Another subnetting section that I need help with. Thank you!! Classical Subnetti
ID: 3585982 • Letter: A
Question
Another subnetting section that I need help with. Thank you!!
Classical Subnetting 200.35.1.0/24 Question 18 (1 point) Assume that you have been assigned the 200.35.1.0/24 network block. You must address up to twenty (20) hosts on each subnet. Specify the minimum number of host bits that are required. (Hint: This is the number of host bits to "keep," not the number of subnet bits to "steal.") ab Save Question 19 (1 point) Given the minimum number of host bits necessary to address twenty (20) hosts, use the formula (2N - 2) to calculate the maximum number of hosts that can be addressed on each subnet. (Hint: "N" is the number of host bits that remain after subtracting the subnet bits.) ab SaveExplanation / Answer
question 18 :
answer is : 5 bits
the minimum number of host bits required are 5 bits.
5^2 = 32 hosts
question 19:
answer is : 3 bits
since, 5 bits required for host. the remaining 3 bits are used
for subnetting.
question 20:
answer is: 8 subnets
2^3 =8
question 21:
binary notation subnet decimal notation
11001000 00100011 00000001 00000000 200.35.1.0
11001000 00100011 00000001 00100000 200.35.1.32
11001000 00100011 00000001 01000000 200.35.1.64
11001000 00100011 00000001 01100000 200.35.1.96
11001000 00100011 00000001 10000000 200.35.1.128
11001000 00100011 00000001 10100000 200.35.1.160
11001000 00100011 00000001 11000000 200.35.1.192
11001000 00100011 00000001 11100000 200.35.1.224
question 26:
6 th subnet is 200.35.1.192
first ip address 200.35.1.193 11001000 00100011 00000001 11000001
second ip address 200.35.1.194 11001000 00100011 00000001 11000010
third ip address 200.35.1.195 11001000 00100011 00000001 11000011
next to last ip address 200.35.1.223 11001000 00100011 00000001 11011111
last ip adress 200.35.1.222 11001000 00100011 00000001 11011110
question 27.
subnet broadcast address is 200.35.1.223 11001000 00100011 00000001 11011111