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Unformatted text preview: MGMT 46000 Operations Management (OM) Ch II: Process Analysis  III Capacity of M1: 6[units/hr] 6[units/hr] Capacity of M2: 10[units/hr] Total for Stage 1: 16 [units/hr] M1 10 min CT for Stage 1 ( = 60 / 16): 3.75 min / unit M3 CT for Stage 2: 4 min / unit 4 min Bottleneck: Stage 2 M2 Stage 2 6 min Process CT : 4 min / unit Process Capacity: 15 units / hour Stage 1 Min TPT: 6 + 4 = 10 min Schedule: Start job every 12 minutes on M1 (0 12 24 Schedule: Start a job every 12 minutes on M1 (0, 12, 24, ..). Start a job every 6 minutes on M2 (0, 6, 12, ..). Example: 7 Design parameters Design M1 M2 Buffer M3 1 2 3 3 1 2 3 4 5 6 6 4 5 6 7 8 9 9 7 8 9 TPT:
Job 1: 6 + 4 = 10 Job 2: 10 + 4 = 14 Job 3: 6 + 2 + 4 = 12
42 6 10 14 18 22 26 30 34 38 Average TPT?
2 ChIIProcessAnalysisIIISp10 1 ChIIProcessAnalysisIIISp10 MGMT 36100 MGMT 36100 Example: 7
M1 10 min M2 6 min
Stage 1 M3 4 min Stage 2 Design parameters Process CT: 4 min / unit Process Capacity: 15 units / hour Min TPT: 10 min
Start Start 1 job every 12 min. on M1 (0, 12, 24, ..). Start 1 job every 6 min. on M2 (0, 6, 12, ..). Applying Little’s formula to processes with variable processing times / arrival times. WIP = Flow rate * Flow time Ex: 8 M1 M1 M2 Buffer M3 1 2 3 3 1 2 3 4 5 6 6 4 5 6 7 8 9 9 7 8 9 200 customers arrive in a twohour period. 45 customers inside. Time spent in the facility? Flow rate = 100 [customers / hr] [customers WIP = 45 [customers] [customers] Flow time = 45 [customers] / 100 [customers / hr] = 0.45 hours = 0.45 [hours] * 60 [min / hr] = 27 minutes 27 minutes Ex: 9
CT: Average TPT: WIP M1: M2: Buffer: M3: Total: 4 [min/unit] 12 [min] 5/6 units 1 units 1/6 units 1 unit 3 units
3 6 10 14 18 22 26 30 34 38 42 WIP = TPT / CT 3 = 12 / 4
ChIIProcessAnalysisIIISp10 6000 claims per year (50 Weeks). Processing time 2 weeks. # of applications in the process? Flow rate = 6000 [claims / year] / 50 [weeks / year] = 120 [claims/ week] Flow time = 2 [weeks] WIP = 120 [claims/ week] * 2 [weeks] = 240 claims 4 ChIIProcessAnalysisIIISp10
MGMT 36100 MGMT 36100 WIP = Flow rate * Flow time WIP = Flow rate * Flow time Ex: 10 $100M per year. Accounts receivables: $15M. Payment days? Flow rate = 100M [$ / year] , WIP = 15M [$] Flow time = 15M [$] / 100M [$ / year] = 0.15 [years] = 0.15 [years] * 360 [days / year] = 54 days 15 customers waiting. Processing 2.5 customers per minute. How long will the customer wait? WIP = 15 [customers] Flow rate = 2.5 [customers / min] [customers Flow time = 15 [customers] / 2.5 [customers/ min] = 6 minutes Ex: 12
Date 12/5 12/7 12/8 Average balance $3000. Money turned over 6 times per year (every 2 months). How many dollars per year? Transaction Deposit Check #123 Check # 124 Amount 1200.00 77.37 100.00 Balance 4533.37 4456.00 4356.00 Ex: 11 11 WIP = 3000 [$] , Flow time = 1/6 [year] Flow rate = 3000 [$] / (1/6) [year] = 18,000 [$] / [year] ChIIProcessAnalysisIIISp10 5 ChIIProcessAnalysisIIISp10 6 MGMT 36100 MGMT 36100 ChIIProcessAnalysisIIISp10 1 Assembly Operations  Ex: 13
Stage 1 M1 9 min AA M2 5 min P1 4 min Stage 4 BB X1 3 min Stage 2 Y1 5 min Stage 3 AA
St BB
Cycle Time King Soopers Bakery
Custom Cake Job shop process. Low volume Low / no automation Pastry Batch process. Moderate volume Bread Continuous process. High volume R1 16 min R2 18 min R3 15 min Stage 5 1 2 3 4 5 3.21 [min/unit] 3.00 [min/unit] 5.00 [min/unit] 4.00 [min/unit] 5.41 [min/unit] Moderate automation More automation Difficult and Difficult and expensive expensive to change capacity Labor: Little skill Bottleneck Minimum TPT = ? From Stage 1, you can get at the end of 5 minutes From Stages 2 and 3, you can get at the end of 8 minutes Easy and inexpensive Moderately difficult Easy and inexpensive Moderately difficult / to to change capacity expensive expensive to change …. capacity Labor: skilled . Labor: Semiskilled Semi Both are required for the assembly. So you can start assembly after 8 minutes. Add 4 minutes on P1 then 15 minutes on R1.
ChIIProcessAnalysisIIISp10 Minimum TPT = 8 + 4 + 15 = 27 minutes
MGMT 36100 7 ChIIProcessAnalysisIIISp10 8 MGMT 36100 Other Process Characteristics
A Buffer B
Y X X X Z Z blocked
X X Z Industrial engineering
• Frederick Taylor (1856 – 1915) is called the father of scientific management. First to perform stop watch studies for process improvement. Frank Gilbreth (18681924) is known for his work in motion studies. His wife, Lillian Gilbreth (18781972) was a professor at Purdue (19351948). Henry Gantt (18611919): known for Gantt charts. 1908 – Pennsylvania State College starts Industrial Engineering. A B Y X Z • • • • A: Utilization 100% WIP = 1.000 B: Utilization < 100% WIP < 1.000 A: Utilization < 100% WIP = 1.000 B: Utilization < 100% WIP < 1.000 idle Utilization: productive use of the resource. WIP contribution: average number of jobs occupying a resource (or inprocess buffer). in • Courses offered in I.E.: Quality Control Work Control, Work Measurements, Production Planning and Control, Simulation, O.R., … • Industrial engineers use process charts with standardized symbols.
9 ChIIProcessAnalysisIIISp10 10 ChIIProcessAnalysisIIISp10 MGMT 36100 MGMT 36100 Subject: Hamburger Assembly Process Hamburger Chart No: Prepared By: Present Method: Proposed Method: Date:
Distance (ft.) Time (m) Symbol Process Description Process Process Chart Example Some Some insights
Capacity of a process is the maximum output rate. If input rate becomes higher than the capacity, the process will not become stable. TPT for incoming jobs will become higher and higher. If we need to increase the capacity of any process, we need to add resources at the bottleneck stage. This may shift bottleneck to another stage. For two schedules it is possible to have same cycle time (output rate) but different WIP and TPT. In a number of cases, a batch of jobs is released and the whole batch moves from station to station. Obviously TPT values will be higher since the whole batch must be completed. Batching of products means more WIP and need for buffer space. Analysis for nondeterministic cases can be complex. Since processing times are variables, bottleneck operations can shift frequently. If an important job is to be completed as fast as possible, we may have to free up fastest resources at stage. The time required to complete this is given by the minimum TPT.
11 ChIIProcessAnalysisIIISp10 12 0.00 1.5 0.05 2.50 0.05 1.0 0.5 0.5 0.05 0.15 0.10 0.20 0.05 Meat patty in storage Transfer to broiler Broiler Inspect visually Transfer to rack Temporary storage Obtain bun, lettuce, etc. Assemble order Transfer to finish rack Place in finish rack ChIIProcessAnalysisIIISp10 MGMT 36100 MGMT 36100 ChIIProcessAnalysisIIISp10 2 ...
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 Spring '10
 panwalker
 Management

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