# Quantitative Review III

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Quantitative Review III

Chapter 6 and 6S Statistical Process Control

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)

x-Bar Computations x-bar = sample average

“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. deviations desired to use to develop the control limits

Assume 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 1 Sample 2 Sample 3 Observation 1 15.8 16.1 16.0 Observation 2 15.9 Observation 3 Observation 4 Sample means 15.875 15.975 18.56, 16.32 16.22, 18.56 17.62, 14.23 19.01, 12.56 18.33, 14.28

Inspectors 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 Means What are the upper and lower control limits for this process? Z=3 = 4.27

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.

Control Chart Factors D3 D4 2 0.00 3.27 3 2.57 4 2.28 5 2.11 6 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 Factors for R-Chart Sample Size (n)

Example Sample 1 Sample 2 Sample 3 Observation 1 15.8 16.1 16.0
15.9 Observation 3 Observation 4 Sample means 15.875 15.975 Sample ranges =0.2 =0.3

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

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. .996, -.320 1.233, 0 1.055, 0 .788, 1.201 1.4, 0

(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

(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

Note: 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. Sample Number of Defective Tires Number of Tires in each Sample Proportion Defective 1 3 20 .15 2 .10 .05 4 5 Total 9 100 .09 Solution: Note: Lower control limit can’t be negative.

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

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

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. Week Number of Complaints 1 3 2 4 5 6 7 8 9 10 Total 22 Solution:

Process Capability Index
Can a process or system meet it’s 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

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 Cp assumes that the process is centered in the specification range Cpk helps to address a possible lack of centering of the process

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

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

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

Computations Cp: Cpk:

Chapter 3 Project Mgt. and Waiting Line Theory

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.

Example

CPM Diagram

Identify Unique Paths

Calculate 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

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

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

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 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 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

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

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

Activity Slack = TLS - TES = TLF - TEF
Activity Slack Time TES = earliest start time for activity TLS = latest start time for activity TEF = earliest finish time for activity TLF = latest finish time for activity Activity Slack = TLS - TES = TLF - TEF

Calculate Activity Slack

Waiting Line Models

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

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

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

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)

Average Utilization

Average Number of Students in the System

Average Number of Students Waiting in Line

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

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

Consider 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? 2.25 3 4 3.25 4.25 60/80-60 = 3