A (not-so-well maintained machine) machine is down roughly 40% of the time. One
ID: 364827 • Letter: A
Question
A (not-so-well maintained machine) machine is down roughly 40% of the time. One operator is assigned to service four identical such machines. Each machine can produce 60 widgets/hr if running (i.e. not completely idle). The operator is paid $10.00/hr and each machine costs $20.00/hr for power and supplies.
a) What is the actual production for the four machines supervised by one operator for an 8hour shift given the above conditions?
b) What is the unit cost for each widget?
c) Since there is so much lost production due to idle time, management is considering hiring another worker to assist the first operator in servicing these four machines.
There are two approaches or choices:
Assign machines #1 and #2 to the first operator and machines #3 and #4 to the second operator
Or have both operators help each other and service all four machines as needed
Which choice is best, i.e. lowest unit cost?
Explanation / Answer
Calculate the probability and idle time (hours)
a) Actual production = (4*8-2.76-2.46-0.61)*60 = 1,570.2 parts
b) Total cost = machine cost + operator cost = 4*8*20+8*10 = $ 720
Unit cost = 720/1570.2 = $ 0.459 per unit
c)
Option 1 - Assign machines #1 and #2 to the first operator and machines #3 and #4 to the second operator
Lost time occurs only if both machines of an operator are down. Probability of this event = 2C0*0.42*0.60 = (2!/0!2!)**0.42*0.60 = 0.16. Total lost hours for each operator (machine 1 and 2 or 3 and 4) = 8*0.16 = 1.28
Production = (2*8-1.28)*60 = 883.2 units
Total cost of each split (machine 1&2 or 3&4) = (2*20+10)*8 = 400
Unit cost = 400/883.2 = $ 0.453 per unit
Option 2 - Both operators help each other and service all four machines as needed
Increase in production = (2.76+0.1536*8+0.0256*8)*60= 251.6 units
New total cost = 720+80 = 800
New unit cost = 800/(1570.2+251.6) = $ 0.439 per unit
Unit cost of option 2 is lower. Therefore the best option is option 2 (both operators help each other and service all four machines as needed)
Up Down Probability Idle time 4 0 4!(.40 )(.64 )/0!4! = 0.1296 0 3 1 4!(.41 )(.63 )/1!3! = 0.3456 0 2 2 4!(.42 )(.62 )/2!2! = 0.3456 0.3456*8*1=2.76 1 3 4!(.43 )(.61 )/3!1! = 0.1536 0.1536*8*2=2.46 0 4 4!(.44 )(.60 )/4!1! = 0.0256 0.0256*8*3=0.61Related Questions
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