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ISQS 3344 Quantitative Review #3

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Control Charts for Measurements of Quality Example Usage: number of ounces per bottle; diameters of ball bearings; lengths of screws 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 (largest - smallest) over time

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X-Bar Chart Computations 1.First, find xbar-bar, the average of the averages 2. Now find, where σ is the standard deviation and n is the size of each sample 3). Find the upper control limit (UCL) and lower control limit (LCL) using the following formulas: Number of sample averages Z is the number of sigma limits specified in the problem. For 3 sigma limits use z = 3, for example.

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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. Sample 1Sample 2Sample 3 Observation 115.816.116.0 Observation 216.0 15.9 Observation 315.8 15.9 Observation 415.9 15.8 Sample means15.87515.97515.9

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Range or R Chart k = # of sample ranges A range chart measures the variability of the process using the average of the sample ranges (range = largest – smallest) The values of D 3 and D 4 are special constants whose values depend on the sample size. These constants will be given to you in a chart.

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Range Chart Factors

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First Example Revisited Sample 1Sample 2Sample 3 Observation 115.816.116.0 Observation 216.0 15.9 Observation 315.8 15.9 Observation 415.9 15.8 Sample means15.87515.97515.9 Sample Ranges0.20.30.2

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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.

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

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1). Find p-bar: 2). Compute 3). Compute UCL and LCL using the formulas: P-Chart Computations Number of observations per sample As with the X-Bar chart, z is the number of sigma limits specified in the problem If LCL turns out to be negative, set it to 0 (lower limit cant be negativewhy?)

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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. Sampl e Number of Defectiv e Tires Number of Tires in each Sample Proportio n Defective 1320.15 2220.10 3120.05 4220.10 5120.05 Total9100.09

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Count charts Used when looking at # of defects Control limits are based on average number of defects, C-Charts

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Number-of-Defectives or C Chart C-Chart Computations 1). Compute c-bar: 2). Compute 3). Compute LCL and UCL using the formulas: As with the X-Bar chart, z is the number of sigma limits specified in the problem As with the P-Bar chart, if the LCL turns out to be negative, set LCL to 0 (LCL cant be negative, why?

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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 13 22 33 41 53 63 72 81 93 101 Total22

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Capability : 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 Cp assumes that the process is centered on the specification range, which may not be the case! To see if a process is centered, we use Cpk: Process Capability

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min = minimum of the two = mean of the process A value of Cpk < 1 indicates that the process is not centered. Cpk Cp=Cpk when process is centered

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Example Design specifications call for a target value of 16.0 +/-0.2 ounces. Observed process output has a mean of 15.9 and a standard deviation of 0.1 ounces. Is the process capable? LSL = 16-0.2 = 15.8 USL =16 + 0.2 = 16.2

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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: Longest path through the process 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. Critical Path Method (CPM)

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Example

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Step 1: Draw the Diagram

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Step 2: Add Activity Durations

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Step 3: Identify All Unique Paths And Path Durations Path Duration = Sum of all task times along the path PathDuration ABDEGHJK40 ABDEGIJK41 ACFGHJK22 ACFGIJK23 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 E=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 Strategy: Find all the ESs and EFs first by moving left to right (start to finish). Then find LF and LS by working backward (finish to start)

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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 If an item is on the critical path, there is no slack!!!!

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Calculate Activity Slack The critical path was ABDEGIJK Notice that the slack for these task times is 0.

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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 Arrival rate and service rate must be in the same units!!

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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: –µ = 20 –λ= 15 Note that both arrival rate and service rate are in hours, so we dont need to do any conversion.

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Average Utilization Average Number of Students in the System 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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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? Convert to hours first!

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