1. For each mask length: 30, 29, 28, 27, 26, 25, answer the following items a. N
ID: 663669 • Letter: 1
Question
1. For each mask length: 30, 29, 28, 27, 26, 25, answer the following items
a. Number of usable IPs
b. Mask number
2. Given a class C network, when we need to arrange 4 subnets (5 host, 20 host, 40 host, 80 host),
a. What are their mask number and length?
b. How many usable IPs are left (excluding arranged subnets)?
3. Given a topology: S1-R1-R2-S2 in a 192.168.1.0/24 network:
S1 need 20 IPs; needs 40 IPs.
Describe IP configure of required interfaces in each router and network ID of all subnets.
Explanation / Answer
1.a &b ) 2^27 (since for mask length 30 we must have mask number or mask bit as 5(since 5 bits gives us 2^5 =32 addresses) so available ip address is 2^(32-mask number))
for case of 29 : again 2^27 for the same reason as above , again here the mask number is 5 .
for case of 28 : again 2^27 ip addresses. and mask number is 5
for case of 27 : again 2^27 ip addresses. and mask number is 5
for case of 26 : again 2^27 ip addresses. and mask number is 5
for case of 25 : again 2^27 ip addresses. and mask number is 5
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