I am trying to understand why the calculating the last delay is \$max(t,mt)+2t\$. I suppose thats derived from \$max(max(t,mt), t)+t\$? But how do you extract the \$t\$ from \$max(..., t)\$ out.


I also dont get the next part ...

whats m? and how is \$S_i=max(t,mt)+t\$ and \$C_{i+1}=max(t,mt)+2t\$ simplified to \$2t\$ and \$3t\$ respectively?

  • \$\begingroup\$ It's about the race between X/Y '0/0' inputs and C 'mt' \$\endgroup\$ – kenny Sep 22 '11 at 11:17
  • \$\begingroup\$ @kenny, what about them? I don't understand ... I also updated my question \$\endgroup\$ – Jiew Meng Sep 22 '11 at 13:01
  • \$\begingroup\$ max(x,y) takes the maximum of x or y and it's algebra \$\endgroup\$ – kenny Sep 22 '11 at 13:16
  • \$\begingroup\$ oh yes, I understand that part. What I dont understand is how I get \$C_{i+1}=max(t,mt)+2t\$. What I got, from my own understanding is \$C_{i+1}=max(max(t,mt)+t, mt)+t\$. How do I simplify down to \$C_{i+1}=max(t,mt)+2t\$? Also I dont get whats \$m\$ in the update. How do I simplify from \$C_{i+1}=max(t,mt)+2t\$ to \$C_5=9t\$ for example? I may know \$i\$ but not \$t\$? \$\endgroup\$ – Jiew Meng Sep 22 '11 at 14:11
  • \$\begingroup\$ @Brian Carlton - I previously removed the homework tag as it's deprecated, so I rolled back to that version. \$\endgroup\$ – stevenvh Sep 22 '11 at 15:50

The answer to the first question is that if you take max(max(t,mt)+t,mt)+t, you can bring the t in the inner brackets, obtaining max(t+t, mt+t): since the latter is always bigger than mt (unless you have a negative t) you can simplify with max(t+t, mt+t)+t and then take out again the +t and obtain the value shown.

For the second part, m is again the coefficient that multiplies t: let's take your case.

m=7, so max(t, 7t)+2t=9t as shown.

It's easy!


With the 'mt' for the C input, this just means that the input isn't ready until mt time after the X and Y inputs. Each gate adds propagation delay t, so this just tells you that Cin was produced by m gates.

About those two gates that produce Cout, and the nested max() -

max(max(mt,t), t) reduces to max(mt,t) :

let f(m,t) = max(max(mt,t),t)
let g(m,t) = max(mt,t)

if mt > t, then
f(mt,t) equals max(max(mt,t), t) equals max(mt, t) equals mt
g(mt,t) equals max(mt, t) equals mt

if mt <= t, then
f(m,t) equals max(max(mt,t), t) equals max(t, t), equals t
g(m,t) equals max(mt, t) equals t

So, for any m and t, functions f() and g() are equivalent. I.e., max(max(mt,t),t) reduces to max(mt,t) !

Now, with that out of the way, you should see that the t wasn't extracted from the max() expression; that t delay was incurred by the rightmost AND gate. It was the inputs to that AND gate that were ready at time max(mt, t), and its output was ready t units after that, for max(mt,t) + t. Likewise, the OR that produces the C output adds another t of propagation delay, giving you the max(mt,t) + 2t result.


You said you have,


Start as far in as possible:

max(t,mt)+t = mt+t

And make your way outward.

max(max(t,mt)+t,mt) => max(mt+t,mt) = mt+t

mt+t +t = mt + 2t

Ci+1 = mt + 2t

Where you could of course retain what you have as the solution:

Ci+1 = max(t,mt) + 2t

But after substitution, max(t, mt) will always evaluate to mt. I have never done this sort of thing before with circuit delays, but I assumed mt was just a constant scaled t. As such, just carry out the algebra where max(a,b) returns the max of the two arguments.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.