35 34 37 31 WC1 WC 2 WC 3 WC4 WC 5 Assembly Line 242 Times at Standard (seconds)

Question

35 34 37 31 WC1 WC 2 WC 3 WC4 WC 5 Assembly Line 242 Times at Standard (seconds) Answer the following questions regarding Assembly Line 242. A. What is the cycle time of the assembly line? seconds B. What is efficiency of the assembly line? C. Assuming 10 hours per day, 5 days per week, 1 worker at each WC, what is the weekly loss due to balance delay on this assembly line, if the average hourly wage is $20.00? D. Assuming the following workers would achieve their given e fficiency at any of the above work centers, at which WC would you assign each worker to maximize nominal daily capacity in units (10 hours per day). What would the nominal daily capacity be? WC 1 WC 2 WC 3 WC 4 WC 5 Will125 oey Nancy 1.15 Tom 110 George 1.20 90 Nominal Capacity Units

Explanation / Answer

a) Cycle time of the assembly line is the time of bottleneck process which is WC3

Hence it is 37 seconds

b) Efficiency = Total task time/(Cycle time*Number of station)*100 = (35+34+37+30+31)/(37*5)*100 = 167/185*100 = 90.27%

c) Assuming 10 hours per day and 5 days per week, Total hours per week = Number of workers*Number of days per week*Number of hours per day = 10*5*5 = 250

Idle % = 100 - Efficiency = 100 - 90.27 = 9.73%

Idle hours = 9.73%*250 = 24.32 hours

Rate = 20 $ per hour

Loss due to balance delay = Idle hours*Rate = 24.32*20 = 486.49$

d) Assign the most efficient worker to workstation with highest time

Capacity per day = 10*60*60/WC time

Nominal capacity = Capacity per day*Efficiency

WC 3 = Will = 36000/37*1.25 = 1216

WC1 = George = 36000/35*1.2 = 1234

WC 2 = Nancy = 36000/34*1.15 = 1218

WC 5 = Tom = 36000/31*1.1 = 1277

WC 4 = Joey = 36000/30*0.9 = 1080

So, Nominal capacity = 1080 units

              

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