Week 8 HW

The following data are the production units from a manufacturing line for the pas Week Monday Tuesday Wednesday Thursday Friday X Bar 1 12 15 8 7 10 10.4 2 6 9 11 13 7 9.2 3 15 11 9 6 12 10.6 4 8 7 6 11 14 9.2 5 20 12 10 5 14 12.2 6 13 10 9 8 12 10.4 7 10 14 11 8 5 9.6 8 10 15 17 11 9 12.4 X2 Bar 10.5 R Bar 8.625 A2 0.58 UCL Xbar 15.5025 LCL Xbar 5.4975 UCL R 18.1988 LCL R 0 D3 0 D4 2.11 Calculate the Averages from the weeks to collect the X bar Calculation Calculate the Ranges of the Data for the R Bar Calculations (MAX - Min) Calculate the Averages of both the Xbar and Rbar data to get X2 Bar and R Bar Calculate the UCL Xbar & LCL Xbar using the A2 which was from the Sample Size Calculate the UCL Rbar & LCL Rbar with the D3 and D4 values from sample size 1 2 0 2 4 6 8 10 12 14 16 18 X Ba

st 8 weeks. The manager wants to know if the process is stable and in control. Pro R Bar UCL XBAR LCL XBARX2 BAR UCL R LCL R 8 15.5025 5.4975 10.5 18.1988 0 7 15.5025 5.4975 10.5 18.1988 0 9 15.5025 5.4975 10.5 18.1988 0 8 15.5025 5.4975 10.5 18.1988 0 15 15.5025 5.4975 10.5 18.1988 0 5 15.5025 5.4975 10.5 18.1988 0 9 15.5025 5.4975 10.5 18.1988 0 8 15.5025 5.4975 10.5 18.1988 0 3 4 5 6 7 8 X BAR CHART ar UCL XBAR LCL XBAR X2 BAR 1 2 0 2 4 6 8 10 12 14 16 18 20

A project manager at Q Corporation has a commitment to deliver 200 cylinde a ) Calculate the process capability index (CP and CPk) and indicate if the p USL (60+1) cm = 61 cm LCL (60-0.5) cm=59.5 cm Standard Deviation (r) 0.5 Mean (m) 60.4 Cp = (USL - LSL) / 6r (61-59.5)/(6*.5) = .5 CPp for USL (USL-m)/(3*r) (61-60.4)/(3*.5) = .4 Cp for LSL (m-LSL)/3*r) (60.4-59.5)/(3*.5) = .6 Cpk = Min {CPu, CPI} (Min{.4,.6} = .4 The Process is not capable because CP is <1 b ) How many cylinders the PM should plan to produce in order for 200 of t % less than 59.5 = LSL-m/r = (59.5-60.5)/0.5 = -1.8 = 3.59% % more than 61 = (USL-m)/r = (61-60.4)/0.5 = 1.2 = 88.49% = 15.1% (100-88.4 Percentage beyond specification limit= (11.51+3.59)% = 15.1% Percentage within specification limit = (100-15.1)% = 84.9% Number of cylinders = t (84.9/100)*t = 200 236 Cylinders need to be produced to meet specification c ) What should be the standard deviation in order for the process to becom For the process to be cables CP >1 {(61 - 59.5)/6r}>1 6r >1.5 r>.25 Standard Deviation (r) needs to be >.25 for the process to be capable

ers to a customer. The internal diameter of these cylinders must be 60 centim process is capable. them to meet specification. 49%) me capable? Note the process is not centered.

duction process is 60.4 and the standard deviation is 0.5,

A product has specification limits of 18 to 20. If the process mean is 19.2 and t a) The process capability for this production line. USL 20 LSL 18 Standard Deviation (r) 0.6 (USL-LSL)/6r (20-18)/6*.6 CP 2/3.6 =.556 b) Is the process Capable? No the process is not Cablable, CP < 1 c) What % of the products are out of specs if any? To solve we must calculate the Z -values for the LSL & USL ZUSL=(USL-process mean)/std deviation (20-19.2)/.6 ZUSL=1.3333 ZLSL=(process mean-LSL)/std deviation (19.2-18)/.6 ZLSL=2 P(Z>ZUSL) =P(Z>1.3333)=.0912 = 9.12% P(Z>ZLSL)=P(Z>2)= .0228= 2.28% Total %= 9.12%+2.28% Total % out of specs is 11.4% d) What should be the process standard deviation in order for the proces We can solve for the standard deviation using the equation for CpK, using 1.333 Process Capability Index= CpK=Minimum(Process mean-LSL)/3*Mean, (USL-Process mean)/3*Mean Minimum(19.2-18)/3*x, (20-19.2)/3*x Minimum (1.2/3X), (.8/3X) Minimum CpK=1.333

An IT organization has a service level agreement (SLA) with a major custome The IT manager has developed the following data: Average Daily volume of tier one calls is 30 with a standard deviation of 8 ca Mean processing time in the center is 10 hours with a standard deviation of 1 a) Calculate the % daily tier 1 problems that misses the SLA. UCL=Mean processing time+(3×Standard deviation of processing ti UCL=10 hours+(3×1.1 hours)=10 hours+3.3 hours=13.3 hours Now, we know that any call taking more than 12 hours will miss the To do this, we need to calculate the z-score for 12 hours: Z= Target time−Mean processing time/ Standard deviation of processing time Z= 12 hours−10 hours / 1.1 hours Using a standard normal distribution table or calculator, we find tha Z = 1.82 This will give us the percentage of daily tier 1 pr P(Z > 1.82) = .0344 This means approximately 3.44% of daily tier 1 problems will miss t b) Estimate the annual penalty cost to the center. Number of calls missing SLA=Average daily volume of tier 1 calls×P = 30*0.0344 = 1.032 Since we can't have a fraction of a call, we'll round up to the nearest whole Annual penalty cost=Number of calls missing SLA per day×Penalty per call× Annual Penalty Cost = 2 x 150 x 250 Estimated Annual Penalty Cost = $75,000 c) If you were the manger, would you add another technician to help d I would conduct a cost-benefit analysis to determine if the cost of hiring ano If the cost of hiring another technician, including salary, benefits, training, a whether or not to add another technician to the help desk to avoid penalty c While penalty costs are certainly a concern, they should be weighed against

er requiring that tier 1 calls/problems be resolved within 12 hours. There is a p alls 1.1 hour ime) e SLA. So, we need to find out how many calls fall into this category. at the proportion of calls taking more than 12 hours can be approximated by roblems that miss the SLA. the SLA. Percentage of calls missing SLA number. Therefore, approximately 2 calls per day will miss the SLA. ×Number of working days per year desk to avoid the penalty cost. Why or why not. Use your own estimate and r other technician is justified by the potential savings from avoiding penalty costs. In this and equipment, is significantly lower than the penalty cost, it might be financially prude costs would depend on a thorough assessment of various factors, including cost-effecti t the broader context of organizational priorities and objectives.

gnment with long-term strategic goals.