Question

1. This study (work sampling*) is focusing on machine operations. Using the following data,

1. Categorize the events into the interesting activities such as machine idle (I), machine servicing (S).

   Observed Data	Categorize         √ S (Service) I (Idle)	Number of Observation
   Machine Running	S or I	222
   Remove Scrap	S or I	48
   Change tool	S or I	17
   Change material	S or I	60
   Adjust die	S or I	23
   Wait for maintenance service	S or I	53
   Wait for next material	S or I	25

2. What % (p) of the time Machine is down (not idle but “serviced by the operator”) roughly?
3. An analyst determined that the machining time (Tm) per piece was 40min. What is the actual production for the one machine supervised by one operator for an 8-hour shift given the above conditions?
4. What is the unit cost ? The operator is paid $10.00/hr and each machine costs $20.00/hr for power and supplies.
5. Since there is so much lost production due to idle time, management is considering assigning another machine to the first operator. There are three approaches or choices: i) Assign Machines #1 to the first operator and Machines #2 to the second operator, or ii) have both operators help each other and service all two machines as needed. iii) Assign Machines #1 and #2 to the first operator and don’t assign a second operator. Which choice is best, i.e. lowest unit cost?

Answer 1

ANSWER :

a)

Machine Idle (I): these are the events during which neither the machine is operating nor it is getting serviced. So there are two processes when machine was ideal.

Machine Ideal events = wait for maintenance service + wait for next material = 53 + 25 = 78 observations

Machine servicing (S): These are the events when machine goes through overhaul activities.

Machine servicing (S) = Remove scrap + change tool + change material + adjust die

Machine servicing (S) = 48 + 17 + 12 + 23 = 100 observations

b)

Machine service (S) = Remove scrap + change tool + change material + adjust die

Machine service (S) = 48 + 17 + 60 + 23 = 148 observations.

Total machine observations = 222 + 48 + 17 + 60 + 23 + 53 + 25 = 448 observations.

Percent of time machine is down (Not Idle but serviced by the operator):

P% = [Machine service (S) / Total machine observations] x 100 percent

P % = [148/448] x 100 %

P% = 33.03 Percent

c)

Machine time per piece = 40 Min

Supervisor time = 8 hour per shift

Actual production for one machine = machine capacity

= operating hours x operating rate x number of machines

= (40/60) Hours x 8 hours per shift x 1

= 5.36

Thus actual production for one machine is approximately 5 units.

d)

Unit cost = [operator’s payment + machine cost for power and supplies] / total units produced per hour

Unit cost = [$10 + $20] / 5.36

Unit Cost = 30 / 5.36 = $5.6

e)

Assign machine 1 to first operator and machine 2 to second operator = unit cost of machine 1 and unit cost of machine 2

= {[$10 + $20] / 5.36} + {[$10 + $20] / 5.36}

Unit cost = $11.2

Have both operators help each other and service all two machines as needed. This choice will result into no production in the event of both the machine breakdown

Unitl cost = [$10 + $20] for operator 1 + [$10 + $20] for operator 2

Unit cost = $60

Assign machine 1 & 2 to the first operator and don’t assign second operator = unit cost of machine 1 and 2 keeping operator constant

= {$10 + ($ 20 x 2)} / 5.36

Unit cost = $9.32

Thus from above calculations, Third choice is the best where machine 1 & 2 is assigned to the first operator and second operator is not assigned with the machine since the unit cost of this choice is $9.32 which is less than the unit cost of other two choices.