HW Sol Due 4-2-2009

# HW Sol Due 4-2-2009 - S's/Mala Problems 6.1 Consider a...

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Unformatted text preview: S's/Mala Problems 6.1. Consider a facility that produces twO products in three workstations. Product 1 follows the probabilistic workstation transition matrix given by From/To 1 2 3 1 0.0 0.3 0.5 2 0.2 0.0 0.8 ‘3 0.4 0.5 0.0 while Product 2 has the transition matrix Problems ————-————_—____' From/"1‘0 i 2 3 1 0.0 0.6 0.4 2 0.3 0.0 0.7 3 0.4 0.1 0.0 The workstation processing time distributions are different by product. For Prod— uct 1, these data are Workstation # E [1}] 5 ————~——_____. 1 - 1.1hr 1.0 2 1.0 hr 1.5 3 0.611r 2.0 For Product 2, these data are . . Workstation # E [m C? _____._.________ 1 0.25 hr -1.0 2 0.35 hr 1.5 3 0.60 hr 2.0 ——_—.______—____ The mean reIease rate for Product 1 is 0.2 jobs per hour and for Product 2 is 0.3 jobs per hour, both Poisson distributed releases into Workstation 1. (a) Determine the minimum number of (identical) machines that must be placed in each workstation so that a steady-state system results. . (b) Using the number of machines détennined in Part (a), ﬁnd the workstation and system performance measures: cycle time, work-in—process, and throughput. 185- Chapter 6 Problem 1 Part a Computer solution follows: External Arrival Stream Mean Rates : G external inflow by product to Steps 1 { 0.200, 0, 0} 2 { 0.300, 0, 0} External Arrival Streams Squared Coef. Var. by step: Cao 1 { 1.000, 0, 0} 2 { 1.000, 0, 0} Steps to Workstations Conversion 1 2 3 1 2 3 Routing Step Matrix by Product: QPRP 0.000 0.300 0.500 0.200 0.000 0.800 0.400 0.500 0.000 0.000 0.600 0.400 0.300 0.000 0.700 0.400 0.100 0.000 Prod mean times 1.100 1.000 0.600 0.250 0.350 0.600 Prod Serv Vars 1.210 1.500 0.720 0.063 0.184 0.720 External Inflow by Product to WorkStations 1 { 0.200, 0.000, 0.000} 2 { 0.300, 0.000, 0.000} External Arrival Squared Coef. Var. by WorkStation: Caows uses asymptotic weighted Cao of steps into workstations 1 { 1.000, 0.000, 0.000} 2 { 1.000, 0.000, 0.000} S — Serv Time Sq. Coef. Var. by Product 1.000 1.500 2.000 1.000 1.500 2.000 QPRP 1 0.000 0.300 0.500 0.200 0.000 0.800 0.400 0.500 0.000 G[i] by step: { 0.200, 0, 0} Lam: l = { 0.619, 0.567, 0.763} QPRP 2 0.000 0.600 0.400 0.300 0.000 0.700 0.400 0.100 0.000 G[i] by step: { 0.300, 0, 0} Lam: 2 = { 0.680, 0.468, 0.600} Lam by Product per Step 0.619 0.567 0.763 0.680 0.468 0.600 Lam by Product and WorkStation 0.619 0.567 0.763 0.680 0.468 0.600 Total wslam by WorkStation 1.299 1.035 1.363 Branching Probabilities by Workstation: Qws 0.000 0.457 0.448 0.245 0.000 0.755 0.400 0.324 0.000 Gam Inflow by Workstation and Product: Gws 0.200 0.000 0.000 0.300 0.000 0.000 wsgam: { 0.500, 0.000, 0.000} total gam: 0.500 mean proc times at Workstations ‘ mtws: { 0.655, 0.706, 0.600} 3 Sq. Coef Serv Times at Workstations 1 Csws: { 1.841, 2.025, 2.000} station loads 1 " util: { 0.851, 0.731, 0.818} Part a) answer is 1 machine per workstation. Continuing with the solution for Part b): Ca.2 equations: C3 (1) = (1 — 0.8512 )C3 (1) + 0.35 12 (1.840) c3 (2) = (1 - 0.7315030) + 0.73 12 (2.026) C3 (3) = (1 — 0.3182)C3(3)+ 0.8182(2) C3 (1) = £352 + 1_-03_15;_1;§15_1[0245C§(2)+ 1— 0.245] + 1.363(0.4) 1.299 1.299(0457) L035 +W[1—0324+0.324C3(3)] i 1.035 1 WP _ 0.448 + 0.448C§(1)] 1.363 + 01.035(0.75 5) 1.363 [0.4G3(3)+1— 0.4] C3 (2) = [1 — 0.457 + 0.45703 (1)] C: (3) = [1— 0.755 + 0.7563(2)] num machs at stations m: { l, 1, l} external arriv sq coef var to stations: asymptotic method Caosws: { 1.000, 0.000, 0.000} I Citer: 0 Ca: Cda: { 1.293, 1.513, 1.262} { 1.000, 1.000, 1.000} Citer: 1 Ca: { 1.138, 1.252, 1.353} Cd: { 1.608, 1.548, 1.669} Cda: { 1.272, 1.413, 1.217} Citer: 2 Ca: { 1.164, 1.278, 1.411} Cd: { 1.647, 1.665, 1.786} Cda: { 1.289, 1.502, 1.254} Citer: 3 Ca: { 1.168, 1.283, 1.418} Cd: { 1.654, 1.677, 1.805} Cda: { 1.293, 1.511, 1.261} Citer: 4 Ca: { 1.168, 1.283, 1.419} Cd: { 1.655, 1.679, 1.807} I Cda: { 1.293, 1.513, 1.261} l , | Citer: 5 Ca: { 1.168, 1.283, 1.419} Cd: { 1.655, 1.679, 1.807} . Citer: 6 Ca: { .168, 1.283, 1.419} Cd: { .655, .679, 1.808} Cda: { 1.293, 1.513, 1.262} H H |_| asymptotic method - Ca: { 1.168, 1.283, 1.419} Cd: { 1.655, .679, 1.808} Cda: { 1.293, 1.513, 1.262} I-1 rho: { 0.851, 0.731, 0.818} mtws: { 0.655, 0.706, 0.600} Results by Workstation W81: 1 CTqi: 5.605 WIPqi: 7.281 CTsi: 6.260 WIPsi: 8.132 WSi: 2 CTqi: 3.172 WIPqi: 3.284 CTsi: 3.878 WIPsi: 4.015 WSi: 3 CTqi: 4.602 WIPqi: 6.272 CTsi: 5.202 WIPsi: 7.090 total system inflow (tgam): 0.500 WIPsyS: 19.236 CTsys: 38.473 Individual Product Cycle Times Prod: 1 Prodl TH: 0.200 CTs: 52.408 hours CTs: 2.184 days , Prod: 2 Prod2 TH: 0.300 CTs: 29.183 hours CTs: 1.216 days C3 (1") Computations: Workstation 1 A“ = 0.619, E[S,‘] = 1.1, C2[Sf] = 1.0, E[(S:)2]=(1.1)2(Cz[s:]+1)=2.42 2,2 = 0.680, ELsf} = 0.25, C’ISE] =' 1.0, E[(S,2 )2] = (0.25)2 (C2[Sf] + 1) = 0.125 El:(§1)2:| = 0.619 x2.42 + 0.680x0.125 = 1.219 1.2991 BUSH)? 1219421840 .C2[Sl]: E[§1]Z =0.6552 Workstation 2 212' _= 0.567, E[S;] = 1.0, C2[S;] = 1.5, E[(s;)2] = (1)2 (CZ[S;] + 1) = 2.5 2; = 0.468, E[S§'] = 0.35, C2[S§]=1.5, E[(S§ )2] = (0.35)2 (C2[s§] + 1) 2 0.30625 I E[(§2)2] = W = 1.508 1.035 c2[§,]=E[(~2)2] 1.508 —-‘T= 2 —1=2.0249 ELSE] 0.706 Workstation 2 C2[§3] = 2.0 ...
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