3Control Charts for Variable Data (length, width, etc.) Mean (x-bar) chartsTracks the central tendency (the average or mean value observed) over timeRange (R) charts:Tracks the spread of the distribution over time (estimates the observed variation)
5“n” here equals # of observations in each sample = standard deviation of the sample means“n” here equals # of observations in each sample“k” here = # of samples= standard normal variable or the # of std. deviationsdesired to use to develop the control limits
6Assume the standard deviation of the process is given as 1 Assume the standard deviation of the process is given as 1.13 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.Sample 1Sample 2Sample 3Observation 115.816.116.0Observation 215.9Observation 3Observation 4Sample means15.87515.97518.56, 16.3216.22, 18.5617.62, 14.2319.01, 12.5618.33, 14.28
8Inspectors want to develop process control charts to measure the weight of crates of wood. Data (in pounds) from three samples are:Sample Crate Crate 2 Crate 3 Sample MeansWhat are the upper and lower control limits for this process?Z=3= 4.27
10Range or R Chart k = # of sample ranges Range Chart measures the dispersion or variance of the process whileThe X Bar chart measures the central tendency of the process. Whenselecting D4 and D3 use sample size or number of observations for n.
14Ten samples of 5 observations each have been taken form a Soft drink bottling plant in order to test for volume dispersionin the bottling process. The average sample range was foundTo be .5 ounces. Develop control limits for the sample range..996, -.3201.233, 01.055, 0.788, 1.2011.4, 0
16(P) Fraction Defective Chart Used for yes or no type judgments (acceptable/not acceptable, works/doesn’t work, on time/late, etc.)p = proportion of nonconforming items= average proportion of nonconforming items
17(P) Fraction Defective Chart n = # of observations in each sample= standard normal variable or the # of std. deviationsdesired to use to develop the control limitsK = number of samples
18Note: Lower control limit can’t be negative. P-Chart Example: A Production manager for a tire company has inspected the number of defective tires in five random samples with 20 tires in each sample. The table below shows the number of defective tires in each sample of 20 tires. Z= 3. Calculate the control limits.SampleNumber of Defective TiresNumber of Tires in each SampleProportion Defective1320.152.10.0545Total9100.09Solution:Note: Lower control limit can’t be negative.
19Number-of-Defectives or C Chart Used when looking at # of defectsc = # of defects= average # of defects per sample
20Number-of-Defectives or C Chart = standard normal variable or the # of std. deviationsdesired to use to develop the control limits
21C-Chart Example: The number of weekly customer complaints are monitored in a large hotel using a c-chart. Develop three sigma control limits using the data table below. Z=3.WeekNumber of Complaints13245678910Total22Solution:
22Process Capability Index Can a process or system meet it’s requirements?Cp < 1: process not capable of meeting design specsCp ≥ 1: process capable of meeting design specsOne shortcoming, Cp assumes that the process is centered on the specification range
23Process Capability Product Specifications Preset product or service dimensions, tolerancese.g. bottle fill might be 16 oz. ±.2 oz. (15.8oz.-16.2oz.)Based on how product is to be used or what the customer expectsProcess Capability – Cp and CpkAssessing capability involves evaluating process variability relative to preset product or service specificationsCp assumes that the process is centered in the specification rangeCpk helps to address a possible lack of centering of the process
24min = minimum of the two= mu or mean of the processIfIs less than 1, then the process is not capable oris not centered.
25Capability Indexes Centered Process (Cp): Any Process (Cpk): Cp=Cpk when process is centered
26ExampleDesign specifications call for a target value of /-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
29Critical Path Method (CPM) CPM is an approach to scheduling and controlling project activities.The critical path is the sequence of activities that take the longest time and defines the total project completion time.Rule 1: EF = ES + Time to complete activityRule 2: the ES time for an activity equals the largest EF time of all immediate predecessors.Rule 3: LS = LF – Time to complete activityRule 4: the LF time for an activity is the smallest LS of all immediate successors.
34Calculate Path Durations The longest path (ABDEGIJK) limits the project’s duration (project cannot finish in less time than its longest path)ABDEGIJK is the project’s critical path
35Adding Feeder Buffers to Critical Chains The theory of constraints, the basis for critical chains, focuses on keeping bottlenecks busy.Time buffers can be put between bottlenecks in the critical pathThese feeder buffers protect the critical path from delays in non-critical paths
37Some Network Definitions All activities on the critical path have zero slackSlack defines how long non-critical activities can be delayed without delaying the projectSlack = the activity’s late finish minus its early finish (or its late start minus its early start)Earliest Start (ES) = the earliest finish of the immediately preceding activityEarliest Finish (EF) = is the ES plus the activity timeLatest Start (LS) and Latest Finish (LF) depend on whether or not the activity is on the critical path
38ES=4+6=10EF=10LS=4LF=10ES=16EF=30ES=32EF=34ES=10EF=16D(6)E(14)H(2)B(6)Latest EF= Next ESES=35EF=39ES=39EF=41A(4)K(2)G(2)J(4)ES=0EF=4LS=0LF=4C(3)ES=30EF=32LS=30LF=32F(5)I(3)Calculate EarlyStarts & FinishesES=4EF=7LS=22LF=25ES=7EF=12LS=25LF=30ES=32EF=35LS=32LF=35
43Arrival & Service Patterns Arrival rate:The average number of customers arriving per time periodService rate:The average number of customers that can be serviced during the same period of time
44Infinite Population, Single-Server, Single Line, Single Phase Formulae
45Infinite Population, Single-Server, Single Line, Single Phase Formulae
46ExampleA 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.Based on this description, we know:Mu = 20 (exponential distribution)Lambda = 15 (Poisson distribution)
50Average Time a Student Spends in the System .2 hours or 12 minutes
51Average Time a Student Spends Waiting (Before Service) Too long?After 5 minutes peopleget anxious
52Consider a single-line, single-server waiting line system 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 45 seconds per customer. What is the average number of customers in the system?2.25343.254.2560/80-60= 3