248 = 11111 000 ) This yields up to 30 subnets (2 5 - 2) with up to 6 has (2 3 - 2) per subnet 192.168.1.0 – 192.168.1.7 Discard as Subnet Id = 0! ( 00000 000 – 00000 111 ) 192.168.1.8 – 192.168.1.15 ( 00001 000 – 00001 111 ) (dispose of 1 st and last IP addr as host Id can\'t be all 0\'s or 1\'s) 192.168.1.16 – 192.168.1.23 ( 00010 000 - 00010 111 ) (dispose of 1 st and last IP addr) . . . 192.168.1.240 – 192.168.1.247 ( 11110 000 – 11110 111 ) (dispose of first and last IP addr) 192.168.1.248 – 192.168.1.255 (dispose of subnet Id = every one of the 1\'s) ( 11111 000 – 11111 111 ) Remove first and last subnets and in addition first and last IP addresses for each subnet The last legitimate scope of locations are as per the following: 192.168.1.9 – 192.168.1.14 192.168.1.17 – 192.168.1.22 192.168.1.241 – 192.168.1.246 Instructor: Sam Nanavaty

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Practice illustration IP address : 192.168.1.0 Subnet Mask: 255.255.255.224 This yields up to ________ subnets with up to _____ has per subnet Now compute the scope of locations for this subnet cover Next decide the substantial IP addresses in this range Instructor: Sam Nanavaty

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IP Network address : 192.168.1.0 Subnet Mask: 255.255.255.224 (224 = 111 00000) This yields up to 6 subnets with up to 30 has per subnet Range : 192.168.1.0 – 192.168.1.31 ( 000 00000 – 000 11111 ) (dispose of as subnet ID =0) 192.168.1.32 – 192.168.1.63 ( 001 00000 - 001 11111 ) (dispose of addr w/have =0 and 255) 192.168.1.64 – 192.168.1.95 ( 010 00000 - 010 11111 ) (dispose of addr w/have =0 and 255) 192.168.1.96 – 192.168.1.127 ( 011 00000 - 011 11111 ) (dispose of addr w/have =0 and 255) 192.168.1.128 – 192.168.1.159 ( 100 00000 - 100 11111 ) (dispose of addr w/have =0 and 255) 192.168.1.160 – 192.168.1.191 ( 101 00000 - 101 11111 ) (dispose of addr w/have =0 and 255) 192.168.1.192 – 192.168.1.223 ( 110 00000 - 110 11111 ) (dispose of addr w/have =0 and 255) 192.168.1.224 – 192.168.1.255 ( 111 00000 - 111 11111 ) (dispose of as subnet ID = 255) Instructor: Sam Nanavaty

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Address Mask A = (D & M) Where: A = 32 bit IP address M = 32 bit address veil D = Destination address Instructor: Sam Nanavaty

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CIDR Notation 128.10.0.0/16 The number after the/demonstrates the quantity of 1\'s in the subnet cover. Used to segment addresses: 128.211.0.0/16 ISP 128.211.0.16/28 Network A (128.211.0.17-128.211.0.30) 128.211.0.32/28 Network B IP saves have address zero (means arrange) IP saves have address all 1\'s for communicate) IP holds organize prefix 127/8 for loopback Instructor: Sam Nanavaty

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172.16.210.0/22 Which subnet does this deliver have a place with? Educator: Sam Nanavaty

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201.100.5.68/28 Find the subnet this deliver has a place with. Educator: Sam Nanavaty

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If you are utilizing a subnet veil of 255.255.255.224, can the accompanying IP deliver be alloted to a host utilizing the given subnet cover. 217.168.166.192 Instructor: Sam Nanavaty

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ISP 172.16.1.1/24 192.168.100.17/28 Identify the legitimate IP address task for the workstation 192.168.100.20 255.255.255.240 DG 172.16.1.1 192.168.100.30 255.255.255.240 DG 192.168.100.17 192.168.100.19 255.255.255.248 DG 172.16.1.1 Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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Instructor: Sam Nanavaty

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