# Goodparts

Submitted By smokinjoe
Words 312
Pages 2
Goodparts Case

A. Assembling A & B is the Process with the smallest hourly production (160/hr), there for the maximum weekly production would be (5*8*160 = 6400 components produced per week)

B. Because the drilling process is only occurring 8 hours a day and with 4 machines, the maximum amount of part c that can be drilled in one day is 1920 or (8*240). Assemble line 1 and the final assembly can produce 2560 (320 * 8) and 2880 (360 * 8) respectively. There will not be enough drilled part C for assembly line 1 and the final assembly to run at full efficiency. Therefore, the daily production will be limited to 1920 or 9600 per week. Based on this, the drilling operation limits the capacity.

C. Assembly line 1: 2560/day or (160*16)
Part C drilling: 2400/day or (60*5*8)
Final assembly 2160/day or (180*12)

Since the final assembly operation can only produce a maximum of 2160 a day, the maximum weekly capacity would be 10800 or (2160*5). Based on this, the final assembly operation limits the weekly capacity.

D. From question B
Part A: 9600 * .55 = 5280
Part B: 9600 * .45 = 4320
Part C: 9600 * .35 = 3360
Assembly 1: 9600 * .4 = 3840
Drilling: 9600 * .25 = 2400
Fin Assembly: 9600 * .4 = 3840
Cost of Elec: 9600 * .01 = 96
Depreciation: 75
Total Costs: 26311
Div: Parts Assembled 9600
Cost per unit: 2.74

From question C
Part A: 10800 * .55 = 5940
Part B: 10800 * .45 = 4860
Part C: 10800 * .35 = 3780
Assembly 1: 10800 * .4 = 4320
Drilling: 10800 * .25 = 2700
Fin Assembly: 10800 * .4 = 4320
Cost of Elec: 10800 * .01 = 108