System Design and Operation - Term Paper Example

System Design and Operation System Design Line Balancing Line balancing is workload leveling across all the operations with the aim of eliminating excess capacity and bottlenecks as defined by the six-sigma materials. Considering mass production, products are often produced in sets or lines of machines as opposed to single production that uses a single machine (Stevenson 188). Therefore, a line may be modular line, assembly line or a set of lines with online finishing and packing. In a system design, a line usually includes multiple workstations that have varied work contents. The production time of each workstation varies depending on the work content of the station, machine capacity, operators’ skills, and work force within the station. For instance, having five operation units or machines lead to the expectation that these machines have different work capacity and efficiencies; thus, line processes are expected to take different times within these machines along the line. Assuming that these machines take 0.45, 1.35, 1.8, 0.6, 0.3 minutes of their individual operations consecutively and operating 15 hours a day that translates to 1080 minutes a day. Using the same data it is apparent that the minimum cycle time = 1.8 minutes and the maximum cycle is obtained by summing all the time taking in conducting the task = A + B + C + D + E = 4.5 minutes. The output rate for the above systems = operating time per day/ minimum cycle time =1080/1.8 = 600 units per day The output rate of this system = operating time per day/maximum cycle time = 1080/4.5 = 240 units per days The above calculations indicate that the selected output for the system’s line is within the range of 240 and 600 units per day. Thus, cycle time = operating time per day/desired output rate = 1080/600 = 1.8 minutes per day Nmin = ∑t / cycle time = 4.5/1.8 = 3 stations The above information can be presented in a time diagram below The idling time per operation is tabulated below: WorkStation Time remaining Eligible Assign task Revise time remaining Station idle time 1 1.8 A,C A 1.35 1.35 B B 0.0 2 1.8 C C 0.0 3 2.85 D D 1.2 1.2 E E 0.9 0.9 From this table, the percentage of idle time can be deduced as: Percentage Idle time = (item time per cycle/ (N-actual * cycle time)) * 100 = (0.9/ (4.5*1.8))*100 = 25.92592592592593 = 25.92% Efficiency = 100 - 25.92 = 74.18 % with the efficiency of the system being more than 50%, it implies that the system has less idling time. Continuous Multiple Facility Placement There are usually problems with the continuous location problems especially when a firm wants to start two or more facilities in a highly competitive environment. The facilities in same service or product lines often exist in such areas; therefore, there are needs to locate a position properly and offer quality services and products at these new facilities in order to maximize profit in the new facilities. Nonetheless, the Continuous Multiple Facility Placement is a global optimization problem that affects many local optima (Stevenson 125). However, there are any ways of approach towards solving this problem including heuristics or the three-multi start location search, evolutionary algorithm variance, and multi start simulated annealing algorithm. Of the three possible solutions pathways formulae, the evolutionary algorithm provides the best heuristics for the Continuous Multiple Facility Placement. For instance, the same applications can be used to solve the location problem and customers’ needs and the same can be understood by using the five customers with different location and demands presented in the table below. Customers Demand Coordinates (x, y) 1 140 50, 40 2 110 20, 60 3 90 10, 90 4 100 40, 50 5 70 30, 70 Scaling the Demands as per the individual customer Customers Scaled 1 2.3 2 1.4 3 1.5 4 1.6 5 1.2 The third step to working out the continuous location problem is drawing a point graph towards determining the minimum cost constants The forth step is calculating the distance between the points 36.45 * 2.3 = 84.84 38.45 * 1.4 = 53.83 59.82 * 1.5 = 89.73 74.38 * 1.6 = 119.01 85.68*1.2 = 102.82 From the calculations above, it is apparent (38.45 * 1.4 =) 53.83 translates to the minimum cost constant. Therefore, the investor involved in this business may opt for the same location to maximize its operations despite the existed of other operators in his or her product line. However, for him/her to attract customer on the same demand on his/her facility he/her must provide quality products and or services. Work Cited Stevenson, William J. Operations Management. New York: McGraw-Hill/Irwin, 2012. Print. Read More