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Danila Fedorin
2021-03-16 16:31:00 -07:00
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HW3.tex
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@@ -80,7 +80,7 @@ For all inverters, then, we get the following:
\pagebreak
\section*{Q4}
First, to compute stage effort $\hat{f}$.
First, to compute total effort $F$.
\begin{equation*}
\begin{aligned}
@@ -92,29 +92,27 @@ First, to compute stage effort $\hat{f}$.
\end{aligned}
\end{equation*}
Assuimng a $p_\text{invs}$ of 1, and thus $\rho = 3.59$, we get:
Since we are using the same inverter as the one in Q3, we are
once again using $p_\text{inv}=5$, and thus, have that $\rho=6.138$.
From this, we can determine the ideal number of stages:
\begin{equation*}
\log_\rho F = 7.2 \approx 7
\log_\rho F = 5.13 \approx 5
\end{equation*}
Since we currently have 3 stages, we should insert 4 inverters.
It appears as though inserting inverters only at the end makes it
too difficult for the first-stage NAND gate to drive the 3-branched
NOR gates (we end up with an optimal size less than 1). Instead,
I will insert two inverters right after the NAND2 gate, and two more
inverters at the end. We can now compute gate sizes:
Since we currently have 3 stages, we should insert 2 inverters.
I will insert these at the end of the path in question. From
there, we once again compute stage effort $\hat{f}$, and work
backwards to determine the optimal sizes for all of the stages.
\begin{equation*}
\begin{aligned}
\hat{f} &= \sqrt[7]{11111} \\
\text{sz}_7 &= 1000/\hat{f}^1 = 264 \\
\text{sz}_6 &= 1000/\hat{f}^2 = 69.8 \\
\text{sz}_5 &= 1000/\hat{f}^3 * \left(\frac{5}{3}\right) = 30.8 \\
\text{sz}_4 &= 1000/\hat{f}^4 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right) = 13.5 \\
\text{sz}_3 &= 1000/\hat{f}^5 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)3 = 10.7 \\
\text{sz}_2 &= 1000/\hat{f}^6 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)3 = 2.84 \\
\text{sz}_1 &= 1000/\hat{f}^7 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)3 = 1 \\
\hat{f} &= \sqrt[5]{11111} \\
\text{sz}_\text{inv1} &= 1000/\hat{f}^1 = 155 \\
\text{sz}_\text{inv2} &= 1000/\hat{f}^2 = 24.1 \\
\text{sz}_\text{nand3} &= 1000/\hat{f}^3 * \left(\frac{5}{3}\right) = 6.23 \\
\text{sz}_\text{nor2} &= 1000/\hat{f}^4 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right) = 1.61 \\
\text{sz}_\text{nand2} &= 1000/\hat{f}^5 * \left(\frac{5}{3}\right)\left(\frac{5}{3}\right)\left(\frac{4}{3}\right)3 = 1
\end{aligned}
\end{equation*}