Assembly_Diagram Product Structure Tree Level 0 Chair 1 Short Top Rails (2) Legs

Question

Assembly_Diagram Product Structure Tree Level 0 Chair 1 Short Top Rails (2) Legs (8) Screws (20) 2Lumber (5) 2by2 (5) 2by2 (25) Bill of Materials and Inventory Records Quantity for Currently Schedule Receipt (inLead OrderSafety Time Quantity Stock Lot-for-Lot Lot-for-Lot 360 1200 120 Lot-for-Lot 2500 Weeks)* 80 (w1) 90 (w1) Parent 1 of Parent On-Hand Item Chair Short Top Rail Leg 30 t n Chair Chair Chair 2 160 550 1200 (wl) Lumber Short Top 2by2 ** Rail, Leg 0.5, 0.25 0.5 120 (w1) & 120 (w3)3 480 250 (w2) 2500 (w1) Screw Chair 20 2400 * Under the Scheduled Receipt column wi, w2, and w3 indicate that the related orders will be completed/received at the beginning of weeks 1, 2, and 3 respectively. ** Each rail requires 0.5 2by2s and each leg requires 0.25 2by2s.

Explanation / Answer

SOLUTION:

1. How many screws are required to assemble N chairs

Answer: Screw quantity for 1 chair which is its parent is 20. Hence for N chairs, we need (N*20 = 20N screws)

2. How many 2 by 2s are required to assemble N chairs

Answer: In order to assemble N chairs, we need N 2by2 for Rail and 2N 2by2 for Leg

(for N chairs, we need 2N Rail and 8N legs. Hence for 2by2, (2N*0.5 Rail = N for Rail) and (8N*0.25 Leg = 2N for Legs)

3. What is the Chair's planned order release for week 4

Answer: Planned order release of chairs for week 4 is 80. Since the Net Inventory before production is -40 as the opening inventory was -40 and there was no requirement in week 4 we go with the production lot size which is 80.

4. What is the short top net requirement for week 2

Answer: Short top net is equal to the quantity of parent which is chair is 1 (1:1) Hence for week 2, we need 50+N top nets which are the same quantity of chairs which is 50+N.

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