Constructing Higher MUX using Lower MUX derivation of formula

To construct a {\color{Red} M\times 1} using {\color{Red} N\times 1} 


no. of levels (K) =     \left \lceil log _{n}^{m}\right \rceil

and Total number of MUX needed(n)=   {\color{Red} \sum_{p=1}^{k}\left \lceil M/N^{p} \right \rceil}  



1. Construct 32\times 1 using 2\times 1


Total number of levels= \left \lceil log_{2}^{32} \right \rceil = 5

Total number of MUX= \left \lceil 32/2^{1} \right \rceil +\left \lceil 32/2^2 \right \rceil+\left \lceil 32/2^3 \right \rceil +\left \lceil 32/2^4 \right \rceil+\left \lceil 32/2^5 \right \rceil

                                            = 16+8+4+2+1=31



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