It takes 8 minutes to load and 6 minutes to unload a machine. Inspection, packing, and...
It takes 8 minutes to load and 6 minutes to unload a machine. Inspection, packing, and travel between machines totals 6 minutes. Machines run automatically for 40 minutes. Operators cost $30 per hour; machines cost $55 per hour.
a. What is the maximum number of machines that can be assigned to an operator without creating machine idle time?
b. What is the number of machines that minimizes the cost?
c. If two machines are assigned to an operator, what will be the cost per unit produced?
d. If the cost per operator is unknown, for what range of values Co will the optimum assignment remain the same?
Homework Answers
Time to load = 8 minutes
Time to unload = 6 minutes
Inspection, packing and travel time = 6 minutes
Run time = 40 minutes
Operator cost = $ 30 per hour
Machine cost = $ 55 per hour
1)
Total worker time per machine = 8+6+6 = 20 minutes
Total machine time per cycle = loading and unloading time + run time = 8+6+40 = 54 minutes
Number of machines that can be assigned to a worker without causing machine idle time = total machine time per cycle / total worker time per machine
= 54/20
= 2.7
2.7 is a fraction, if we round it up to 3, then total worker time per cycle (i.e. for 3 machines) will be 20*3 = 60 minutes, which is more than the cycle time for machine (54 minutes). This will cause machine idle time.
Therefore, maximum number of machines that can be assigned to an operator without creating machine idle time
= 2 machines
b)
If two machines are assigned to a operator, then the machine cycle time = 54 minutes
which means 2 units are produced every 54 minutes
Productivity per hour = 2 machines * 60/54
Total cost per hour = operator cost + machine cost = 30+2*55
= $ 140
Cost per unit produced = Total cost per hour / Productivity per hour
= 140 / (2*60/54)
= $ 63 per unit produced
c)
Optimum assignment is 2 machines per operator
The range of values of Co are determined by equating the cost of 2 machines with operator cost
So, maximum of Co is 2*55 = $ 110 per hour
Range of Co is from $ 0 to $ 110 per hour
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