Significance of Operations Management - Math Problem Example

Project management Name of the student Name of the Institution Date of submission Question one (a) Productivity = {Output of a specific Product}/ {Input of a specific Resource} Total jackets produced = 100 Good quality product = 80%*100 = 80 jackets per month Hours of input = 11 hours per jacket 100 jackets = (11*100) = 1100hours Productivity = (80*450)/1100hours = 32.73$/hr Multifactor productivity ratio= [Quantity at standard cost]/ [Labour cost + Materials cost + Overhead cost] Overhead cost = $ 6000 Material cost = (150*100) = $ 15000 Labour cost = (1100*9) = $9900 Standard cost =36,000 / (6000 + 15000 +9900) Multifactor productivity ratio = 36,000 /30900 Multifactor productivity ratio = 1.17 Operations resources consist of what we term the five Ps; People, plants, parts, processes, planning & Control. İn order to establish the productivity of a given process, all the fives p’s must be considered and the higher the productivity ratio the better the process. From the process, the company output is 32.73 dollars per hour meaning within one hour company is able to produce an output of 32.73. using this figure, the management is able to manupulate the various factors in order to improve the productivity of its workforce at any given time. Using mltifactor productivity ratio of 1.17, he is able to validate ist cause and influence the output either by manipulating the labor cost or sales price per unit. (b) After some study, Nick is considering three options to improve the multifactor productivity ratio (i) Increasing sales by 10% = (36000*110%) = 39600 Multifactor productivity ratio = 39600/30900 = 1.28 (ii) Improve quality so that only 15% are defective That means 85% will be good Sales = (85*450) = $ 38,250 Multifactor productivity ratio =$ 38,250/30900 = 1.24 (iii) Reduce labour, material, and overhead costs by 10%. Total cost = 30900 10% reduction = 27,810 Multifactor productivity ratio =36,000/27,810 = 1.29 Using the first multifactor productivity ratio as the base, when sales are increased by 10%, the percentage change in multifactor productivity will be given by (1.28-1.17)/ 1.17 = 9.4% increase in the multi-productivity factor. If the management decides to improve production efficiencies to 85%, then the multifactor productivity will increase by (1.24 -1.17)/1.17 = 5.98% increase. And lastly if the management decides to reduce the labour cost by 10% then the multifactor productivity will increase by (1.29-1.17)/1.17 = 10.26%. Therefore, if the management decides to reduce the production cost then it will have the greatest impact on the multifactor productivity ratio (Duncan 2006). (c) Significance of Operations Management to this (or any) business Operation management in most cases is considered as a multi-disciplinary filed that focuses mostly in all aspects of the organization operations. It involves overseeing, designing, controlling the production process and designing of various business operations to aid in the production of goods and services. Operation management helps in efficient planning and forecasting of production process. Running an efficient production process requires a great deal of planning and includes long range of decision making process and all decisions are aimed at meeting customer need. This can only be achieved through efficient planning which is a function of operation management. Operation management also help in managing the system, it involve working with people to encourage them to improve organization performance. For instance, in the example above, the management gives specific target to each employee and throughout they are encouraged to work hard and meet their targets. Lastly operation management helps in understanding business operations and helps in contributing towards organization success. Question two From the case study, the Roberts Auto Sales and Services RASAS operates four major divisions of their company which initially started as a Ford dealership that owner, Vicky Roberts inherited from her father and decided to expand the business. The three business sections include; Two auto parts stores Three car dealership that specialize in the sales and services of American and Japanese cars Body shop that deals with painting and does other bodywork Auto salvage yard These are the Competitive Priorities aspects and frontiers of the business. Robert has purchased a 1965 Shelby Mustang that she is tasking her director of service operations to assemble a restoration schedule and budget that would bring the total cost of acquisition and restoration to no more than $70,000 without spending more than $3600 per week and in order for it to be completed it should take 45 days. Activity Description time Cost Immed.Pred Ers.start Ers Finish Latest Start Latest Finish time Slack variables A order all needed materials 2 $ 100   0 2 0 2 0 B Receive uphostry 30 $ 2,100 A 2 32 2 32 0 C Receive windshield 10 $ 800 A 2 12 17 27 15 D Receive carburetor and oil pump 7 $ 1,750 A 2 9 14 21 12 E Remove chrome 1 $ 200   0 1 15 16 14 F Remove body from frame 1 $ 300 E 1 2 23 27 14 G Repair fenders 4 $ 1,000 F 2 6 16 22 21 H Repair doors, trunk and hood 6 $ 1,500 F 2 8 16 17 14 I Pull engine from chassis 1 $ 200 F 2 3 19 22 14 J Remove rust from frame 3 $ 900 I 3 6 17 22 16 K Regrind engine valves 5 $ 1,000 I 3 8 21 22 14 L Replace carb and oil pump 1 $ 200 I,D 9 10 35 38 12 M Rechrome the chrome 3 $ 210 E 1 4 22 23 34 N Reinstall engine 1 $ 200 K,L 10 11 22 23 12 O Put doors, hood, trunk on 1 $ 240 H,J 8 9 23 14 14 P Rebuild transmission 4 $ 2,000 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 $ 1,700 R,Q 16 20 28 32 12 T Reupholster 7 $ 2,400 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 $ 1,000 U,T 39 41 39 41 0     Total cost $ 18,100.00             Based upon the scheduled activities, we have identified the following Earliest Start, Earliest Finish, Latest Start and Latest Finish times for each activity as well as identified their preceding activities. Initial cost calculations reveal that the restoration process will cost approximately $18,100 including transportation of the mustang to the Detroit auto show, well within her budget of $70,000 including cost of acquisition. Based upon the network diagram, restoration would be completed within 39 (Earliest Finish) or 41 (Latest Finish) days. The Critical Path of A-B-T-V has been identified as the longest path with no slack. In terms of the restoration and delivery of the Shelby Mustang to the Detroit Auto show, RASAS Director of Service Operations and his team can assuredly meet the deadline within the projected budget and cost constraints. However, in terms of the feasibility of launching a vintage auto restoration line of business, there would be a significant shift in operations in terms of developing new timetables for her employees for restoring vintage parts and cars. Based upon the network diagram, the majority of the activities were dependent upon the sourcing and delivery of parts. Sourcing NOS parts, therefore, is a key component in the ability of RASAS to restore vintage cars in a timely manner as well as selling these parts to DIY vintage car enthusiasts. Mint restoration is risky and not as feasible given the uncertainty in the supply chain of NOS parts. While RASAS may be a first mover in the market for vintage restoration, they cannot guarantee the timely delivery of NOS parts. Therefore, RASAS jeopardizes their competitive priorities of quality and customer service (Chapman & Ward 2006). If RASAS only provides exterior and cosmetic restoration through their body and paint shop, which is feasible given the existing facilities and employee skill set, they can avoid an unstable supply chain. However, the price of the finished product will drop dramatically, which brings into the question of the cost benefit to expanding into the vintage restoration market. Question three Activity Predecessor Normal Time Normal cost Crash time Crash Cost NT NC CT CC Weeks ($, 000) (Weeks) ($, 000) A 5 2 3 4 B 8 5 7 8 C A 10 10 8 12 D A,B 4 3 3 7 E B 3 4 2 5 F D 9 8 6 14 G E,F 2 2 2 2 H G 8 6 5 9 I C,F 9 7 7 15 We need to first identify critical path 10 9 5 4 8 4 9 9 3 2 2 8 A-C-I = 24 A-D-F-I = 27 A-D-F-G-H = 28 B-D-F-I = 30 B-D-F-G-H = 31 B-E-G-H = 21 From the analysis, the critical path is B-D-F-G-H= 31 To shorten a project, crash only activities that are critical Crash from least expensive to most expensive From B-D-F-G-H; G is the least or the cheapest hence we crash it maximum days is 2 so if we crash it by 1crashing cost in this case is 2000 Critical path remain B-D-F-G-H; and If we crash D, by 2, the number of weeks will reduces to 28, and the crashing cost will be 10,000 Still B-D-F-G-H is the critical path, we crash B by 3, will have a total of 25 and the crashing cost will be 3*8000= 24,000. The total crashing cost = 2000 + 10,000+ 24,000= $ 36,000 Activity Predecessor Normal Time Normal cost Crash time Crash Cost Max Cost crash cost Cost to crash NT NC CT CC Red per period ($ 000) Weeks ($, 000) (Weeks) ($, 000) A 5 2 3 4 2 2 1.0 B 8 5 7 8 1 3 0.3 C A 10 10 8 12 2 2 1.0 D A,B 4 3 3 7 1 4 0.3 E B 3 4 2 5 1 1 1.0 F D 9 8 6 14 3 6 0.5 G E,F 2 2 2 2 0 0 0.0 H G 8 6 5 9 3 3 1.0 I C,F 9 7 7 15 2 8 0.3 The above table clearly shows the crash cost and savings which can be done if the critical path above is followed. I recommend crashing the critical path From B-D-F-G-H; G is the least or the cheapest hence we crash it maximum days is 2 so if we crash it by 1crashing cost in this case is 2000. Critical path remain B-D-F-G-H; and if we crash D, by 2, the number of weeks will reduces to 28, and the crashing cost will be 10,000. Still B-D-F-G-H is the critical path, we crash B by 3, will have a total of 25 and the crashing cost will be 3*8000= 24,000. The total crashing cost = 2000 + 10,000+ 24,000= $ 36,000 In crashing project, the project completion times may need to be shortened because: Different deadlines Penalty clauses Need to put resources on a new project Promised completion dates This is only made possible due to crashing hence the benefits of crashing project. On the other hand, crashing have several limitations including high cost which sometimes may be associated ith crashing of the projects. Question four Weighted-distance method: A mathematical model used to evaluate flexible-flow layouts based on proximity factors. Euclidean distance is the straight-line distance, or shortest possible path, between two points. Rectilinear distance: The distance between two points with a series of 90 degree turns, as along city blocks. dAB = √ (24-9)2 + (27-16)2 = √225+121 √346 = 18.60 = (30-23) + (12-8) = 7+4 = 11 From the results, we find different due to excessive relocation costs, Shipping and Receiving must remain in its current location. This can be used by management to help in making decision Reference Chapman, C. and Ward, S., 2006. Project risk management: processes, techniques and insights. John Wiley. Duncan, W. R. 2006 A guides to the project management body of knowledge Appendices Read More