Operations Management - Assignment Example

Topic: Operations Management Name: Institution: Course Code Date: Introduction Operation management is an activity that has been considered important in management of resources of a business in a number of ways. This is because, effective operation management ensures a businessman invests in the right manner as well as manages both human and capital resources. This is because, the goal of any business undertaking is to ensure profit is generated and cost of production is kept as low as possible. While most businesses have implemented operation management strategies in a number of ways, some businesses have not been effective in the manner in which resources should be managed. This paper presents a course of actions that can be followed in order to manage resources and activities effectively (Krajewski, Ritzman & Malhotra, 2013). The main areas of focus include management of a process to ensure a process is in control. It also includes determination of various methods of making investment choices such as the use of Net Present Value (NPV) as well as determining the impact of value stream planning as a way of managing activities in an operational environment. Q 1: The Case of industrial Soap Company The following data has been used to compute the X-bar and R charts for this case Sample No. Bar 1 Bar 2 Bar 3 Bar 4 Bar 5 1 167.0 159.6 161.6 164.0 165.3 2 156.2 159.5 161.7 164.0 165.3 3 167.0 162.9 162.9 164.0 165.4 4 167.0 159.6 163.7 164.1 165.4 5 156.3 160.0 162.9 164.1 165.5 6 164.0 164.2 163.0 164.2 163.9 7 161.3 163.0 164.2 157.0 160.6 8 163.1 164.2 156.9 160.1 163.1 9 164.3 157.0 161.2 163.2 164.4 10 156.9 161.0 163.2 164.3 157.3 11 161.0 163.3 164.4 157.6 160.6 12 163.3 164.5 158.4 160.1 163.3 13 158.2 161.3 163.5 164.6 158.7 14 161.5 163.5 164.7 158.6 162.5 15 163.6 164.8 158.0 162.4 163.6 16 164.5 158.5 160.3 163.4 164.6 17 164.9 157.9 162.3 163.7 165.1 18 155.0 162.2 163.7 164.8 159.6 19 162.1 163.9 165.1 159.3 162.0 20 165.2 159.1 161.6 163.9 165.2 21 164.9 165.1 159.9 162.0 163.7 22 167.6 165.6 165.6 156.7 165.7 23 167.7 165.8 165.9 156.9 165.9 24 166.0 166.0 165.6 165.6 165.5 25 163.7 163.7 165.6 165.6 166.2 Ms Excel was used to process the above data so that the mean, variance, range and other statistical characteristics could be obtained. It was found that X-bar was 162.7 the mean bar value was 9.8. This is illustrated in the figure below. x-bar 162.9 162.2 162.6 162.2 163.5 162.7 R 12.7 9 9 8.9 8.9 9.8 Ms Excel was also used to compute the standard deviation for the data and it was found that SD =2.8 a. Computation of the R Chart Upper Controlled limit and lower control limits were as follows: UCL =D4R, According to R charts, D4 = 2.246 Hence, UCL = 9.8*2.246 = 22.018 Centerline = R = 9.8 Lower Control Limit LCL = D3R, According to R charts, D3 = 0, Hence, LCL = 9.8*0 = 0 The resulting R Chart. Computation of the X-bar Chart Centerline = X bar = 162.7 , UCL= Xbar + A2R = 162.7 + 0.734*9.7 = 169.82 LCL = X bar – A2R = 172.7 – 0.734*9.8 = 155.6. when X-bar Chart is computed, we obtain From the the Xbar and R charts, it is can be observed that the process was in a statistical control. This is because the values of the plots were within the control limits. b. Computation of the values of Cp and Cpk from the initial capability study. Process capability, Cp refers to process capability and establishes the ability of the project to achieve a particular output while it is being performed In the initial capability, Upper Control Limit = 170 Lower Control Limit = 162 From calculation, mean or X bar = 166 while Standard deviation = 5.68 Cp is computed by (USL-LSL)/6*SD = (170-162)/(6*5.68) = 0.2347 Cpk is determined by (Mean – LSL)/3*SD = (166-162)/(3*5.68) = 0.2347 Based on the outcome of Cp and Cpk as shown above, it is concluded that the process had a very low capability to achieve the objectives of producing soaps. Thus, there was the need to improve the process. c. Values of Cp and Cpk after improvements have been made will be as follows When improvements have been made, UCL = 169.8 and LCL = 155.6 and mean = 162.7 Cp = (UCL-LCL)/6*SD. Substituting the above values into the equation, we get Cp = (169.8-155.6)/6*2.8 = 0.8452 Cpk is obtained by (Mean – LSL)/3*SD = (162.7 -155.6)/(3*2.8) =0.8452 When the process is based on 166grams, respective values of Cp and Cpk were obtained as follows. The values of SD was 1 gram, X-bare was 163 grams and R-bar was 2.326 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 Interpretation of the results: It is found that the process centered values of Cp and Cpk are lower than the improvement process but higher than the initial process capability. This implies that the best strategy among these cases is the process improvement strategy. Question 2: TS Henderson & Company Ltd Company investment Case Study. When the company makes the option of either buying Machine 1 or Machine 2, the Net Present Value after 5 years will be obtained as follows: Machine 1 at an interest rate of 15% Year 0 Cost =0 NPV0 = - (500000)/ (1+ 0.15)0 = -500000 Year 1 Cost = 1000000, less 12% discount = 880000 NPV1= (880000-500000)/ (1+0.15)1 = 330434 Year 2 Cost = 1350000 less 12% discount =1188000 NPV2 = (1188000-500000)/ (1+0.15)2 = 520226 Year 3 Cost = 1400000 less 12% discount = 1232000 NPV3 = (1232000-500000)/ (1+0.15)3 = 481301 Year 4 Cost =1450000, less 12% discount = 1276000 NPV4= (1276000-500000)/(1+0.15)4 = 443680 Year 5 Cost = 2550000, less 12% discount = 2244000 NPV5 = (2244000-500000)/(1+0.15)5 = 867076 Total = NPV0 +NPV1 + NPV2 + NPV3 + NPV4 + NPV5 = -500000 + 330434 + 520226 + 481301 + 443680 + 867076 = 2142720 Machine 2 Year 0 Cost =0 NPVo = - (900000)/(1+ 0.25)0 = -900000 Year 1 Cost = 1000000 less 12% discount = 880000 NPV1 = - (900000)/ (1+ 0.25)0 = -16000 Year 2 Cost = 1350000, less 12% discount = 1188000 NPV2 = (1188000-900000)/(1+0.25)2 = 184320 Year 3 Cost = 1400000, less 12% discount = 1232000 NPV3 = (1232000-900000)/(1+0.25)3 = 169984 Year 4 Cost = 1450000, less 12% discount = 1276000 NPV4 = (1276000-900000)/(1+0.25)4 = 154009 Year 5 Cost = 2550000, less 12% discount = 2244000 NPV5 = (2244000-900000)/(1+0.25)5 = 112721 Total = NPV0 + NPV1 + NPV2 + NPV3 +NPV4 + NPV5 = -900000 -16000 + 184320 + 169984 + 154009 + 112721 = 294964 I would recommend that machine 1 should be bought because it has a high NPV after 5 years compared with machine 2. b. A case when the discount is reduced to 8%. In this case, considering: Machine 1 at an interest rate of 15% Year 0 Cost =0 NPV0 = - (500000)/ (1+ 0.15)0 = -500000 Year 1 Cost = 1000000, less 8% discount = 920000 NPV1= (920000-500000)/ (1+0.15)1 = 365217 Year 2 Cost = 1350000 less 8% discount = 1242000 NPV2 = (1242000-500000)/ (1+0.15)2 = 561058 Year 3 Cost = 1400000 less 8% discount = 1288000 NPV3 = (1288000-500000)/ (1+0.15)3 = 518123 Year 4 Cost =1450000, less 8% discount = 1334000 NPV4= (1334000-500000)/(1+0.15)4 = 476842 Year 5 Cost = 2550000, less 8% discount = 2346000 NPV5 = (2346000-500000)/(1+0.15)5 = 917788 Total = NPV0 +NPV1 + NPV2 + NPV3 + NPV4 + NPV5 = -500000 + 365217 + 561058 + 518123 + 476842 + 917788 = 2339028 The case of Machine 2at 25% savings Machine 2 Year 0 Cost =0 NPVo = - (900000)/ (1+ 0.25)0 = -900000 Year 1 Cost = 1000000 less 8% discount = 920000 NPV1 = (920000-900000)/(1+0.25)1 = 16000 Year 2 Cost =1350000 less 8% discount =1242000 NPV4 = (1242000-900000)/ (1+0.25)2 = 218880 Year 3 Cost = 1400000 less 8% discount = 1288000 NPV3 = (1288000-900000)/ (1+0.25)3 = 198656 Year 4 Cost =1450000, less 8% discount = 1334000 NPV4 = (1334000-900000)/ (1+0.25)4 = 177766 Year 5 Cost = 2550000, less 8% discount = 2346000 NPV4 = (2346000-900000)/ (1+0.25)5 = 473825 Total = Total = NPV0 +NPV1 + NPV2 + NPV3 + NPV4 + NPV5 = -900000 + 16000 + 218880 + 198656 + 177766 + 473825 = 185127 This option shows that Machine 1 will have a high NPV after 5 years compared with Machine 2. This implies that it would still be beneficial to buy Machine 1 rather than machine 2. Question 3: a. The following will be the earnings of 10 workers in a week = 10* 25*18*6 = £ 27000 The total overhead cost per week is = £ 35000 Fixed costs to be incurred each week = £ 62000 The case of Model A: Quantity of model A o be produced in a day = (18*60)/90 = 20 units Quantity of Model A produced in a week = 20*6 = 120 units Weekly material costs = 50*120 = £ 6000 Amount of earning from the sales of Model A in a week = 450*100 = £ 45000 The case of Model B: Quantity of Model B produced in a day = (18*60)/90 = 20 units Quantity of model B produced in a week = 20*6 = 120 Material costs during production of model B = 40*120 = £ 4800 Amount of income as a result of the sales of Model B in a week = 400*75 = £ 30000 The case of Model C: Amount of production of model C in a day = (18*60)/130 = 8 units Amount of production of Model C in a week = 8*6 = 48 units Material cost during production of Model C in a week = 110*48 = £ 528000 Amount of earning as a result of sales of Model C in a week = 40*110 = £ 4400 b. It is suggested that the production process should use combinations of Model A and B in order to generate high profits. This is because, this combination results into a higher income compared to any other combination. This is illustrated by: Weekly sales - Total Costs in each week Weekly Sales = £ 30000 + 45000 = £ 75000 Weekly costs= 62000 + 6000 +4800 = £ 72500 Profit generated each week = £ (75000 -72500) = £ 2500 Question 4:Value Stream Mapping at Antonio Meirels Co. a) Present inventory level of the cell In the cutting process, time taken to produce 400 units will be obtained by the product of the number of units and the cycle time. That is 120*400 = 48000 seconds or 800 minutes The available time during the cutting shift = 8 hour with 45 minutes of break. This results into available time of 7 hours 15 minutes, when the setup time of 15 minutes is subtracted, we obtain 7 hours 12 minutes or 642 minutes, thus, the number of units produced = 642*400/800 = 321 units. Thus, the maximum amount of units that the cell can cut in a particular shift = 321 units. This implies that the number of units that will be transferred to the next shift are 400-321 = 79 units. In process 2 which is a bending process, the number of units that need to be bent are 500 while the cycle time is 250 seconds. This implies that the total time required to bend all the units is (500*250)/60 = 2083 minutes. The available time for production of the units in a single shift = 7 hours 15 minutes – set up time of 5 minutes = 7 hours 10 minutes or 640 minutes. The number of units bent in each shift = 640*500/2083 = 153 units. Sine there will be 2 people during the shift. The total number of units produced = 306 units. In process 3 which involves punching, the time available will be 7 hours 15 minutes or 645 minutes. In order to produce 200 units, the time required will be 200*140/60 = 467 minutes. Thus, all the units will be punched before the end of the available time. b) Takt time is the rate at which a worker completes a particular task during a shift to ensure products are completed so that customer demands can be met. Takt time for each process can be found as follows: Process 1: Cutting process Takt time = (7 hours 15 minutes – 3 minutes)/ (400 units = 1.6 minutes per The takt time for process 2: bending will be 640/500 = 1.28 The takt time for process 3 will be 645 minutes/200 units = 3.225 minutes per unit c) Lead time is the time when there is a delay during management of inventory. It is found by adding delay time and the productivity time. It can also be obtained by use of the formula: lead Time = Cycle Time * WIP The lead time for Process 1= (120*400) = 800 minutes Lead time for process 2: bending process = (250*500)/60 = 2083 minutes. Lead time for process 3: Punching = (140*200) = 28000 seconds or 467 units d) Processing time is the time taken to produce a particular component so that it can be bought by customers. During process 1, the processing time = 8 hours -45 minutes break -3 minutes for set up = 7 hours 12 minutes. During process 2, the processing time = 8 hours – 45 minutes break – 5 minutes et up time = 7 hours 10 minutes. In the punching process, processing time = 8 hours – 45 minutes = 7 hours 15 minutes e) Cell capacity refers to the amount of products it can produce to completion without being subjected to strenuous working conditions. The capacity of the cell in this case is = 327+ 306 +1000 = 1633 units Application of Streamline Mapping to this problem There are various ways in which streamline mapping can be applied to this process. It can be applied during the process of scheduling the tasks to be performed in the cell such as the sequence in which tasks will be performed and the tools and set ups that will need to be followed so that the a particular work being performed in the cell is successfully completed. Furthermore, streamline mapping method will be used in assessing areas where the worker is inefficient so that the person involved can take measures to acquire competent skills applicable in a particular situation. The benefits of applying streamline mapping method in performing operation tasks in a production department is that it ensures a person is organized and performs work in an organized manner so that errors and accidents are prevented (Hill, 2012). It is also a method that contributes to improved output when properly implemented. There are a number of limitations associated with application of streamline mapping method such as its detailed approach to tasks that makes some tasks not to be completed. It is also associated with high cost of implementation and cannot be effectively implemented when an employee is not provided with the right guidance before the employee can learn it effectively. This has resulted into a number of recommendations in the use of streamline mapping methods. For instance, it is suggested that during the process of learning streamline mapping method, it should involve cross-functional approach. This is where the leaner is taught a number of skills while it is being implemented. It is also suggested that streamline mapping method can be understood better by starting with simpler tasks after which a person can implement it in more complex tasks. 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. Read More