1. Chapter 6 Network-Based Scheduling
Prepared by: Dr. Tsung-Nan Tsai

2. Introduction Job sequence is important. Cutting → milling → drilling.
The predecessor-successor relationships can be made by a network representation.
Critical path method (CMP，要徑法) and Project Evaluation and Review Technique (PERT，計畫評核術)
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3. 6.1 Critical Path Method (CPM)
Two basic elements to construct a network:
Activity (活動) : represented by a arrow, the time- and resource-consuming portion of the operations.
Event (事件): represented by a node, indicates start or completion of the activity.
The logical connection of nodes and activities,”→” the precedence relationships.
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4. 6.1 Critical Path Method (CPM)
同時進行
同時完成
Activity a is designed as
1-2,
starts in node 1 and
complete in node 2
duration i-j is denoted
as di-j. 虛擬
Resource: time
Node 3 to start,
a, b must be completed
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5. Time elements - CPM
Earliest start time (ETi): Earliest possible time when the processing of activities emerging from i can commence.
Earliest expected project completion time (CE): The maximum time required in the network to complete all jobs.(最早完成時間)
Latest start time from node i (LTi): The latest time that processing of an activity emerging from node can commence such that the expected project completion time CE is not delayed. (最遲開始時間)
Slack (S): The permissible delay in the completion of a node such that the earliest expected completion time CE is not delayed.(浮時/寬放), slack = 0 is the critical node.
See page 115 for more explanations.
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6. 6.1 Critical Path Method (CPM) - Example
ET LT
For Node 3 is Max(ET1 + d1-3, ET2 + d2-3)
= max(0+3, 2+0) = 3
For node 4 is max(0+2, 3+3) = 6
For node 5 is max(2+3, 3+5, 6+1) = 8
Si = LTi – ETi
The Early Start Time
- Forward pass (前向計算)
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7. 6.1 Critical Path Method (CPM) - Example
The latest completion times, follows a backward pass (後推計算).
For end node: LT5 = CE = 8
Node 2 and 3 have two emerging activity
for node 2, nodes 3 and 5 must be resolved
for node 3, nodes 4 and 5 must be known
Start with node 4.
LT4 = LT5 – d4-5 = 8-1 = 7
LT3 = (min(LT5 - d3-5, LT4 - d3-4))
= min(8-5, 7-3) = 3
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8. 6.2 Scheduling a Network of jobs on a Specified Number of Parallel Processors
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9. 6.2 Scheduling a Network of jobs on a Specified Number of Parallel Processors
ddd
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10. 6.2 Scheduling a Network of jobs on a Specified Number of Parallel Processors
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11. 6.2 Scheduling a Network of jobs on a Specified Number of Parallel Processors
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12. 6.2 Scheduling a Network of jobs on a Specified Number of Parallel Processors
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13. 6.2 Scheduling a Network of jobs on a Specified Number of Parallel Processors
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14. 6.4 Assembly Line Balance
The assembly line balancing needs the precedence relationships of related stations.
If a line is perfectly balanced, each station required identical time to complete all its assigned tasks.
A perfect balance is rarely possible, and the largest station time become cycle time for the assembly line.
Breaking the total job content into small tasks or work elements and determine the precedence relationship among the tasks is the first step to line balancing.
The 2nd step is to distribute the total work content among the stations and have a balanced load to each station.
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15. 6.4 Assembly Line Balance - LCR
Largest Candidate Rule – LCR
List the tasks in descending order of task times and list the corresponding immediate predecessor task(s) for each task.
Start with 1st station, number the remaining stations consecutively. See step #3
Beginning at the top the task list, assign the first feasible task to the station under consideration. Once the task is assigned, all reference to it is removed from the predecessor task list (with --). The cumulative time must be smaller than cycle time.
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16. 6.4 Assembly Line Balance - LCR
Delete the task that is assigned from the first column of the task list. If the list is empty then go to step #6, otherwise, return to step #3.
Create a new station (Station = Station +1). Return to Step #3.
All jobs are assign, and the present station no. reflects the number of stations required. The largest cumulative time is the cycle time.
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17. 6.4 Assembly Line Balance – LCR Example
A job is broken into 9 tasks.
The desired cycle time is 0.95 minutes, or production rate is 500/0.95 = 526 for a 500-minute working shift.
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18. 6.4 Assembly Line Balance – LCR Example
Step #1: ranking the tasks in descending order of their time requirements.
前置作業
** Assignment has been made.
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19. 6.4 Assembly Line Balance – LCR Example
Step #2: Starting with the first station and applying with step #3 (no predecessor is required).
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20. 6.4 Assembly Line Balance – LCR Example
Maxim, CT
4.40
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21. 6.5 Mixed-Model Assembly (MMA) Line Balancing
A number of similar products or a number of different models of an entity are manufactured on the same assembly line is called mixed-model assembly (MMA).
Generally, there are many common elements between the models and workers have the same skills and knowledge to make the development of a MMA feasible.
Two stages to solve the problem.
Balancing: determining which products/models should be grouped together. Using knowledge and judgment of the problem.
Scheduling: design the assembly line for the group.
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22. 6.5 Mixed-Model Assembly (MMA) Line Balancing - Example
See Page 131. example for a diesel engine factory. (柴油引擎)
VA: air-cooled engines, VA4, VA6, and VA6TC.
VA: water-cooled engines, VW2 and VW3.
Shift production requirement: VA4: 2 units, VA6: 3 units, VA6TC: 1 unit
VW2: 2 units, VW3: 2 units.
Total time to produce the required
Production of each model:
2(271)+3(246)+1(437)+2(464)+2(526)
= 3,697 minutes/per shift
assuming the total available time
per shift: 500 minutes/per operator
3,697/500 = 7.39 ≈ 8
efficiency is 3,697/(8 × 500) = 92.42%
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23. 6.5 Mixed-Model Assembly (MMA) Line Balancing - Example
Step #1: Grouping the similar works (開始群聚相似作業). Guideline is:
1) models requiring exactly the same tasks should be grouped together.
2) the groups should be well balanced regarding the work content.
See page 132.
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24. 6.5 Mixed-Model Assembly (MMA) Line Balancing - Example
Largest Candidate Rule – LCR method to both groups
Available time/per operator: 500 minutes → per shift
Group #1: 1, 7, 4 → 8, 5 → 9, 10 (500 minutes)
Group #2: 1, 2, 3 → 4, 5, 6, 7 → 8, 9, 10
1280
2417
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25. 6.5 Mixed-Model Assembly (MMA) Line Balancing - Example
Cul. time
1-[(500 × 3) - (3 × 471)]/3 × 500
= 94.2%
471
316
493
1-[(500 × 3) - (3 × 493)]/3 × 500
= 98.6%
Cul. time
1-[(1000 × 3) - (3 × 975)]/3 × 1000
= 94.2%
975
462
980
1-[(1000 × 3) - (3 × 980)]/3 × 1000
= 98%
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26. 6.5 Mixed-Model Assembly (MMA) Line Balancing - Example
3 operators
5 operators
Total efficiency: [3(85.33)+5(95.96)]/8 = (in %)
Group #1
Group #2
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