Solution :
                                 A        B       C        D
The Activity of time per unit(min) (X) Â Â 0.25 Â Â Â Â Â 0.33 Â Â Â 0.2 Â Â Â Â Â 0.5
Capacity per worker(units/min) (Y) Â Â Â Â Â 4 Â Â Â Â Â Â Â 3 Â Â Â Â Â Â 5 Â Â Â Â Â Â 2
Number of workers (Z) Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2 Â Â Â Â Â Â Â 3 Â Â Â Â Â Â Â 2 Â Â Â Â Â Â 4
Capacity (Units/min) (ZxY) Â Â Â Â Â Â Â Â Â Â Â 8 Â Â Â Â Â Â Â 9 Â Â Â Â Â Â 10 Â Â Â Â Â 8
A and D re the bottlenecks with a minimum capacity of 8 units/min
Hence, initial system capacity = 8 units/min
It is given that activity time per unit in D step is [tex]$\text{reduced by 0.25}$[/tex] Â min per unit
Capacity/worker = 1/0.25 = 4 per min
Number of worker for D = 4
New capacity of D per min = 4 x 4 = 16 units
Steps                    A      B       C        D  Â
New capacity(units/min) Â Â Â Â 8 Â Â Â Â Â 9 Â Â Â Â Â Â 10 Â Â Â Â Â Â 16
Therefore now, step A is the bottleneck as it has the lowest capacity of 8 units/min.
Capacity of the entire process = 8 units/min