Resistors for electronic circuits are manufactured on a high

Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is set up to produce a large run of resistors of 1,000 ohms each.

       To set up the machine and to create a control chart to be used throughout the run, 15 samples were taken with four resistors in each sample. The complete list of samples and their measured values are as follows: Use three-sigma control limits. Refer Exhibit 10.13.

Calculate the mean and range for the above samples. (Round \"Mean\" to 2 decimal places and \"Range\" to the nearest whole number.)

Determine the UCL and LCL for R-chart. (Leave no cells blank - be certain to enter \"0\" wherever required. Round your answers to 3 decimal places.)

SAMPLE NUMBER	READINGS (IN OHMS)
1	990	983	1016	979
2	1011	978	1023	976
3	974	1003	996	998
4	981	988	1001	1007
5	992	976	978	1001
6	988	1030	1030	1014
7	1022	1010	1021	990
8	1010	998	981	975
9	997	1018	983	1024
10	1028	974	989	989
11	972	1010	1007	992
12	1013	981	1002	1009
13	1005	1021	996	1014
14	995	973	971	975
15	1011	1001	987	1015
a. Calculate the mean and range for the above samples. (Round \"Mean\" to 2 decimal places and \"Range\" to the nearest whole number.) Sample Number      Mean      Range 1        2        3        4        5        6        7        8        9        10        11        12        13        14        15        b. Determine and . (Round your answers to 3 decimal places.)       c. Determine the UCL and LCL for a chart. (Round your answers to 3 decimal places.)   UCL   LCL d. Determine the UCL and LCL for R-chart. (Leave no cells blank - be certain to enter \"0\" wherever required. Round your answers to 3 decimal places.)   UCL   LCL

Solution

Ans-

. Michael’s Engineering, Inc. manufactures components for the ever-changing notebook computer business. He is considering moving from a small custom design facility to an operation capable of much more rapid design of components. This means that Michael must consider upgrading his CAD equipment. Option 1 is to purchase two new desktop CAD systems at $100,000 each. Option 2 is to purchase an integrated system and the related server at $500,000. Michael’s sales manager has estimated that if the market for notebook computers continues to expand, sales over the life of either system will be $1,000,000. He places the odds of this happening at 40%. He thinks the likelihood of the market having already peaked to be 60% and future sales to be only $700,000. What do you suggest Michael do and what is the EMV of this decision? ANS: The EMV for the desktop systems is $620,000 vs. $320,000 for the integrated system. Therefore, Michael should purchase the desktop systems. 3. Westover Electrical, Inc., is a medium-size manufacturer of wire windings (used in transformers) in making electric motors. Joe Wilson, VP operations, has experienced an increasing problem with rejected product found during the manufacturing operation. “I’m not sure where to begin,” admitted Joe at the weekly meeting with his boss. “Rejects in the Winding Department have been killing us the past two months. Nobody in operations has any idea why. I have just brought in a consultant, Roger Gagnon, to take a look at the situation and make recommendations about how we can find out what is going on. I don’t expect Roger to make technical recommendations— just see if he can point us in the right direction.” Gagnon’s first stop later that day was the production floor. His discussions with the production supervisors in the Winding Department indicated that they had no real grasp of what the problem was or what to do to correct it. A tour of the winding operation indicated that there were three machines that wound wire onto plastic cores to produce the primary and secondary electric motor windings. After inspection by quality control (QC), these windings then went to the Packaging Department. Packaging personnel, Gagnon found, inspect their own work and make corrections on the spot. The problem is that too many windings are found to be defective and require reworking before they can be packaged. Gagnon’s next stop was the Quality Control Department, where he obtained the records for the past month’s Winding Department rejects. TABLE 1. Transformer Reject Log: Winding Process Winder Bad Wind Twisted Wire Broken Leads Abraded Wire Wrong Core Wrong Wire Failed Electrical Test fraction defects 1 100 1 1 0 4 1 0 0 1 0.08 2 100 2 2 1 0 0 1 5 0 0.11 3 100 3 0 0 0 1 0 0 3 0.07 4 100 1 0 1 3 0 0 0 0 0.05 5 100 2 3 1 0 0 2 3 0 0.11 6 100 3 0 0 1 2 0 0 0 0.06 7 100 1 1 0 0 2 0 0 0 0.04 8 100 2 0 0 0 0 0 3 0 0.05 9 100 3 0 0 1 2 0 0 3 0.09 10 100 1 0 0 1 0 0 0 0 0.02 11 100 2 0 0 0 0 0 2 0 0.04 12 100 3 0 0 0 3 1 0 3 0.10 13 100 1 0 1 5 0 0 0 0 0.07 14 100 2 0 0 0 0 0 2 1 0.05 15 100 3 0 0 0 3 0 0 2 0.08 16 100 1 0 0 2 0 0 0 0 0.03 17 100 2 0 0 0 0 0 1 0 0.03 18 100 3 0 0 0 3 0 0 3 0.09 19 100 1 0 1 5 0 0 0 0 0.07 20 100 2 0 0 0 0 0 1 0 0.03 21 100 3 0 0 0 1 0 0 4 0.08 22 100 1 0 0 6 0 0 0 0 0.07 23 100 2 1 0 1 0 1 0 0 0.05 24 100 3 0 0 6 1 0 0 4 0.14 25 100 1 0 0 4 0 0 0 0 0.05 No. of Reject Units by Cause No Inspected Inspections a. Prepare an outline for Shambhu Pradhan’s report. What charts or graphs might be included in the report? You could use z=2 confidence level. b. Prepare Shambhu’s recommendation, with justification, on one page. (12+6) p-Chart p = 166/2500 = 0.066 p = sq rt [(p (1 – p ))/n] = sq rt [(0.066x0.934)/100] = 0.025 UCL = p + zp = .066 + 2x.025 = 0.116 LCL = p - zp = .066 - 2x.025 = 0.016 From the p-chart it can be seen that since one of the points (sample no 24) lies outside the upper control limit, this is due to some assignable cause and therefore we can conclude that the process is not within control. Pareto Chart Type of Defect No of defects % of defects Winder 49 29.52% Broken Leads 39 23.49% Faiuled Electric test 24 14.46% Abraded Wire 19 11.45% Wrong wire 17 10.24% Bad wind 8 4.82% Twisted wire 5 3.01% Wrong Core 5 3.01% 166 100.00% From the Pareto chart, it is clear that the major causes of the defects are winder and broken leads. If these two problems are taken care of, the total number of defects will reduce by about 53%. 4. Chicago Supply Company manufactures paper clips and other office products. Although inexpensive, paper clips have provided the firm with a high margin of profitability. Samples of 200 were taken and the numbers of defects in the last 10 samples were as follows: 5, 7, 4, 4, 6, 3, 5, 6, 2 and 8. Establish upper and lower control limits to reflect a 99.73% confidence level? Is the process in control? ANS: Sample No No of defects Fraction defective 1 5 0.025 2 7 0.035 3 4 0.020 4 4 0.020 5 6 0.030 6 3 0.015 7 5 0.025 8 6 0.030 9 2 0.010 10 8 0.040 50 p = Total no of defects ÷ Total no of pcs inspected = (5+7+4+4+6+3+5+6+2+8) ÷ (10 * 200) = 0.025 p = p (1 – p) / n = ((0.025*(1 – 0.025)) / 200) = 0.011 A 99.73% confidence level would reflect a z value = 3 UCL = p + 3p = 0.025 + 3*0.011 = 0.058 LCL = p - 3p = 0.025 - 3*0.011 = - 0.008 = 0 (no negative UCL\'s permitted) Plotting the fraction defectives (no of defects in each sample divided by the sample size i.e. 200), we see that all points fall within these two control limits, and hence we can conclude that the process is in control