Operations Management in Shuzworld - Case Study Example

(Boss) Re: Shuzworld case study analysis and recommended solutions A. Develop a distribution pattern that meets availability and demand constraints and minimizes total shipping costs for Shuzworld The current distribution plan developed should have a Shangai shipping 1300 units to the second warehouse, Shuzworld H shipping units to the first warehouse, 200 units to warehouse 2 , 1800 units to the third warehouse and Shuzworld F shipping 2200 units to the first warehouse. The transportation plan results into an optimal transportation total cost of USD13600. The updated production plan leads to an upsurge in the production of the plant in Shanghai from 1300 to 2800 units which is quite advantageous given that it leads to an improvement in the controlling the cost of transportation of USD 13,400 translating into USD200 savings. This result from the ability of the company to ship the additional 200 units from the Shangai plant to the second warehouse at USD 3 as opposed to shipping the units from Shuzworld H at USD4. The new updated transportation schedule states that Shangai is shipping a total of 1500 units to the second warehouse, Shuzworld H shipping 1800 units to the third warehouse and 300 units to the first warehouse and Shuzworld F shipping 2200 units to the first warehouse. The plan d leads to 1500 units of the excess capacity with Shangai being able to produce 1300 excess of the units and Shuzworld H being able to produce extra 200 units. Copy of output The original shipping plan | Warehouse 1 | Warehouse2 | Warehouse3 | Supply | Shanghai | 4 | 3 | 3 | 1300 | Shuzworld H | 3 | 4 | 2 | 1500 | Shuzworld F | 2 | 4 | 6 | 1800 | Demand | 2500 | 1500 | 1800 |   |   | | | |   | Optimal cost = $13,600 | Warehouse1 | Warehouse2 | Warehouse3 |   | Shanghai |   | 1300 |   |   | Shuzworld H | 300 | 200 | 1800 |   | Shuzworld F | 2200 |   |   |   | New shipping plan   | Warehouse1 | Warehouse2 | Warehouse3 | Supply | Shanghai | 4 | 3 | 3 | 2800 | Shuzworld H | 3 | 4 | 2 | 1500 | Shuzworld F | 2 | 4 | 6 | 1800 | Demand | 2500 | 1500 | 1800 |   |   | | | |   |   | | | |   | Optimal cost = $13,400 | Warehouse1 | Warehouse2 | Warehouse3 | Excess Production | Shanghai |   | 1500 |   | 1300 | Shuzworld H | 300 |   | 1800 | 200 | Shuzworld F | 2200 |   |   |   | Explain why you chose the decision analysis tool you used. The transportation tool is one of the linear programming models that is used in the calculation of the lease cost means of shipping suppliers from different origins and destinations. It is necessary when calculating the optimal shipping schedule and can be used in the calculation of the minimization and maximization problems and also in finding the best distribution of inventory. (Heizer & Render 2010) The tool is important for the Shuzworld given that it enables the senior management to adopt the efficient production from the current warehouse facilities in terms of shipping cost. The application of the data obtained can help the management reduce the costs and increase their profits. B. Analyze the reliability of the computer-driven shoe machines process in the Shuzworld Shanghai plant 1. Recommend ways to increase the reliability of the system, using an appropriate decision analysis tool. From this recommendations perspective, reliability refers to the probability that a machine can properly function for a specific period of time. It can be improved through the improvement of specific individual components or by the provision of redundancy. The reliability of the three step process is as below R1 x R2 x R3 = Rs whereby Rn is the reliability of the component and Rs is the reliability of the process. In the Shuzworld’s case; R1 =0.84, R2 = 0.91 and R3 =0.99 Therefore Rs = 0.84 x 0.91 x 0.99 = 0.757 or 75.7% Therefore the current system reliability is 75.7%. The process that has low reliability should be backed up and in the case the first machine with 84% should be backed. To calculate this process the calculations which are required includes; The probability of the second machine working, probability of needing the second machine, and probability of the first machine working Therefore in our case; (0.84) + ({0.84} x {1 - 0.84}) = 0.84 +0.1344 = 0.9744 Therefore the new reliability for the combination is 0.9744. The reliability equation is R1 x R2 x R3 = Rs Therefore Rs1 = 0.97 x 0.91 x 0.99 = 0.873 If machine 2 is backed then similar calculation applies R1 = 0.84, R2 = 0.91, and R3 = 0.99 R2 Redundant = (0.91) + ({0.91} x {1 - 0.91}) = 0.99 The new reliability of the system with the redundant machine 2 = Rs2= 0.84 x .99 x 0.99 = 0.823 If machine 3 is backed then R1 = 0.84, R2 = 0.91, and R3 = 0.99 R3 redundant = (0.99) + ({0.99} x {1 - 0.99}) = 1.0 The new machine reliability with redundant machine 2 is Rs3= 0.84 x 0.91 x 1.0 = 0.764 or 76.4% The data indicates an upsurge in the reliability that is gained through backing up the process which is least reliable, in this scenario, the first machine. This paper recommends that the first machine should be backed as soon as possible. The output is as indicated in the above calculations Explain why the decision analysis tool you chose is the best fit for this problem The paper picked the reliability decision tool given that the use of the tool is centered on redundancy. Redundancy is the application of components in a parallel manner to enhance reliability. The tool makes it possible for the calculation of the new probability of a system to function. It makes it possible for the management to measure the combined systems functioning probability. It also allows the calculation of the new probability of the functioning of the new system with one or more redundant components. In Shuzworld, it enables the management to calculate the effect of backing any of their production machines besides assisting in enhancing production reliability. Provide the optimum number of shoelaces to order for the Shuzworld Factory, considering appropriate cost balancing, using an appropriate decision analysis tool, The optimum number of the Shuzworld is 27,387. This is also the maximum number of shoe laces that should always be in the inventory at any particular time or maximum optimal inventory number. Averagely, there should be 13,694 within the inventory at any given time. To ensure the achievement of the inventory numbers, there should be 10.94 orders made annually. The annual setup and the associated holding costs should be $1369.31 and $1369.31 respectively. The entire costs on annual basis should be $2738.61. Explain how an economic order quantity (EOQ) amount relates to the problem. The aim is to get the total units that will ensure there is the minimization of the total cost connected with the purchase, delivery and storage. When optimizing the inventory, it is usually a common practice to utilize inventory control techniques such as the quantity discount model, production order quantity or the basic economic order quantity. Neither QD nor POQ is applicable for the Schuzworld case though EOQ fist it well. The model’s main purpose is to ensure the minimization of the total inventory holding costs and ordering costs. (Heizer & Render 2010) When the economic order quantity model is used as the decision tool, it is shown that the optimal order quantity will stand at a total of 27,386 pairs of the shoelaces. This is the amount that must be ordered if they do not want any shortage. This should be placed roughly 11 times per annum. The yearly holding cost and set up for the entire laces is a total of $1,366.31 at a rate of $10 per lace with the average inventory being 13,693 laces. Explain why you chose the decision analysis tool you used. It makes it possible for the user to see the determination of the optimal order quantity besides the optimal number of orders annually, total holding costs and total setup costs for the inventory item. The tool is able to perform inventory calculations using POQ, ABC and, Quantity Discount and EOQ models. In the case, it was possible to calculate the optimal number of the shoelaces to be ordered that will see holding and setup costs as low as possible and ensuring that the demand is met. Compare the one-cashier and two-cashier waiting line systems In the single cashier system, the average number of customers in the system in 1 while in the multiple cashier model with two cashiers, the number reduces to 0.53 customers. What is the average time a customer will spend in the system (waiting in line and being served)? In the single model, the data indicates that the average time is 10 and when the multiple cashier model is used, the number reduces to 5.33 minutes. How many customers will be in line on average? There will be 0.5 customers for the single cashier at anytime while in the multiple cashier model, the number reduces to 0.03. How long will customers wait in line on average? The data calculated indicate that for the single cashier model, the average waiting time will be 5 minutes while for the multiple cashier model, the waiting time will fall to 20 seconds. What is the probability of no one being in line or being served? The probability of no one being on the line is 0.5 for the single cashier systems while the probability of no one being in the line will be 0.6 when second cashier is added to the system. Recommend a one-cashier or two-cashier waiting line system, using an appropriate decision analysis tool. The best action is to ensure the deployment of a two cash system waiting line in the system. As shown above, the additional cashier reduces the waiting time for the customer from 5 minutes to 20 seconds. Given the objectives as explained, this is the best option for the business. Two cashier waiting line Explain why the decision analysis tool you chose is the best fit for this problem. The waiting line decision analysis tools makes it possible for the management and users to estimate the performance of the system by projecting the average number of customers in the system to be served, the average system utilization, average time spent by the customers in the system, average number o customers in the system,, probability of n customers being in the system and the average time the customer waits in the line. (Heizer & Render 2010) The main benefit of the tool is the fact that the output is that it gives the management with the necessary information to decide whether the changes are required. The tool can be used in changing the performance of the waiting line system by changing; arrival rate of the customers, efficiency of the server, number of the service facilities, priority rule and the total number of lines within the system. The Shuzworld data can be used to modify system to ensure better performance. Reference Heizer, J. & Render, B. (2010). Operations Management (10th Ed). New Jersey: Pearson. Read More