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Optimal Distribution Pattern That Minimizes Shipping Costs - Essay Example

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As the paper "Optimal Distribution Pattern That Minimizes Shipping Costs" tells, transportation costs fall to $13,400 in the process. That translates to a $200 savings on the transportation cost, as more products from Shanghai can be transported to the warehouses at a lower shipping cost…
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Optimal Distribution Pattern That Minimizes Shipping Costs
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Optimal Distribution Pattern that Minimizes Shipping Costs The current distribution plan has an optimal shipping cost of $13,600. In the plan, Shangai ships 1300 units to warehouse 2. Shuzworld H ships 100 units to the first warehouse, 200 units to the second warehouse and 1800 units to the third warehouse. Shuzworld F ships 2200 units to warehouse 1. The new production plan causes an increase in production at the plant in Shanghai from 1300 to 2800 units. Transportation costs fall to $13,400 in the process. That translates to a $200 savings on the transportation cost, as more products from Shanghai can be transported to the warehouses at a lower shipping cost. The company is able to ship the additional 200 units from Shangai to warehouse 2 at $3. That would cost $4 if the units were to originate from Shuzworld H. under the new transportation schedule, Shangai ships 1500 units to warehouse 2, Shuzworld H ships 1800 units to the third warehouse and 300 units to the first warehouse while Shuzworld F ships 2200 units to warehouse 1. The plan utilizes the excess capacity of 1500 units Shangai to produce more than the initial 1300 units. Shuzworld H also produces an extra 200 units. Copy of Output The Original Shipping Plan Name Warehouse 1 Warehouse 2 Warehouse 3 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 Name Warehouse 1 Warehouse 2 Warehouse 3 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 Reasons for the Use of Transportation Given that the problem involved several sources and several destinations for the products, the use of transportation became necessary. Transportation as a tool is a linear programming model applicable where one wants to calculate the least cost for the shipping of supplies from different sources and destined for different locations. Through calculating the optimal shipping schedule, one is able to solve minimization and maximization problems. That leads to the most feasible distribution of inventory, (Heizer & Render 2010). For Shuzworld, the transportation technique would enable management determine the most efficient distribution patterns. That would influence production from each production facility as the company seeks to maximize on low shipping costs between the facilities and the warehouses. It would depend on the capacity of each facility and the demand of each warehouse. By maximizing production in facilities that involve low shipping costs to warehouses, management would reduce costs and increase profit. Analysis of the Reliability of the Computer-Driven Shoe Machines Process in the Shuzworld Shanghai Plant Reliability is the probability that a machine can function well for a given period. Improvement of reliability is possible through improving individual components of a system of through reducing redundancy. For the three-step process, reliability would be as follows: R1 x R2 x R3 = Rs, Where Rn represents the reliability of the component and Rs represents 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, reliability of the current system is 75.7%. Where the process has little security, it should be backed up and in this case; the first machine with 84% should be backed up. The calculations for this process are as follows; The probability that the second machine works, the likelihood 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, 0.9744 is the new reliability for the combination. 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, the calculations would be similar. 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 engine 3 is supported 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 device 2 is Rs3= 0.84 x 0.91 x 1.0 = 0.764 or 76.4% From the data, there is an indication that reliability goes up when there is backing up of the component with the least reliability. For this set, machine 1 has the lowest reliability. Backing it up raises reliability from 75.7% to 97.44%. The company should therefore support the first machine as early as possible to improve functionality of the whole process and avoid situations where there is a break down due to the machine not functioning well. Suitability of the Decision Analysis Tool The basis of the use of the reliability decision tool is redundancy. Redundancy involves the duplication of components in a parallel manner that enhances reliability. With the tool, it is possible to calculate the new probability of a system to function. Management is able to determine how possible it is for the combined systems operate. It also allows the calculation of the new probability of the functioning of the new system with one or more redundant components. In Shuzworld, the management is able to use it to calculate the impact backing any of their production machines would have besides assisting in the enhancement of production reliability. Optimum Number of Shoelaces for the Shuzworld Factory The optimum number of shoelaces for Shuzworld to order is 27,387. That is as per the economic order quantity, which is the analysis tool in this case. It tries to balance all costs to arrive at this figure. The figure is also the maximum number of shoe laces that Shuzworld should always stock in the inventory at any particular time. On average, the laces should amount to 13,694 at any given time. To ensure the company maintains these values within the inventory, it should make 10.94 orders annually. That would lead to annual setup and the associated holding costs of $1369.31 and $1369.31 respectively. The total annual costs should be $2738.61. Economic Order Quantity (EOQ) The economic order quantity aims at determining the number of units a company can purchase to minimize costs in relation to the purchase, delivery and storage of inventory. EOQ fits well with Schuzworld as the problem involves ordering costs and holding costs for the inventory. As the model seeks to minimize costs related to inventory, it may be applicable by management (Heizer & Render 2010). Applying the economic order quantity gives an optimal order quantity of 27,386 pairs of shoelaces. The company must order this quantity every time it places an order. That would be the most economical quantity, as it would minimize costs. With the EOQ at 27,386, the company would make approximately 11 orders per year. Total holding and set up costs for each year would amount to $1,366.31. Each lac would go for $10 with average inventory being 13,693. The Choice of the Decision Analysis Tool Used With EOQ, the user is able to determine not just the optimal order quantity, but also the optimal annual number of orders, total holding costs and total setup costs for inventory. The model makes it possible to determine inventory calculations through the use of POQ, ABC and, Quantity Discount and EOQ models. For the case, calculation of the optimal number of the shoelaces per order was possible. The optimal number minimizes holding and setup costs to the lowest possible amount and ensures that the demand is met. Comparison Between the One Cashier and Two Cashier Waiting Line Systems Under the single cashier system, the average number of customers in the system is 1 while the number reduces to 0.53 customers in the multiple cashier model with two cashiers. The Average Time a Customer Will Spend in the System Under the single cashier model, the data indicates that the average duration is 10 minutes, while the number reduces to 5.33 minutes under the multiple cashier model. Average Number of Customers in Line There will be 0.5 customers for the single cashier model at any time while, in the multiple cashier model, the number reduces to 0.03. Average Amount of Time Customers Will Wait in Line From the data, customers will have to wait for 5 minutes on average under the single cashier model. In the multiple cashier model, the average waiting time falls to 20 seconds. Probability That No One is in Line or Being Served Under the single cashier system, the likelihood that there is no one in the line is 0.5 while for the multiple cashier system is 0.6. Recommendation on the System to Adopt The best strategy would to adopt the two-cashier system. It is evident that an additional cashier would reduce the amount of waiting time for customers from 5 minutes to 20 seconds. The system would increase efficiency and improve customer satisfaction, as they do not spend a lot of time waiting in line. The Choice of the Decision Analysis Tool With queuing, management and users may estimate the efficiency of a system. The tool projects the average number of customers in a system, the average time the customers spend in the system and the average utilization of the system. It also projects the average number of customers in the system, the probability of the customers being in the system as well as the average waiting time for the customer in the line (Heizer & Render 2010). Of importance in the model is the fact that it informs management on whether changes are necessary in the system. With information from the analysis, performance may improve through implementation of some changes. The arrival rate of the customers may be altered, improve efficiency of the server, increase the number of the service facilities, priority rule as well as increasing the total number of lines within the system. Modification to the system could help Shuzworld improve performance through efficiency. Reference Heizer, J. and Render, B., (2010). Operations Management (10th Ed). New Jersey: Pearson. Read More
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