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Quantitative Review III. ISQS 3344 Introduction to Production and Operations Management Spring 2014. Waiting Line Theory. Infinite population, single-server, single line, single phase formUlae. Infinite population, single-server, single line, single phase formulae.
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Quantitative Review III ISQS 3344 Introduction to Production andOperations ManagementSpring 2014
Infinite population, single-server,single line, single phase formUlae
Infinite population, single-server,single line, single phase formulae
A help desk in the computer lab serves students on a first-come, first served basis. On average, 15 students need help every hour. The help desk can serve an average of 20 students per hour. Help Desk Example
Based on the description, we know: • = 20 (Exponential distribution) • = 15 (Poisson distribution) • Average System Utilization • Average Number of Students in the System Help Desk Example
Average Number of Students Waiting in Line Average Time a Student Spends in the System Average Time a Student Spends Waiting (Before Service) Help Desk Example
Probability of n Students in the System Help Desk Example
Consider a single-line, single-server waiting line system. The arrival rate is 100 people per hour, and the service rate µ is 150 people per hour. What is the probability of having 3 people in the system? Pn = (1 - ) n = / = 100/150 = 0.67 P3 = (1 - ) 3 = (1 – 0.67)(0.67)3 = 0.0992 Waiting Line PROBLEM
Consider a single-line, single-server waiting line system. Suppose that customers arrive according to a Poisson distribution at an average rate of 60 per hour, and the average (exponentially distributed) service time is 50 seconds per customer. What is the average number of customers in the system? L = / ( - ) = 60 / (72 – 60) = 5 Waiting Line PROBLEM
Consider a single-server queuing system. If the arrival rate is 30 customers per hour and it takes 3 minutes on average to serve a customer, what is the average waiting time in the waiting line in minutes? Waiting Line PROBLEM
Waiting Line PROBLEM (Solution) Mean arrival rate = 30 customers per hour Mean service rate = 3 minutes per customer Average time spent waiting in line Waiting time continually increases!
Quality Control and Statistical Process Control (SPC) CHAPTER 6 & Supplement 6
For Variable Data: • Mean ( ) Chart • Range (R) Chart • For Non-variable Data: • Defective (C) Chart • Fraction Defective (P) Chart Control Charts
Mean ( ) Chart: Tracks the central tendency (the average or mean value observed) over time • Range (R) Chart: Tracks the spread of the distribution over time (estimates the observed variation) Control Charts for Variable Data
= sample average k = # of samples n = # of observations in each sample z = standard normal variable or the # of standard deviations desired to use to develop the control limits Mean ( ) Chart
Assume the standard deviation of the process is given as 1.33 ounces.Management wants a 3-sigma chart (only 0.26% chance of alpha error). Observed values shown in the table are in ounces. Calculate the UCL and LCL. Mean ( ) Chart PROBLEM
Range chart measures the dispersion or variance of the process while the mean chart measures the central tendency of the process. R = range of each sample k = # of samples When selecting D4 and D3 use number of observations for n. Range (R) Chart
Factors for R-Chart Sample Size (n) D3 D4 2 0.00 3.27 3 0.00 2.57 4 0.00 2.28 5 0.00 2.11 6 0.00 2.00 7 0.08 1.92 8 0.14 1.86 9 0.18 1.82 10 0.22 1.78 11 0.26 1.74 12 0.28 1.72 13 0.31 1.69 14 0.33 1.67 15 0.35 1.65 Factor for Range (R) Chart
Four samples of 10 observations each have been taken form a Soft drink bottling plant in order to test for volume dispersion in the bottling process. The average sample range was found to be .4 ounces. Develop control limits for the sample range. Range (R) Chart Problem
Range (R) Chart PROBLEM Develop control limits for the sample ranges.
Defective (C) Chart: Used when looking at # of defects • Fraction Defective (P) Chart: Used for yes or no type judgments (acceptable/not acceptable, works/doesn’t work, on time/late, etc.) Control Charts for NON-Variable Data
c = number of defects = average number of defects per sample z = standard normal variable or the # of standard deviations desired to use to develop the control limits Defective (C) Chart
The number of weekly customer complaints are monitored in a large hotel using a C-chart. Develop three sigma (z=3) control limits using the data table. Defective (C) Chart PROBLEM Note: Lower control limit can’t be negative.
p = proportion of nonconforming items = average proportion of nonconforming items z = standard normal variable or the # of standard deviations desired to use to develop the control limits Fraction Defective (P) Chart
A Production manager for a tire company has inspected the number of defective tires in four random samples with 25 tires in each sample. The table shows the number of defective tires in each sample of 25 tires. z=3. Calculate the control limits. Fraction Defective (P) Chart PROBLEM
Fraction Defective (P) Chart PROBLEM Note: LCL can’t be negative.
Can a process or system meet its requirements? • Product Specifications • Preset product or service dimensions, tolerances, e.g., bottle fill might be 16 oz. ± .2 oz. • Based on how product is to be used or what the customer expects • Process Capability • Assessing capability involves evaluating process variability relative to preset product or service specifications Process capability
Cp assumes that the process is centered in the specification range. • Cp < 1: process not capable of meeting design specs • Cp≥ 1: process capable of meeting design specs Process capability indexes (Cp & Cpk)
Cpkhelps to address a possible lack of centering of the process. • min = minimum of the two • = mu or mean of the process • Cpk < 1: process not capable or not centered • Cpk≥ 1: process capable or centered Cp=Cpkwhen process is centered. Process capability indexes (Cp & Cpk)
Design specifications call for a target value of 16.0 +/- 0.2 ounces (USL = 16.2 & LSL = 15.8) Observed process output has a mean of 15.9 and a standard deviation of 0.1 ounces Process capability example
Cp Cpk Process capability example
Project Management CHAPTER 3 Reference: Erik Larson and Clifford Gray, 2011, Project Management: The Managerial Process, McGraw Hill.
Forward Pass – Earliest Times • Early Start (ES) – How soon can the activity start? • Early Finish (EF) – How soon can the activity finish? • Backward Pass – Latest Times • Late Start (LS) – How late can the activity start? • Late Finish (LF) – How late can the activity finish? • Slack (SL) – How long can the activity be delayed? • Critical Path (CP)– The longest path in the network which, when delayed, will delay the project Network Computation Process
You add activity times along each path in the network (ES + Duration = EF). • You carry the early finish (EF) to the next activity where it becomes its early start (ES), unless • The next succeeding activity is a merge activity. In this case, you select the largest early finish number (EF) of all its immediate predecessor activities. Forward Pass Computation
You subtract activity times along each path starting with the project end activity (LF - Duration = LS). • You carry the late start (LS) to the next preceding activity to establish its late finish (LF), unless • The next preceding activity is a burst activity. In this case, you select the smallest late start number (LS) of all its immediate successor activities to establish its late finish (LF). Backward Pass Computation
Slack for an activity is simply the difference between the LS and ES (LS – ES) or between LF and EF (LF – EF). Slack tells us the amount of time an activity can be delayed and yet not delay the project. When the LF = EF for the end project activity, the critical path can be identified as those activities that also have LF = EF or a slack of zero (LF – EF = 0 or LS – ES = 0). Determining Slack
1 B 6 0 Research Topic 1 5 6 0 A 1 9 D 11 9 9 11 F G E 10 10 12 0 0 0 1 1 Identify Topic Create Graphics Final Draft References Edit Paper 0 1 1 9 2 11 10 10 11 1 1 1 11 11 12 6 C 9 0 Draft Paper Legend 6 3 9 ES ID EF Description SL LS DUR LF Critical path method network Group Term Paper
1 B 6 Research Topic 5 0 A 1 9 D 11 9 11 9 F E G 10 12 10 Identify Topic Final Draft References Create Graphics Edit Paper 1 2 1 1 1 6 C 9 Draft Paper Legend 3 ES ID EF Description SL LS DUR LF Forward Pass Computation Always start at 0 EF = ES+DUR EF = ES+DUR EF = ES+DUR EF = ES+DUR Group Term Paper
1 B 6 0 Research Topic 1 5 6 0 A 1 9 D 11 9 11 9 F E G 12 10 10 1 Identify Topic Create Graphics References Final Draft Edit Paper 0 1 1 9 2 11 11 10 10 1 1 1 12 11 11 6 C 9 Draft Paper Legend 6 3 9 ES ID EF Description SL LS DUR LF Backward Pass Computation Group Term Paper LS = LF - DUR LS = LF - DUR LS = LF - DUR LS = LF - DUR EF=LF
1 B 6 0 Research Topic 1 5 6 0 A 1 9 D 11 9 11 9 G E F 12 10 10 0 0 1 1 0 Identify Topic Create Graphics Final Draft References Edit Paper 0 1 1 9 2 11 10 11 10 1 1 1 11 12 11 6 C 9 0 Draft Paper Legend 6 3 9 ES ID EF Description SL LS DUR LF Determining Slack SL = LS – ES or LF - EF SL = LS – ES or LF - EF SL = LS – ES or LF - EF CRITICAL PATH? Group Term Paper
Compute the early, late, and slack activity times Determine the planned project duration Identify the critical path What should you do if the Doors activity is going to take two extra days? Garage Problem
2 Erect Frame 4 1 3 8 10 7 9 5 4 Pour Foundation Windows Door Opener Paint Clean-up Doors Rough-in Frame Roof 3 4 2 2 1 1 1 1 6 Electrical Legend 3 ES ID EF Description SL LS DUR LF Garage Problem Project Duration: ______ days Critical Path: _______________
3 2 7 0 Erect Frame 3 4 7 0 1 3 7 3 11 7 13 15 11 7 10 4 10 7 9 8 5 13 8 11 16 15 8 0 0 3 0 2 0 3 0 Pour Foundation Doors Door Opener Clean-up Rough-in Frame Windows Paint Roof 0 3 3 7 4 11 13 10 15 10 11 12 1 1 2 2 1 1 11 11 15 13 16 13 7 6 10 1 Electrical Legend 8 3 11 ES ID EF Description SL LS DUR LF Garage Problem Project Duration: ___16___ days Critical Path: __1237910__ If activity 5 is going to take two extra days, you probably do not have to do any thing because this activity has three days of slack – no effect on project duration.
FINAL EXAM • When: • Friday May 9, 2014 • 7.30p.m. – 10.00p.m. • Where: • TBA