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   :apa: スプリットショックウイルス † (splitshockvirus@mstdn.starnix.network)'s status on Tuesday, 25-Feb-2025 02:51:45 JST :apa: スプリットショックウイルス †

   • かき@GNUsocialJP

   I'm finna (going to) get some Orange Juice and sausage from the grocer and then get coffee and donuts.

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     :apa: スプリットショックウイルス † (splitshockvirus@mstdn.starnix.network)'s status on Tuesday, 25-Feb-2025 02:51:41 JST :apa: スプリットショックウイルス †

     in reply to

     • かき@GNUsocialJP
     • idk anymore, man

     @benis_redux Oh yeah and I guess I forgot to say what that subnet is. It's 255.255.255.240

     How you find that is again with the last octet.

     11110000

     An easy way to figure this out would but to understand the following

     10000000 = 128
     01000000 = 64
     00100000 = 32
     00010000 = 16
     00001000 = 8
     00000100 = 4
     00000010 = 2
     00000001 = 1
     00000000 = 0

     So for 11110000 you can just add the following

     10000000 = 128
     +
     01000000 = 64
     +
     00100000 = 32
     +

     00010000 = 16

     11110000 = 240

     Which is how you get 255.255.255.240 from /28

     The subnet mask in binary is
     11111111 11111111 11111111 11110000

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     :apa: スプリットショックウイルス † (splitshockvirus@mstdn.starnix.network)'s status on Tuesday, 25-Feb-2025 02:51:42 JST :apa: スプリットショックウイルス †

     in reply to

     • かき@GNUsocialJP
     • idk anymore, man

     @benis_redux I typed six bits but meant four.

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     :apa: スプリットショックウイルス † (splitshockvirus@mstdn.starnix.network)'s status on Tuesday, 25-Feb-2025 02:51:43 JST :apa: スプリットショックウイルス †

     in reply to

     • かき@GNUsocialJP
     • idk anymore, man

     @benis_redux @benis_redux 14 addresses, and the broadcast is 192.168.8.143.

     Reasoning a /28 has 16 addresses avaliable in the but the first one 128 goes to the network, the last one 143 is the broadcast. so 192.168.8.129-142 is the usable range.

     How you find out how many address there are in a CIDR.

     An IPv4 address is 32 bits
     00000000 00000000 00000000 00000000

     A /28 CIDR implies that the network gets 28 bits out of the 32, so the host bits are 4 bits. You could express the mask in binary like so.

     11111111 11111111 11111111 11110000 in the case of the network
     or
     00000000 00000000 00000000 00001111 in the case of the hosts

     Focusing on the last octet 00001111

     That number in binary is 16. A quick way to calculate that number is you know that four bits are going to the hosts so you can do 24 which equals 16, you always minus two off the end result for network and broadcast address which is why it's 14. You can apply this logic for the remainder of calculating CIDR addresses. And they will have you doing this in the CCNA exam.

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     :apa: スプリットショックウイルス † (splitshockvirus@mstdn.starnix.network)'s status on Tuesday, 25-Feb-2025 02:51:44 JST :apa: スプリットショックウイルス †

     in reply to

     • かき@GNUsocialJP
     • idk anymore, man

     @benis_redux

     How many usable host addresses are on the network192.168.8.128/28 and what is its broadcast address :gunhand:

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     idk anymore, man (benis_redux@cawfee.club)'s status on Tuesday, 25-Feb-2025 02:51:44 JST idk anymore, man

     in reply to
     @splitshockvirus eight addresses and 192.168.8.255 (I haven't started the books I showed you so I reserve the right to be wrong)
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     idk anymore, man (benis_redux@cawfee.club)'s status on Tuesday, 25-Feb-2025 02:51:45 JST idk anymore, man

     in reply to
     @splitshockvirus thanks for the update 👍
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     :apa: スプリットショックウイルス † (splitshockvirus@mstdn.starnix.network)'s status on Tuesday, 25-Feb-2025 08:30:05 JST :apa: スプリットショックウイルス †

     in reply to

     • かき@GNUsocialJP
     • Stephen Brooks 🦆

     @sjb I forgor the sauce age, because I remember I had venison in yhe freezer.

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     Stephen Brooks 🦆 (sjb@mstdn.io)'s status on Tuesday, 25-Feb-2025 08:30:06 JST Stephen Brooks 🦆

     in reply to

     @splitshockvirus