AGENDA Duncan Multiple Ranges Test Non-terminating analysis Report and presentation expectations DUNCAN MULTIPLE RANGE TEST Run after ANOVA Ho has been rejected Know one or more means are statistically different Duncan test indicates which means are statistically different from the others Uses a least significant range value to determine differences for a set of means If the range of means is larger than the least significant range value there is a difference DUNCAN MULTIPLE RANGE TEST PROCEDURE Sort the replication means for each alternative in ascending order left to right. Calculate the least significant range value for all of the possible sets of adjacent means. Compare each set of possible adjacent means with the corresponding least significant range value in descending order with respect to the set size. Mark the non-significant ranges. LEAST SIGNIFICANT RANGE CALCULATIONS STANDARD DEVIATION OF X BAR CALCULATIONS COMPARISON OF ADJACENT MEANS Begin with the largest number of adjacent means If the range is less than the least significant range value there is no difference The range is underlined to represent all the data being the same If the range is greater than the least significant range value there is a difference The range must be split into smaller ranges and examined individually EXAMPLE Number of oil changes in an hour 4 alternatives 12 replications of each alternative ANOVA null hypothesis rejected MSE = 3.857 NUMBERS RANGES FOR 2,3, AND 4 ADJACENT MEANS 2 adjacent means A to B, B to C, C to D 3 adjacent means A to C, B to D 4 adjacent means A to D LEAST SIGNIFICANT RANGE ALPHA = 0.05, DF = 44 2 means = 0.567*2.86=1.622 3 means = 0.567*3.01=1.707 4 means = 0.567*3.10=1.757 RANGES A to D = 2.05 A to C = 1.17, B to D = 1.88 A to B = 0.17, B to C = 1.00, C to D = 0.88 DUNCAN ANALYSIS Actual ranges A to D = 2.05 A to C = 1.17, B to D = 1.88 A to B = 0.17, B to C = 1.00, C to D = 0.88 Least significant ranges 2 adjacent 1.622 3 adjacent 1.707 4 adjacent 1.757 INTERPRETATION No difference between A, B, and C No difference between C and D Difference between D and A or B What do you recommend? ECONOMIC ANALYSIS Statistical analysis means more when operating costs are considered Changes In Operating Policy If no extra costs, recommend best performing Changes in Resources Least expensive similarly performing More expensive must justify increased costs TYPES OF NON-TERMINATING SYSTEMS Most manufacturing systems Service systems that do not close Service systems where the customer leaves something to be picked up at another time Examples Automobile plant Repair facilities Hospitals NON-TERMINATING SYSTEM TIME TACTICAL CONSIDERATIONS Starting conditions… Determining steady state… Autocorrelation… Length of replication… Batch method… STARTING CONDITIONS Begin with the system empty Preferred Begin with the system loaded How many to load with? DETERMIING STEADY STATE Must eliminate initial transient Graphical approach… Linear regression approach… GRAPHICAL APPROACH Visually determine when the slope of the initial transient approaches 0 Highly subjective and influenced by individual interpretation Not recommended LINEAR REGRESSION Uses the least squares method to determine where the initial transient ends If the observations’ slope is not zero, advance the range to a later set of observations Eventually the range of data will have an insignificant slope coefficient Steady state behavior has been reached AUTOCORRELATION Correlation between performance measure observations in the system Possible issue with non-terminating systems Problem if not addressed Practitioner may underestimate variance Results in the possibility of concluding that there is a difference when this actual is not Methods to address Can be accounted for by complex calculations Can be avoided by special techniques BATCH METHOD Identify the non-significant correlation lag size Make a batch 10 times the size of the lag Make the steady state replication run length 10 batches long EXAMPLE Run single replication for 10,000 minutes Observe 4000 entities Initial transient requires 2000 minutes Steady state for 8000 minutes EXAMPLE DETERMINE RUN LENGTH Non-significant lag occurs at 200 observations Batch size 200 x 10 = 2000 observations to remove autocorrelation 2000 x 10 batches = 20000 total observations Time for each observation 10000 / 4000 = 2.5 minutes per observation Total simulation length must be 2000 + 20000 x 2.5 = 52000 minutes FINAL ANALYSIS Remove initial transients Split remainder into 10 batches / replications Process just like terminating simulations REPORT EXPECTATIONS Follow 10 step process Computer generated, no handwriting Bound or in pressboard prong binder (no three ring binders) Title page One page executive summary Computer generated table of contents WBS, LRC, Gantt Chart High level flow chart Screen captures of program and statistics Equations Summary statistics in the text, raw data in the appendix CD PRESENTATION EXPECTATIONS Total of 15 minutes Rehearsed Dress appropriately Follow 10 step process Only summary statistics Preload both presentation and model Test your files Demonstration of base alternative Do not keep flipping back and forth between programs