Production system optimization - Coursework Example

Production System Optimization Production System Optimization In this production system project, the aim is to produce calculations that will maximize the efficiency and the productivity of the assembly lines (Gregory & Kuzmich, 2004). This evaluation is done with an emphasis on the LV Doors as the assembly line where production level can be evaluated and increased.The efficiency of the assembly lines is calculated as: E = ((A + B + C + D) / (N * Cd)) * 100Where A, B, C and D are the Cycle times, N is the minimum number of work stations required and Cd = production time available / the desired output unitsAt present, the efficiency is 94 %.

It can be improved by either increasing the cycle times at every LV Door or by reducing the number of workstations at every LV Door for all LV Doors available in the system. For example, we can increase the cycle times in every LV Door while retaining N and Cd as constants. The new Efficiency will be:E = ((A + B + C + D) / (N * Cd) * 100E = ((195 + 740 + 680 + 680) / (16 * 144)) * 100E = (2295 / 2304) * 100 = 99.609375 Alternatively, we use the results of analysis which indicate that the more the number of LV Doors, the more the number of workstations required for production (Burke, 2005).

It therefore means that the number of work stations needed in the assembly lines has to be reduced (Rutherford, 2002). The required minimum number of the workstations can be calculated as:N = ((A + B + C + D) / Cd) This requires that we reduce the cycle times for every LD Door to obtain the smallest number of workstations required. In the original situation where there are 4 assembly doors, the number of workstations required is 15.10 which are rounded off to 16. However, the reduction of cycle times gives the calculation of the number of work stations as follows:N = (170 + 700 + 550 + 550) / 144 = 13.

68 ≈ 14 Workstations As we maintain the minimum number of workstations at 14, we have balanced the line to its optimal efficiency (Chapman & King, 2008). This efficiency must be maximized since it corresponds to the percentage utilization of the workstations available. Taking a different sample of assembly line specificationWork ElementCycle TimeAComponents Fixation170 minBFlange Wiring700 minCWires Termination540 minDDoor Wiring550 minEfficiency E = ((170 + 700 + 540 + 550) / (144 * 13.68)) * 100E = 1960 / 1969.92E = 99.49 % The high value of Efficiency E reflects a high level of productivity.

In the current situation, the production system produces about 3 sets of MC every day, when the operation runs continuously for 12 hours in a day (Brookhart, 2006). It means that the efficiency has to be increased until the production shifts from 3 to 5. This requires that we measure the percentage by which the production will have increased. This is calculated as follows:The percentage increase in production (PI) =PI = ((5 – 3) / 3) * 100PI = 2 / 3 * 100PI = 67 % Since the existing system has been optimized and the existing efficiency exceeds 90%, it is above average and can be retained at this level.

ReferencesBrookhart, S., 2006. Formative Assessment Strategies for Every Classroom. Alexandria, VA: Association for Supervision and Curriculum Development.Burke, K., 2005. How to Assess Authentic Learning, Fifth Edition. Thousand Oaks, CA: Corwin Press.Chapman, C., & King, R., 2008. Differentiated Instructional Management: Work Smarter, Not Harder. Thousand Oaks, CA: Corwin Press.Gregory, G., & Kuzmich, L., 2004. Data Driven Differentiation in the Standards- Based Classroom. Thousand Oaks, CA: Corwin Press.

Rutherford, P., 2002. Why Didn’t I Learn This in College? Teaching & Learning in the 21st Century. Alexandria, VA: Just ASK Publications & Professional Development.