Sugar content. Ratings. Rich Taste Smooth Texture Distinct flavor Sweet taste

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1 Course title : Operations and Service Management Semester : Fall 2011 MBA Term 4 Assignment : One Due Date : Friday, September 23, 2011 Total Marks : = You wish to compete in the super premium ice cream market. The task is to determine the wants of the super premium market and the attributes/hows to be met by their firm. Use the house of quality concept. Market research has revealed that customers feel four factors are significant in making a buying decision. A rich taste is most important followed by smooth texture, distinct flavor, and a sweet taste. From a production standpoint, important factors are the sugar content, the amount of butterfat, low air content, and natural flavors. Assign customer importance ratings from conducting a small survey in your own classroom or in other semesters. High relationship = 5, medium = 3 and low = 1. Conclude on your findings from this House of Quality tool. One possible solution for this problem is: Ratings Sugar content High butterfat Low air content Natural flavor Rich Taste Smooth Texture Distinct flavor Sweet taste Looking at the house of quality tool, we can see that high butterfat level and sugar content score the highest. So the company should focus on these two product attributes in order to satisfy customer needs. Therefore most importance should be given to these two factors to get a competitive edge. 2. 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

2 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.

3 TABLE 1. Transformer Reject Log: Winding Process No. of Reject Units by Cause No Inspections Inspected Winder Bad Wind Twisted Wire Broken Leads Abraded Wire Wrong Core Wrong Wire Failed Electrical Test fraction defects 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 = σp = sq rt [(p (1 p ))/n] = sq rt [(0.066x0.934)/100] = UCL = p + zσp = x.025 = LCL = p - zσp =.066-2x.025 = 0.016

4 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 % Broken Leads % Faiuled Electric test % Abraded Wire % Wrong wire % Bad wind % Twisted wire % Wrong Core % % 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?

5 ANS: p Sample No No of defects Fraction defective = Total no of defects Total no of pcs inspected = ( ) (10 * 200) = σp = p (1 p) / n = ((0.025*( )) / 200) = A 99.73% confidence level would reflect a z value = 3 UCL = p + 3σp = *0.011 = LCL = p - 3σp = *0.011 = = 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

6 5. 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 1000 ohms each. To set up the machine and to create a control chart to be used throughout the run, 10 samples were taken with 4 resistors in each sample. The complete list of samples and their measured values are as follows: Sample No Readings in ohms i. Is this process in control? ANS (i) S.No Readings in ohms Mean X R X = /10 = ohms R = 193/10 = 19.3 ohms UCLX = X + A2R LCLX = X A2R UCLR = D4R LCLR = D3R From the table A2 = 0.73 D4 = 2.28 D3 = 0 UCLX = *19.3 = ohms

7 LCLX = *19.3 = ohms UCLR = 2.28*19.3 = 44 ohms LCLR = 0 The X-chart will look like below The R-chart will look like below Since all points fall within the control limits of both the X and R charts, we can safely conclude that the process is within control. 6. Lackey s, a local bakery, is worried about increased costs particularly energy. Last year s records can provide a fairly good estimate of the parameters for this year.

8 Charles Lackey, the owner, does not believe things have changed much, but he did invest an additional $3,000 for modifications to the bakery s ovens to make them more energy-efficient. The modifications were supposed to make the ovens at least 15% more efficient, but extra labor-hours were required for workers to become familiar with the process changes. Lackey has asked you to check the energy savings of the new ovens and also look over other measures of the bakery s productivity to see if the modifications were beneficial. He has the following data representing a month from last year and an equivalent month this year:: Last year Now Production (dozen) 1,500 1,500 Labor (hours) Capital Invested ($) $15,000 $18,000 Energy (BTU) 3,000 2,750 Determine the multifactor productivity using the following costs: - labor $8 per hour - capital 0.83% per month of investment - energy $0.60 per BTU You are to provide Mr Lackey with a multifactor indication of the productivity increase or decrease between last year and this year. Energy Change: Last year : 1500*12/3000 = 6.0 loaves/btu Now : 1500*12/2750 = 6.55 loaves/btu Percent Change: (6.55-6)/6*100= 9.17% There has definitely been an increase in the energy productivity which means there have been savings in energy. Labor Change: Last year : 1500*12/350 = loaves/labor hour Now : 1500*12/325 = loaves/labor hour Percent Change: ( )/51.43*100= 7.68% There is also an increase in labor productivity Investment Change: Last year : 1500*12/15000 = 1.2 loaves/$ investment Now : 1500*12/18000 = 1.0 loaves/$ investment Percent Change: ( )/1.2*100= %

9 However as we can see there has been a decrease in the investment productivity. This can be however be substantiated with the fact that due to this investment, the energy savings is now possible and that this has cause an increase in the energy productivity. Now looking at the total multifactor productivity figures Last year : 1500*12/(350*8+3000* *.0083) = 3.81 Now : 1500*12/(325*8+2750* *.0083) = 4.09 Percent Change: ( )/3.81*100= 7.35% So as can be seen the modification could only make the ovens 7.35% more efficient and not 15% more efficient.