The following data are available for the four steps: A B C D Activity time per unit (min) 0.25 0.33 0.2 0.5 Capacity per worker (units/min) 4 3 5 2 Number of workers 2 3 2 4 Suppose the steps in activity D are made easier, so the activity time per unit in step D is reduced by 50% (to 0.25 min per unit). If the assignment of workers to steps remains the same, what is the capacity of the entire process in units per min after the activity time per unit in step D is reduced by 50%

Solution :

                                                                  A              B            C              D

The Activity of time per unit(min) (X)     0.25          0.33       0.2          0.5

Capacity per worker(units/min) (Y)          4              3             5             2

Number of workers (Z)                            2               3              2            4

Capacity (Units/min) (ZxY)                       8               9             10           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  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    

New capacity(units/min)        8           9            10             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

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answer: b. false

the staff is a set of five horizontal lines and four spaces which represent a different musical pitch. the lines and the spaces are numbered from bottom to top, the bottom line is the first line and the fifth line is the line on the top. the notes on the top of the staff (fifth line) have a higher pitch than a note at the bottom of the staff.