Report

ISQS 3344 Quantitative Review #3 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 X-Bar Chart Computations 1. First, find “xbar-bar”, the average of the averages x x 1 x 2 ... x n Number of sample averages k 2. Now find x , where σ is the standard deviation and n is the n size of each sample 3). Find the upper control limit (UCL) and lower control limit (LCL) using the following formulas: UCL x z Z is the number of sigma limits specified in the problem. For “3 sigma limits” use z = 3, for example. n LCL x z n 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 16.0 16.0 15.9 Observation 3 15.8 15.8 15.9 Observation 4 15.9 15.9 15.8 Sample means 15.875 15.975 15.9 x (15 . 875 15 . 975 15 . 9 ) / 3 15 . 917 1 . 13 (from the problem); UCL 15 . 917 3 1 . 13 n 4; z 3 17 . 612 4 LCL 15 . 917 3 1 . 13 4 14 . 222 Range or R Chart UCL R D4R, LCL k = # of sample ranges R D3R R R k A range chart measures the variability of the process using the average of the sample ranges (range = largest – smallest) The values of D3 and D4 are special constants whose values depend on the sample size. These constants will be given to you in a chart. Range Chart Factors Sample Size (n) 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Factors for R-Chart D4 D3 0.00 3.27 0.00 2.57 0.00 2.28 0.00 2.11 0.00 2.00 0.08 1.92 0.14 1.86 0.18 1.82 0.22 1.78 0.26 1.74 0.28 1.72 0.31 1.69 0.33 1.67 0.35 1.65 First Example Revisited Sample 1 Sample 2 Sample 3 Observation 1 15.8 16.1 16.0 Observation 2 16.0 16.0 15.9 Observation 3 15.8 15.8 15.9 Observation 4 15.9 15.9 15.8 Sample means 15.875 15.975 15.9 Sample Ranges 0.2 0.3 0.2 R ( 0 . 2 0 . 3 0 . 2 ) / 3 0 . 2333 n 4 , so D 3 0 ; D 4 2 . 28 UCL 2 . 28 ( 0 . 2333 ) 0 . 5319 LCL 0 ( 0 . 2333 ) 0 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. R 0 . 5 (from the problem) n 5 , so D 3 0 ; D 4 2 . 11 UCL 2 . 11 ( 0 . 5 ) 1 . 055 LCL 0 ( 0 . 5 ) 0 P-Charts P Fraction Defective Chart • “Proportion charts” • Used for yes-or-no type judgments (acceptable/not acceptable, works/doesn’t work, on time/late, etc.) • p = proportion of nonconforming items • Control limits are based on p = average proportion of nonconforming items P-Chart Computations 1). Find p-bar: total number p total number 2). Compute p of defects of units sampled(" k" times" n" ) p (1 p ) n Number of observations per sample 3). Compute UCL and LCL using the formulas: UCL p z p (1 p ) As with the X-Bar chart, z is the number of sigma limits specified in the problem n LCL p z p (1 p ) n If LCL turns out to be negative, set it to 0 (lower limit can’t be negative—why?) 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 1 3 20 .15 2 2 20 .10 p # Defectives Total Inspected p (1 p ) 9 .09 100 (.09)(.91) 3 1 20 .05 σp 4 2 20 .10 5 1 20 .05 UCL p p z σ .09 3(.064) .282 Total 9 100 .09 n 0.064 20 LCL p p z σ .09 3(.064) .102 Since LCL is negative, LCL 0 UCL 0.282 set LCL 0. The limits are C-Charts • “Count charts” • Used when looking at # of defects • Control limits are based on average number of defects, c Number-of-Defectives or C C-Chart Computations Chart 1). Compute c-bar: c number number 2). Compute c of defects observed of samples taken c 3). Compute LCL and UCL using the formulas: UCL c z c LCL c z c 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 can’t be negative, why? 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 2 _ 3 3 c 4 1 5 3 6 3 7 2 LCL c c z c 2.2 3 2.2 2.25. 8 1 Since LCL is negative, 9 3 10 1 Total 22 # complaints # of samples 22 2.2 10 UCL c c z c 2.2 3 2.2 6.65 The limits are then LCL 0 UCL 6.65 set LCL to 0. Process Capability • “Capability” : Can a process or system meet its requirements? Cp product' s design specificat 6 standard deviations ion range of the production USL " Upper Specificat ion Limit" LSL " Lower Specificat ion Limit" USL - LSL system 6 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: USL μ μ LSL Cpk min , 3σ 3σ Cpk USL μ μ LSL Cpk min , 3σ 3σ min = “minimum of the two” = mean of the process A value of Cpk < 1 indicates that the process is not centered. Cp=Cpk when process is centered 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 Cp 16 . 2 15 . 8 0 . 667 6 (. 1) 16 . 2 15 . 9 15 . 9 15 . 8 Cpk min , 3 (. 1) 3 (. 1) min( 1, 0 . 333 ) 0 . 3333 Both Cp and Cpk 1, so process is not capable. Chapter 3 Project Mgt. and Waiting Line Theory Critical Path Method (CPM) 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. Example Ac tiv ity A B C D E F G H I J K D e s c rip tio n D e ve lo p p ro d u c t s p e c ific a tio n s D e s ig n m a n u fa c tu rin g p ro c e s s S o u rc e & p u rc h a s e m a te ria ls S o u rc e & p u rc h a s e to o lin g & e q u ip m e n t R e c e ive & in s ta ll to o lin g & e q u ip m e n t R e c e ive m a te ria ls P ilo t p ro d u c tio n ru n E va lu a te p ro d u c t d e s ig n E va lu a te p ro c e s s p e rfo rm a n c e W rite d o c u m e n ta tio n re p o rt T ra n s itio n to m a n u fa c tu rin g Im m e d ia te D u ra tio n P re d e c e s s o r N one A A B D C E & F G G H & I J (w e e k s ) 4 6 3 6 14 5 2 2 3 4 2 Step 1: Draw the Diagram Step 2: Add Activity Durations Step 3: Identify All Unique Paths And Path Durations Path Duration = Sum of all task times along the path Path Duration ABDEGHJK 40 ABDEGIJK 41 ACFGHJK 22 ACFGIJK 23 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 noncritical paths ES=10 ES=16 EF=16 EF=30 LS=16 LS=10 E Buffer LF=16 LF=30 ES=4 EF=10 LS=4 LF=10 B(6) D(6) E(14) A(4) ES=0 EF=4 LS=0 LF=4 ES=32 EF=34 LS=33 LF=35 H(2) G(2) C(3) ES=4 EF=7 LS=22 LF=25 F(5) ES=7 EF=12 LS=25 LF=30 ES=30 E=32 I(3) ES=32 EF=35 LS=32 LF=35 ES=35 EF=39 LS=35 LF=39 ES=39 EF=41 LS=39 LF=41 J(4) K(2) Critical Path 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=10 EF=16 B(6) D(6) ES=16 EF=30 E(14) Latest EF H(2) = Next ES ES=35 EF=39 ES=39 EF=41 G(2) J(4) K(2) A(4) ES=0 EF=4 LS=0 LF=4 C(3) ES=32 EF=34 F(5) ES=30 EF=32 LS=30 LF=32 ES=4 ES=7 EF=7 EF=12 LS=22 LS=25 Strategy: Find all the ES’s and EF’s first LF=25 LF=30 by moving left to right (start to finish). Then find LF and LS by working backward (finish to start) I(3) ES=32 EF=35 LS=32 LF=35 Calculate Early Starts & Finishes ES=16 EF=30 LS=16 LF=30 ES=10 EF=16 LS=10 LF=16 ES=4 EF=10 LS=4 LF=10 B(6) D(6) E(14) H(2) 39-4=35 ES=35 EF=39 LS=35 LF=39 Earliest LS G(2) = Next LF J(4) A(4) ES=0 EF=4 ES=32 EF=34 LS=33 LF=35 C(3) ES=4 EF=7 LS=22 LF=25 F(5) ES=7 EF=12 LS=25 LF=30 ES=30 EF=32 I(3) LS=30 LF=32 ES=32 EF=35 LS=32 LF=35 ES=39 EF=41 LS=39 LF=41 K(2) Calculate Late Starts & Finishes 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 If an item is on the critical path, there is no slack!!!! Calculate Activity Slack Ac tiv ity A B C D E F G H I J K L a te E a rly S la c k F in is h 4 10 25 16 30 30 32 35 35 39 41 F in is h 4 10 7 16 30 12 32 34 35 39 41 (w e e k s ) 0 0 18 0 0 18 0 1 0 0 0 The critical path was ABDEGIJK Notice that the slack for these task times is 0. 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 • Arrival rate and service rate must be in the same units!! Infinite Population, Single-Server, Single Line, Single Phase Formulae lambda mean arrival rate mu mean service rate L average system utilizatio n average number of customers in system Infinite Population, Single-Server, Single Line, Single Phase Formulae L Q L average number of customers W 1 average time in system in line including service W Q W average time spent waiting in line 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 don’t need to do any conversion. Average Utilization 15 0 . 75 or 75 % 20 Average Number of Students in the System L 15 20 15 3 Average Number of Students Waiting in Line L Q L 0 . 75 3 2 . 25 students Average Time a Student Spends in the System W 1 .2 hours or 12 minutes 1 20 15 Average Time a Student Spends Waiting (Before Service) W Q W 0 . 75 0 . 2 0 . 15 hours or 9 minutes Too long? After 5 minutes people get anxious 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! 1 customer 45 secs 3600 secs 1hr 1hr Now , L 60 80 60 80 customers 3