Cost and delay of MUX

Theory
1

If I have a MUX with two inputs: A[n-1:0], B[n-1:0] and a select string: sel[0] (only one bit)
when calculating the cost and delay, should I treat it as a MUX(2:1) or MUX(n:1)?

2

I might be wrong but I’m pretty sure it’s implemented as n 2:1-Muxes, which will make it c(C) = theta(n), d(C) = theta(1)

1 Like
3

this is correct. MUX(2n:n) is implemented by n, MUX(2:1) in parallel