Answer:
G order Quantity = 320 no's to meet demand of 251 nos. in multiples of 80s.
G Order timing = 2nd week (beginning)
Explanation:
Solution:
Subassembly G order quantity (in units):
Based on the bill of material structure.
G sub assemblies needed for making one robot = 6+1 = 7
G Sub Assemblies needed making 38 robots = 38 x 7 = 266
Inventory of G sub assemblies = 15
G sub assemblies order needed = Demand - inventory
G sub assemblies order needed = 266 - 15
G sub assemblies order needed = 251
G sub assemblies order needed in multiple of 80's
Since, order quantities in multiple of 80's and more than equal to 251
G assemblies order quantity = 80x4 = 320
The multiple of 80's will meet the demand of 251
Less than 320, we cannot order, because it will not meet the demand of 251. As 80x3 = 240.
So,
320 is the minimum order quantity in multiple of 80's to meet the demand of 251.
b.
Sub assembly G order should start at beginning of the week.
Note: Lead time structure with respect to BOM structure for robot is attached below in the attachment.
Since, G order timing = 2nd week.
