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3.15. Determine the impact of an arrival rate of 5 per day in Example 3.4 (a = 5, u = 3,7= 2 in Eq. 3.13) as it reflects on the system parameters. (a) Write the system of equations for the steady-state probabilities. (b) Obtain the system performance measures: CTs, CT7, WIPS, WIPq, utilization u, mean service time E(Ts), and throughput re. Example 3.4. An overhaul facility for helicopters is open 24 hours a day, seven days a week and helicopters arrive to the facility at an average rate of 3 per day according to a Poisson process (i.e., exponential inter-arrival times). One of the areas within the facility is for degreasing one of the major components. There is only room in the facility for 4 jobs at any one time and there are two machines that do the degreasing. The newer of the two degreasing machines takes an average of 8 hours to complete the degreasing and the older machine takes 12 hours for the degreasing operation. Because of the large variability in helicopter conditions, all times are exponentially distributed. Thus, we have a = 3 per day, u = 3 per day, and y= 2 per day. The system of equations given by (3.13) become 84 3 Single Workstation Factory Models 3po 3p1f-2p1s =0 3p1f-2p2 2pis = 0 3pf +3ps 5p2 = 0 3p2 5p3 = 0 p3 5p4 = (0) po + Pif+pis + p2 + p3 + 24 = 1. The solution to this system of equations is po = 0.288, Pf = 0.209, pis = 0.118, p2 = 0.196, p3 = 0.118, P4 = 0.071. a pif = a po = Upif+ Ypis Yp2 + ypis pif+apis = (y+u)p2 2p2 = (y+up3 2 pz = (y+u)p4 . (3.13) 3.15. Determine the impact of an arrival rate of 5 per day in Example 3.4 (a = 5, u = 3,7= 2 in Eq. 3.13) as it reflects on the system parameters. (a) Write the system of equations for the steady-state probabilities. (b) Obtain the system performance measures: CTs, CT7, WIPS, WIPq, utilization u, mean service time E(Ts), and throughput re. Example 3.4. An overhaul facility for helicopters is open 24 hours a day, seven days a week and helicopters arrive to the facility at an average rate of 3 per day according to a Poisson process (i.e., exponential inter-arrival times). One of the areas within the facility is for degreasing one of the major components. There is only room in the facility for 4 jobs at any one time and there are two machines that do the degreasing. The newer of the two degreasing machines takes an average of 8 hours to complete the degreasing and the older machine takes 12 hours for the degreasing operation. Because of the large variability in helicopter conditions, all times are exponentially distributed. Thus, we have a = 3 per day, u = 3 per day, and y= 2 per day. The system of equations given by (3.13) become 84 3 Single Workstation Factory Models 3po 3p1f-2p1s =0 3p1f-2p2 2pis = 0 3pf +3ps 5p2 = 0 3p2 5p3 = 0 p3 5p4 = (0) po + Pif+pis + p2 + p3 + 24 = 1. The solution to this system of equations is po = 0.288, Pf = 0.209, pis = 0.118, p2 = 0.196, p3 = 0.118, P4 = 0.071. a pif = a po = Upif+ Ypis Yp2 + ypis pif+apis = (y+u)p2 2p2 = (y+up3 2 pz = (y+u)p4 . (3.13)