ISQS 3344 Quantitative Review #3

1. ISQS 3344Quantitative Review #3

2. 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

3. X-Bar Chart Computations 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.

4. Assume the standard deviation of the process is given as 1.13 ouncesManagement 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.

5. 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 D3 and D4 are special constants whose values depend on the sample size. These constants will be given to you in a chart.

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

7. First Example Revisited

8. 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.

9. 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 = average proportion of nonconforming items

10. P-Chart Computations 1). Find p-bar: 2). Compute 3). Compute UCL and LCL using the formulas: 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 can’t be negative—why?)

11. 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.

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

13. C-Chart Computations Number-of-Defectives or C Chart 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 can’t be negative, why?

14. 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.

15. Process Capability • “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 • Cpassumes 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:

16. Cpk min = “minimum of the two” = mean of the process A value of Cpk < 1 indicates that the process is not centered. • Cp=Cpkwhen process is centered

17. 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

18. Chapter 3Project Mgt. and Waiting Line Theory

19. 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.

20. Example

21. Step 1: Draw the Diagram

22. Step 2: Add Activity Durations

23. Step 3: Identify All Unique Paths And Path Durations Path Duration = Sum of all task times along the path Critical path

24. 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

25. 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=39 EF=41 LS=39 LF=41 ES=35 EF=39 LS=35 LF=39 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 E=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

26. Some Network Definitions • All activities on the critical path have zero slack • Slack defines how long non-criticalactivities can be delayedwithout 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

27. 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 Strategy: Find all the ES’s and EF’s first by moving left to right (start to finish). Then find LF and LS by working backward (finish to start)

28. ES=16 EF=30 LS=16 LF=30 ES=32 EF=34 LS=33 LF=35 ES=10 EF=16 LS=10 LF=16 39-4=35 ES=4 EF=10 LS=4 LF=10 ES=39 EF=41 LS=39 LF=41 ES=35 EF=39 LS=35 LF=39 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

29. 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!!!!

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

31. Waiting Line Models

32. 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!!

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

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

35. 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.

36. Average Utilization Average Number of Studentsin the System Average Number of StudentsWaiting in Line

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

38. Average Time a StudentSpends Waiting (Before Service) Too long? After 5 minutes people get anxious

39. 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!