In: Operations Management
Dumfries Industries is generating a materials requirement plan for the next six weeks. Here is its most current information:
Master Production Schedule
Week |
1 |
2 |
3 |
4 |
5 |
6 |
Item A |
10 |
0 |
20 |
20 |
10 |
10 |
Other information
Item |
Item’s Parent |
Lead time |
Lot size |
Current inventory |
A |
None |
1 week |
LFL |
20 |
B |
A |
1 week |
20 |
20 |
C |
B |
1 week |
50 |
20 |
How many levels does a product structure for Dumfries item A have?
Using MRP methodology, how much of Parts A, B, and C should be ordered, and when should they be ordered?
Which item will experience the highest average inventory, following this plan?
Consider the following three jobs: Job Alpha requires two weeks of work and is due in four weeks. Job Beta requires three weeks of work and is due in five weeks. Job Carlos requires one week of work and is due in three weeks. If these jobs are completed in alphabetical order, how late will Job Beta be?
Suppose you have three jobs waiting to be completed, and you must determine the sequence in which these jobs should be worked on. You calculated the critical ratio for each of the jobs and the results were that Job A has a critical ratio of 1.27, Job B has a critical ratio of -0.89, and Job C has a critical ratio of 0.15. Which of these jobs is already late? Why?
(1) MRP Problem of Dumfries Industries.
MPS for End-Item A:
Week number |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
Quantity |
0 |
10 |
0 |
20 |
20 |
10 |
10 |
Item A OH =20 LT=1 SS=0 Q=LFL |
Gross reqmts. |
0 |
10 |
0 |
20 |
20 |
10 |
10 |
Scheduled receipts |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
Projected on hand |
20 |
20 |
10 |
10 |
0 |
0 |
0 |
|
Net reqmts. |
0 |
0 |
0 |
10 |
20 |
10 |
10 |
|
Planned-order receipts |
0 |
0 |
0 |
10 |
20 |
10 |
10 |
|
Planned-order releases |
0 |
0 |
10 |
20 |
10 |
10 |
0 |
|
Item B OH =20 LT=1 SS=0 Q=20 |
Gross reqmts. |
0 |
0 |
10 |
20 |
10 |
10 |
0 |
Scheduled receipts |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
Projected on hand |
20 |
20 |
20 |
10 |
10 |
0 |
10 |
|
Net reqmts. |
0 |
0 |
0 |
10 |
0 |
10 |
0 |
|
Planned-order receipts |
0 |
0 |
0 |
20 |
0 |
20 |
0 |
|
Planned-order releases |
0 |
0 |
20 |
0 |
20 |
0 |
0 |
|
Item C OH =20 LT=1 SS=0 Q=50 |
Gross reqmts. |
0 |
0 |
20 |
0 |
20 |
0 |
0 |
Scheduled receipts |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
Projected on hand |
20 |
20 |
20 |
0 |
0 |
30 |
30 |
|
Net reqmts. |
0 |
0 |
0 |
0 |
20 |
0 |
0 |
|
Planned-order receipts |
0 |
0 |
0 |
0 |
50 |
0 |
0 |
|
Planned-order releases |
0 |
0 |
0 |
50 |
0 |
0 |
0 |
Formula:
Net requirements = Gross requirements – On hand inventory – Scheduled receipts.
Product Structure for Dumfries Item A has three levels with Item A being at Level 0.
The number of units of Parts A, B and C that should be ordered at particular points of time is given in the preceding MRP Schedule.
Item C will experience the highest average inventory as per the given plan.
(2) Scheduling Problem: Jobs Alpha, Beta, Carlos.
Job Sequence |
Processing Time (in weeks) |
Due Date (in weeks) |
Flow Time (in weeks) |
Days Tardy = Flow Time - Due Date (in weeks) (0 if negative) |
Alpha |
2 |
4 |
0 + 2 = 2 |
0 |
Beta |
3 |
5 |
2 + 3 = 5 |
0 |
Carlos |
1 |
3 |
5 + 1 = 6 |
3 |
There will be no delay for Job Beta as its tardiness is zero.
(3) Critical Ratio Problem.
Job B and Job C are already late as their Critical Ratios are less than 1, with Job B being most late (having lowest value of Critical Ratio).
The Product Tree Structure is given below: