Reliability of the Computer-Driven Shoe Machines Process in the Shuzworld Shanghai Plant - Term Paper Example

 Task 1: Transportation model Reduction of transportation costs is key to cost management within the organization (Martin, 2009, p. 79). The model provided assesses the best options that Shuz World can use in its warehousing in order to minimize costs associated with transportation. According to the findings, the following table offers the best model that will ensure lowest cost in a feasible manner. The cost of transportation based on this model will amount to $ 13, 400. DEMAND 2500 1500 1800 Warehouse 1 Warehouse 2 Warehouse 3 CAPACITY Shanghai $ 4.00 $ 3.00 $ 3.00 2800 Shuzworld H $ 3.00 $ 4.00 $ 2.00 2300 Shuzworld F $ 2.00 $ 4.00 $ 6.00 2200 Shipments Warehouse 1 Warehouse 2 Warehouse 3 Shanghai 0 1500 0 Shuzworld H 300 0 1800 Shuzworld F 2200 0 0 Received Demand 2500 1500 1800 Total Cost $ 13,400.00 The distribution system is based on the principle of allocating quotas to the option that offers the least possible costs for as long as it is possible. Once the option offering the least costs has been fully exploited, the remaining allocation is transferred to the next option that offers least cost amongst the remaining options. The reason this tool is chosen is because it is simple to understand and use in addition to allow the user the freedom of playing around with allocations as appropriate. Task b: Improving reliability as well as maintenance of the three machines required for production of deck shoes no doubt requires critical analysis.   The three machines need to work alongside each other in order to ensure high reliability standards, failure of a single machine results into a decline in productivity. The whole systems reliability is obtained by the equation (Rx) = (R1) x (R2) x (R3) where R1, R2, and R3 represent the reliabilities of machine 1, 2, and 3.   The overall past reliability figures are therefore given as .84 *. 91* .99. Each machine has individual tasks which vary from each other’s; nonetheless, it requires all the three machines working appropriately and effectively so as to produce the finished products. Each machine’s reliability is unique to the respective machine and has no effect on the reliability of the others.   To raise reliability levels, the company might incorporate the idea of investing in a replacement/back up machine for machine 1 to raise the system’s reliability rating.   This has the potential of raising the overall reliability of the whole system. Redundancy can also increase the process of raising reliability (Martin, 2009, p. 87). This involves backing up of the existing components additional components.   To be able to determine the system reliability in instances where redundancy is applicable, the general network tool is applicable.  If a single machine fails, production duration is not lost as a result of the presence of back-up machine.   Reliability via a parallel process results into redundancy of the two less reliable machines, which is 1 and 2. In undertaking this, the system reliability is raised from .757 to .957.   The general network tool is chosen due to component availability as well as the desire to raise reliability of the system. Task C: Queuing system Queuing system is important in ensuring maximum staff utilization as well client satisfaction (Resing & Adan, 2001). It involves creation of a balance between client satisfaction and staff utilization. The tables below shows the results of using one server compared to two servers. Single server The table below shows the result for the single server system. Basic Inputs: Number of Servers, S = 1 Queue Capacity, M = 10 Arrival Rate,  = 10 Service Rate Capacity of each server,  = 5 Arrivals: Average Rate Joining System (Lambda-Bar) = 4.99878 Average Rate Leaving Without Service (balking) = 5.00122 Customers who Balk: Probability that System is Full (Pfull) = 50.01% The Waiting Line: Average Number Waiting in Queue (Nq) = 9.003 Average Waiting Time (Tq) = 1.80107 Q: Probability of more than 0 customers waiting = 99.93% Service: Average Utilization of Servers = 99.98% Average Number of Customers Being Served (Ns) = 0.99976 The Total System (waiting line plus customers being served): Average Number in the System (N) = 10.003 Average Time in System (Tq + Ts) = 2.00107 Probability Distribution: n = total number of customers in the system q = number of customers in the waiting line n P(n) Cumulative q P(q) Cumulative 0 0.0002 0.0002 1 0.0005 0.0007 0 0.0007 0.0007 2 0.0010 0.0017 1 0.0010 0.0017 3 0.0020 0.0037 2 0.0020 0.0036 4 0.0039 0.0076 3 0.0039 0.0075 5 0.0078 0.0154 4 0.0078 0.0154 6 0.0156 0.0310 5 0.0156 0.0310 7 0.0313 0.0623 6 0.0313 0.0622 8 0.0625 0.1248 7 0.0625 0.1248 9 0.1250 0.2498 8 0.1250 0.2498 10 0.2501 0.4999 9 0.2501 0.4998 11 0.5001 1.0000 10 0.5001 1.0000 The information in the table indicates that on average ten (10) customers will be in the system. The average time that a customer will spend in the system is 2 minutes and on average nine (9) customers will be in the waiting line. On average customers will wait in the line for 1.8 minutes. The probability of no one being in the line and being served is 0.07%. Two server system Basic Inputs: Number of Servers, S = 2 Queue Capacity, M = 10 Arrival Rate,  = 10 Service Rate Capacity of each server,  = 5 Arrivals: Average Rate Joining System (Lambda-Bar) = 9.2 Average Rate Leaving Without Service (balking) = 0.8 Customers who Balk: Probability that System is Full (Pfull) = 8.00% The Waiting Line: Average Number Waiting in Queue (Nq) = 4.400 Average Waiting Time (Tq) = 0.47826 Q: Probability of more than 0 customers waiting = 80% Service: Average Utilization of Servers = 92.00% Average Number of Customers Being Served (Ns) = 1.84 The Total System (waiting line plus customers being served): Average Number in the System (N) = 6.240 Average Time in System (Tq + Ts) = 0.67826 Probability Distribution: n = total number of customers in the system q = number of customers in the waiting line n P(n) Cumulative q P(q) Cumulative 0 0.0400 0.0400 1 0.0800 0.1200 2 0.0800 0.2000 0 0.2000 0.2000 3 0.0800 0.2800 1 0.0800 0.2800 4 0.0800 0.3600 2 0.0800 0.3600 5 0.0800 0.4400 3 0.0800 0.4400 6 0.0800 0.5200 4 0.0800 0.5200 7 0.0800 0.6000 5 0.0800 0.6000 8 0.0800 0.6800 6 0.0800 0.6800 9 0.0800 0.7600 7 0.0800 0.7600 10 0.0800 0.8400 8 0.0800 0.8400 11 0.0800 0.9200 9 0.0800 0.9200 12 0.0800 1.0000 10 0.0800 1.0000 The table above indicates that when the number of servers is increased to two (2) on average the number of customers in the system will decline from ten customers to six (6.24) customers. The average time that a customer will spend in the system will also decline from 2 minutes to 0.68 minutes and on average the number of customers in the waiting line will decline from nine to four (4.4). The average time that customers wait in the line will also decline from 1.8 minutes to 0.48 minutes. The probability of no one being in the line and being served will on the other hand increase from a marginal 0.07% to 20% which suggests that both servers will not be occupied 20% of the time. The graph below shows additional information. According to Stevenson (2012) waiting lines have several implications – cost of waiting space, the possibility of losing business as well as goodwill; and the possibility of a reduction in customer satisfaction. The results indicate that there is a slim chance that there will be no customer waiting while in the two server system no customer will be in the line 20% of the time. It therefore means that the servers may be underutilized. The decision on whether to employ one or two servers will depend on the response received from customers and careful consideration of the implications of waiting lines. However, the fact that when the two server model is employed there is the possibility of a 20% underutilization of staff; it means that there will be capacity for an increase in customers. Additionally, if the arrival rate of customers’ increase then the utilization of the servers will be increased. Furthermore, using the staff at or close to 100% may result in errors which can cause the company as much as or even more than poor customer service. In fact, it can lead to other customer service issues. Therefore, the foregoing suggests that a two server system would be more beneficial to the company in the long run. Task 4: Ordering system More often than not, inappropriate ordering decisions cost organizations lots of money which would have otherwise been avoided if an alternative approach had been used (Howard, 2002). In order to make a determination of the optimal number of units to order at any point in time, the economic order quantity (EOQ) model is very useful. This model is used to determine the will minimize the costs that are associated with the purchase of inventory (Stevenson 2012). These costs are holding costs and ordering costs. The formula for calculating the optimum or economic order quantity which is shown below takes these two costs into consideration. Qo = √2DS/H Where, D = Demand in units per year; S = Ordering costs per order; and H = holding cost per unit per year. Shuzworld is interested in finding out the optimum number of pairs of shoelaces to order at any one time. The company’s annual demand for shoelaces is 300,000 pairs, order cost is $125 per order and the cost of holding each pair in inventory for a year is 10 cents. When these figures are substituted in the model the result is: Qo = √(2 x 300,000 x 125)/0.1 = √750,000,000 = 27,386 The result indicates that 27,386 pairs of shoe laces should be ordered at any one time in order to minimize costs. This suggests a total of 11 orders per annum. This model assumes that demand is evenly spread out throughout the year; that lead time is known and remains the same throughout the year, that each order involves a single delivery; and that no quantity discounts are received (Stevenson 2012). Generally, the decision tools used in this paper aim to ensure that the most appropriate and feasible decisions are adopted and hence the operations at Shuzworld are maximized while at the same time the cost of production is not unnecessarily inflated. The model is chosen due to its ability to provide information that will facilitate a reduction in the annual costs of holding and ordering inventory. References Howard, R. A. (2002). The Evolution of Decision Analysis,’’ in The Principles and Applications of Decision Analysis, ed. R. A. Howard and J. E. Matheson, SRI and Strategic Decisions Group (1983). Martin, H. (2009). Systems modeling and simulation. Decision Analysis Journal, 12 (3), p. 423-446. Resing, J & Adan, I. (2001). Queuing Theory. Eindhoven University: Department of mathematics and computing. Stevenson, W.J. (2012). Operations Management. 11th ed. New York, NY: McGraw Hill/Irwin Read More