Let me answer without formula first
4 Full adders are there
F$_1$-->F$_2$-->F$_3$-->F$_4$
to calculate sum are of Full adders need two bit to sum and a carry bit from previous Full adder
so F$_1$ takes 5ns to calculate carry and then passes it on to F$_2$ during same time calculates sum also
- at 6th ns F$_2$ starts to calculate carry and sum and in the same time F$_1$ is working on its sum
after 10ns F$_1$ produces its sum and F$_2$ produces its carry
- at 11th ns F$_3$ starts to calculate carry and sum and in the same time F$_2$ is working on its sum
after 15ns F$_2$ produces its sum and F$_3$ produces its carry
after 20ns F$_3$ produces its sum and F$_4$ produces its carry
- at 25th ns F$_4$ finishes its sums also
so it took 25ns to complete 1 sum
1 sum time = 25 * 10-9 seconds
=>in 1 sec 1/25 * 10-9 sums will get executed
=> 109 /25 = 4 * 107
Method 2
let T(n) is the time taken by nth full adder as it has to wait for 5ns for previous adder to calculate its carry
so we can form a recurrence
T(n) = T(n-1) + PD$_C$ (Propagation delay of carry)
T(1) = (PD$_s$) Propagation delay of sum (as for only 1 full adder time taken = 10ns(=(PD$_s$)
solving by back substitution
T(n) = T(n-2) + (PD$_c$)+ (PD$_c $)
T(n) = T(n-2) + 2*(PD$_c$)
T(n) = T(n-k) + k*(PD$_c$)
if n-k=1 ,k=n-1
T(n) = T(n-(n-1)) + (n-1)*(PD$_c$)
T(n) = T(1) + (n-1)*(PD$_c$)
T(n) = (PD$_c$) + (n-1)*(PD$_c$) we can derive this formula by ton a ways
here n=4 (PD$_c$) = 5ns (PD$_s$) =10ns
T(4) = 10 + (4-1)*(5)
T(4) =25ns
1 sum time = 25 * 10-9 seconds
=>in 1 sec 1/25 * 10-9 sums will get executed
=> 109 /25 = 4 * 107