Process Analysis - Class Notes - Mod 2 - Part 2(1)

8/13/2019 Process Analysis - Class Notes - Mod 2 - Part 2(1)

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MGMT 361Operations Management

OM)

Process Analysis part 2(multiple stage processes Littles Law)

Prof Julia A Kalish


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Every Day Starts theDay Before

Mgmt 361 2Module 2 - Process Analysis


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Last Section : Precedence diagrams and flow charts

Design parameters/ Run-Time parameters

Analysis of Single Stage Processes

This Section :

Littles Law Analysis of Multiple Stage Processes

Overview of Class:

Mgmt 361 Module 2 - Process Analysis 3


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Quick Repetition of last ClassSequential versus Parallel Machines

Capacity determined by slowest stage:Bottleneck: Stage 1 - Capacity : 120 [units/hour]((1/30) * 60 * 60) = 120

1

30

4

5

3

8

2

2

5

3

Capacities add:Capacity of M1 = 1/10 [units / sec]

= 360 [units / hour]Capacity of M2 = 225 [units / hour]Capacity of M3 = 180 [units / hour]

System capacity = 765 [units / hour]

Time in Seconds

M110

M320

M216

Time in Seconds

DoNOT

add oraverage CT:

CT = 1 / Capacity

Sequential Machines:

Parallel Machines:

Mgmt 361Module 2 - Process Analysis

4

CT = 1 / Capacity= 3600 / 765 = 4.706 [sec / unit] or

765/60 = 12.75, 12.75/60 =.2125, 1/.2125 = 4.706 sec/unit

CT = 1 / CapacityCapacity = 1/CT


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Dimensional Analysis:Quantity [Q], Time [ T ] and Cost [$]Capacity, output rate, input rate: [Q/T]

Cycle time: [T/Q], WIP [Q] Throughput Time (TPT) : [ T ]

__________ displaytasks (activities) that have adefined starting and endingtime on a resource .

M1

M2

M3

Buffer

6 10 14 18 22 26 30 34 38 42

5

4

4

5

6

6

6

2

1

1

2

3

3

3

8

7

7

8

9

9

9

Alternateterminology Operations tasks, activities.

____________ ( jobs ) materials, work orders, customers, patients, products,projects, cash, claims forms.Output rate flow rate (FR) , production rate, throughput rate

__________ flow time (FT) , manufacturing lead time (MLT)Stage station

Idle time ignored in WIP calculations



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Work in Process & Throughput ______________ (WIP) includes the set of unfinished

items for products in a production process. These items arenot yet completed but either just being fabricated or waitingin a queue for further processing or in a buffer storage.

Optimal production management aims to minimize work inprocess. Work in process requires storage space, represents

bound capital not available for investment and carries aninherent risk of earlier expiration of shelf life of the products. Average WIP Avg. number of jobs in the process .

______________ (TPT) aka flow time (FT) Average TPT Avg. time for a transaction to go through a

stable process. Minimum Throughput Time (TPT) How fast a transaction

goes through the process when the process is empty.Mgmt 361 Module 2 - Process Analysis 6


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Parameter relationships

When a process becomes stable, there is a relationship betweenprocess parameters _______________________.

WIP = Flow time * Flow rate

= FT * FR[ Q ] = [ T ] * [ Q / T ]

WIP = Flow Time / Cycle Time= FT / CT = TPT / CT

[ Q ] = [ T ] / [ T / Q ]

Inventory = Throughput rate * Flow time

Quantity [Q]Time [ T ]Cost [$]Capacity, output rate, input rate: [Q/T] Cycle time: [T/Q]WIP [Q] Throughput Time (TPT) : [ T ]

Mgmt 361Module 2 - Process Analysis 7


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5 min

Stage 1

4 min.

Stage 2

Example: 2.5

5 min / unit4 min / unit

Stage 15 min / unit12 units / hour5 + 4 = 9 min

CT for Stage 1:CT for Stage 2:

Bottleneck:Process CT :

Process Capacity:Min TPT:

Part (a)Design parameters

Stage 1

Stage 2

M1: 5 min

M2: 5 min

M3: 4 min




Example: 2.6

2.5 min / unit4 min / unit

Stage 24 min / unit15 units / hour5 + 4 = 9 min


DoNOT add or

average CT:

CT = 1 / Capacity


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5 min

Stage 1

4 min.

Stage 2

Example 2.5 Two-Stage Serial Process




Part (a)Design parameters

Bottleneck: largest CT or smallest Capacity

Process CT=Bottleneck CTProcess Capacity=Bottleneck Capacity



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Part (b): Schedule a new job every 5 minutesstarting at 0.

Design parametersCT = 5 min / unitBottleneck: Task 1Capacity = 12 units / hour

Minimum TPT = 9 min

M1, Task 15 m

M2, Task 24 m

Littles formula: WIP = TPT/ CT 1.8 [units] = 9 [min] / 5 [min / unit]

CT:

Average TPT:

WIP M1:M2:

Total :

5 min / unit

9 min

1.00 units0.80 units

1.80 units

14

5

9

0

19

10

24

15

Example: 2.5 Continued

1 2 3 4 5 6 7 8 9



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Stage 1

Stage 2

M1: 5 min

M2: 5 min

M3: 4 min

CT for Stage 1:

CT for Stage 2:Bottleneck :Process CT :


1/24 h / u = 2.5 min / unit

4 min / unitStation 24 min / unit1/4 u/min = 15 units / hour5 + 4 = 9 min

Example 2.6-A Hybrid Process

Capacity for stage 1 :

M1: 12 [units / hour]M2: 12 [units / hour]Total: 24 [units / hour]

Design parameters


(1/4) * 60 = 15

(1/5 = .2 .2 + .2 = .4 1/.4 = 2.5)


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Example 2.6 - Schedule 1

Send 2 jobs every 8 min. One to M1, one

to M2. Input rate = 15 units/hr(2/8) * 60 = 15

M1

M2

Bu

M3

Cycle time: 4 min / unitOutput rate: 15 units/hr

Average TPT: (9 + 13) / 2

= 11 min.

WIP

M1M2BuM3

Total

5/8 = 0.6255/8 = 0.6254/8 = 0.5008/8 = 1.000

2.750

Littles Formula

WIP = TPT / CT2.75 = 11 / 4

M34 mM2

5 m

M15 m

Input rate = Capacity


Bottleneck


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Example 2.6 - Schedule 2

Send a job to M1 at 0,8,16, .. And to M2 at 4, 12,20, Input rate = 15 units/hr. M34 mM2

5 m

M15 m

M1

M2

M3

Cycle time: 4 min / unit

Output rate: 15 units/hr

Average TPT: 9 min.WIP

M1

M2M3 (4/4)

Total

5/8 = 0.625

5/8 = 0.6258/8 = 1.000

2.250

Littles Formula

WIP = TPT / CT2.25 = 9 / 4

Input rate=Capacity



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Example 2.6 - Schedule 3: Input rate


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Example 2.6 - Comparison of Schedules

Littles Formula: WIP=TPT/CT

Which schedule is better: Schedule 1 or 2 ?

Schedule 2 : Same output rate (CT) with lower WIP & TPT

M34 mM25 m

M15 m

Schedule 1: CT = 4, Avg. TPT = 11, WIP = 2.75



Design parameters:

CT = 4 min/unit

Capacity = 15 units/hr

Min TPT = 9 min


Documents

Process Analysis - Class Notes - Mod 2 - Part 2(1)