( video ) n
Process Analysis n
Process Flowcharting n
Types of Processes n
Process Performance Metrics and Examples n n
Process Throughput Time Reduction
第 3 章專案管理：重要路徑法 ( 要徑法 ,
Critical Path Method, CPM)
– 點子來自 Anupindi, Managing Business Process
Flows
1

n n
Process: Transforms inputs into outputs (with greater value to the organization)
Cycle ( 週期 ) Time: Average time between completions of successive units
– Mechanical ( 機械 ) slot machine ( 吃角子老虎 ):
Assume payback to customer 95%, casino ( 賭場 ) keeps 5%. 賭本 $100
•
• Average player feeds coins: 15 sec / coin.
1
15
$ 60 s 1 s m
60
1 m h
×
0 .
05
=
12
$ h
⇒
100 $
12
$ h
=
8 .
3 h
2
– Electronic ( 電子式 ): 10 sec / coin. Lose 18 $/h.
Finish in 5.5 h

3
n
Utilization ( 使用率 ):
Time that a resource is actually activated / Time that it is available for use
– Machine used 12 out of 24 hours: 12 / 24 = 0.5 = 50%
– Revenue
• 機械式： 12 $/h * 12 h/d = 144 $ per day
• 電子式： 18 $/h * 12 h/d = 216 $ per day

Process Flowcharting ( 流程圖 )
Tasks or operations
Decision points
Storage areas
4

1 2 n n n
Station 1 cycle time 30 sec / unit.
Station 2 cycle time 45 sec / unit
– 1 st stage blocked ( 阻塞 ) for 15 seconds
– 2nd stage bottleneck ( 瓶頸 )
Station 1 cycle time 45 sec / unit.
Station 2 cycle time 30 sec / unit
– 2nd stage starved ( 停工待料 ) for 15 seconds
– 1 st stage bottleneck ( 瓶頸 )
第 18 章限制理論或 E. Goldratt 著，目標
5

n n n
Make-to-order (MTO ，接單生產 )
– Only activated in response to an actual order
– Both work-in-process and finished goods inventory kept to a minimum
– Before fast food. Slow response time
Make-to-stock (MTS ，存貨生產 ) (next slide A)
– Process activated to meet expected or forecast demand
– Customer orders are served from target stocking level
– Quicker response time (for small children)
Configuration to order (CTO ，客製化生產，接單組
裝 ): PC
6

Exhibit 5 .2 Making Hamburgers at McDonald’s,
Burger King, and Wendy’s
(B) Hybrid ( 混合 )
(c) 肉片先烤好，
如果太久，當湯
(d) 1999 新方
法， hybrid ，新
技術 (45 秒煮好
肉， 9 秒烤好麵
包 )
7
• 不同的目標顧
客群

8

9
n
Operation time (OT ，作業時間 ) = Setup time ( 整備時間
)
+ Run time ( 生產時間 )
– Cereal-boxing machine: (1) Produce 30 boxes / min.
Then Run time 2 sec. (2) Switch the machine from
16-ounce boxes to 12-ounce boxes: Setup time 30 min
(3) OT (for 10,000 12-ounce boxes) = 1800 sec + 2 sec
/ box
×
10,000 = 21,800 sec n
Throughput time (TT ，產出時間 ) = Average time for a unit to move through the system
– Six-machine assembly line with a cycle time of 30 sec.
Then TT = 30 sec / station
×
6 station = 3 min

10 n
Throughput rate ( 產出率 ) : Output rate that the process is expected to produce over a period of time n
– Above assembly line: 1 / Cycle time = 1 / 30 sec / unit
= 2 units per min
Value-added ( 附加價值 ) time: Time that useful work is actually being done on the unit n
Process velocity (or Throughput ratio)
= Throughput time / Value-added time
– Assume 10 additional buffer positions with 30 sec each. Then PV = (10
×
30 + 180) / 180 = 2.66

n
Little’s Law: Throughput time
= Work-in-process / Throughput rate
11
WIP TR
– Assembly line with 1 unit of WIP at each station.
– T. Time = 6 Units / 2 Units / min = 3 min n
Cash Flow Example: (Anupindi, Managing Business Process
Flows, p. 44)
– Steel company processes $400 million of raw materials/year. Cost of processing: $200 mil./year
– Average inventory: $100 million
– Average flow time T = 100 m / 600 m / year = 1/6 years =
2 months. On average, a dollar spends 2 months in the process

12

5
n
Exhibit 5 .4: Bread-making (BM), Packing (Pa) n
For the original operation (A)
– BM is the bottleneck.
– Bakery’s capacity 100 loaves / hr
– Pa idle for ¼-hour periods with 75% utilization n
For the new operation (B)
– Cycle time for the 2 BM lines: 0.5 hr per 100 loaves
– Pa is the bottleneck
– BM 2 (8-hr) shifts: 16 hr / 1 (hr / 200 loaves) = 3200 loaves
– Pa 3 shifts: 24 hr / ¾ (hr / 100 loaves) = 3200 loaves
13

14
Example 5 .1 Bread Making: Process Analysis (1/2) n
For the original operation (A)
– No inventory buildup between BM and Pa
(Wrong, next slide)
– Throughput time = 1.75 hr
– Throughput rate = 100 loaves / hr n
For the new operation (B)
– The 1st 100 loaves move into Pa while the 2nd 100 loaves wait
– During the first 2 shifts, inventory builds from 0 to
(200 – 400/3)16 = 1067 loaves. Correct: 200
×
16 –
15
×
400/3 = 1200
– Average WIP = 533. Correct 600

15 n n
製作麵包：批 (100 條 / 小時 ) 。
包裝：條 (100 條 / 45 分 ⇒
0.45 分 / 條 ) 。 n n n
平均庫存 = 面積 / 時間
Left: (100
×
0.45 + 99
×
0.45 + … + 1
×
0.45) / 60 = 37.875
Right: (45
×
100)/ (2
×
60) = 37.5 (Good approximation)

5
– During the 3 rd shift, drop from 1200 to 0.
– Process cycle time = 0.75 hr / 100 loaves. Then throughput rate = 100 / 0.75 = 133.3 loaves / hr
– Average WIP time = 600 loaves /
133.3 loaves / hr = 4.5
hr (By Little’s Law)
– Total throughput time = 1 (BM) + 4.5
(inven) +
¾ (Pa) = 6.25
hr
16

5
n
Design our buffet so customers take an average of
30 minutes to get their food and eat n
Assume they eat in groups of 2 or 3 to a table n
Restaurant has 40 tables. Each table can accommodate 4 people. n
Maximum capacity?
17

18
5
n
Average customer party: 2.5 individuals, so average seat utilization (under full capacity)
= 2.5 seats / party
÷
4 seats / table = 62.5 % n
Restaurant’s cycle time
= 30 min / table
÷
40 tables = 0.75 min, so on average a table becomes available every n
45 sec.
Could handle 60 min
÷
0.75 min / party =
80 customer parties per hour

19
Time
11:30-11:45
11:45-12:00
12:00-12:15
12:15-12:30
12:30-12:45
12:45-1:00
Total parties
Parties arriving
15
35
30
15
10
5
110

Time
11:30-11:45
Arri
(Cum)
15
Dep
(cum)
At table or wait
Tabl e used
0 15 15
Waiting party
Waiting time
(min)
11:45-12:00 35 (50) 0 50 40 10 7.5
12:00-12:15 30 (80) 15 65 40
12:15-12:30 15 (95) 20 (35) 60 40
12:30-12:45 10 (105) 20 (55) 50 40
25
20
10
18.75
15
7.5
12:45-1:00 5 (110) 20 (75) 35 35
1:00-1:30 0 (110) 35
(110)
Waiting time = cycle t.
*WP = 0.75*WP

n n n n n
Perform activities in parallel ( 執行平行的作業 )
– Concurrent engineering ( 同步工程 ) in product development
– 同時煮湯和炒菜
改變作業執行的順序：洗澡和寫功課
Reduce interruptions ( 減少干擾因素 ) ：期限前交作業
刪除不必要的步驟與重複的作業：公文與電話旅行
( 下 3 張 ) 要徑法
21

22

1
2
CPM 網路
23
• LS: Late start, ES: Early start
• Slack time = LS
−
ES

n n n n n
可以明確定義專案中的活動。
可以明確定義專案活動進行的順序，並形成網路
圖。
重點應放在重要路徑上。 If any activity along the critical path is delayed, then the entire project will be delayed.
可以建立成線性規劃 (Linear programming) 的模型
強調時間 ( 速度 )
– 可以花錢，加快速度
– 當活動時間不確定，也可以處理
24

( 謝謝郭幸民教授 )
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