Let there be m bits in sequence number.
We have 2$^{m}$ $\geq$ W$_{s}$ + W$_{r}$
Case 1) Sliding window protocol:
Here, W$_{s}$ = 1+2a; W$_{r}$=0
(we do not have any buffer at the receiver end. The receiver simply sends a cumulative acknowledgement.)
2$^{m}$ $\geq$ W$_{s}$ + W$_{r}$
2$^{m}$ $\geq$ 1+2a
m $\geq$ $\left \lceil log (1+2a) \right \rceil$
Case 2) Go back n protocol:
Here, W$_{s}$ = 1+2a; W$_{r}$=1
2$^{m}$ $\geq$ W$_{s}$ + W$_{r}$
2$^{m}$ $\geq$ 1+2a + 1
m $\geq$ $\left \lceil log (2+2a) \right \rceil$
Case 3) Selective repeat:
Here, W$_{s}$ = W$_{r}$ = 1+2a;
2$^{m}$ $\geq$ W$_{s}$ + W$_{r}$
2$^{m}$ $\geq$ 1+2a + 1 + 2a
m $\geq$ $\left \lceil log (2+4a) \right \rceil$
Someone please let me know if this is right. :)