How to calculate the broadcast address?
How to calculate the broadcast address?
I handle the network address well, and the initial host address is also acceptable. To find the last host address, you need a broadcast address and then reduce the fourth octet by one. This gives the correct last host address. In addresses with all zeros, it always shows 255, which works. However, when the third octet has 128, the broadcast address appears as 135 instead of the expected 128. The fourth octet is 255 because it’s zero, and that’s why it appears correctly.
the subnet is 255.255.255.252 or /30, which means each network uses only 4 IP addresses—just 2 bits are missing, resulting in 2^2=4 IPs. then subtracting the two reserved ones leaves 2 usable IPs. the network address is 70.195.99.132. the first usable IP in this range is 70.195.88.133, the last is 70.195.99.134, and the broadcast is 70.195.99.135. i hope this matches what you needed.
Your mask is /30, meaning you're using the last two bits as IP addresses, giving you four IPs: 70.195.99.132, 70.195.99.133, 70.195.99.134, and 70.195.99.135. The final one is the broadcast address.
this involves the subnet mask, the range for 142.34.128.0 is 255.255.128.0 or /17. the /17 indicates you'll use half of the third block in each network, which means not all addresses are usable—some are reserved for IPs. on this subnet, the network would look like 142.34.0.0 to 142.34.127.255, then 142.34.128.0 to 142.34.255.255, and so forth. the network address is 142.34.128.0, starting at 142.34.128.1, and when you reach 142.34.128.255 it jumps to 142.34.129.255. this pattern continues until you hit 142.34.255.255, which is the broadcast address.
I approached it another way, yet the outcome remained consistent. I flipped the mask in reverse, swapping 1s and 0s. After converting the binary mask to decimal, I combined it with the network address. The same process was followed in the second case. Appreciate the assistance!
Additional data is placed in the first host address. To determine the last host's address, reduce one from the fourth octet of the broadcast address. The example shows 142.34.255.254, which equals 142.34.255.253 after subtraction. This process applies only to the fourth octet.
Yes, the final host is 142.34.255.254. If you meant the last usable IP is found by summing the IPs and subtracting one from the last block, that's correct.
You're looking for a more straightforward way to determine the number of subnets based on the mask. One alternative is to count the number of '1's in the last octet after converting the binary mask to decimal. Each group of 8 bits indicates a potential subnet boundary. If you have a pattern like 11111111, it means 8 ones, which equals 2 subnets. This method avoids complex calculations and directly gives you the result.
well, you kinda have a mistake; the last block only has 7 bits, so you're missing a 0 the last block for /28 would be 11110000, and you can determine the size of the network by the value of first 1 after the 0s, in this case that's 16 networks will be 15.40.17.0 15.40.17.16 . . . 15.40.17.128 15.40.17.144 the last exercise is exaclty the same concept except that you're dealing with the 3rd block and not the 4th