Lean Six Sigma Black Belt Mock Exam Name of the Black Belt: ……………………………
Please check you are taking the correct exam. This is the Black Belt Exam for Services/Industry. This is an open-book exam. exam. You are allowed to use books books and notes. You are no t allowed to use a calculator, telephone, telephone, tablet or computer. Please circle your answer. Calculation of points Score Score = 10(#correct 10(#correct – 10) / 30 Your Your scor score e will will be roun rounde ded d to halv halves es,, with with the the excep excepti tion on of 5.5. 5.5.
This This mate materi rial al is inte intellllec ectu tual al prop proper erty ty of the Instit Institute ute for Busine Business ss and Indust Industria riall Stat Statist istics ics of the University of Amsterdam Amsterdam (IBIS UvA) http://www.ibisuva.nl.
1 Process Capability of Gespreksduur (min.)
USL
Process Data LSL Target USL Sample Sample Mean Mean Sample N StDev(With StDev(Within) in) StDev(Over StDev(Overall) all)
Within Overall
* * 5.00000 3.0332 3.03327 7 250 0.92945 0.92945 0.91889 0.91889
Potential (Within) Capability Cp * CPL * C PU PU 0 .7 .7 1 C pk pk 0 .7 .7 1 CCpk CCpk 0. 0.71 71
Observed Performance % < LSL * % > USL 1.60 1.60 % Tota Totall 1. 1.60 60 Exp. Within Performance % < LSL * % > USL 1.72 1.72 % Tota Totall 1. 1.72 72
Overall Capability Pp PPL P PU PU Ppk Cpm
Exp. Overall Performance % < LSL * % > USL 1.62 1.62 % Tota Totall 1. 1.62 62
0.75
1.50
2.25
3.00
3.75
4.50
* * 0 .7 .7 1 0.71 *
5.25
This is a process capability analysis of the duration of telephone calls in a call center. What can be concluded from this analysis? a) 1.62% of the calls is longer than the norm of 5 minutes. b) The The aver averag age e dura durati tion on of a call call is too too long long.. The call call dura durati tion on does does not not foll follow ow a norm normal al dist distri ribu buti tion on.. c) The d) A call center employee can handle around 12 calls per hour.
2
Incoming requests
Step 1
Step 2
Step 3
Takt time: 3 min. 20 jobs / hour
Cycle time: 5 min.
Cycle time: 2 min.
Cycle time: 11 min.
Requests are processed in three steps. The cycle time of each process step is given (the cycle time is the processing time per employee per request). The takt rate can be assumed constant (so, no fluctuations in workload), and the variation in processing times is marginal. How many employees are required in total for these three steps? a) 3. b) 5. c) 7. d) 18.
3 USL
90% 90%
50% 50%
95% 95%
75% 75%
0
2
4
6
99% 99%
8
10
12
14
Throughput times: 50% shorter than 2.6 days 75% shorter than 5.1 days 90% shorter than 8.3 days 95% shorter than 10.6 days 99% shorter than 16.0 days
16
The graph above indicates that 50% of throughput times are shorter than 2.6 days, 75% is shorter than 5.1 days, et cetera. By means of which techniques has the BB determined these percentages? a) b) c) d)
Descriptive statistics (Basic Statistics > Display Descriptive statistcs). Process capability analysis (Quality tools > Capability analysis). Empirical CDF (Graph > Empirical CDF). Normal distribution (Calc > Probability distributions > Normal).
4
Why was the name ‘Six Sigma’ chosen for the programme?
a) Because all processes are allowed to fluctuate in a range of 6 around the mean. b) Because it was invented by statisticians. c) For normally distributed variables, a distance of six times the (long-term) standard deviation between the specification limits and the process’s average ensures a defect rate of 3.4 PPM. d) It is symbolic for the systematic pursuit of extremely low defect rates in all processes.
5
The length of a certain component has a normal distribution with mean 12.3 mm. and standard deviation 0.21. We produce 1000 components and measure their length. Which of the following statements is false?
a) Of the 1000 components, around 25 have a length smaller than 11.88 mm. b) Around 500 have a length smaller than 12.3 mm. c) Around 950 have a length smaller than 12.72 mm. d) The average of the 1000 lengths is about equal to their median.
6
A BB has performed a process capability analysis with the Stat > Quality tools > Capability Analysis (Normal) procedure in Minitab. Suppose that the data have a strongly non-normal distribution. Which ones of the results of the Minitab analysis are likely to be wrong ?
a) b) c) d)
Only the estimated StDev (both within and overall). The estimated StDev (within and overall) and the observed performance. Only the estimated “Within” (= short term) performance and the Cp and Cpk. The estimated “Within” performance and the Cp and Cpk, but also the estimated “Overall” (= long term) performance and the P p and Ppk.
7
A BB has done a regression analysis to study the effect of an X onto his CTQ. The result is displayed below.
Regression Analysis: CTQ versus X The regression equation is CTQ = 0.675651 + 0.695515 X S = 1.39516 R-Sq = 34.8 % R-Sq(adj) = 32.5 %
Analysis of Variance Source DF SS Regression 1 29.1469 Error 28 54.5010 Total 29 83.6478
MS F P 29.1469 14.9743 0.001 1.9465
Which is no t true? a) b) c) d)
8
X has a significant effect on the CTQ. The predicted value for X = 6 equals CTQ = 4.85. The model gives precise predictions about the CTQ. For fixed X, the CTQ has a standard deviation of about 1.40.
Typical questions for Black Belt for Industry Optimization of the production of decaf Questions 8 through 24 all relate to this case.
We consider a project at a manufacturer of coffee. The process under study is the extraction of caffeine from raw coffee beans in order to make decaffeinated coffee (“decaf”). According to food laws, the caffeine percentage of decaf should be below 0.1%. The objective of the project is to reduce the cycle time of decaffeinating a batch of coffee, thereby increasing the capacity of the production facility (that is, increasing the number of batches that the facility can process in a week). Processing one additional batch per week would represent extra revenues of k€15. Of course, the final product should meet the norms as stated in the food laws. What is a suitable operational definition of the CTQ(s) in this project? a) CTQ-1: cycle time (measurement procedure: elapsed time registered in process log on the computer / unit: batch / requirement: as low as possible). CTQ-2: caffeine% (measurement procedure: chemical analysis / unit: batch / requirement: USL=0.1). b) CTQ-1: cycle time (measurement procedure: lapsed time registered in process log on the computer / unit: batch / requirement: as low as possible). c) The project is successful if it succeeds in reducing the cycle time sufficiently to allow the processing of one additional batch per week, thus creating k€15 in extra revenue. d) The CTQ is cycle time. The relevance of the CTQ is that a reduction allows an increased capacity without additional investments (k€15 for each additional batch).
9 Optimization of the production of decaf The BB does a gauge R&R study. The set-up is as follows: 10 batches are randomly selected, and from each batch a sample is taken. The 10 samples are measured three times (on three different occasions), in a randomized order, by a single operator.
What is a consequence of the chosen set-up? a) This is the standard set-up of a gauge R&R study, and the results will allow a good assessment of the measurement system. b) With the chosen set-up, it is not possible to estimate the part-to-part spread. c) It is not wise to select 10 batches randomly. As a consequence, the estimated GR&R% is unreliable. d) With the chosen set-up, it is not possible to determine the reproducibility of the measurements.
10 Gage R&R
Sour ce Tot al Gage R&R Repeat abi l i t y Par t - To- Par t Tot al Var i at i on
VarComp 0. 0000294 0. 0000294 0. 0000605 0. 0000899
%Cont r i but i on ( of VarComp) 32. 74 32. 74 67. 26 100. 00
Upper process tol erance l i mi t = 0. 1 Study Var Source StdDev ( SD) ( 6 * SD) Tot al Gage R&R 0. 0054252 0. 0325515 Repeat abi l i t y 0. 0054252 0. 0325515 Par t - To- Par t 0. 0077763 0. 0466581 Tot al Var i at i on 0. 0094818 0. 0568909
%Study Var %Tol erance ( %SV) ( SV/ Tol er) 57. 22 45. 59 57. 22 45. 59 82. 01 65. 35 100. 00 79. 68
Number of Di st i nct Categori es = 2
How precise are the caffeine% measurements? a) The 99.7% confidence margins on the true caf% are: ± 0.0000294. b) The 99.7% confidence margins on the true caf% are: ± 0.0000882. c) If a single person measures a single batch multiple times, the standard deviation of the results is 0.0054252. d) If a single person measures a single batch multiple times, the standard deviation of the results is 0.0094818.
11 Optimization of the production of decaf The BB does a process capability study. She has the caffeine% determined for 50 batches. The resulting analysis is on the next slide.
What should the BB conclude from the analysis? a) The Cp cannot be calculated because the data do not have a normal distribution. b) In the long run, around 0.00% total of all batches will be nonconforming (meaning: caffeine% > 0.1%). c) Over a longer period, the standard deviation of the caffeine percentages is around 0.013342. d) This is a 0.040026 sigma process.
Process Capability of Caf% USL Process Data LSL * Target * USL 0.1 Sample Mean 0.0562153 Sample N 50 StDev(Within) 0.010803 StDev(Overall) 0.013342
Within Overall Potential (Within) Capability Cp * CPL * CPU 1.35 Cpk 1.35 Overall Capability Pp * PPL * PPU 1.09 Ppk 1.09 Cpm *
0.024 0.036 0.048 0.060 0.072 0.084 0.096 Observed Performance % < LSL * % > US L 0 .00 % Tot al 0.00
Exp. Within Performance % < LSL * % > U SL 0 .0 0 % T ot al 0.00
Exp. Overall Performance % < LSL * % > U SL 0 .0 5 % T ot al 0.05
12 Optimization of the production of decaf The BB wishes to establish whether the caffeine percentages have a normal, lognormal or Weibull distribution, or maybe even an altogether different distribution.
What is the best tool to study the distribution of the data? a) b) c) d)
The control chart. The histogram. The empirical CDF (cumulative distribution function). The probability plot.
13 Optimization of the production of decaf Suppose the BB is worried about possible contaminations in the dataset, and wants to screen the data for outliers.
What is the best tool to identify outliers in the dataset? a) b) c) d)
The histogram. The empirical CDF (cumulative distribution function). The probability plot. The 1-sample t-test.
14 Optimization of the production of decaf The largest component of the batch cycle time, is the extraction time ( ET). In normal production, ET is set by the operators at 400 minutes per batch. To investigate the effect of ET on the caffeine percentage, the BB does an experiment, in which ET is varied from 200 to 700 minutes. The results are analyzed using regression analysis (see next slide).
Is this analysis acceptable? a) b) c) d)
No: the BB should remove the outliers from the dataset. No: the BB should try to fit a transfer function with a quadratic term. No: 87.1% is not an acceptable value for the R-sq. No: the value S = 0.0161376 indicates that there is too much random scatter in the dataset.
Regression Analysis: Caf% versus ET
The r egr ess i on equat i on i s Caf% = 0. 1788 - 0. 000261 ET S = 0. 0161376
R- Sq = 87. 1%
R- Sq( adj ) = 86. 3%
Anal ysi s of Var i ance
Fitted Line Plot Plot Sour ceLine DF
SS
Regress 1 0.CT 0315351 0.1788i -on 0.000261 Caf% = = 0.000261 CT Er r or Tot al
0.16
18 19
0. 0046876 0. 0362227
0.14
Sequent i al Anal ysi s of
0.12
Sour ce Li near Quadr at i c
0.10
DF 1 1
S R-Sq R-Sq Vari ance R-Sq(adj)
SS 0. 0315351 0. 0031627
% % f 0.08 f a a C C 0.06
0.04 0.02 0.00 200
300
400
500 CT ET
600
MS 0. 0315351 0. 0002604
700
F 121. 09 35. 26
F 121. 09 0.0161376 0.0161376 87.1% 86.3%
P 0. 000 0. 000
P 0. 000
15 Optimization of the production of decaf Assume for the sake of this question that the analysis on the previous slide is correct and acceptable.
Based on the fitted line plot on the previous slide, what would be a suitable extraction time ET? a) b) c) d)
Caf% = 0.1%. ET = 230 min. ET = 300 min. ET = 430 min.
16 Optimization of the production of decaf In the regression analysis of Caf% versus ET (two slides back) it said: Anal ysi s of Sour ce Regr essi on Err or Tot al
Var i DF 1 18 19
ance SS 0. 0315351 0. 0046876 0. 0362227
MS 0. 0315351 0. 0002604
F 121. 09
P 0. 000
The value for P is 0.000
What does this mean? a) It means that the slope of the fitted line differs significantly from zero (that is, the fitted line is not horizontal). b) It means that the effect of Caf % is significant. c) It means that the extraction time determines almost 100% of the caffeine percentage. d) It means that the dataset is not large enough to prove that there is a relationship between Caf% and ET.
17 Optimization of the production of decaf The BB runs an experiment in which she investigates the following factors: - Temperature - Soaktime - Heattime
Besides the Extraction Time ET, the Soak-time and heat-time are two other components of the total batch cycle time. Temperature is a setting on the machine. The CTQ in the experiment was the Caffeine%. (see next slide). What experimental design did the BB use? a) b) c) d)
Box-Behnken design with four center points. 25 factorial design with center points 25–1 fractional factorial design with center points 24 factorial design
StdOrder RunOrder PtType
Blocks
Temperat Soaktime Heattime Caffeine%
9
1
2
1
77.5
30
60
0.11
6
2
2
1
85
40
60
0.18
14
3
0
1
77.5
40
75
0.06
5
4
2
1
70
40
60
0.17
8
5
2
1
85
40
90
0.08
12
6
2
1
77.5
50
90
0.03
7
7
2
1
70
40
90
0.08
2
8
2
1
85
30
75
0.06
11
9
2
1
77.5
30
90
0.03
3
10
2
1
70
50
75
0.12
15
11
0
1
77.5
40
75
0.07
10
12
2
1
77.5
50
60
0.15
16
13
0
1
77.5
40
75
0.06
13
14
0
1
77.5
40
75
0.08
4
15
2
1
85
50
75
0.09
1
16
2
1
70
30
75
0.12
18 Optimization of the production of decaf The results of the first analysis are given on the next slide.
1. There are insignificant effects that should be dropped. 2. The R2 (R-sq) indicates a poor fit. 3. The analysis should have been done in uncoded units.
Which of the claims 1., 2. or 3. are valid? a) Only 1. b) Only 3. c) Both 1. and 3. d) All three 1., 2. and 3.
Response Surface Regression: Caffeine% versus Temperature; Soakti me; Heattime The anal ysi s was done usi ng coded uni t s. Est i mated Regressi on Coef f i ci ent s f or Caff ei ne% Ter m Const ant Temper at ur e Soakt i me Heatt i me Temper at ur e*Temper at ur e Soakt i me*Soakt i me Heatt i me*Heatt i me Temper at ur e*Soakt i me Temper at ur e*Heat t i me Soakt i me*Heatt i me
Coef 0. 067500 - 0. 010000 0. 008750 - 0. 048750 0. 038750 - 0. 008750 0. 021250 0. 007500 - 0. 002500 - 0. 010000
SE Coef 0. 008036 0. 005683 0. 005683 0. 005683 0. 008036 0. 008036 0. 008036 0. 008036 0. 008036 0. 008036
T 8. 399 - 1. 760 1. 540 - 8. 579 4. 822 - 1. 089 2. 644 0. 933 - 0. 311 - 1. 244
P 0. 000 0. 129 0. 175 0. 000 0. 003 0. 318 0. 038 0. 387 0. 766 0. 260
S = 0. 0180728 PRESS = 0. 0208889 R- Sq = 94. 96% R- Sq( pr ed) = 32. 05% R- Sq( adj ) = 87. 40%
19 Optimization of the production of decaf Consider this analysis: Ter m Const ant Temper at ur e Heatt i me Temper at ur e*Temper at ur e Heatt i me*Heatt i me Temper at ur e*Heat t i me S = 0. 0195891
Coef 0. 063125 - 0. 010000 - 0. 048750 0. 038750 0. 021250 - 0. 002500
SE Coef 0. 007616 0. 006219 0. 006219 0. 008795 0. 008795 0. 008795
T 8. 288 - 1. 608 - 7. 839 4. 406 2. 416 - 0. 284
P 0. 000 0. 139 0. 000 0. 001 0. 036 0. 782
R- Sq = 89. 94% R-Sq( adj ) = 84. 91%
Besides the ‘constant’, which terms should be included in the transfer function? a) Only Heat t i me and Heatt i me*Heatt i me b) Only Heat t i me, Heatt i me*Heatt i me, and Temper at ur e* Temper at ur e c) Heat t i me, Heat t i me*Heat t i me, Temper at ur e, and Temper at ur e* Temper at ur e d) All the terms given in the table.
20 Optimization of the production of decaf Remember the objective of the project: batch cycle time should be reduced, while keeping the caffeine percentage safely below 0.1%. Heat-time is a component of the total batch cycle time.
The results of the experiment are summarized in the contour plot (next slide).
Is (heat-time = 66; temperature = 78.5) a recommendable process setting? a) No: (heat-time = 90; temperature = 78.5) is the optimal setting. b) No: with that setting, around 50% of the batches will have an unacceptable caffeine%. c) No: (heat-time = 60; temperature = 70) gives maximal caffeine percentages of more than 0.16%. d) Yes: the given setting meets the project’s objectives.
Contour Plot of Caffeine% vs Heattime; Temperature 90
0.04 85
0.08
80 e m i t 75 t a e H
70
0.06
0.12 0.10
65
0.14 60 70
72
74
0.14 76 78 80 Temperature
82
84
21 Optimization of the production of decaf The current process settings are:
Temperature = 73 Heat-time = 85 Suppose it is acceptable if at most 2.5% of the batches is above 0.1% (in caffeine%).
Based on the results of the experiment, how much can the batch cycle time be reduced? a) b) c) d)
No reduction in batch cycle time possible based on the results of the experiment. By about 9 minutes. By about 15 minutes. By about 19 minutes.
22 Optimization of the production of decaf
The following principle is called Robust Design: If a nuisance variable has an interaction effect with a control variable, choosing a suitable setting for the control variable may reduce the effect of the nuisance variable onto the CTQ, and thus, the CTQ is less sensitive to variability in the nuisance variable. The following principle is called Tolerance Design: If much of the variation of a CTQ is caused by a nuisance variable, then variability of the CTQ may be reduced by setting tolerance limits on the nuisance variable, which limit the variability that is allowed in the nuisance variable. Based on the results of this experiment, can the BB develop further improvement actions? a) Robust design: by selecting a suitable value for heat-time, the caffeine percentages can be made less sensitive to variation in temperature. b) Tolerance design: by setting suitable tolerance limits to the caffeine percentages, variability in temperature may be reduced. c) Robust design: by selecting a suitable value for caffeine%, the temperature can be made less sensitive to variability in heat-time. d) None of these three ideas is feasible.
23 The BB has executed an experiment with 2 factors at 2 levels with cp’s. Below is the analysis of the data. So far, the BB has only removed the interaction term. Ter m Const ant A B
Ef f ect - 0. 2723 0. 3533
Coef 42. 1251 - 0. 1362 0. 1767
SE Coef 0. 2173 0. 2173 0. 2173
T 193. 86 - 0. 63 0. 81
Anal ysi s of Vari ance f or Response (coded uni t s) Sourc e DF Seq SS Adj SS Mai n Ef f ect s 2 0. 39799 0. 39799 Curvat ure 1 0. 33031 0. 33031 Resi dual Err or 8 3. 02182 3. 02182 Lack of Fi t 1 0. 02496 0. 02496 Pure Err or 7 2. 99687 2. 99687 Tot al 11 3. 75011
P 0. 000 0. 548 0. 440
Adj MS 0. 19899 0. 33031 0. 37773 0. 02496 0. 42812
F 0. 53 0. 87
P 0. 610 0. 377
0. 06
0. 816
Here are 4 explanations for the analysis above. 1) Curvature has ruined the analysis. 2) The pure error is overestimated, due to irregulaties in the experiment. 3) A and B were studied in too small a range. 4) A and B have no or minor effects. Which possible explanations make sense? a) 4 b) 2 and 4 c) 2, 3 and 4 d) 1, 2, 3 and 4
24 Ter m Const ant A B C A*B A*C B*C
Ef f ect 3. 7509 - 0. 0413 0. 8467 - 0. 3650 1. 1588 - 0. 4096
Coef 49. 7576 1. 8754 - 0. 0207 0. 2734 - 0. 1825 0. 5794 - 0. 2048
SE Coef 0. 2656 0. 2656 0. 2656 0. 2656 0. 2656 0. 2656 0. 2656
T 187. 32 7. 06 - 0. 08 1. 93 - 0. 69 2. 18 - 0. 77
P 0. 000 0. 000 0. 940 0. 130 0. 509 0. 057 0. 460
Anal ysi s of Vari ance f or Response (coded uni t s) Sourc e Mai n Ef f ects 2- Way I nter acti ons Resi dual Err or Lack of Fi t Pure Err or Tot al
DF 3 3 9 1 8 15
Seq SS 57. 478 6. 576 10. 161 1. 173 8. 988 74. 214
Adj SS 57. 478 6. 576 10. 161 1. 173 8. 988
Adj MS 19. 159 2. 192 1. 129 1. 173 1. 123
F 16. 97 1. 94
P 0. 000 0. 193
1. 04
0. 337
Having collected the results of an experiment, the BB runs the ‘Analyze factorial design’ procedure for the first time. The Minitab output is shown above. What is a sensible next step? a) Conduct steepest ascent. b) Do additional runs to fit a quadratic model. c) Eliminate the highly insignificant terms (p > 0.2). d) Draw contour plots and find optimal settings.
Typical questions for Black Belt for Services Reducing turnaround time for mortgage offers Questions 25 through 40 all relate to this case.
We consider an administrative process in a bank. The process handles applications for mortgages, and in case the application is accepted, produces an offer. On average 40 applications enter the process each day. In the first stage, they are handled by 4 coordinators, who check applications for completeness, and collect additional information from the Real Property Registration (Kadaster). Next, the coordinators send the application to one of 3 rate officers. They judge wether the applicant qualifies for a loan, and they determine the interest rate that is offered. Applications involving a higher loan sum (25% of the applicactions) are then sent to an underwriter, who assesses the risk and decides on a rejection or acceptance. In the end, 10% of the applications are rejected, in which case a secretary writes a rejection letter. For the accepted applications (90%), the secretary prepares the offer and sends it to the applicant. The company is losing business due to the rather long turnaround times (total throughput time) of the process. If applicants do not receive an offer within a few working days, they simply give up and apply at another bank. This results in a poor conversion rate (= the percentage of offers accepted by applicants).
25
Revenue
Operational cost
Conversion rate (= % offers accepted by clients)
Turnaround Time (TT)
Man-hours per application
Waiting times (WTs) in between process steps
Processing times ( PTs) for the process steps
Is the CTQ flowdown above acceptable? a) Yes, the flowdown is good as it is. b) No, Waiting Times (WTs) should not be connected to Man-hours per application. c) No, Processing Times (PTs) should not be connected to Man-hours per application . d) No, Conversion Rate should be replaced with Quality and Efficiency.
26
What is a practical and effective way to measure the PTs and WTs? a) “Travel-sheet”: follow a sample of applications through the process, while employees place time-stamps when they start or complete a task. b) “Day-in-the-life-of” (“multi-moment-opname”): follow a number of employees over a day. Each 15 minutes, they indicate on a check sheet what sort of task they are performing. c) “Time-study”: Have an employee perform the first task in the process for 10 applications and measure the processing times with a stopwatch. The same for the second step in the process, and so on. d) “Process-study”: Measure on what moments an application arrives in the first process step, then in the second step, and so on.
27
DMAIC 2: Validate the measurement procedures The BB wants to validate the measurement methods and measurement plan that she has devised for determining the WTs and PTs.
Which of these techniques is not relevant for validating the measurement plan? a) Gauge R&R study. b) Do a test measurement. c) After the data have been collected: check face validity of the results. d) Control chart.
28 Gage R&R
%Cont r i buti on ( of Var Comp) 6. 20 4. 79 1. 41 1. 41 93. 80 100. 00
Sour ce Tot al Gage R&R Repeat abi l i t y Repr oduci bi l i t y Measur er Par t - To- Par t Tot al Var i at i on
Var Comp 0. 32292 0. 24965 0. 07327 0. 07327 4. 88506 5. 20798
Sour ce Tot al Gage R&R Repeat abi l i t y Repr oduci bi l i t y Measur er Par t - To- Par t Tot al Var i at i on
St dDev ( SD) 0. 56826 0. 49965 0. 27068 0. 27068 2. 21022 2. 28210
St udy Var ( 5. 15 * SD) 2. 9265 2. 5732 1. 3940 1. 3940 11. 3826 11. 7528
%St udy Var ( %SV) 24. 90 21. 89 11. 86 11. 86 96. 85 100. 00
Given the results above, what are the 99% uncertainty margins (precision) for the time measurements? a) ± 0 .323 b) ± 0 .568 c) ± 1.46 d) ± 2.28
29 I Chart of PT (coordinator) 35 UCL=32.71 30
e 25 u l a V l a 20 u d i v i d n 15 I
_ X=19.26
10 LCL=5.80
5 1
4
7
10
13 16 19 Observation
22
25
28
On the basis of 30 measurements, the BB makes a control chart for the PT of the coordinators (see above). What can she conclude? a) The processing times of the coordinators are acceptable. b) The PTs of the coordinators are no t in statistical control. c) There is no evidence for trends or outliers (the PTs are stable). d) All of the PTs of the coordinators are within the norms as agreed in the SLA (Service Level Agreement).
30
The BB wishes to determine whether the WTs have a normal, lognormal or Weibull distribution, or maybe even an altogether different distribution.
What is the best tool to study the data’s distribution? a) b) c) d)
The empirical CDF (cumulative distribution function). The histogram. The control chart. The probability plot.
31
The BB makes an empirical CDF graph of 50 observations of the Turnaround Time (TT); see next slide.
What can she conclude from this graph? a) About 4% of applications are longer than 10 work days in the process. b) About 24% of the applications has a turnaround time of 5 days. c) The Weibull distribution is not a good fit, as the data are not on a straight line. d) The effect of TT is not linear.
Empirical CDF of TT Weibull Shape 3.575 Scale 7.100 N 50
100
80
t 60 n e c r e P 40
20
0 2
3
4
5
6 7 8 TT (work days)
9
10
11
32
DMAIC 4: Identify potential influence factors. Trying to get the TTs down to more acceptable levels, the BB considers doing an FMEA with the employees working in the process.
Is it useful, in this project, to do an FMEA? a) No, since the objectives of the project are not about risks. b) No: an FMEA is used to identify failure modes (possible malfunctions) in products. c) Yes, provided that part of the long TTs are caused by errors and mistakes. d) Yes, as it helps to identify the bottleneck in the process.
33
DMAIC 4: Identify potential influence factors. The BB organizes a brainstorming meeting with the employees in the process. The goal of the meeting is to identify the causes of long TTs and to devise possible solutions. The meeting is quite successful and results in 64 ideas.
According to the DMAIC model, what should the BB do with these ideas? a) Participants vote about the value of these ideas, and the ideas with the most votes are implemented (“multi-voting”). b) Find evidence for the effects of these candidate causes and ideas. c) Delegate: let the participants decide which ideas they want to implement. d) Brainstorming sessions are not based on data. Therefore, it is not a very suitable technique, and the BB should be very suspicious about the results.
34
Based on the measurements, the BB makes a value stream map (next page).
What are the capacities of the coordinators and the secretary? a) Coordinators: Cap = 3 applications/day; Secretary: Cap = 6. b) Coordinators: Cap = 21; Secretary: Cap = 6. c) Coordinators: Cap = 21; Secretary: Cap = 42. d) Coordinators: Cap = 84; Secretary: Cap = 42.
Applications
Client
Taktrate: 40/day
(offer or rejection letter) 75%
Coordinator
WT
PT: 20 min #: 4 Cap:
Rate officer
WT
PT: 15 min #: 3 Cap:
Secretary
WT 25%
Under ‐ writer
WT
Work hours: 7 hrs / day
PT: 15 min #: 1 Cap:
PT: 10 min #: 1 Cap:
35
Which resource is the bottleneck in this process? (That is, which bottleneck will have the highest utilization?) a) The coordinators. b) The rate officers. c) The underwriter. d) The secretary.
36 The arrival rate of 40 applications per day is an average. Arrivals are not spread evenly over a day and over a week. Also the given PTs are averages, but there is some variability in the processing times of applications.
Will there be long waiting queues in the process (large WTs)? a) No, as the capacity of each of the process’s resources is larger than the workload. b) No; some resources have a capacity lower than the workload, other resources have a capacity higher than than the workload. But the average capacity is large enough to handle the workload. c) Yes, there will be a long waiting time especially before the bottleneck resource. d) Yes, there will be long waiting times before all resources (all WTs are the same order of magnitude).
37 Two- sampl e T f or Of f er vs Rej ect i on Of f er Rej ecti on
N 6 6
Mean 8, 754 12, 063
St Dev 0, 820 0, 846
SE Mean 0, 33 0, 35
Di f f erence = mu ( Of f er) - mu ( Rej ect i on) Est i mate f or di f f erence: - 3, 309 95% CI f or di f f erence: (- 4, 396; - 2, 221) T- Test of di f f er ence = 0 ( vs not =) : T- Val ue = - 6, 88 P- Val ue = 0, 000 DF = 9
The BB does a small study. For six accepted applications, she records how long it takes (PT) the secretary to compose the offer. Also, for six rejected applications, she records how long it takes to write the rejection letter. Assuming a normal distribution for the data, what can she conclude from the analysis above? a) The time it takes to compose an offer is significantly shorter than the time to write a rejection letter. b) The sample sizes in the study are too small to make reliable conclusions. c) The means and the variances differ significantly between the two groups. d) No conclusions are possible, as the BB should have used the ANOVA technique.
38 The BB wants to determine the r elationship between the waiting time WT in the workqueue before the coordinators, and the workload WL (the number of applications submitted per day). On 20 different days, she records the number of applications that have been submitted. And she also records, for a randomly selected application, how long it spends in the coordinators’ queue. See the regression analysis on the next slide. Is this a good fit? a) Yes, there is no evidence for a poor fit. b) No, a quadratic fitted line is not needed, a linear line would have been just as good. c) No, the relationship is not significant. d) No, there are outliers.
Pol ynomi al Regr essi on Anal ysi s: WT ver sus WL The r egr ess i on equat i on i s WT = 24, 64 - 1, 091 WL + 0, 01276 WL**2 S = 1, 15449
R- Sq = 96, 3%
Anal ysi s of Var i ance Sour ce DF SS Regress i on 2 308, 243 Err or 9 11, 996 Tot al 11 320, 238
R- Sq( adj ) = 95, 4%
MS 154, 121 1, 333
F 115, 63
P 0, 000
Fitted Line Plot
14 Sequent i al Anal ysi s of Vari ance Sour ce DF SS F P Li near 1 278, 632 66, 97 0, 000 12 Quadr at i c 1 29, 611 22, 22 0, 001 10 8
) s r 6 u o h ( T 4 W 2
S R-Sq R-S q(ad j)
0 40
45
50 55 60 65 WL (applications per day)
1,13149 94,1% 93,4%
70
39
Is the workload (WL) the most important influence factor for the waiting time (WT) before the coordinators?
a) Yes, the small p-value indicates that this is the vital influence factor having the largest impact. b) The high R2-value indicates that this factor explains almost all differences in WT observed in the regular process. c) No, there is considerable random scatter of the observations around the fitted curve. d) To make that conclusion, one should include other potential influence factors in the regression analysis as well.
75
40
Suppose that the workload is WL = 62.5 applications per day. What can you say about the waiting time in the coordinators’ queue?
a) In that case, 95% of all applications will spend 6 hours in the queue before the coordinators. b) 95% of the applications will have a waiting time below 8.3 hours. c) There is a 50% chance that WT will be above 6 hours. d) 95% of the applications have a waiting time in between 4.85 and 7.15 hrs.
Answers Question Answer
Question
Answer
Question
Answer
Question
Answer
1
A
11
C
21
B
31
A
2
C
12
D
22
D
32
C
3
C
13
C
23
C
33
B
4
D
14
B
24
C
34
D
5
C
15
D
25
B
35
D
6
D
16
A
26
A
36
C
7
C
17
A
27
D
37
A
8
A
18
A
28
C
38
A
9
D
19
C
29
C
39
B
10
C
20
B
30
D
40
C
