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Operations & Process Analysis - Assignment Example

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The paper "Operations & Process Analysis" is an outstanding example of a management assignment. Labor productivity is a measure indicating the output that is productive for an hour within a specified period of time (Krajewski et al., 2013, p.37). From the formula of calculating labor productivity, we should have the total output as well as the total productive hours…
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Extract of sample "Operations & Process Analysis"

Title: Operations & Process Analysis Name: Institutional Affiliation: Tutor: Date: QUESTION 1 A.) Labor productivity is a measure indicating the output that is productive for an hour within a specified period of time (Krajewski et al., 2013, p.37). From the formula of calculating the labor productivity, we should have the total output as well as the total productive hours. In this case, Total Output: Here we shall consider the costs incurred in the production of the total units in the one month against the total number of hours worked throughout the one month period. Total Output: 185 each at £250 = £21250 Total number of input hours per month = 100 x 10 =1000 Labor Productivity = 21250/1000 = £21.25 per hour Calculation of the multifactor productivity ration basically entails incorporating the total input costs incurred in the manufacturing process with respect to the total output (Mantel et al., 2007, p.68). In the case of Nick’s race shop, we shall have as follows: Total output = £21250 Inputs: Direct material = 40 x 100 = 4000 Direct labor = 10 x 10 x 100 = 10000 Overheads = 5000 Total Inputs = 19000 Therefore, the multifactor productivity gives 21250/19000 = 1.118 B.) Let us consider each option: i) Increasing sales by 10%. This will result to 110% x 250 x 85 = 23375/19000 = 1.23 ii) Increasing efficiency to only 10 defective: 250 x 90 = 22500/19000 = 1.184 iii) Reducing material, overhead, and labor costs by 10%: 21250/ (19000 -1900) = 1.243 From the above calculations, it is quite clear that the third (iii) option will yield the greatest impact on the multifactor productivity of the firm as it provides the largest value for the ratio as compared to the other two available options. C. Operations management is crucial in optimizing productivity in an organization. As can be observed in Nick’s Race jacket shop, the various factors involved in the manufacturing process impact on the final productivity of the organization. The primary aim of operation management, therefore, is make sure that the delicate balance between the factors of production is maintained (Nicholas & Herman, 2008, p.59). An organization should price its product prudently in the market of it is to maintain financial leverage. Moreover, operations management aids in designing and readjusting input costs accordingly to achieve desired levels of productivity (Waller, 2003, p.42). As evidenced in the calculations above, by carefully reducing the costs of the various inputs at the production level, the shop was able to achieve a much higher multifactor productivity ratio. Fundamentally, the whole process of operation management is pivotal in the decision making process. QUESTION 2. A.) For the successful implementation of any project, there should be critical consideration of the core competitive priorities. The fact that we are working under a timeline of 45 days means that these competitive factors should be carefully analyzed to ensure that we strike a balance between high output and the strict schedule. As such, the competitive priorities will be taken to consideration as follows: 1. Cost – For RASAS to achieve a competitive pricing and business for the mustang, the key consideration that should be put in mind is focusing on condition restoration of the mustang which would help in improving the value and the overall price of the automobile. If they conduct a partial restoration, the overall value of the mustang would be negatively affected. The organization should at all times ensure that it delivers goods at the lowest cost possible (Krajewski et al., 2013, p.33). 2. Quality – The issue of quality deals with determining the availability of the needed parts for the vehicle. Moreover, it is important for RASAS to make a great impression to the customers. Therefore, the aim of quality consideration is to determine if all the needed arts are available. Moreover, it sets to check whether in the event these parts are not available, the company is in a position to manufacture them on its own. 3. Customer Service – the issue of customer service is very critical in any organization as it ensures stability of business is maintained (Flynn et al., 2011, p.46).. Faced with the difficulty in finding and maintaining a constant supply of the necessary parts, it is important that the issue of obtaining parts be solved so as to provide the needed parts when need arises. 4. Flexibility – It is quite clear that at the moment, RASAS is not in a favorable condition to engage in vintage car restoration. This is due to the fact the organization does not have in place a department dealing with these types pf cars. AS a result, it is necessary that RASAS devotes considerable time, financial resource, and research on the viability of this project before kicking it off B.)From the Appendix 1, we can clearly determine the critical path to be falling on the path A-B-T-V. Further, the diagram shows the estimated slack times for each activity with the estimated times required for the earliest start, finish, late starts and finish. In the implementation, the company should be active in ensuring that they are able to obtain a steady supply of the needed materials for the vintage cars. Moreover, they should also ensure that they are in a position to constantly deliver the parts when requested by the customers. C.)From the project schedule shown on Appendix 2, it is possible to very easily check how the earliest start, finish and the latest start and finish were obtained. Further, the calculations obtained indicate that the entire process of restoration accrues a total cost of £18,100. This, according to part C of this question, means that it is indeed possible for the project to be completed within the allocated budget of £20,000. On the other hand, with the consideration that each week has five days, it is clear that each week will have an average spending of £2,011, which is way below the anticipated spending of £3,600. From these calculations, it is evident that there should not be any major issues within the budget. However, in the event that some activities in the restoration process would more expensive than expected, this issue can be dealt with by using the surplus cash (£1,900) to cater for such expenses. QUESTION 3. From the above critical analysis diagram, it is possible to calculate the completion time it would take (normal) as shown below B-D-F-G-H-I. This is the critical path. From this path, the completion time will be 40 weeks. Further, in seeking to shorten the duration it would take to complete the project, it is essential that we consider the possible decreases that can be taken into consideration in order to come up with considerable savings on the project. We also realize that as a result of the delay in completion of the project there is a penalty of £1,000 calculated weekly if the project takes longer than 25 weeks. In our case above, the penalty shall be of (40-25) x1000 = £15,000. The aim is therefore to reduce the duration it takes to carry out the project to a maximum of 25 weeks. First, we shall consider the possible decreases that we can make by considering the amount of time that can be saved from the crash time as shown below:   Required Time (weeks) Time Difference (weeks) Activity Normal Time Crash Time NT-CT Cost Per Week £’ A 5 3 2 1000 B 8 7 1 3000 C 10 8 2 1000 D 4 3 1 4000 E 3 2 1 1000 F 9 6 3 2000 G 2 2 0 0 H 8 5 3 1000 I 9 7 2 4000 The column on the cost per week is obtained by dividing the difference between crash cost and normal cost by the (NT-CT) column. From the above table, we are in a position to evaluate the possible activities from which we can reduce the number of weeks and hence obtain a total reduction in the costs. To begin with, we are going to crash B by one week. This is informed by the fact that it incurs the lowest crash cost. The cost of this is £3000. Although activity G is not expensive, there is no indication of any time saved by crashing it. Next we crash activity D at a cost of £4,000. We are now left with activities FH and I. From the activities above, we choose the activity that is the cheapest. Activity I is the cheapest so we crash this activity by one week at a cost of £4000. Therefore, the activities reduced will appear as follows: Procedure Activity Crashed Cost Weeks Reduced 1 B £3000 1 2 I £4000 1 3 D £4000 1 TOTAL = £11,000 TOTAL = 3 With this proposal, the company will be in a position to make a total savings of the following: i.) The number of weeks will be reduced from the initial 40 weeks to 37 weeks. ii.) Further, the company will make a reduction on the crash costs by a total of £11,000. This will be a substantial reduction for the company considering it will be attained by making a reduction of only 3 weeks. iii.) From the 3 weeks reduction, the company will have reduced the amount of penalty charged from the £15,000 to £12,000. iv.) The recommended completion week for the project will be in the 37th week. Crash analysis is majorly used in project management. There are various benefits attributed to the use of crash analysis. Ideally, when an organization is running out of time on a project, the use of crash analysis addresses the need to match productivity with the limited time (Boyer & Velma, 2010, p.54). Moreover, it helps in bringing on board more resources on a project and in as a result helps in effectively reducing the time required on a project significantly. Another major benefit of using crash analysis in project management is the important role it plays in reducing the initial costs anticipated in a project. This is because crash analysis involves selecting the alternative that incurs the lowest costs on a project. On the other hand, the use of crash analysis also has its negative aspects. One the major limitation of using crash analysis is the risk that even after utilizing a lot of resources on a project, there is no assurance that the project will prove to have better results (Flynn et al., 2011, p.46). As crash analysis involves using more resources in a bid to match the demanding project needs, an organization may be tempted to use so much resources only to fail to get equally high returns (Waller, 2003, p.73). Additionally, the use of crash analysis has the greatest risk on the quality of work produced. The project manager’s core objective in any project is to ensure that the project is completed within the given duration and achieve the set standards for the project. With the limitation of time in mind, there is a risk of substandard output should the project be carried within a very limited timeline. QUESTION 4. a.) Department Pair Closeness Factor Distance Weighted Distance score 1,2 16 3 48 1,3 6 1 6 1,5 18 2 36 1,6 11 1 11 2,4 5 1 5 3,5 20 3 60 3,6 15 2 30 4,6 5 1 5 5,6 8 1 8 TOTAL 209 b.) On exchanging the positions of the inspection areas and the tool store, we obtain a new diagram of the block plan and hence we shall have a new layout taking the form shown below 3 4 2 1 5 6 As such the weighted distance score obtained will bear slight difference with our current weighted distance as shown: Department Pair Closeness Factor Distance Weighted Distance score 1,2 16 3 48 1,3 6 1 6 1,5 18 1 18 1,6 11 2 22 2,4 5 1 5 3,5 20 2 40 3,6 15 2 30 4,6 5 2 10 5,6 8 1 8 TOTAL 187 c.) In redesigning the new block plan, there are several factors that should be put in consideration. First, the question requires that shipping and Receiving department remain where it is. This therefore means that while the other departments are highly flexible, these two will remain immobile. Secondly, emphasis has to be placed on the closeness factor of the departments. Therefore, the departments with the highest closeness factor will be located within the closest distances to each other. Department 3 and 5 should be close Departments 1 and 5 should be close Departments 1 and 2 should be close. Initializing the block plan, we shall have only department 3 with a sure location. 3 Then, by trial and error, we shall consider immediate closeness between 3, 5 1, 5 and 1, 2. 3 5 2 1 This therefore leaves out 4 and 6. Further, the closeness between 1 and 6 is considerably high thus putting them close. Therefore, a good block plan would take the arrangement below: 3 5 4 2 1 6 d.) The layout is essential in determining the arrangement of resources in an organization with the primary aim of improving efficiency, sharing resources and improving communication. What the layout design above intimates is that for the organization to continue performing effectively, then there should be close relative proximity between certain departments. In particular, the layout indicates that the Burr and grind, Tool Store, Shipping and receiving, and NC Machines departments should be located close to each other. Further, the whole process of structuring the manufacturing process is critical in enhancing flexibility of an organization’s workforce (Cousins et al., 2006, p.108). In addition to this, the use of the above layout design helps in proper allocation of the different process in an organization, categorizing the batch processes, job processes, continuous and line processes. Therefore, through proper application of the layout design, it is possible for organizations to strike the delicate balance between maintaining a relatively high output levels and reduction on production cost per unit. Equally, as a result of the reduced cost, an organization is in a position to increase the volume of produced goods (Boyer & Velma, 2010, p.87). In addition to this, an organization is in a position to engage highly specialized labor in each department. For the case of the organization shown above, the management would be in a position to make the best use of the equipment and the labor in each department with such a layout. REFERENCES. Boyer, K. K., & Verma, R., 2010. Operations & supply chain management for the 21st century. Mason, Ohio, South-Western/Cengage Learning. Cousins, P. D., Lawson, B., & Squire, B., 2006. Supply chain management theory and practice the emergence of an academic discipline? Bradford, England. Flynn, B. B., Morita, M., & Machuca, J., 2011. Managing global supply chain relationships: operations, strategies and practices. Hershey Pa, Business Science. Krajewski L., Ritzman, L. and Malhotra, M., 2013. Operations Management Processes and Supply Chains. 10th ed. Pearson Education Limited. Mantel Jr., Samuel J., Jack R. Meredith, Scott M. Shafer, and Margaret M. Sutton., (2007) Project Management in Practice, 3rd ed. New York: John Wiley & Sons. Nicholas, John M, and Herman Stein., 2008 Project Management for Business, Engineering, and Technology, 3rd ed. Burlington, MA: Butterworth-Heinemann. Waller, D. L., 2003. Operations management: a supply chain approach. London [u.a.], Thomson Learning. APPENDICESAPPENDIX 1 NB. The network diagram above can be ‘saved as picture’ to obtain a clearer view of the content. ACTIVITY TYPE OF ACTIVITY TIME (Days) COST PREDECESSOR EARLIEST START EARLIEST FINISH LATEST START LATEST FINISH SLACK A Order the needed materials 2 100 0 2 0 2 0 B Receive Upholstery 30 2100 A 2 32 2 32 0 C Receive Windshield 10 800 A 2 12 17 27 15 D Receive Oil Pump and Carburetor 7 1750 A 2 9 14 21 12 E Remove Chrome 1 200 0 1 14 15 14 F Remove Body from frame 1 300 E 1 2 15 16 14 G Repair the fenders 4 1000 F 2 6 23 27 21 H Repair trunk, doors and hood 6 1500 F 2 8 16 22 14 I Pull engine from chassis 1 200 F 2 3 16 17 14 J Remove rust from frame 3 900 I 3 6 19 22 16 K Regrind engine Valves 5 1000 I 3 8 17 22 14 L Replace oil pump and Carb 1 200 I, D 9 10 21 22 12 M Rechrome the chrome 3 210 E 1 4 35 38 34 N Reinstall Engine 1 200 K, L 10 11 22 23 12 O Put doors, hood, and trunk 1 240 H,J 8 9 22 23 14 P Rebuild transmission and replace brakes 4 2000 N,O 11 15 23 27 12 Q Replace windshield 1 100 C 12 13 27 28 15 R Put fenders back on 1 100 G, P 15 16 27 28 12 S Paint car 4 1700 R, Q 16 20 28 32 12 T Reupholster 7 2400 B, S 32 39 32 39 0 U Put Chrome back on 1 100 S, M 20 21 38 39 18 V Auto show 2 1000 U, T 39 41 39 41 0 TOTAL COST 18,100 APPENDIX 2 Read More
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