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Quantitative Review III

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Chapter 6 and 6S Statistical Process Control

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Control Charts for Variable Data (length, width, etc.) Mean (x-bar) charts –Tracks the central tendency (the average or mean value observed) over time Range (R) charts: –Tracks the spread of the distribution over time (estimates the observed variation)

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x-Bar Computations x-bar = sample average

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= standard deviation of the sample means = standard normal variable or the # of std. deviations desired to use to develop the control limits n here equals # of observations in each sample k here = # of samples

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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. 0 Sample 1Sample 2Sample 3 Observation Observation Observation Observation Sample means A.18.56, B.16.22, C.17.62, D.19.01, E.18.33, 14.28

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Inspectors want to develop process control charts to measure the weight of crates of wood. Data (in pounds) from three samples are: Sample Crate 1 Crate 2 Crate 3Sample Means What are the upper and lower control limits for this process? Z=3 = 4.27

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Range or R Chart k = # of sample ranges Range Chart measures the dispersion or variance of the process while The X Bar chart measures the central tendency of the process. When selecting D4 and D3 use sample size or number of observations for n.

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Control Chart Factors

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Example Sample 1Sample 2Sample 3 Observation Observation Observation Observation Sample means Sample ranges = = =0.2

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R-chart Computations (Use D3 & D4 Factors: Table 6-1)

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Ten samples of 5 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.5 ounces. Develop control limits for the sample range. A..996, B.1.233, 0 C.1.055, 0 D..788, E.1.4, 0

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(P) Fraction Defective Chart Used for yes or no type judgments (acceptable/not acceptable, works/doesnt work, on time/late, etc.) p = proportion of nonconforming items = average proportion of nonconforming items

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(P) Fraction Defective Chart n = # of observations in each sample = standard normal variable or the # of std. deviations desired to use to develop the control limits K = number of samples

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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 Tires Number of Tires in each Sample Proportion Defective Total Solution: Note: Lower control limit cant be negative.

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Number-of-Defectives or C Chart Used when looking at # of defects c = # of defects = average # of defects per sample

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Number-of-Defectives or C Chart = standard normal variable or the # of std. deviations desired to use to develop the control limits

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C-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 Complaints Total22 Solution:

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Process Capability Index Can a process or system meet its requirements? Cp < 1: process not capable of meeting design specs Cp 1: process capable of meeting design specs One shortcoming, Cp assumes that the process is centered on the specification range

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Process Capability Product Specifications –Preset product or service dimensions, tolerances –e.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 expects Process Capability – Cp and Cpk –Assessing capability involves evaluating process variability relative to preset product or service specifications –C p assumes that the process is centered in the specification range –C pk helps to address a possible lack of centering of the process

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min = minimum of the two = mu or mean of the process IfIs less than 1, then the process is not capable or is not centered.

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Capability Indexes Centered Process (C p ): Any Process (C pk ): Cp=Cpk when process is centered

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Example Design 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

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Computations C p : C pk :

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Chapter 3 Project Mgt. and Waiting Line Theory

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Critical 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 activity Rule 2: the ES time for an activity equals the largest EF time of all immediate predecessors. Rule 3: LS = LF – Time to complete activity Rule 4: the LF time for an activity is the smallest LS of all immediate successors.

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Example

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CPM Diagram

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Add Activity Durations

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Identify Unique Paths

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Calculate Path Durations The longest path (ABDEGIJK) limits the projects duration (project cannot finish in less time than its longest path) ABDEGIJK is the projects critical path

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Adding 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 path These feeder buffers protect the critical path from delays in non- critical paths

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B(6) D(6) A(4) C(3) F(5) E(14) G(2) I(3) H(2) J(4) K(2) ES=0 EF=4 LS=0 LF=4 ES=4 EF=10 LS=4 LF=10 ES=10 EF=16 LS=10 LF=16 ES=16 EF=30 LS=16 LF=30 ES=32 EF=34 LS=33 LF=35 ES=35 EF=39 LS=35 LF=39 ES=39 EF=41 LS=39 LF=41 ES=32 EF=35 LS=32 LF=35 ES=30 EF=32 LS=30 LF=32 ES=7 EF=12 LS=25 LF=30 ES=4 EF=7 LS=22 LF=25 Critical Path E Buffer

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Some Network Definitions All activities on the critical path have zero slack Slack defines how long non-critical activities can be delayed without delaying the project Slack = the activitys late finish minus its early finish (or its late start minus its early start) Earliest Start (ES) = the earliest finish of the immediately preceding activity Earliest Finish (EF) = is the ES plus the activity time Latest Start (LS) and Latest Finish (LF) depend on whether or not the activity is on the critical path

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B(6) D(6) A(4) C(3) F(5) E(14) G(2) I(3) H(2) J(4) K(2) ES=0 EF=4 LS=0 LF=4 ES=4+6=10 EF=10 LS=4 LF=10 ES=10 EF=16 ES=16 EF=30 ES=32 EF=34 ES=35 EF=39 ES=39 EF=41 ES=32 EF=35 LS=32 LF=35 ES=30 EF=32 LS=30 LF=32 ES=7 EF=12 LS=25 LF=30 ES=4 EF=7 LS=22 LF=25 Calculate Early Starts & Finishes Latest EF = Next ES

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B(6) D(6) A(4) C(3) F(5) E(14) G(2) I(3) H(2) J(4) K(2) ES=0 EF=4 ES=4 EF=10 LS=4 LF=10 ES=10 EF=16 LS=10 LF=16 ES=16 EF=30 LS=16 LF=30 ES=32 EF=34 LS=33 LF=35 ES=35 EF=39 LS=35 LF=39 ES=39 EF=41 LS=39 LF=41 ES=32 EF=35 LS=32 LF=35 ES=30 EF=32 LS=30 LF=32 ES=7 EF=12 LS=25 LF=30 ES=4 EF=7 LS=22 LF=25 Calculate Late Starts & Finishes Earliest LS = Next LF 39-4=35

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Activity Slack Time T ES = earliest start time for activity T LS = latest start time for activity T EF = earliest finish time for activity T LF = latest finish time for activity Activity Slack = T LS - T ES = T LF - T EF

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Calculate Activity Slack

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Waiting Line Models

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Arrival & Service Patterns Arrival rate: –The average number of customers arriving per time period Service rate: –The average number of customers that can be serviced during the same period of time

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Infinite Population, Single-Server, Single Line, Single Phase Formulae

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Example 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. Based on this description, we know: –Mu = 20 (exponential distribution) –Lambda = 15 (Poisson distribution)

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Average Utilization

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Average Number of Students in the System

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Average Number of Students Waiting in Line

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Average Time a Student Spends in the System.2 hours or 12 minutes

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Average Time a Student Spends Waiting (Before Service) Too long? After 5 minutes people get anxious

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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? A.2.25 B.3 C.4 D.3.25 E /80-60 = 3

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