For FCFS only D and $\Delta$ are required.
The first process will complete at D, Second at 2D+$\Delta$, Third at 3D+2$\Delta$..., the nth process at nD+n-1$\Delta$
The arrival time of all process is 0
So Turn around will be Completion - Arrival= D+2D+$\Delta$+3D+2$\Delta$...
$\rightarrow$ D + 2D + 3D ... + nD + $\Delta$ + 2$\Delta$ + ...+ (n-1)$\Delta$
$\rightarrow$ D$\frac{n(n+1)}{2}$+ $\Delta \frac{n(n-1)}{2}$
so average TT will be D$\frac{n+1}{2}$+$\Delta \frac{n-1}{2}$
Waiting time = TT- burst time = D$\frac{n(n+1)}{2}$+ $\Delta \frac{n(n-1)}{2}$ - nD
Average waiting time = $\frac{D\frac{n(n+1)}{2}+ \Delta \frac{n(n-1)}{2}-nD}{n}$
$\rightarrow$ D$\frac{n-1}{2}+ \Delta \frac{n-1}{2}$
Average Waiting time will be = $\frac{(n-1)(D+\Delta)}{2}$
For RR
let D=kd
Completion time for i$^{th}$ process can be given by following equation
$p_{i}=n*(k-1)*(d+\delta)+(i-1)(d+\Delta)+d$
so Turn around can be given by = $D(\delta+n)+\frac{n-1}{2}*(d+\Delta)$
Avg waiting time will be = $D(\delta+n-1)+\frac{n-1}{2}*(d+\Delta)$