A manufacturing plant is trying to determine standard production per day for an incentive program. Suppose that the incentive program will pay $1 per unit produced above standard rates. A worker is timed at an average of 280 seconds per unit, with a performance rating of 105%. The plant has an allowance factor of 13%. If the worker later produces 100 units in an 8 hour shift, how much would he receive under the incentive plan

Respuesta :

Answer:

He would receive $15 under incentive plan.

Explanation:

The given values are:

Average observed time

= 280 seconds per unit

Performance rating

= 105%

i.e.,

= 1.05

Allowance factor

= 13%

i.e.,

= 0.13

So,

⇒  [tex]Standard \ time = \frac{(Average \ observed \ time\times Performance \ rating)}{1-Allowance \ factor}[/tex]

On putting the estimated values, we get

                             [tex]=\frac{(280\times 1.05)}{(1-0.13)}[/tex]

                             [tex]=\frac{294}{0.87}[/tex]

                             [tex]= 337.93 \ seconds[/tex]

The available time will be:

= [tex](8 \ hours\times 60 \ min/hr\times 60 \ sec/min)[/tex]

= [tex]28800 \ seconds[/tex]

Now,

The Standard production per day will be:

= [tex]\frac{Available \ time}{Standard \ time}[/tex]

= [tex]\frac{28800}{337.93}[/tex]

= [tex]85.22 \ units[/tex]

Since he generates 100 units, he consumes about 15(00-85,22) units per day well above normal production.  

So that he's going to get:

= [tex]15\times 1[/tex]

= [tex]15[/tex] ($)