ChII-ProcessAnalysisII - Quantity Dimensional Analysis:...

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Unformatted text preview: Quantity Dimensional Analysis: Quantity [Q], Time [ T ] and cost [$] MGMT 46000 Operations Management (OM) Ch II: Process Analysis - II Alternate Alternate terminology Capacity, Capacity, output rate, input rate: [Q/T] rate: Cycle time: [T/Q], WIP [Q] Throughput Time - TPT : [ T ] Operations: tasks, activities. Transactions ( jobs ): materials, work orders, customers, materials, patients, products, projects, cash, claims forms. Output rate: flow rate (FR), production rate, throughput rate flow manufacturin TPT: TPT: flow time (FT), manufacturing lead time (MLT) time Stage: station M1 2 1 3 3 1 2 3 4 5 4 5 6 6 6 7 8 7 8 9 9 9 Gantt charts display display tasks (activities) that have a defined starting and ending time on a resource. M2 Buffer M3 6 10 14 18 22 26 30 34 38 42 1 Idle time ignored in WIP calculations 2 Parameter relationships When a process becomes stable, there is a relationship between process parameters – Little’s formula. WIP = Flow time * Flow rate = FT * FR [Q ] = [ T ] * [ Q /T ] WIP WIP = Flow Time / Cycle Time Flow Cycle = FT / CT = TPT / CT [ Q ] = [ T ] / [ T /Q ] Example: 5 M1: 5 min Example: 6 M3: 4 min M2: 5 min 5 min Stage 1 4 min. Stage 2 Stage Stage 1 2.5 min / unit 4 min / unit Stage 2 4 min / unit 15 units / hour 5 + 4 = 9 min 4 Stage 2 Part (a) (a) Design parameters Little’s original formula : L = * W 3 CT for Stage 1: 5 min / unit CT for Stage 1: CT for Stage 2: 4 min / unit CT for Stage 2: Bottleneck: Stage 1 Bottleneck: Process CT : 5 min / unit Process CT : Process Capacity: 12 units / hour Process Capacity: Min TPT: 5 + 4 = 9 min Min TPT: Example: 5 M1, Task 1 5m M2, Task 2 4m Part (b): Schedule a new job every 5 minutes starting at 0. 0 5 10 15 Design parameters CT = 5 min / unit Bottleneck: Task 1 Capacity = 12 units / hour Minimum TPT = 9 min Example 6 M1 5m M2 5m M3 4m CT Design parameters: CT = 4 min/unit, Capacity = 15 units/hr, Min TPT = 9 min. Schedule 1: Send 2 jobs every 8 min. One to M1 , one to M2. Input rate = 15 units/hr CT: 5 min / unit Average TPT: 9 min WIP M1: 1.00 units M2: 0.80 units Total: 1.80 units 9 123456789 M1 M2 Bu M3 14 19 24 Little’s formula: WIP = TPT / CT 1.8 [units] = 9 [min] / 5 [min / unit] 5 Cycle time: 4 min / unit M1 5/8 = 0.625 Output rate: 15 units/hr M2 5/8 = 0.625 Average TPT: (9 + 13) / 2 WIP Bu 4/8 = 0.500 1.000 M3 = 11 min. Total 2.750 Little’s Formula WIP = TPT / CT 2.75 = 11 / 4 6 1 Example 6 M1 5m M2 5m M3 4m Design parameters: CT = 4 min/unit, CT Capacity = 15 units/hr, Min TPT = 9 min. Schedule 2: Send a job to M1 at 0,8,16, .. And to M2 at 4, 12, 20, … Input rate = 15 units/hr. Example 6 M1 5m M2 5m M3 4m Design parameters: CT = 4 min/unit, CT Capacity = 15 units/hr, Min TPT = 9 min. Schedule 3: Not enough work. Send a job to M1 at 0,10, 20, .. And to M2 at 4, 14, 24, … Input rate = 12 units/hr. M1 M2 M3 M1 M2 M3 Cycle time: 4 min / unit M1 5/8 = 0.625 Output rate: 15 units/hr M2 5/8 = 0.625 1.000 Average TPT: 9 min. WIP M3 2.250 Total Little’s Formula WIP = TPT / CT 2.25 = 9 / 4 Cycle time: 5 min / unit M1 5/10 = 0. 5 Output rate: 12 units/hr M2 5/10 = 0. 5 8/10 = 0.8 Average TPT: 9 min. WIP M3 Total 1.8 Little’s Formula WIP = TPT / CT 1.8 = 9 / 5 7 8 Example 6 M1 5m M2 5m M3 4m Shouldice Hospital Why Shouldice? Schedule 1: CT = 4, Avg. TPT = 11, WIP = 2.75 Low cost. Better process (low rate of recurrence). Non – hospital like atmosphere. Patients with similar ailment. Faster recovery time. Annual checkup checkup. Design parameters: CT = 4 min/unit min/unit, Capacity = 15 units/hr, Min TPT = 9 min. Schedule 2: CT = 4, Avg. TPT = 9, WIP = 2.25 2: Avg TPT WIP Capacity calculations: Where is the bottleneck? • About 7000 patients / year. • Average stay 3.5 days. • Patients admitted on 5 days each week, operations performed on 5 days each week (M-F). (M• 12 doctors on staff. Perform 3 to 4 operations / day. • 5 operating rooms, 7 operations scheduled each day in a room. • Patients examination takes 15 to 20 minutes, exams sessions are conducted in 6 examination rooms for 3 hours each day. • 89 beds. 9 10 Schedule 3: CT = 5, Avg. TPT = 9, WIP = 1.8 Capacity calculations: Where is the bottleneck? 7000 patients / year 140 patients / week Average stay = 3.5 days (50% leave after 3 days, 50% after 4 days) Examination room: 15 to 20 minutes per patient (average 17.5) {6 rooms * 180 [min /(day * room)] * 5 [days / week] } {17.5 [min / patient]} 300+ [patients / week] Operating room: 5 rooms * 7 [operations / (day * room)] * 5 [days / week] rooms [operations (day room)] [days w eek] = 175 [operations / week] Doctors: 10[doctors] * 3.5 [operations/(day * doctor)] * 5 [days / week] = 175 [operations / week] Beds: 89[beds] 89[beds] * 7 [days / week)] / 3.5 [ bed -days / patient] = 178 [patients / week] 11 More refined calculations for bed capacity: 140 patients admitted Sunday through Thursday or 28 per day. Assume that we start with no patients in the hospital. Week 1 Day In Out Beds In Week 2 Out Beds In Week 3 Out Beds Sun Mon Tue Wed Thu Fri Sat 28 28 28 28 28 0 0 14 28 28 28 28 56 84 98 98 70 42 28 28 28 28 28 0 0 28 14 0 14 28 28 28 42 56 84 98 98 70 42 28 28 28 28 28 0 0 28 14 0 14 28 28 28 42 56 84 98 98 70 42 Only 89 beds, how are extra people accommodated? 12 2 Applying Little’s formula In Week 2 Day Sun Mon Tue Wed Thu Fri Sat Out I 28 28 28 28 28 0 0 Flow rate = 140 [patients / week] / 7 [days/week] Patients = 20 [patients / day] Flow time = 3.5 [days] P 42 56 84 98 98 70 42 O 28 14 0 14 28 28 28 Week 3 P 42 56 84 98 98 70 42 I 28 28 28 28 28 0 0 O 28 14 0 14 28 28 28 WIP = average no. of patients = (42+56+84+98+98+70+42)/7 = 490 / 7 = 70 [patients] WIP = Flow rate * Flow time 70 [patients] = 20 [patients / day] * 3.5 [days] 3.5 13 3 ...
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This note was uploaded on 01/31/2011 for the course MGMT 361 taught by Professor Panwalker during the Spring '10 term at Purdue University-West Lafayette.

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