Product Line Case Study

1163 Words3 Pages

1. A product line has two models: X and Y. Model X consists of 4 components: a, b, c, and d. The number of processing operations required to produce these four components are 2, 3, 4, and 5, respectively. Model Y consists of 3 components: e, f, and g. The number of processing operations required to produce these three components are, 6, 7, and 8 respectively. The annual quantity of Model X is 1000 units and of Model Y is 1500 units. Determine the total number of (a) components and (b) processing operations associated with these two models, (c) workers that will be needed to accomplish the processing operations if each worker works 800 hr/yr. Solution: (a) The total number of components: npf = PQnp 1000(4) + 1500(3) = 4000 + 4500 = 8500 components …show more content…

Total work content for a product assembled on a manual production line is 33.0 min. Production rate of the line must be 47 units/hr. Base parts are attached to a moving conveyor whose speed = 2.2 m/min. Repositioning time per worker is 6 sec, and line efficiency is 94%. Due to imperfect line balancing, the number of workers needed on the line must be two more workers than the number required for perfect balance. Manning level = 1.6, excluding utility workers. Determine (a) the number of workers and (b) the number of workstations on the line. (c) What is the balance efficiency for this line? (d) If the workstations are arranged in a line, and the length of each station is 3.3 m, what is the tolerance time in each station? (e) What is the elapsed time a work unit spends on the line? Solution: (a) Tc = 60(0.94)47= 1.2 min, Ts = 1.2 - 0.1 = 1.1 min For perfect balance, Eb = 1.0. w = Minimum Integer ≥ 33.0 1.0(1.1)= 30 workers. With 4 more workers than required for perfect balance, w = 30 + 2 = 32 workers (b) n = 32 1.6 w M = 20 stations (c) Eb = 33.0 32(1.1)= 0.9375 = 93.75% (d) Tt = Ls/vc = (3.3 m)/(2.2 m/min) = 1.5 …show more content…

A ten-station rotary-indexing machine performs machining operations at nine workstations, and the tenth station is used for unloading and loading parts. The longest process time on the line is 1.75 min and the loading/unloading operation requires less time than this. It takes 9.0 sec to index the machine between workstations. Stations break down with a frequency of 0.006, which is considered equal for all ten stations. When breakdowns occur, it takes an average of 8.0 min to diagnose the problem and make repairs. The starting work part costs $2.50 per unit. Operating cost of the indexing machine is $96/hr, and tooling cost is $0.38 per piece. Determine (a) line efficiency, (b) average hourly production rate, and (c) completed part cost. Solution: (a) F = np = 10(0.006) = 0.06 Tc = 1.75 + 9/60 = 1.90 min Tp = 1.90 + 0.06(8) = 1.90 + 0.48 = 2.38 min/pc E = 1.90/2.38 = 0.798 = 79.8% (b) Rp = 60/2.38 = 25.2

Open Document