TC 1830107 - its my exam note PDF

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its my exam note...

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BGC TRUST UNIVERSITY BANGLADESH

ASSIGNMENT ON TELECOMMUNICATION ENGINEERING

Submitted to:

Submitted by:

Md. Abdul Wahab

Shuveccha Barua

Lecturer

Roll – 1830107 Semester – 7th

Department of CSE Sec – B Bgc trust university,Bangladesh Batch – 30th

Date of submission: 23rd March,2021. A.N.S TO THE QUESTION N0-1(a) Group 1: For this group, each customer needs 256 addresses. This means that (log2 256)= 8 bits are needed to define each host. The prefix length is then ,32− 8 = 24. The addresses are: 1st Customer:190.100.0.0/24 - 190.100.0.255/24 2nd Customer:190.100.1.0/24 - 190.100.1.255/24 3rd Customer:190.100.2.0/24 - 190.100.2.255/24 4th Customer:190.100.3.0/24 - 190.100.3.255/24 ………. 63th Customer:190.100.62.0/24 - 190.100.62.255/24 64th Customer:190.100.63.0/24 - 190.100.63.255/24 Total usage of address= 64 x 256 = 16,384

A.N.S TO THE QUESTION N0-1(b) Group 2: For this group, each customer needs 128 addresses. This means that (log 2 128)= 7 bits are needed to define each host. The prefix length is then ,32 - 7 = 25. The addresses are, 1st Customer: 190.100.64.0/25 - 190.100.64.127/25 2nd Customer:190.100.64.128/25 - 190.100.64.255/25 3rd Customer:190.100.65.0/25 - 190.100.65.127/25 4th Customer:190.100.65.128/25 - 190.100.65.255/25

…….. 127th Customer:190.100.127.0/25 - 190.100.127.127/25 128th Customer:190.100.127.128/25 - 190.100.127.255/25 Total usage of address = 128 x 128 = 16,384

A.N.S TO THE QUESTION N0-1(c) Group 3 : For this group, each customer needs 64 addresses. This means that (log 2 64)= 6 bits are needed to each host. The prefix length is then ,32 - 6 = 26. The addresses are, 1st Customer: 190.100.128.0/26 - 190.100.128.63/26 2nd Customer: 190.100.128.64/26 - 190.100.128.127/26 3rd Customer: 190.100.128.128/26 - 190.100.128.191/26 4th Customer: 190.100.128.192/26 - 190.100.128.255/26 …….. 127th Customer: 190.100.159.128/26 - 190.100.159.191/26 128th Customer: 190.100.159.192/26 - 190.100.159.255/26 Total usage of address = 128 x 64 = 8,192 Number of granted addresses to the ISP= 65,536 Number of allocated addresses by the ISP=(16,384+16,384+8,192)=40,960 Number of available addresses= (65,536-40,960)=24,576

A.N.S TO THE QUESTION N0-2(a) Group 1: For this group, each customer needs 256 addresses. This means that (log2 256) =8 bits are needed to define each host. The prefix length is then ,32− 8 = 24. The addresses are: 1st Customer:194.168.0.0/24 - 194.168.0.255/24 2nd Customer:194.168.1.0/24 - 194.168.1.255/24 ………. 127th Customer:194.168.126.0/24 - 194.168.126.255/24 128th Customer:194.168.127.0/24 - 194.168.127.255/24 Total usage of address= 128 x 256 = 32,768 Group 2: For this group, each customer needs 128 addresses. This means that (log2 64)= 7 bits are needed to define each host. The prefix length is then ,32− 7 = 25. The addresses are: 1st Customer:194.168.128. 0/25 - 194.168.128.127/25 2nd Customer:194.168.128.128/25 - 194.168.128.255/25 ……….

63th Customer:194.168.159.0/25 - 194.168.159.127/25 64th Customer:194.168.159.128/25 - 194.168.159.255/25 Total usage of address= 64 x 128= 8,192....