The daily demand for parts from a machining workcell JMB to an assembly workcell is 1,600 units. The average processing time is 25 seconds per unit. A container spends, on average, 6 hours waiting at JMB before it is processed. Each container holds 250 parts. Currently 10 containers are being used for the part.

What percent safety margin of stock is being carried?

If you wanted to retain the safety margin, but remove one container, to what value must the waiting time be reduced?

What would happen if the demand for the parts increased to 1,900 with K = 10?

Daily demand for parts = 1600 units

Avg processing time = 25 seconds.

Capacity of a container = 250 units

Waiting time of a container before the JMB = 6 hours

Number of containters currently deployed = 10

Solution:

Total number of parts carried by the containter = 10 x 250 = 2500 parts.

Total numbers carried in excess as safety stock = 2500-1600 = 900

Percentage of safety stock = (900/1600) x 100 = 56.25%

If one container is removed total parts carried = 2250 units

Processing time at machine for 250 parts after loading in JMB = 250 x 25 seconds = 1.736 hours, but it has to wait for another 4.26 hours for the next container to be loaded because of the waiting time of the containter is 6 hours.

Processing time at machine for 1600 parts = 1600 x 25 seconds = 11.11 hours

Number of containers required for 1600 parts = 1600/250= 6.4 ~ 7 containters

Waiting time of 7 containers = 7 x 4.26 = 29.82 hours.

If one of the containers has to be removed 1600 parts has to be carried in 6 containers ie., 29.82hours - 6 hours = 23.82hours

Waiting time per container before loading = 23.82/6 = 3.97 hours.

Reduced Total waiting time of the container = 3.97 hours + the processing time of 250 parts = 3.97 + 1.736 =5.7hours ~ 5 hours and 42 minutes

answer; /// i believe the correct answer is first in, first out(fifo); //