Jean is the plant manager of Asus laptop computer in Asia an

Jean is the plant manager of Asus’ laptop computer in Asia and she wants to determine the best option proposed by her supervisors to increase the productivity performance of the manufacturing plant.Currently, the plant has 50 staff members working 8 hrs/day at a rate of $15.00/hr. There are two machines and each consume 200 kilowatt-hrs(KWH) of power per day at a cost of $1.00/KWH. The plant produces 20 units per day and the material cost is per unit $6.00. Option 1 is to reduce the number of staff members by 22% and the payroll cost per hour by 20% but raise the power cost per KWH by 100%. Option 2 is to reduce the number of staff members to 20% and the material cost per unit by 50% but raise the number of machines by 150%. Option 3 is to raise the working hours per day by 25%, KWH used per machine by 10%, and units produced per day by 70%. Which of the three options can provide the highest productivity increase in terms of labor only, power only or both factors?

Solution

The obvious first step should be to find the present level of productivity.

Labor productivity = 20 per day / $6000 per day = 0.00333 units/$

Power productivity = 20 per day / $400 per day = 0.05 units/$

Multifactor Productivity = Output / Input = 20 per day / ($6000+$400+$120) per day = 0.003067 units/$

Option 1

Reduce the number of staff members by 22% i.e. from 50 to (50 - 50*0.22) = 39
Reduce the payroll cost per hour by 20% i.e from $15 to ($15 - $15*0.20) = 12
Raise the power cost per KWH by 100% i.e. from $1 to ($1 + $1*1) = $2

Labor productivity = 20 per day / $3744 per day = 0.005342 units/$

Power productivity = 20 per day / $800 per day = 0.025 units/$

Multifactor productivity = Output / Input = 20 per day / ($3744+$800+$120) per day = 0.004288 units/$

So,

% increase in labor productivity = (0.005342 - 0.00333) / 0.00333 = 60.4%

% increase in power productivity = (0.025 - 0.05) / 0.05 = -50%

% increase in multifactor productivity = (0.004288 - 0.003067) / 0.003067 = 39.8%

Option 2

Reduce the number of staff members to 20% i.e. from 50 to (50 - 50*0.20) = 40
Reduce the material cost per unit by 50% i.e. from $6 to ($6 - $6*0.50) = $3
Raise the number of machines by 150% i.e. from 2 to (2 + 2*1.5) = 5

Labor productivity = 20 per day / $4800 per day = 0.004167 units/$

Power productivity = 20 per day / $1000 per day = 0.02 units/$

Multifactor productivity = Output / Input = 20 per day / ($4800+$1000+$60) per day = 0.003413 units/$

So,

% increase in labor productivity = (0.004167 - 0.00333) / 0.00333 = 25.14%

% increase in power productivity = (0.02 - 0.05) / 0.05 = -60%

% increase in multifactor productivity = (0.003413 - 0.003067) / 0.003067 = 11.28%

Option 3

Raise the working hours per day by 25% i.e. from 8 to (8 + 8*0.25) = 10
Raise the KWH used per machine by 10% i.e. from 200 to (200 + 200*0.10) = 220
Raise the units produced per day by 70% i.e. from 20 to (20 + 20*0.7) = 34

Labor productivity = 20 per day / $7500 per day = 0.002667 units/$

Power productivity = 20 per day / $440 per day = 0.045 units/$

Multifactor productivity = Output / Input = 34 per day / ($7500+$440+$204) per day = 0.004175 units/$

So,

% increase in labor productivity = (0.002667 - 0.00333) / 0.00333 = -19.9%

% increase in power productivity = (0.045 - 0.05) / 0.05 = -10%

% increase in multifactor productivity = (0.004175 - 0.003067) / 0.003067 = 36.12%