a) That question is worded weirdly. Basically if we divided our /17 in 16 smaller subnets, what would the mask be. First thing to do is find the size of our initial network. The /17 is in the number of network bits out of 32. The remaining number of bits is for host addressing, in this case 15. 2^15 = 32768. Divide that by the desired number of subnets and we have our subnet size. If the number subnets is an odd number you will end up with a size between that doesn't fit into a subnet size, go up to the next larger subnet. In this case our resulting subnet does line up on a bit boundary. I'm going to assume you can divide 32768 by 16 and get 2048. So I will leave you with this, what subnet mask allows for 2048 hosts?
b) The first IP in a subnet reserved as the subnet (or network address) and can't be used to address a host. In which case the second IP in the network is the first host address. Also, the last IP in the network is reserved as the broadcast address and can't be used for a host address (for future questions).
c) We found that in step a, so you got a freebie.
Suppose that you, the network administrator, had to assign the IP address 178.71.128.0/17 to a network. However, the network needs to be first divided into 16 subnets from the available addresses.
a. Determine the subnet mask in decimal notation for a host on one of 16 subnets that will result from creating 16 subnets out of the available 178.71.128.0/17.
b. Suppose that we are numbering the subnets and hosts within subnets sequentially. What will be the IP address, in decimal notation, of the first host of the subnet numbered 1?
c. What is maximum number of hosts that one of the 16 subnets can support?
