ipv4 exercices (part 3 & 4)
Calculate the following ipv4-configurations, given ip-address and netmask,
- the net-ID
- the broadcast address
- CIDR-notation and decimal notation of the netmask
- maximum possible hosts in this network ...
a. 189.14.105.22 netmask 255.255.248.0
max #hosts: 32 -/21 = 11 >> 2power11-2=2046
b. 19.200.219.137/19
netmask decimal = 255.255.224.0 (/19) power = 32-19=13
max hosts = 2power13-2=8190
c. 55.114.166.148/17
d. 102.132.178.38/25
e. 186.82.147.248 mask 255.255.252.0
f. 208.24.133.47/26
g. 5.200.232.114 netmask 255.255.254.0
h. 10.237.12.23 mask 255.255.224.0
g. 11.195.17.26 netmask 255.248.0.0
max hosts: power is 32 - 13 = 19 -->> max hosts is 2power19 - 2 = 524286
h. 181.194.88.186/9
i. 172.18.253.143/12
j. 88.190.124.208/15
e. 83.86 .186.22/31
f. 112.180.205.95/10
g. 192.168.1.84/16
h. 10.198.17.3/8
i. 195.157.120.88/30
j. 192.83.5.116/19
k. 206.219.37.30/26
l. 25.184.112.165/29
m. 170.104.71.189/22
n. 94.161.222.76/28
o. 115.51.203.69/12
p. 70.167.13.25/27
q. 5.72.141.171/26
r. 40.216.148.29/18
s. 175.179.89.127/16
t. 16.167.138.46/23
u. 32.136.47.180/11
v. 100.46.200.134/13
w. 45.42.34.185/22
x. 7.181.65.168/10
y. 40.174.210.85/22
z. 125.199.164.46/10
aa. 182.126.163.23/26
ab. 64.14.18.113/11
ac. 11.221.209.83/29
ad. 106.23.31.79/18
ae. 135.172.94.138/15
af. 102.187.216.194/25
ag. 196.162.183.205/28
ah. 24.118.219.202/22
ai. 94.39.179.73/17
aj. 193.162.77.21/21
ak. 172.196.209.50/26
al. 116.206.141.60/14
am. 182.87.176.124/10
an. 189.56.181.125/26
ao. 147.175.144.95/18
ap. 214.31.89.58/11
aq. 84.170.183.160/12
ar. 52.14.35.131/23
as. 146.26.115.38/26
at. 62.220.84.185/29
au. 11.200.169.99/12
av. 141.12.144.18/22
aw. 171.204.97.90/16
ax. 125.151.188.83/30