Work Center A produces parts that are then processed by Work Center B. Kanban co

Question

Work Center A produces parts that are then processed by Work Center B. Kanban containers used by the work centers hold 100 parts. The overall rate of demand is 4.5 parts per minute at Work Center B. The table below shows setup, move and wait times for parts at each of the work centers. ?? Work Center ??? A. ?B Setup??? 4? 3 Run time per unit? 0.1? 0.4 Move time?? 2. 6 Wait time?? 10? 20 a. What is the minimum number of containers needed between these two work centers? T is the time required to go through both work centers and back to its starting point. Therefore, T = ___+ ___ = ____minutes of setup time, _____ x (____ + ____) = ____ minutes of run time, ___ + ___ minutes of move time, and ____ + ____ minutes = ___ minutes of wait time. T = ___ + ____ + ___ + ____ = _____ minutes N = DT / C = (_____ x (____)) / ______ = ____._____ containers à round up to ____ containers b. Assume that two extra containers are available at no extra cost. If these work centers use the two containers, what is the maximum parts per minute that could be expected to flow through these two work centers? n = DT / C ___ + ___ = (D x (____)) / _____ Revise formula, as shown below, to: D = n x (C / T) to find D. D = ____ x (_____ / _____) = ____._____ parts per minute c. Could the work centers handle a demand of 8.5 parts per minute? Work Center A produces parts that are then processed by Work Center B. Kanban containers used by the work centers hold 100 parts. The overall rate of demand is 4.5 parts per minute at Work Center B. The table below shows setup, move and wait times for parts at each of the work centers. ?? Work Center ??? A. ?B Setup??? 4? 3 Run time per unit? 0.1? 0.4 Move time?? 2. 6 Wait time?? 10? 20 a. What is the minimum number of containers needed between these two work centers? T is the time required to go through both work centers and back to its starting point. Therefore, T = ___+ ___ = ____minutes of setup time, _____ x (____ + ____) = ____ minutes of run time, ___ + ___ minutes of move time, and ____ + ____ minutes = ___ minutes of wait time. T = ___ + ____ + ___ + ____ = _____ minutes N = DT / C = (_____ x (____)) / ______ = ____._____ containers à round up to ____ containers b. Assume that two extra containers are available at no extra cost. If these work centers use the two containers, what is the maximum parts per minute that could be expected to flow through these two work centers? n = DT / C ___ + ___ = (D x (____)) / _____ Revise formula, as shown below, to: D = n x (C / T) to find D. D = ____ x (_____ / _____) = ____._____ parts per minute c. Could the work centers handle a demand of 8.5 parts per minute? Work Center A produces parts that are then processed by Work Center B. Kanban containers used by the work centers hold 100 parts. The overall rate of demand is 4.5 parts per minute at Work Center B. The table below shows setup, move and wait times for parts at each of the work centers. ?? Work Center ??? A. ?B Setup??? 4? 3 Run time per unit? 0.1? 0.4 Move time?? 2. 6 Wait time?? 10? 20 a. What is the minimum number of containers needed between these two work centers? T is the time required to go through both work centers and back to its starting point. Therefore, T = ___+ ___ = ____minutes of setup time, _____ x (____ + ____) = ____ minutes of run time, ___ + ___ minutes of move time, and ____ + ____ minutes = ___ minutes of wait time. T = ___ + ____ + ___ + ____ = _____ minutes N = DT / C = (_____ x (____)) / ______ = ____._____ containers à round up to ____ containers b. Assume that two extra containers are available at no extra cost. If these work centers use the two containers, what is the maximum parts per minute that could be expected to flow through these two work centers? n = DT / C ___ + ___ = (D x (____)) / _____ Revise formula, as shown below, to: D = n x (C / T) to find D. D = ____ x (_____ / _____) = ____._____ parts per minute c. Could the work centers handle a demand of 8.5 parts per minute? Work Center A produces parts that are then processed by Work Center B. Kanban containers used by the work centers hold 100 parts. The overall rate of demand is 4.5 parts per minute at Work Center B. The table below shows setup, move and wait times for parts at each of the work centers. ?? Work Center ??? A. ?B Setup??? 4? 3 Run time per unit? 0.1? 0.4 Move time?? 2. 6 Wait time?? 10? 20 a. What is the minimum number of containers needed between these two work centers? T is the time required to go through both work centers and back to its starting point. Therefore, T = ___+ ___ = ____minutes of setup time, _____ x (____ + ____) = ____ minutes of run time, ___ + ___ minutes of move time, and ____ + ____ minutes = ___ minutes of wait time. T = ___ + ____ + ___ + ____ = _____ minutes N = DT / C = (_____ x (____)) / ______ = ____._____ containers à round up to ____ containers b. Assume that two extra containers are available at no extra cost. If these work centers use the two containers, what is the maximum parts per minute that could be expected to flow through these two work centers? n = DT / C ___ + ___ = (D x (____)) / _____ Revise formula, as shown below, to: D = n x (C / T) to find D. D = ____ x (_____ / _____) = ____._____ parts per minute c. Could the work centers handle a demand of 8.5 parts per minute?

Explanation / Answer

a)T= 4+3= 7 minutes of setup time

100 *(0.1+0.4)= 50 minutes of run time

2+6=8 minutes of move time

10+20=30 minutes of wait time

Thus T= 7+50+8+30=95 MINUTES

N= DT/C =4.5*95/100=4.275 rounded upto 5 containers

b) N= DT/C

5+2=D*95/100

D=7*100/95= 7.37 Parts per minute

So the it cannot handle the demand rate of 8.5 units per minute if they have seven containers.

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