Given a Class A network with the /20 prefix, determine the subnet mask in binary.?
I'll answer your questions one by one 1) Subnet mask in binary - 11111111.11111111.11110000.00000000 How it is calculated CIDR is 20 so mark that number from the left xxxxxxxx.xxxxxxxx.xxxx[20th bit]xxxx.xxxxxxxx each octet is 8 bits which I represent with x. Everything before the marking is 1 and everything after that is 0 2) Subnet mask in decimal - 255.255.240.0 How it is calculated Convert the binary to decimal let me show the first octet 11111111 = 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = 128+64+32+16+8+4+2+1 The result is 255 Take the 3rd octet now 11110000 = 2^7 + 2^6 + 2^5 + 2^4 + 0 + 0 + 0 + 0 = 128+64+32+16 The result is 240 3) Maximum number of hosts - 4094 (usable IP addresses) How it is calculated Host bit = 32 - CIDR = 32 - 20 = 12 No. of IP addresses = 2^12 = 4096 First IP is network IP and last IP is broadcast IP so 4096 - 2 = 4094 - Jesin ( http://jesin.tk )
Given Class A network 19.0.0.0 with a subnet mask of 255.240.0.0.?
Subnet portion is 11 bits. (8 in the first byte, 3 in the second byte) Increment is 0, 16, 32, 48, 64, 80, 96, 112, ...etc First six subnets: 19.0.0.0, 19.16.0.0, 19.32.0.0, 19.48.0.0, 19.64.0.0, 19.80.0.0 Host Range of the forth subnet: 19.48.0.1 - 19.48.255.254 Broadcast of the forth subnet: 19.48.255.255 good luck
Cisco Router and Subnets?
The problem with the first IP address was that it was a network address, if you were to use .1 for the fourth octet it would work. For address assignment you cannot use the network and broadcast addresses, the network is the first IP in the subnet while the broadcast is the last. The reason the second address didn't work was because you used a /12 subnet mask and not a /16, it would have worked if it was a /16 and wasn't a network address. The range for a /16: * 172.24.0.0 - 172.24.255.255 The range for a /12: * 172.16.0.0 - 172.31.255.255 So it would've worked if the second interface had an IP address of 172.32.0.1. Also the reason it worked even though you used 24.0.0 was because 16.0.0 was the network address so it was okay. Given that the IP address listed should be /16 I would recommend using either .1 at the end or .254, that's usually what people set for gateways. Same for the /24 addresses, except append to the third octet instead of the second, so: 192.168.0.1 192.168.1.1 192.168.2.1 etc. My advise would be to learn how to subnet. :) Also for the static routes you're going to want to do this: B1: * 0.0.0.0 0.0.0.0 s0/0/0 10.0.0.1 HQ: * 172.24.0.0 255.252.0.0 s0/0/0 10.0.0.2 * 192.168.0.0 255.255.252.0 s0/0/1 10.0.0.6 B2: * 0.0.0.0 0.0.0.0 s0/0/0 10.0.0.5 # I set the HQ as .1 and the branches as .2 The 0.0.0.0 0.0.0.0 thing is a default route, it basically means if I don't know where to send this I will resort to this and send it out that way. Because directly-connected routes have a lower administrative distance (0) they're chosen before static routes (1). So it knows how to get to it's own networks already. The 255.252.0.0 and 255.255.252.0 thing is summarized, basically includes all of the networks because it's a larger subnet. Hope this helps, Cheers @SiriusParadox I don't think he has a choice, he's taking CCNA Exploration so it's probably due or something tomorrow xD