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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
8/13/2019 Process Analysis - Class Notes - Mod 2 - Part 2(1)
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Every Day Starts theDay Before
Mgmt 361 2Module 2 - Process Analysis
8/13/2019 Process Analysis - Class Notes - Mod 2 - Part 2(1)
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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:
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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
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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 ]
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8/13/2019 Process Analysis - Class Notes - Mod 2 - Part 2(1)
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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
CT for Stage 1:CT for Stage 2:
Bottleneck:Process CT :
Process Capacity:Min TPT:
Example: 2.6
2.5 min / unit4 min / unit
Stage 24 min / unit15 units / hour5 + 4 = 9 min
Mgmt 361 Module 2 - Process Analysis 8
DoNOT add or
average CT:
CT = 1 / Capacity
8/13/2019 Process Analysis - Class Notes - Mod 2 - Part 2(1)
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5 min
Stage 1
4 min.
Stage 2
Example 2.5 Two-Stage Serial Process
CT for Stage 1:CT for Stage 2:
Bottleneck:Process CT :
Process Capacity:Min TPT:
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 :
Process Capacity:Min TPT:
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
Mgmt 361 Module 2 - Process Analysis 11
(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
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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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8/13/2019 Process Analysis - Class Notes - Mod 2 - Part 2(1)
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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
Schedule 2: CT = 4, Avg. TPT = 9, WIP = 2.25
Schedule 3: CT = 5, Avg. TPT = 9, WIP = 1.8
Design parameters:
CT = 4 min/unit
Capacity = 15 units/hr
Min TPT = 9 min
Mgmt 361 Module 2 - Process Analysis 15