172.18.129.0/24
172.18.130.0/24
172.18.132.0/24
172.18.133.0/24
The four segments of the same parts we don't move it,
172.16.? .?

10000001, 129,
10000010, 130,
10000100, 132,
10000101, 133,
The same top five binary
As we know, have a 32-bit IP address, four quarter
No section 8, in 172.16 the two did not change,
Can I add it is concluded that the network number 8 + 8 + 5=21
Because the previous five same, behind three different is change
1 s and 0 s, in turn, change, remove the three changes, namely the three into a 0000 - plus in front of the same five 10000
10000000=128
The summary route for 172.16.128.0/21
In the nets broken front four can pass
Thus reducing the routing entry, reduce the burden of router
Enter the 10000000