POM 601
1) A network consists of the activities in the following list. Times are given in weeks.
| Activity | Preceding | Time |
| A | — | 8 |
| B | — | 3 |
| C | A | 7 |
| D | A, B | 3 |
| E | C | 4 |
Draw the network diagram.
8 weeks
7 weeks 4 weeks
3 weeks
3 weeks
Calculate the ES, EF, LS, LF, and Slack for each activity.
| Activity | Preceding | Time | ES | EF | LS | LF | Slack |
|---|---|---|---|---|---|---|---|
| A | — | 8 | 0 | 8 | 0 | 8 | 0 |
| B | — | 3 | 0 | 3 | 13 | 16 | 13 |
| C | A | 7 | 8 | 15 | 8 | 15 | 0 |
| D | A, B | 3 | 8 | 11 | 16 | 19 | 8 |
| E | C | 4 | 15 | 19 | 15 | 19 | 0 |
What is project completion time? Project completion time = A-C-E = 8+7+4= 19 weeks
2) A toy manufacturer makes its own wind-up motors, which are then put into its toys. While the toy manufacturing process is continuous, the motors are intermittent flow. Data on the manufacture of the motors appears below.
Annual demand (D) = 50,000 units
Daily subassembly production rate = 1,000
Setup cost (S) = $85 per batch
Daily subassembly usage rate = 200
Carrying cost = $.20 per unit per year
To minimize cost, how large should each batch of subassemblies be?
Q*P = √ 2DS = √ 2 * 50000 * 85 = 7288.68 = 7289 units
H – (1 –d/p) .2 (1 – 200/1000)
Approximately how many days are required to produce a batch? 7289/1000 = 7.289 = 7.3 days
How long is a complete cycle? 7289/200 = 36.445 = 36 days
What is the average inventory for this problem? (Solve for maximum inventory first)
Q*{1/d-p} = 7288.7*{1-200/1000} = 5830.96 = 5831 units; Average inventory = 5831/2 = 2915.5
What is the total inventory cost (rounded to nearest dollar) of the optimal behavior in this problem?
TC= 50000 * 85 + 5831 *.2 = 583.07 + 583.10 = $1166.17
7289 2
3) A very simple product (A) consists of a base (B) and a casting (C). The base consists of a plate (P) and three fasteners (F). There are currently 30 castings and 100 bases on hand. Final assembly takes one week. The casting has a lead time of three weeks. All other parts have one week lead times. There are no scheduled receipts. All components are lot-for-lot. The MPS requires 80 units of product A in week 5 and 120 in week 8. Produce the MRP for the upcoming eight weeks. Produce a list of all planned order releases.
Indented Bill of Material (BOM)
| Component Level # per Parent Indented BOMA 0 1 AB 1 1 BP 2 1 PF 2 3 FC 1 1 CTotal Components 5 |
|---|
| Gross Requirements | 0 | 0 | 0 | 0 | 80 | 0 | 0 | 120 | ||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Scheduled Receipts | ||||||||||||
| On-hand Inventory | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ||||
| Net Requirements | 0 | 0 | 0 | 0 | 80 | 0 | 0 | 120 | ||||
| Planned Order Receipts | 0 | 0 | 0 | 0 | 80 | 0 | 0 | 120 | ||||
| Planned Order Releases | 0 | 0 | 0 | 80 | 0 | 0 | 120 | 0 |
| Gross Requirements | 0 | 0 | 0 | 80 | 0 | 0 | 120 | 0 | ||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Scheduled Receipts | ||||||||||||
| On-hand Inventory | 100 | 100 | 100 | 100 | 100 | 20 | 20 | 20 | 0 | |||
| Net Requirements | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 100 | 0 | |||
| Planned Order Receipts | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 100 | 0 | |||
| Planned Order Releases | 0 | 0 | 0 | 0 | 0 | 0 | 100 | 0 | 0 |
| Gross Requirements | 0 | 0 | 0 | 80 | 0 | 0 | 120 | 0 | ||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Scheduled Receipts | ||||||||||||
| On-hand Inventory | 30 | 30 | 30 | 30 | 30 | 0 | 0 | 0 | 0 | |||
| Net Requirements | 0 | 0 | 0 | 50 | 0 | 0 | 120 | 0 | ||||
| Planned Order Receipts | 0 | 0 | 0 | 50 | 0 | 0 | 120 | 0 | ||||
| Planned Order Releases | 50 | 0 | 0 | 120 | 0 | 0 | 0 | 0 |
| Gross Requirements | 0 | 0 | 0 | 0 | 0 | 100 | 0 | 0 | ||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Scheduled Receipts | ||||||||||||
| On-hand Inventory | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ||||
| Net Requirements | 0 | 0 | 0 | 0 | 0 | 100 | 0 | 0 | ||||
| Planned Order Receipts | 0 | 0 | 0 | 0 | 0 | 100 | 0 | 0 | ||||
| Planned Order Releases | 0 | 0 | 0 | 0 | 100 | 0 | 0 | 0 |
| Gross Requirements | 0 | 0 | 0 | 0 | 0 | 300 | 0 | 0 | ||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Scheduled Receipts | ||||||||||||
| On-hand Inventory | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ||||
| Net Requirements | 0 | 0 | 0 | 0 | 0 | 300 | 0 | 0 | ||||
| Planned Order Receipts | 0 | 0 | 0 | 0 | 0 | 300 | 0 | 0 | ||||
| Planned Order Releases | 0 | 0 | 0 | 0 | 300 | 0 | 0 | 0 |
4) Fred’s Fabrication has the following aggregate demand requirements and other data for the upcoming four quarters.
| Quarter | Demand | Previous quarter’s output | 800 units | |
| 1 | 700 | Beginning inventory | 0 units | |
| 2 | 900 | Stockout cost | $100 per unit | |
| 3 | 1200 | Inventory holding cost | $10 per unit at end of quarter | |
| 4 | 600 | Hiring workers | $20 per unit | |
| Laying off workers | $40 per unit | |||
| Subcontracting cost | $200 per unit | |||
| Unit cost | $100 per unit |
Which of the following production plans is better: Plan A: chase demand by hiring and layoffs; Plan B: pure level strategy; or Plan C: 700 level with the remainder by subcontracting?
Production Plan B is best as it has the lowest costs.
Plan A
| Demand | Regular Time Capacity | Regular Time Production | Units Increase | Units Decrease | |
| Initial Inventory | |||||
| Q 1 | 700 | 1200 | 700 | 0 | 100 |
| Q 2 | 900 | 1200 | 900 | 200 | 0 |
| Q 3 | 1200 | 1200 | 1200 | 300 | 0 |
| Q 4 | 600 | 1200 | 600 | 0 | 600 |
| Total Units | 3400 | 4800 | 3400 | 500 | 700 |
| @$100/unit | @$20/unit | @$40/unit | |||
| Subtotal | 340,000 | 10,000 | 28,000 | ||
| Total | $378,000 |
Plan B
| Demand | Regular Time Capacity | Regular Time Production | Inventory (end) | Shortage (end) | Units Increase | ||||||
| Initial Inventory | 0 | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Q 1 | 700 | 1200 | 850 | 150 | 0 | 50 | |||||
| Q 2 | 900 | 1200 | 850 | 100 | 0 | 0 | |||||
| Q 3 | 1200 | 1200 | 850 | 0 | 250 | 0 | |||||
| Q 4 | 600 | 1200 | 850 | 0 | 0 | 0 | |||||
| Total Units | 3400 | 4800 | 3400 | 250 | 250 | 50 | |||||
| @$100/unit | @$10/unit | @$100/unit | @$20/unit | ||||||||
| Subtotal | 340,000 | 2,500 | 25,000 | 1,000 | |||||||
| Total | $368,500 | ||||||||||
| Demand | Regular Time Production | Subcontracting Production | Regular Time Production | Subcontracting | Inventory (end) | Units Decrease | |||||
| Initial Inventory | 0 | ||||||||||
| Q 1 | 700 | 700 | 0 | 700 | 0 | 0 | 100 | ||||
| Q 2 | 900 | 700 | 200 | 700 | 200 | 0 | 0 | ||||
| Q 3 | 1200 | 700 | 500 | 700 | 500 | 0 | 0 | ||||
| Q 4 | 600 | 700 | 0 | 700 | 0 | 100 | 0 | ||||
| Total Units | 3400 | 2800 | 700 | 2800 | 700 | 100 | 100 | ||||
| @$100/unit | @$200/unit | @$10/unit | @$40/unit | ||||||||
| Subtotal | 280,000 | 140,000 | 1,000 | 4,000 | |||||||
| Total | $425,000 | ||||||||||
5) A Methods and Measurements Analyst needs to develop a time standard for a certain task. The task involves use of ruler, square, and portable electric saw to mark and cut the “notch” in a rafter (a standard carpentry task of home construction). In a preliminary study, he observed one of his workers perform this task five times. The observations were made in an air-conditioned, well-lighted training facility, at ground level, with all tools and equipment clean and readily available.
| Observation: | 1 | 2 | 3 | 4 | 5 |
| Task time (seconds): | 82 | 74 | 80 | 83 | 76 |
What is the actual average time for this task? 82+74+80+83+76/5 = 79 seconds
What is the normal time for this task if the employee worked at a 20% faster pace than is typical for adequately trained workers? 79*1.20= 94.8 seconds
What is standard time for this task if allowances are 8% constant and 6% variable?
d) If the analyst then thought more carefully about his experiment, and decided that the variable allowances needed to be increased to match the real (outside, un-air-conditioned) work environment, and that the proper variable allowance was not 6% but 12%, what is the revised standard time? 94.8 / (1-.20) = 118.5 seconds
- 94.8 / (1-.14) = 110.23 seconds
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Operations Mgmt POM 601.docx
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