A flexible machining system is being planned that will consist of four workstati

Question

A flexible machining system is being planned that will consist of four workstations plus a

part-handling system. Station 1 will be a load/unload station. Station 2 will consist of

horizontal machining centers. Station 3 will consist of vertical machining centers. Station 4 will be an inspection station. Four the part mix that will be processed by the FMS, the workloads at the four stations are as follows: WL1 = 7.5 min, WL2 = 22.0 min, WL3 = 18.0 min, and WL4 = 10.2 min. The workload of the part-handling system WL5 = 8.0 min. The FMS will be operated 16 hours per day, 250 days per year. Maintenance will be performed during nonproduction hours, so uptime proportion (availability) is expected to be 97%. Annual production of the system will be 50,000 parts. Determine (a) the number of machines (servers) of each type (station) required to satisfy production requirements and (b) the utilization of each station. (c) What is the maximum possible production rate of the system if the bottleneck station were to operate at 100% utilization?

Explanation / Answer

a. RP =50000/(16*250*0.97)

=12.8866 pc/hour =0.2147 pc/Minutes

S1=Station 1 = Minimum Integer (0.2147 (7.5) = 1.61) = 2 servers (load/unload workers)

S2=Station 2 = Minimum Integer (0.2147 (22.0) = 4.73) = 5 servers (horizontal machining centers)

S3=Station 3= Minimum Integer (0.2147 (18.0) = 3.87) = 4 servers (vertical machining centers)

S4=Station 4 = Minimum Integer (0.2147 (10.2) = 2.19) = 3 servers (inspection stations)

S5=Station 5= Minimum Integer (0.2147 (7.5) = 1.61) = 2 servers (work carriers)

b. U Denotes utilization

U1= 1.61/2 = 0.805 = 80%

U2= 4.73/5 = 0.945 =94.5%

U3= 3.87/4= 0.966 =96.6%

U4= 2.19/3= 0.730 = 73%

U5=1.61/2= 0.805 =80.5%

c. Maximum value of utilization is at station 3. If station 4 were to operate at 100% utilization, R p * = 12.8866/0.966 = 13.34 pc/hr

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