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I'm web programmer and i'm not familiar with networking , i have this next question i have searched many articles on the internet but it was not helpful and i found difficulties with understanding subnetting .. i preach it anyone who can help me to solve this question :)

The question is :

You have been allocated a class B network address: 165.10.0.0/16

Departments   Number of employees
D1                   800
D2                   200
D3                    50
D4                   1600
D5                   210
D6                    60
D7                    60
D8                    36
D9                    210

How to distribute this addresses over the departments with FLSM Method ?

Greg Hewgill
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Nasouh Almrstani
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    where did you get this block from? This is a public block - and since you say - you are not aware of networking, I am just curious, whether you indeed want to expose this public block to the Internet. – gabhijit Jul 01 '15 at 03:49
  • its uni small project i have and its important to me – Nasouh Almrstani Jul 01 '15 at 08:28
  • O ok. So this is just a project where you are not going to configure it on a 'real router'. Then the simplest is - you can fit all of them into different /20 networks Because highest is - 1600 that should easily fit in 12 bits - so create multiple /20 networks for each of them. – gabhijit Jul 01 '15 at 09:54

2 Answers2

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•   Max host in D4 which is 1600 and will take it to calculate subnet 
•   Address Size :  2^11-2=2046
•   Subnet Mask : 255.255.248.0
•   To calculate new subnet mask: 256 - 248 = 8
•   8 between networks and each one has 2046 addresses.




  Dep   Number of Employees Address Size    Subnet Address
        Subnet Mask Start - End IP  Broadcast Address
    D4  1600    2046    165.10.0.0/21   255.255.248.0   165.10.0.1 - 165.10.0.254   165.10.0.255
    D1  800 2046    165.10.8.0/21
        255.255.248.0   165.10.8.1 - 165.10.8.254   165.10.8.255
    D5  210 2046    165.10.16.0/21
        255.255.248.0   165.10.16.1 - 165.10.16.254 165.10.16.255
    D9  210 2046    165.10.24.0/21
        255.255.248.0   165.10.24.1 - 165.10.24.254 165.10.24.255
    D2  200 2046    165.10.32.0/21
        255.255.248.0   165.10.32.1 - 165.10.32.254 165.10.32.255
    D6  60  2046    165.10.40.0/21
        255.255.248.0   165.10.40.1 - 165.10.40.254 165.10.40.255
    D7  60  2046    165.10.48.0/21
        255.255.248.0   165.10.48.1 - 165.10.48.254 165.10.48.255
    D3  50  2046    165.10.56.0/21
        255.255.248.0   165.10.56.1 - 165.10.56.254 165.10.56.255
    D8  36  2046    165.10.64.0/21
        255.255.248.0   165.10.64.1 - 165.10.64.254 165.10.64.255

sorry for unstyled formatted text i couldn't paste my table here ..

Nasouh Almrstani
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Host bits will be 11... Means each n/w will have 2^11 hosts N/w bits will be 21... Means there will be (16-11) ie 5.. ie 2^5 n/w ie 32 n/w SM = 255.255.248.0

Rudrani
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