Acknowledgment size for STOP-AND-WAIT protocol

I can help to address the first part of your question. As an abstraction of a single switching device inside a processor, imagine a MOSFET connected to ground with a load resistor to the power supply (in a real CMOS part there wouldn't be a load resistor, but another transistor, but this distinction is not important for the analysis). Connected from the junction of the resistor and the transistor is a capacitor, representing all the input capacitances of the transistor that the transistor under discussion is driving. When the first transistor switches off, this capacitance will be charged up through the load resistor. When the first transistor switches back on, the charge stored on the capacitor will be discharged through the first transistor.

It can be shown that when a capacitor is charged through a load resistor, 1/2 of the energy used in charging the capacitor is lost in the resistor, for a total energy dissipation of \$\frac{1}{2}C{V_s}^2\$, where \$V_s\$ is the supply voltage. When the switch then turns on, assuming the resistance of the switch is much less than the load resistance, that same energy will be dissipated in the switch, for a total energy of \$C{V_s}^2\$. Dividing this by the switching period gives you the dynamic power dissipation of the switch/capacitance combo, \$C{V_s}^2f\$. Shrinking the die reduces the junction capacitances of the MOSFETS, so if you know the supply voltage, switching frequency, number of transistors and the approximate junction capacitances of a certain process, you could calculate a ballpark figure of what kind of power savings a process shrink entails, all other things being equal.

What if we start with:

$$U_{a,eff,max} = \frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C}$$

Then we get:

$$P_L = \frac{(U_{a,eff,max})^2}{R_L}=\frac{(\frac{U_0}{2\sqrt{2}}\frac{R_L}{R_L+R_C})^2}{R_L} = \frac{U^2_0R^2_L}{8R_L(R_L+R_C)^2}=\frac{U^2_0R_L}{8(R_L+R_C)^2}$$

And with \$R_C=R_L\$ we get for\$P_{L,max}\$:

$$P_{L,max}=\frac{U^2_0}{32\cdot R_L}$$

For the circuit power we have to consider the power loss of \$R_C\$ and the losses of the transistor itself.

$$P_{R_C} = P_{R_C,DC}+P_{R_C,AC}$$ $$P_{R_C,DC} = \frac{U_{R_C,DC}^2}{R_C}=\frac{(\frac{1}{2}(U_{R_C,max}+U_{R_C,min}))^2}{R_C}=\frac{(\frac{1}{2}(U_0+U_{a,max}))^2}{R_C}$$

If we set again \$R_C=R_L\$ then \$U_{a,max} = \frac{1}{2}U_0\$ and \$U_{R_C,DC}=\frac{3}{4}U_0\$ we get:

$$P_{R_C,DC} = \frac{9U^2_0}{16R_C}$$

For the AC-Part follows:

$$P_{R_C,AC}=\frac{(U_{R_C,eff})^2}{R_C}=\frac{(\frac{U_{R_C,max}-U_{R_C,DC}}{\sqrt{2}})^2}{R_C}=\frac{(\frac{U_{0}-\frac{3}{4}U_{0}}{\sqrt{2}})^2}{R_C}=\frac{U^2_0}{32R_C}$$

Finlay we get:

$$P_{R_C} = \frac{9U^2_0}{16R_C} + \frac{U^2_0}{32R_C} = \frac{19U^2_0}{32R_C}$$

Next how do we get \$P_{transistor}\$ ?