GATE Overflow Test Series | Computer Networks | Test 1 | Question: 25

Question

In the $IPv4$ addressing format, if the number of networks allowed under $\text{Class A, Class B and Class C}$ respectively are $2^a,2^b$ and $2^c,$ the value of $2a + 3b + 5c$ = __________

Answer 1

In $IPv4$, classes are identified by the leading bits.
    
    Leading bit $0\;\text{- class A}$. Network field size $=8\;\text{bits}$. Number of networks $=2^{8-1} = 2^7.$
    
    Leading bits $10\;\text{- class B}$. Network field size $=16\;\text{bits}$.  Number of networks $=2^{16-2} = 2^{14}.$
    
    Leading bit $110\;\text{- class C}$. Network field size $=24\;\text{bits}$.  Number of networks $=2^{24-3} = 2^{21}.$
    
    So, $a=7,b=14,c=21$
    
    $\implies 2a+3b+5c = 161.$

Comments

Sir, No. of networks for Class A should be from 1 to 126 as 0 and 127 are reserved, right?

Ref :

https://www.ciscopress.com/articles/article.asp?p=330807&seqNum=2

https://nptel.ac.in/content/storage2/courses/106105080/pdf/M6L2.pdf