Engineering Asset Management - Assignment Example

Topic: Operations Management Name: Registration No: Institution: Course Code: Date Due: Question 1. a. Control Charts for the data in the initial study Sample No. Bar 1 Bar 2 Bar 3 Bar 4 Bar 5 Averages 1 167 159.6 161.6 164 165.3 163.5 2 156.2 159.5 161.7 164 165.3 161.34 3 167 162.9 162.9 164 165.4 164.44 4 167 159.6 163.7 164.1 165.4 163.96 5 156.3 160 162.9 164.1 165.5 161.76 6 164 164.2 163 164.2 163.9 163.86 7 161.3 163 164.2 157 160.6 161.22 8 163.1 164.2 156.9 160.1 163.1 161.48 9 164.3 157 161.2 163.2 164.4 162.02 10 156.9 161 163.2 164.3 157.3 160.54 11 161 163.3 164.4 157.6 160.6 161.38 12 163.3 164.5 158.4 160.1 163.3 161.92 13 158.2 161.3 163.5 164.6 158.7 161.26 14 161.5 163.5 164.7 158.6 162.5 162.16 15 163.6 164.8 158 162.4 163.6 162.48 16 164.5 158.5 160.3 163.4 164.6 162.26 17 164.9 157.9 162.3 163.7 165.1 162.78 18 155 162.2 163.7 164.8 159.6 161.06 19 162.1 163.9 165.1 159.3 162 162.48 20 165.2 159.1 161.6 163.9 165.2 163 21 164.9 165.1 159.9 162 163.7 163.12 22 167.6 165.6 165.6 156.7 165.7 164.24 23 167.7 165.8 165.9 156.9 165.9 164.44 24 166 166 165.6 165.6 165.5 165.74 25 163.7 163.7 165.6 165.6 166.2 164.96 x-bar 162.892 162.248 162.636 162.168 163.536 162.696 R 12.7 9 9 8.9 8.9 9.7 sd 3.7817456 2.686869 2.474685 2.925138 2.45117  2.8639 Max 167.7 166 165.9 165.6 166.2   Min 155 157 156.9 156.7 157.3   R-Chart Centerline RA = 9.7 Upper Control Limit =UCL =D4RA =2.282*9.7= 22.1354 Lower Control Limit =LCL= D3RA =0*9.7 = 0 x-bar chart Centerline = X bar =162.696 Upper Control Limit = X bar +A2R bar = 162.696 + 0.729*9.7 =162.696 + 7.0713 = 169.7673 Lower Control Limit = X bar – A2Rbar = 162.696 – 7.013 = 155.6647 I can be explained that the process was in control. This is because most values of the x-bar chart are within the control regions. Despite the existence of some sections of the R Chart out of control, there is a significant section of the chart under control that leads to the conclusion that the process was under control. b. Values of Cp and Cpk for the initial capability study USL = 170 LSL = 162 Mean = 166 SD = 5.657 Cp = (USL-LSL)/6*SD = (170-162)/(6*5.657) = 0.2357 Cpk = (Mean – LSL)/3*SD = (166-162)/(3*5.657) = 0.2356 The values of Cp and Cpk are considerably low, thus illustrating the process capability was also low. This implies that there was the need to improve the values of Cp and Cpk so that a higher process capability could be achieved. c. Values of Cp and Cpk for the improvement Cp = (USL-LSL)/6*SD = (169.7673-155.664)/6*2.8639 = 14.1033/ 17.1834 = 0.82075 Cpk value from Initial Values Cpk = (Mean – LSL)/3*SD = (162.696 -155.664)/(3*2.8639) = 7.032/8.5917 = 0.79804 d. The final values of Cp and Cpk when the process is centered at 166 grams When  N= 18 samples LSL = mean – R3σ = 163 – 2.326*1 = 160.674 USL = mean + R3σ = 163 + 2.326 = 165.326 Cp = (USL – LSL)/6*SD = (165.326 – 160.674)/6*1 = 0.7753 Cpk = (Mean – LSL)/3*SD = (163-160.674)/3*1= 0.7753 Comment: It is observed that the value of Cp and Cpk is lower in the process cantered at 166grams compared with the other processes. Question 2.TS Henderson a. Calculation of Cash flow Year Cost (£) Less 12% Discount Machine 1 Net Present Value of 1 Machine 2 Net Present Value of 2 0 -(500000)/(1+ 0.15)0 -500000 -(900000)/(1+ 0.25)0 -900000 1 1000000 880000 (880000-500000)/(1+0.15)1 330434.78 (880000-900000)/(1+0.25)1 -16000 2 1350000 1188000 (1188000-500000)/(1+0.15)2 520226.84 (1188000-900000)/(1+0.25)2 184320 3 1400000 1232000 (1232000-500000)/(1+0.15)3 481301.88 (1232000-900000)/(1+0.25)3 169984 4 1450000 1276000 (1276000-500000)/(1+0.15)4 443680.51 (1276000-900000)/(1+0.25)4 154009.6 5 2550000 2244000 (2244000-500000)/(1+0.15)5 867076.22 (2244000-900000)/(1+0.25)5 112721.92 2142720.23 -294964.48 Based on the above Net Present values, it is observed that Machine 1 has a NPV of £ 2142720.23 while Machine B, has a NPV of £ -294964.48. Thus it is recommended that Machine 1 should be purchased. b. Calculation of Cash flow when the discount rate is reduced to 8%. The cash flow will be as illustrated below. Year Cost (£) Less 8% Discount Machine 1 Net Present Value of 1 Machine 2 Net Present Value of 2 0 -(500000)/(1+ 0.15)0 -500000 -(900000)/(1+ 0.25)0 -900000 1 1000000 920000 (920000-500000)/(1+0.15)1 365217 (920000-900000)/(1+0.25)1 16000 2 1350000 1242000 (1242000-500000)/(1+0.15)2 561058 (1242000-900000)/(1+0.25)2 218880 3 1400000 1288000 (1288000-500000)/(1+0.15)3 518123 (1288000-900000)/(1+0.25)3 198656 4 1450000 1334000 (1334000-500000)/(1+0.15)4 476842 (1334000-900000)/(1+0.25)4 177766 5 2550000 2346000 (2346000-500000)/(1+0.15)5 917788 (2346000-900000)/(1+0.25)5 473825 2339028 185127 It is observed that when the discount rate is reduced to 8%, the NPV of Machine 1 is 2339028 compared to 185127 for Machine 2. Thus, the decision will still be made to acquire Machine 1. Question 3. Motec Data Acquisition a. Total payment for 10 workers in a week = 10* 25*18*6 = £ 27000 Overhead Costs per week = £ 35000 Total Fixed Costs per week = £ 62000 Considering Model A: No of units produced in a day = (18*60)/90 = 20 units No of units produced in a week = 20*6 = 120 units Costs of materials in a week = 50*120 = £ 6000 Income generated by the model in a week = 450*100 = £ 45000 Considering Model B: No of units produced in a day = (18*60)/90 = 20 units No of units produced in a week = 20*6 = 120 units Costs of materials in a week = 40*120 = £ 4800 Income generated by the model in a week = 400*75 = £ 30000 Considering Model C: No of units produced in a day = (18*60)/130 = 8 units No of units produced in a week = 8*6 = 48 units Costs of materials in a week = 110*48 = £ 528000 Income generated by the model in a week = 40*110 = £ 4400 b. Based on the above comparisons, it is observed that combinations of Model A and B generates the highest profit. This is obtained by: Total sales per week –Total Costs per week Total sales = £ 30000 + 45000 = £ 75000 Total costs per week = 62000 + 6000 +4800 = £ 72500 Profit = £ (75000 -72500) = £ 2500 Question 4. Antonio Meireles Co. - Value Stream Mapping a) Current inventory level of the cell In process 1, the Cycle time = 120 seconds, thus 400 units will take (400*120) seconds. This is equal to 800 minutes or 13 hours 20 minutes Time allocated = 7 hours, 15 minutes or 645 minutes The number of units to be cut in a particular day = 654*400/800 = 327 units This implies that only 327 units will be cut in a particular day. Thus the available units to be cut are more than the inventory level of the cell. The number of units that will be transferred to the next shift is 400-327 = 77 units. In process 2, the total time required to bend 500 units will be (500*250/60) minutes. This is equal to 2083 minutes The available time is 7 hours, 15 minutes which is equivalent to 645 minutes The number of units bent in a day = 645*500/2083 = 154 units. Since two operators are involved in the shift, the number of units produced is assumed to double, i.e 154*2 = 308 units. This implies that the maximum number of units that can be bent by the cell in a shift is 308, thus, the available number of units to be bent is higher than the shift capacity of the cell. The number of units that will need to be transferred to the next shift is 500- 308 = 192 units. Considering process 3, the available time is similar to other shifts, i.e 7 hours 15 minutes or 645 minutes. The time required for 200 units is 200*140/60 = 467 minutes. This implies that all the units in the punching process will be punched before the end of the shift. b) Takt time is defined as the rate at which a finished product needs to be completed so that customer demands can be met, due t high demand for the product when its production is completed. The takt time of this cell will be defined as the ratio of available time for production/required units of production In process 1: Cutting Process Takt time = (7 hours 15 minutes- 3 minutes)/ 400 units = 642 minutes/400 units = 1.605 minutes per unit In process 2: Bending process Takt time = 640 minutes/500 units = 1.284 minutes per unit In process 3: Punching process Takt time = 645 minutes/200 units = 3.225 minutes per unit c) Lead time refers to the delay applicable for inventory management purposes. It represents the sum of delay such as time when production has to take place plus the time that productivity has taken place. It is obtained by the formula: Lead time = Cycle Time * WIP In process 1: Cutting Process Lead Time = (120*400) = 4800 seconds or 800 minutes In process 2: Bending process Lead Time = (250*500) = 125000 seconds or 2083.3 minutes In process 3: Punching process Lead Time = (140*200) = 28000 seconds or 467 minutes d) Processing time is the time taken to produce a particular component. In this case, it includes that time taken during actual cutting, punching and bending activities. In the cutting process Processing time = available time- set up time –break time = 8 hrs – 3 minutes – 45 minutes = 7 hours 12 minutes In the bending process Processing time = 8 hours – 5 minutes – 45 minutes = 7 hours 10 minutes In the punching process Processing time = 8 hours – 0 minutes – 45 minutes = 7 hours 15 minutes. e) Capacity of the cell is the amount of products it can produce in a particular day. This is determined by the sum of the amount of the product cut, bent and punched. In this case, the daily capacity of the cell is 327 + 308 + 1000 units = 1635 units. Discussions There are many ways in which value stream mapping method can be applied in this problem. For instance, it can be used during the process of improving processes such as motions and flow of materials so that there is efficiency in the manner in which materials are handled (Hill 2012, p. 44). In addition, value stream mapping method enables interaction between a number of functions in a manufacturing process in addition to ancillary functions and will be important in managing of this cell by enabling planning, scheduling and management of materials so that wastes are eliminated. This will result into a saving in costs of performing tasks such as disposal of waste materials. In addition, it will enable the operator of the machine to be organized so that possibilities of making mistakes that can result into injury is eliminated. The benefits include highlighting areas where problems are likely to be experienced as well as areas where employees are likely to show inefficiencies within a complex system. The strategy integrates information and the flow of materials in addition to the sequence in which tasks are performed such as cycle times and lags in times (Mahadevan, 2010). In addition, the method has the advantage that it ensures the worker is able to identify restrictions, bottlenecks in addition to a number of factors which have an impact on the efficiency of the task performed. Another advantage of this method is that it ensures the operator of the cell implement countermeasures in a significantly visual manner so that culture change can be achieved in an organization. In addition, the property of implementing graphical depictions of limiting factors that can be understood by all stakeholders makes the practice most relevant in management of operations in a production department. Furthermore, value stream mapping method ensures direction is achieved for lean improving of teams, supervisors and the management by encouraging their participation towards improvement of processes in the organization. The limitations include the complexity of tasks that it incorporates by detailing the approach towards a task. This can prevent activity of the worker thus reducing productivity. Another limitation that is associated with value stream mapping method is that it is costly to use. This is because this method involves the risk of opportunity cost and it is usually not easy to quantify (Krajewski, Ritzman & Malhotra, 2013). Another limitation of value stream mapping is that it cannot be easily implemented to identify the areas where practices are not effective if the user does not conduct a detailed analysis of a particular situation. It also requires high commitment and willingness to change to the demands of the working environment which is not usually easy to adapt to. There are a number of improvements that have been suggested so that value stream method can be implemented effectively in this case. It has been suggested that during implementation of the method, the person involved should first learn, control the project in a cross-functional manner. This should be followed by evaluating and mapping simpler activities in processes in a particular area in an organization. It is expected that some early wins will be achieved. This should be followed by advancing the implementation of value-stream mapping such as during the process of addressing activities in cross-functional areas. It is also recommended that mapping process should be detailed so that obvious opportunities of improvement can be identified. When these recommendations are implemented, it is expected that even organizations that are using the method for the first time can be successful in their operational activities. References Hill, A. V. 2012. The encyclopedia of operations management: A field manual and glossary of operations management terms and concepts. Upper Saddle River, N.J: FT Press. Krajewski, L. J., Ritzman, L. P., & Malhotra, M. K. (2013). Operations management. Upper Saddle River, N.J: Pearson. Mahadevan, B. 2010. Operations management: Theory and practice. Upper Saddle River: Pearson. Mathew, J. 2006. Engineering Asset Management: Proceedings of the 1st World Congress on Engineering Asset Management (WCEAM) 11-14 July 2006. S.l.: Cieam/mesa. Read More