關於IPv4各class所占比例及計算可用位址，路由器的classless和classful routing

ghost999991

我自己練習有不懂的地方，因為也沒解答，所以請求各位協助!

A. Assume the total available number of IPv4 addresses to be 2^32=4.29 billion addresses.
       Based on this number, calculate the followings:
     - What percentage of the available addresses are class A?
     - What percentage of the available addresses are class B?
     - What percentage of the available addresses are class C?

我想法是
class A => 2^31 / 2^32 = 1/2 = 50%
class B => 2^30 / 2^32 = 1/2^2 = 25%
class C => 2^29 / 2^32 = 1/2^3 = 12.5%

    B. Assume that in the current allocation of IPv4 space, the efficiency of using IPv4
       addresses are as follows:
       Class A: 65% of total available IP addresses are used
       Class B: 72% of total available IP addresses are used
       Class C: 83% of total available IP addresses are used

      With the above values,
      a. Find the percentage of IPv4 addresses which are currently in use?
      b. What percent of total available IPv4 addresses are used for multicast (or Class D)?
      c. What percent of total addresses in IPV4 space does not belong to classes, A, B, C or D??
      d. You can see the status of IPv4 Address Space Registry in the following link:
         http://www.iana.org/assignments/ ... 4-address-space.xml
         As seen in the link, many Class A addresses were either controlled or assigned to large            corporations. For example, IBM, HP, Xerox Corporation, and Ford Motor Company each
         were assigned a full class A network.
         If the total usable available address space is the sum of addresses available in classes A, B,
         and C, what percentage of this total useable space was assigned to just four companies?
      e. Is assignment of addresses justified? Why?

我想法是
a. 0.5*0.65 + 0.25*0.72 + 0.125*0.83 = 60.875%

b. the percentage of class D is 2^28 / 2^32 = 0.0625

   0.0625 : 0.875 = x : 0.60875  => x = 0.04348 ~= 0.04 = 4%

c. 0.60875+ 0.04 = 0.64875  ==>  1 - 0.64875 = 0.35125 ~= 0.35 = 35%

  d和e我就完全看不懂，所以嘗試寫了下，我也不知題目給那連結到底用不用的到

d. 0.60875 / 4   因為只能分四家公司?

e.  不會

另一題是關於路由器的classful和classless routing      

   In this question, we study the effect of classless routing on the IPv4 routing table size.
    Assume that a router has the following entries in its routing table:

    Destination     Mask              Next hop
    129.2.128.0     255.255.128.0     11.10.25.9
    128.8.192.0     255.255.192.0     12.10.29.13
    129.2.0.0       255.255.240.0     12.10.29.13
    129.2.32.0      255.255.240.0     11.10.25.9
    12.10.20.0      255.255.255.0     142.15.12.49
    129.2.48.0      255.255.240.0     142.15.12.49
    12.10.21.0      255.255.255.0     13.49.25.17
    210.32.240.0    255.255.248.0     142.15.26.11
    215.10.42.0     255.255.255.0     125.42.37.23
    215.0.0.0       255.255.254.0     125.42.37.23

    A. If we do not use classless method and switch the router to a full classful method,
       exactly, how many entries will the router need? Fill those entries in the first two
       columns of the table below.
       (Note: you do not need to fill out the “next hop” column in the table below because
       the field would not be known with the give information when switching back to classful
       method. Also,you may not need all the entries of the table.)

    B. Is it correct to say that using classless routing, the size of routing table
       always increases?

我想法是

A.

我算法是(以第一個舉例)

Destination => 129.2.128.0 = 10000001 . 00000010 . 10000000 . 00000000
     AND
Mask  => 255.255.128.0  = 11111111 . 11111111. 10000000 . 00000000

結果   => 10000001 . 00000010 . 10000000 . 00000000
           =   129.2.128.0


所以做完發現跟題目表格一樣....

    Destination     Mask              
    129.2.128.0     255.255.128.0     
    128.8.192.0     255.255.192.0     
    129.2.0.0       255.255.240.0     
    129.2.32.0      255.255.240.0     
    12.10.20.0      255.255.255.0   
    129.2.48.0      255.255.240.0   
    12.10.21.0      255.255.255.0   
    210.32.240.0    255.255.248.0     
    215.10.42.0     255.255.255.0     
    215.0.0.0       255.255.254.0   

B. no ,或許一樣?  因為A.算出來的size跟題目一樣