Mgmt 361 - Ch 2 II0[1]

# Mgmt 361 - Ch 2 II0[1] - MGMT 361 Operations Management...

This preview shows pages 1–7. Sign up to view the full content.

MGMT 361 Operations Management (OM) Ch II: Process Analysis - II Mgmt 361 Chapter 2 Process Analysis 1 Justin Jia

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
A Framework for Process Analysis 1. Draw a Precedence Diagram 2. Draw a Process Flowchart 3. Determine Capacity for Each Resource in the Process and identify the Bottleneck 4. Calculate Relevant Process Measures Review 5. Understand Process Economics (Little’s formula) 6. Consider Changes to Improve System Performance Mgmt 361 Chapter 2 Process Analysis 2
Process Parameters: Design Capacity : Maximum rate of transaction processing. (Design) Cycle Time = 1/( capacity). Minimum T hrough p ut T ime ( TPT ) – How fast a transaction goes through the process when the process is empty. Run time Input Rate : transaction arrival rate Output Rate : Current rate of processing Actual Cycle Time = 1 / (output rate) Average TPT : Avg. time for a transaction to go through a stable process. Average WIP : Avg. number of transactions in the process . Design parameters indicate what a process can achieve. Run time parameters indicate what the current values are for a specific schedule . These values change with the schedule. Mgmt 361 Chapter 2 Process Analysis 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Run Time parameters Design parameters Example. 3: Process Parameters for a Single Machine M1 4 min (15 units/ hour) Part A: A new component is scheduled every 4 minutes M1 Job 1 4 Job 3 12 Job 4 16 Job 2 8 Input rate Output rate Actual CT AverageTPT WIP = 1/4[u/min]*60[min/h] = 15 [units / hour] = 1/4[u/min]*60[min/h] = 15 [units / hour] = 1/15[hour/unit] *60[min/h] = 4 [min/unit] = 4 [min] = 1.0 [units] (TPT /CT) so 4/4 = 1 Cycle time = Capacity = Minimum TPT = 4 [min. / unit] ¼ [units/min.] * 60 [min / hour] = 15 [units/hour] 4 [min.] Gantt Chart Mgmt 361 Chapter 2 Process Analysis 4
Run Time parameters Example 3 continued M1 4 min Input rate Output rate Actual CT TPT WIP Gantt Chart Part B: A new component is scheduled every 10 minutes M1 Job 2 10 14 Job 1 4 = 1/10 [u/ min]*60[min/h] = 6 [units / hour] = 1/10 [u/ min]*60[min/h] = 6 [units / hour] = 1/6 [hours/unit]*60[min/h] =10 [min/unit] = 4 [min] = TPT/CT = 4/10 = 0.4 [units] Run Time parameters depend on schedule! Input rate = output rate IF process is stable. Mgmt 361 Chapter 2 Process Analysis 5

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Design Parameters Capacity = 15 [units /hr] Cycle time = 4 [min/unit] Minimum TPT = 4 [min] Part C: A new component is scheduled every 3 minutes. Input rate
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/11/2012 for the course MGMT 361 taught by Professor Panwalker during the Spring '10 term at Purdue.

### Page1 / 28

Mgmt 361 - Ch 2 II0[1] - MGMT 361 Operations Management...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online