Computer Science - Networking - Discussion
Discussion Forum : Networking  - Section 2 (Q.No. 33)
                   
                                       
                                33.
You are working with a network that has the network ID 192.168.10.0. What subnet should you use that supports up to 25 hosts and a maximum number of subnets?
 
                                    Discussion:
14 comments Page 1 of 2.
                
                        Patty said: 
                         
                        3 years ago
                
                B is correct.
2^n-2> = number of host.
2^n-2> = 25 then what number exponent to 2 and near reach of 25 or above
2^5-2> = 25 were 5 match is remaining host bits on host portion, then the borrowed bits from host portion are 3bits...then 3 bits match to 224 in binary ..11100000 last 0 take 0, second 0 from 1, p to n 0 then take 2^5+2^6+2^7 = 224.
                2^n-2> = number of host.
2^n-2> = 25 then what number exponent to 2 and near reach of 25 or above
2^5-2> = 25 were 5 match is remaining host bits on host portion, then the borrowed bits from host portion are 3bits...then 3 bits match to 224 in binary ..11100000 last 0 take 0, second 0 from 1, p to n 0 then take 2^5+2^6+2^7 = 224.
                        Kaa said: 
                         
                        5 years ago
                
                The formula for finding the host is 2n  (where n indicates the number of host bits)
We required 5 host bits (2n = 25) = 32
Now, let's find the subnet mask.
Hence, 5 bits are reserved for the host. There are 3 bits remaining for the network. Count the value of all the remaining three bits (27 + 26 + 25 = 224).
Therefore, the new subnet mask you will get is 255.255.255.224.
                We required 5 host bits (2n = 25) = 32
Now, let's find the subnet mask.
Hence, 5 bits are reserved for the host. There are 3 bits remaining for the network. Count the value of all the remaining three bits (27 + 26 + 25 = 224).
Therefore, the new subnet mask you will get is 255.255.255.224.
                        Gee said: 
                         
                        7 years ago
                
                Please, anyone,explain the correct answer.
                
                        Tharindu said: 
                         
                        7 years ago
                
                B is correct.
It is about subnet @Sibasish.
                It is about subnet @Sibasish.
                        Geetha said: 
                         
                        7 years ago
                
                Yes, D is the correct answer.
                
                        Kailas Naik said: 
                         
                        8 years ago
                
                Right answer is D.
                
                        Jumth said: 
                         
                        8 years ago
                
                The correct answer is option D.
                
                        Jyothi said: 
                         
                        9 years ago
                
                Yes, Option D is correct.
                
                        Sibasish said: 
                         
                        9 years ago
                
                We required 25 subnets.
2^4<100<2^5 OR 16<100<32.
no of subnets are 2^5 = 32 (no of 1's are 5).
Since default subnet mask of class c is 255.255.255.0.
i.e. 11111111.11111111.11111111.00000000 so 5 more 1 will get add so it will be;
11111111.11111111.11111111.11111000 i.e. 255.255.255.248.
                2^4<100<2^5 OR 16<100<32.
no of subnets are 2^5 = 32 (no of 1's are 5).
Since default subnet mask of class c is 255.255.255.0.
i.e. 11111111.11111111.11111111.00000000 so 5 more 1 will get add so it will be;
11111111.11111111.11111111.11111000 i.e. 255.255.255.248.
                        Abhishek said: 
                         
                        9 years ago
                
                Nice explanation, Thank you all.
                Post your comments here:
 
            
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