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This Lean Six Sigma Project done by student from Advance Innovation group which is posted to provide for benchmarking and best practices sharing purposes.

This is a sample project on improvement of productivity in BPO back office process

Additionally, it is advisable that you also visit and subscribe Advance Innovation Group Blog (http://advanceinnovationgroup.com/blog) for more Lean Six Sigma Project, Case Studies on Lean Six Sigma, Lean Six Sigma Videos, Lean Six Sigma Discussions, Lean Six Sigma Jobs etc.

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- 1. Indexing ProcessSix Sigma Project onSix Sigma Project onImproving Productivity
- 2. Understanding VOCCustomer Customer Comments Customer (CTQs)Champion – Assistant VicePresident“The objective is to minimize process level variations,raising the overall productivity of the process withoutcompromising on quality parameters. I expect at least 20%productivity increase at a process level.”Improvement in process productivityReduce variationsQualityProcess Manager I am looking towards a productivity improvement of 20%-25% in the indexing process. This project will lead toincreasing the productivity/efficiency/agentutilization/client satisfactionImprovement in productivityClient satisfactionEfficiencyAgent utilization
- 3. Project CharterdDefineProject Leader: Binny Arora Team MembersBusiness Case:Advance Innovation Group India Ltd (AIG) is a Business processOutsourcing Company, operating out of Noida. Aviva Life InsuranceCompany UK has outsourced its Claims Indexing process to AIG.Claims Indexing Process: The Claims Indexing process was migrated toAIG India in January 2009. The process has not been meeting therequired productivity expectations and as a result the backlog hasincreased and transactions are missing turn around time. Customers arecalling and complaining. As per state laws a claim has to be processedwithin 30 days of receiving it - Claims are not being processed within 30days and Aviva is paying huge fines to the State Government. AIG is alsopaying financial penalties for not meeting the SLA target for last threemonths. Improving productivity will increase Business end-end TAT andboth Aviva and AIG will benefit from improved productivity. This also willresult in reduced operating cost for AIG.Stakeholders Business LeaderChampion Vice PresidentSponsor Assistant Vice PresidentMBB Pranay KumarLBB Jai KapoorTeam MemberSME,QCA, 4 Associates, MI team, AM PE & AMOperationsProblem Statement:For the period March 10 to May 10 the average performance onproductivity for the process was 48.56 documents per hour. Against atarget on 58 documents per hour. The backlog has increased by 10,000documents and TAT% is at 85% not met.Goal Statement:To improve process productivity from 48.56 documents per hour to 58documents per hour by 21st November 2010Project In Scope:1. . Associates in production effective October 2009Project Out of Scope:1. Associates in training as on June 20102. Any new work or queue added effective May 20103. Indexing Process at AIG USATimelines/Milestones/PhasesStart Date End DateStart date: 5thJune 2010 -DEFINE 15thJune 2010 10thJulyMEASURE 11thJuly 15thAugustANALYZE 17thAugust 30thSeptemberIMPROVE 5th October 2010 20thNovemberCONTROL 25thNovember 15thDecember 2010
- 4. ARMI and Communication PlanKey Stakeholders ARMI WorksheetDefine Measure Analyze Improve ControlStakeholders I I I I ISponsor I I I I IChampion I & A I & A I & A I & A I & AMBB A & I A & I A & I A & I A & ILBB I & R I & R I & R I & R I & RProcess Manager I & M I & M I & M I & M I & MBinny Arora R & M R & M R & M R & M R & MTeam Members M M M M MA – Approval of team decisions I.e., sponsor, business leader, MBB.R – Resource to the team, one whose expertise, skills, may be needed on an ad-hoc basis.M – Member of team – whose expertise will be needed on a regular basis.I – Interested party, one who will need to be kept informed on direction, findings.Communication PlanInformation Or Activity Target Audience Information Channel Who WhenProject Status Leadership E-mails Binny Arora BI-WeeklyTollgate Review BB,LBB,MBB & Champion E-mails or Meetings Binny Arora As per Project PlanProject Deliverables or Activities Members Emails, Meetings Binny Arora WeeklydDefine
- 5. CTQ TreeImprovementin processproductivityImprovementin processproductivityCTQsDocuments per hour(Project Y Metric)Documents per hour(Project Y Metric)58 documents perhour(Target)58 documents perhour(Target)58 documents perhour(Lower SpecificationLimit)58 documents perhour(Lower SpecificationLimit)Any day whendocuments per houris less than 58would beconsidered adefective day(Defect Definition)Any day whendocuments per houris less than 58would beconsidered adefective day(Defect Definition)Project YProject Y = Number of documents processed in a dayTotal no. of actual hours worked in a daydDefine
- 6. SIPOCSuppliersSuppliers InputsInputs Process OutputsProcess Outputs CustomersCustomersSSSS IIII PPPP OOOO CCCCIndexeddocumentsClaim OfficerClaims Departmentsin AIG(US & India)Paper MailElectronicallyImportedMailAIG US ScanningTeamClaims OfficerStartLog intoApplicationSelect QueueNew Docs orOld MailOpen ScannedImageTakeAppropriateActionComplete theDocumentEnddDefineMissing infodocuments
- 7. Data Collection PlanmMEASUREProject-Y Operational Definition Defect Def Performance StdSpecification LimitOpportunityLSL USLNumber of DocumentsProcessed in a dayVolumes of Documents processed in aday at a process level as generated by theBOSS applicationDocuments perhour below 5858 documents perhour58 NADailyTotal numbers ofworked hours in a dayActual worked hours (time on system –excluding breaks and other auxes) in aday at a process level as generated byPROTRACDailyProject-Y Data TypeData ItemsNeededFormula to beusedUnitPlan to collect DataPlan tosampleWhat Databaseor Container willbe used torecord this data?Is this anexistingdatabaseor new?If new, Whenwill thedatabase beready for use?When is theplanned startdate for datacollection?Number ofDocumentsProcessed in adayDiscreteTotaldocumentsProcessed in adayTotal Documentsprocessed in aday/Total workedhours in a dayDocs/per hour Excel yes NAAlreadystartedMarch 2010to May 2010.Total numbers ofworked hours ina dayContinuousWorked Hoursin a dayHours per day Excel yes NAAlreadystartedMarch 2010to May 2010.
- 8. %TolerenceProcess tolerance = 0.3Study Var %Study Var %ToleranceSource StdDev (SD) (5.15 * SD) (%SV) (SV/Toler)Total Gage R&R 0.10056 0.51789 7.70 172.63Repeatability 0.10056 0.51789 7.70 172.63Reproducibility 0.00000 0.00000 0.00 0.00Part-To-Part 1.30154 6.70292 99.70 2234.31Total Variation 1.30542 6.72290 100.00 2240.97Number of Distinct Categories = 18%ContributionSource VarComp (of VarComp)Total Gage R&R 0.01011 0.59Repeatability 0.01011 0.59Reproducibility 0.00000 0.00Part-To-Part 1.69400 99.41Total Variation 1.70411 100.00Validate Measurement System Gage R&R Annova MethodmMEASUREKPI Data TypeDPH Continuous1. % Contribution of Gage R&R variation is less thanpart-part variation2. % Tolerance Gage R&R against total variation is at7.70%3. Number of Distinct Categories = 18Gauge is successful, hence the data is good for furtheranalysis
- 9. StabilityTake away: Patterns suggest that the variation observed is due to "special causes“. further investigation needs to be done to ascertainthe causes of clusters.The data is free fromtrends, mixtures andoscillationsmMEASUREP-Value is greater than 0.05P-Value is greater than 0.05There is clustering inthe data
- 10. Existence of outliers depict special causes .The special causes can be attributed to extra processing done in order to cater to the Diwali holidays.mMEASURECapability
- 11. Normality Normality: P value: 0.053 Shape: Non-Normal Measure of central tendency :data isnon-normal - measure of centraltendency will be median = 48.91 Spread: Stability factor = Quartile 1divided by quartile 3 Stability Factor = 45.12/52.57= 0.85Aim: The project shall focus on shifting the central tendency and reduction in variation. Stability factor is 0.85mMEASURETargetNull and alternate hypothesisHO – Data is normalHA – Data is non normal
- 12. Normality Normality: P value = 0.053 Shape: Non-Normal Measure of central tendency :data isnon-normal measure of centraltendency will be Median = 48.67mMEASURE
- 13. Box Plot Day of the WeekThursday has the highest variation in the process and Monday, Tuesday and Friday have low averages. And Fridayhave low averagesmMEASURETargetmean
- 14. BOX Plot Agent WiseThere is a lot of variation at the agent level.mMEASURE
- 15. Spread of DataMost of the data points are below 58 Docs per hours. Huge variation in Docs Per Hour observed - Focus onreducing variation. Data is skewed – focus on reducing variation and shifting the central tendencyCurrentMeanTargetmean
- 16. Current Capability - Process Sigma LevelmMEASURECurrent process sigma is 0.33DefectOpportunities perunitNumber of unitsTotal number ofdefectsDefects peropportunityDPMO Sigma1 69 61 0.884 884,57 0.33
- 17. Organize Potential CauseFactors identified through brainstormingaAnalyze
- 18. Organize Potential CausesThe Highlighted Causes will be further analyzed. Factors in control to be statistically analyzedusing Hypothesis TestingaAnalyseHigh Medium LowIn ControlProcess KnowledgeQuality ScoreTyping SpeedTyping AccuracyUse of Shortcut KeysTenureTrainerEducational BackgroundPrior ExperienceIdle TimeAgeTraining ScoreIncoming VolumeResidentail LocationWork ExperienceMaritial StatusOut ofControlApplications Slows DownSystem LatencyCustomer Name not availableClaim Number not availableImpactControl
- 19. Data Collection – For Potential X’sData to be collected around the Potential X’saAnalyze
- 20. Statistically Significant X’s aAnalyseRelationship between productivity and tenure in the processChi-Square Test:Expected counts are printed below observed countsChi-Square contributions are printed below expected countsDPH DPH lessabove 50 than < 50 Total1 9 2 115.26 5.742.658 2.4362 2 10 125.74 6.262.436 2.233Total 11 12 23Chi-Sq = 9.763, DF = 1, P-Value = 0.002HO – There is no relationship between tenure in the process andproductivityHA – There is relationship between tenure in the process andproductivityP<.05 , Accept HA :There is a statistically significant relationshipbetween tenure in the process and productivityDPH above 50 DPH less than < 509 2 Tenure>52 10 Tenure <5Regression Analysis: DPH versus AgeThe regression equation isDPH = 71.68 - 0.8007 AgeS = 2.88835 R-Sq = 61.4% R-Sq(adj) = 59.5%Analysis of VarianceSource DF SS MS F PRegression 1 265.890 265.890 31.87 0.000Error 20 166.851 8.343Total 21 432.741Relationship between Productivity and Age of the associateHO – There is no relationship between Age of associate andproductivityHA – There is relationship between Age of associate and productivityP<.05 , Accept HA :There exists a statistically significant relationshipbetween Age of associate and productivity
- 21. Statistically Significant X’saAnalyseHO – There is no relationship between Trainer and productivityHA – There is a relationship between Trainer and productivityP<.05 , Accept HA :there is a statistically significant relationshipbetween tenure in the process and productivityHO – There is no relationship between prior experience andproductivityHA – There is a relationship between prior experience andproductivityP>.05 , Accept Ho there is no statistically significant relationshipbetween prior experience and productivityRelationship between productivity and trainer Relationship between productivity and prior experienceMood Median Test: DPH versus Trainer 1 or 2Mood median test for DPHChi-Square = 5.24 DF = 1 P = 0.022Trainer Individual 95.0% CIs1 or 2 N<= N> Median Q3-Q1 -------+---------+---------+---------1 9 3 47.2 7.6 (-----------*---------)2 3 8 52.0 4.7 (-------*-----)-------+---------+---------+---------45.5 49.0 52.5Overall median = 50.5A 95.0% CI for median(1) - median(2): (-7.2,-0.4)Mood Median Test: DPH versus Previous ExpMood median test for DPHChi-Square = 2.39 DF = 3 P = 0.495Individual 95.0% CIsPrevious Exp N<= N> Median Q3-Q1 -+---------+---------+---------+-----Back Office 3 5 51.6 5.2 (------------*--)Blended 7 3 47.0 10.9 (--------------*----------------)NON-BPO 1 2 52.3 3.9 (----*-----)Voice 1 1 49.6 2.2 (--*--)-+---------+---------+---------+-----42.0 45.5 49.0 52.5Overall median = 50.5
- 22. Statistically Significant X’sRelationship between productivity and use of short cutkeysHO – There is no relationship between use of short cut keys andproductivityHA – There is a relationship between use of short cut keys andproductivityP<.05 , Accept HA :there is a statistically significant relationshipbetween use of short cut keys and productivityMann-Whitney Test and CI:Short Cut Key Used, Short Cut keys not usedN MedianShort Cut Key Used 13 52.008Short Cut keys not used 10 47.351Point estimate for ETA1-ETA2 is 4.36095.6 Percent CI for ETA1-ETA2 is (0.198,7.701)W = 190.0Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0377aAnalyseRelationship between productivity and ProductKnowledgeHO – There is no relationship between Product knowledge andproductivityHA – There is a relationship between Product knowledge andproductivityP<.05 , Accept HA :there is a statistically significant relationshipProduct knowledge and productivityMood Median Test: C5 versus C4Mood median test for C5Chi-Square = 5.49 DF = 1 P = 0.019C4 N<= N> Median Q3-Q1Process Knowledge- score above 4 9 52.01 5.89Process Knowledge- score below 8 2 47.35 4.65Individual 95.0% CIsC4 +---------+---------+---------+------Process Knowledge- score above (--------*-------)Process Knowledge- score below (-------*--------)+---------+---------+---------+------45.0 48.0 51.0 54.0Overall median = 50.50A 95.0% CI for median(Process Knowledge- score above) - median(ProcessKnowledge- score below): (1.01,6.38)
- 23. Statistically Significant X’sRelationship between Productive and Quality scoreHO – There is no relationship between Quality score and productivityHA – There is relationship between Quality score and productivityP<.05 , Accept HA :there is no statistically significant relationship between Quality Score and productivityChi-Square Goodness-of-Fit Test for Observed Counts in Variable: DPHUsing category names in Quality score above and belowTest ContributionCategory Observed Proportion Expected to Chi-SqY 603.837 0.5 582.477 0.783258N 561.118 0.5 582.477 0.783258N DF Chi-Sq P-Value1164.95 1 1.56652 0.211aAnalyse
- 24. Statistically Significant X’saAnalyseRelationship between productivity and Training ScoreHO – There is no relationship between Training Score andproductivityHA – There is a relationship between Training Score andproductivityP<.05 , Accept HA :there is no statistically significant relationshipbetween Training Score and productivityMood Median Test: C5 versus C4Mood median test for C5Chi-Square = 0.05 DF = 1 P = 0.827C4 N<= N> Median Q3-Q1Training - score above below 8 6 5 48.48 5.77Training - score above 85% 6 6 51.22 7.19Individual 95.0% CIsC4 -----+---------+---------+---------+-Training - score above below 8 (--------*--------------)Training - score above 85% (-----------------*---------)-----+---------+---------+---------+-47.5 50.0 52.5 55.0Overall median = 50.50A 95.0% CI for median(Training - score above below 8) - median(Training -score above 85%): (-5.25,4.29)Relationship between productivity and Typing SpeedHO – There is no relationship between Typing Speed andproductivityHA – There is a relationship between Typing Speed andproductivityP<.05 , Accept HA :there is a statistically significant relationshipbetween Typing Speed and productivityMood Median Test: C5 versus C4Mood median test for C5Chi-Square = 5.49 DF = 1 P = 0.019C4 N<= N> Median Q3-Q1Typing Speed above 30 WPM 4 9 52.01 5.89Typing Speed below 30 WPM 8 2 47.35 4.65Individual 95.0% CIsC4 +---------+---------+---------+------Typing Speed above 30 WPM (--------*-------)Typing Speed below 30 WPM (-------*--------)+---------+---------+---------+------45.0 48.0 51.0 54.0Overall median = 50.50A 95.0% CI for median(Typing Speed above 30 WPM)- median(Typing Speed below 30 WPM): (1.01,6.38)
- 25. Statistically Significant X’saAnalyseRelationship between productivity and Typing AccuracyHO – There is no relationship between Typing Accuracy and productivityHA – There is a relationship between Typing Accuracy and productivityP<.05 , Accept HA :there is no statistically significant relationship between Typing Accuracy and productivityMann-Whitney Test and CI: Typing Accuracy above 90%,Typing Accuracy below 90%N MedianTyping Accuracy above 90% 13 52.008Typing Accuracy below 90% 10 47.351Point estimate for ETA1-ETA2 is 4.36095.6 Percent CI for ETA1-ETA2 is (0.198,7.701)W = 190.0Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0377
- 26. Statistically Significant X’saAnalyseRelationship between Volumes Processed Per day and Docs Per HourHO – There is no relationship between Volumes Processed Per Day & productivityHA – There is a relationship between Volumes Processed Per Day and productivityP<.05 , Accept HA :there is a statistically significant relationship between Volumes Processed Per Day & productivityCorrelations: DPH, VolumePearson correlation of DPH and Volume = 0.338P-Value = 0.018
- 27. Statistically Significant X’saAnalyseRelationship between Volumes Processed Per day and Idle Time in HoursHO – There is no relationship between Volumes Processed Per Day & Idle TimeHA – There is a relationship between Volumes Processed Per Day and Idle TimeP<.05 , Accept HA :there is a statistically significant relationship between DPH & Idle TimeCorrelations: DPH, Idle Time in HoursPearson correlation of DPH and Idle Time in Hours = -0.737P-Value = 0.000
- 28. Summary of Statistical AnalysisThe factors highlighted in Red are the vital X’saAnalyseMeasure Type Data Type P-value Take AwayProductivity Y Continuous NA NAProcess Knowledge X Discrete <0.05 Statistically Significant relationship with YTyping Speed X Continuous <0.05 Statistically Significant relationship with YQuality Score X Discrete >0.05 Statistically no Significant relationship with YTyping Accuracy X Discrete <0.05 Statistically Significant relationship with YTraining Score X Discrete >0.05 Statistically no Significant relationship with YVolumes Received X Discrete <0.05 Statistically Significant relationship with YTrainer X Discrete <0.05 Statistically Significant relationship with YIdle Time X Continuous <0.05 Statistically Significant relationship with YAge X Continuous <0.05 Statistically Significant relationship with YTenure X Continuous <0.05 Statistically Significant relationship with YPrior Experience X Discrete >0.05 Statically no Significant relationship with YUse of Shortcut Keys X Discrete <0.05 Statistically Significant relationship with Y
- 29. Summarizing Findings on Root CausesiImproveVital Xs Mode of Analysis FindingsProcess Knowledge Mood MedianAssociates who have better process knowledge have a Significant Impact onProductivity. Assocaiates who scored above 90% have a higher DPHTyping Speed Mood MedianAssociates who have better Typing Speed have a Significant Impact onProductivity.Typing Accuracy Mann-WhitneyAssociates who have better Typing Accuracy have a Significant Impact onProductivity.Volumes Received CorrelationStatistical analysis on volume and DPH signifies that there exists a relationshipbetween both and an increase in volumes also drives an increase in theproductivityTrainer Mood MedianAssociates trained by Trainer 2 perform better as compared to associatestrained by the other trainerIdle Time CorrelationStatistical analysis on Idle time and DPH signifies that there exists arelationship between both and a decrease in idle time also drives an increasein the productivity.Age Regression & Correlation Analysis Associates below 29 years of age have better productivityTenure Chi-SquareTenure of an associate plays an important role in the time taken to processdocuments, more the tenure of the associate better is his/ her productivity.Use of Shortcut Keys Mann-WhitneyAssociates who use the short cut keys while using the application performbetter as compared to associates who do not use the short cut keys
- 30. Implementing Solutions & TimelinesiImprove
- 31. Pilot Results iImproveMood Median Test:Mood median test for C10Chi-Square = 11.80 DF = 1 P = 0.001C9 N<= N> Median Q3-Q1DPH-Post Solutions 0 11 63.0 6.0DPH_1 124 110 48.9 7.1Individual 95.0% CIsC9 -----+---------+---------+---------+-DPH-Post Solutions (-------*---)DPH_1 (--*)-----+---------+---------+---------+-50.0 55.0 60.0 65.0Overall median = 48.9A 95.0% CI for median(DPH-Post Solutions) - median(DPH_1): (10.1,16.4)Pilot resultsThe pilot results shows steep increase in productivity from48.02 DPH to 62.78 DPHPilot resultsThe pilot results shows steep increase in productivity from48.02 DPH to 62.78 DPHHypothesis Testing for validating improvement Null Hypothesis (Ho) = There is no statisticaldifference in the mean performance of DPHbetween baseline and improve Pilot (Ha) = There is a statistical difference in themean performance of DPH between baselineand improve Pilot P < .05 , Accept Ha : Post first stage of Implementingsolutions DPH is statistically greater than baselineperformanceHypothesis Testing for validating improvement Null Hypothesis (Ho) = There is no statisticaldifference in the mean performance of DPHbetween baseline and improve Pilot (Ha) = There is a statistical difference in themean performance of DPH between baselineand improve Pilot P < .05 , Accept Ha : Post first stage of Implementingsolutions DPH is statistically greater than baselineperformance
- 32. Evaluate Improve ResultsiImproveTake AwayThere is a significant difference in productivity variationbefore and after implementation of ‘Improve’ phase.Take AwayThere is a significant difference in productivity variationbefore and after implementation of ‘Improve’ phase.Hypothesis Testing for validating improvement Null Hypothesis (Ho) = There is no statistical difference in themean performance of DPH between baseline and Post improve (Ha) = There is a statistical difference in the mean performance ofDPH between baseline and Post improve P < .05 , Accept Ha : Post first stage of Implementing solutions DPH isstatistically greater than baseline performanceHypothesis Testing for validating improvement Null Hypothesis (Ho) = There is no statistical difference in themean performance of DPH between baseline and Post improve (Ha) = There is a statistical difference in the mean performance ofDPH between baseline and Post improve P < .05 , Accept Ha : Post first stage of Implementing solutions DPH isstatistically greater than baseline performanceMood Median Test: DPHMood median test for DPHChi-Square = 35.73 DF = 1 P = 0.000Individual 95.0% CIsDate N<= N> Median Q3-Q1 --+---------+---------+---------+----Improve 6 28 75.0 8.0 (---*)Measure 27 2 49.4 16.2 (---*-------------)--+---------+---------+---------+----48.0 56.0 64.0 72.0Overall median = 67.0A 95.0% CI for median(Improve) - median(Measure): (14.0,29.4)56% productivity improvementfrom baseline 48.78 DPH to76.31 DPH (above project targetof 58 DPH)
- 33. iImprove
- 34. iImprove
- 35. iImprove
- 36. iImprove
- 37. iImprove
- 38. iImprove
- 39. iImprove

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