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Homework Sol Due 10-4 Nonserial Set 2

Homework Sol Due 10-4 Nonserial Set 2 - 5M Homework Due...

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Unformatted text preview: 5M . Homework Due 10—4-20} 1 Nonserial Set 2 1 \- 3/4 as 3 114 —*I 213 Let the arrivals to the factory be 3 jobs per hour to workstation 1, Poisson distributed (C;E =1), and 1 job per hour to workstation 2, with squared coefﬁcient of variation of 2 (C: = 2) . Let the four workstations have processing time data of: W51: EESI]:O.25hrs C2[Sl]:4 W52: E[Sz}=0.40hrs C2[S2}:3 WS3: E[S3]=0.30hrs Cz{S3]=2 W54: E[Sd]:0.22hrs C2{82]22.5 Compute the factory performance measures of throughput, cycle time and WIP. X58 u.:%(.25> = .75 CTC!) = (LE WWIXOLQ / ).25 +. 25 22/25 4616 75 \$1590): 3C3? [2‘5 )=‘- é 3'75 frﬁ—c M50) : 3 . Cf“): (I—JSZXCDJ—QEYL/r 26:88 W52 (gm: W3:- A??? a ‘\ CEO->93 = égam +25. = 25427 )3: :2 3 Caz(z):/.79/ 55555 -E ill a: CEO) + {—93 Cal—163: %(2.¢89)+ #3 = 2.:253 1 n.— Cq(2-¢3)=2if.(2.6(ol3+% _. [.3903 c0303): N3 Cwas) + mama?» I\\""5 + [\7—3 CQQ‘C‘EE) = ('91) 2.1753 +Q/z)(tuzczo:?): L978? 52+ ’52. -""'"'""'_—-—-—-- )3 : 2+*/-2,= 2.5 Maw». 3 ~ (Ag—saga) 2.75 g €75.63): 0,973+2 [LE—23¢ 5)+.3 '= 3.0?0 W. 3‘ ~1 ' . marge): 3.5(2099): 5215 70—8—4 —H\\$C5) = :25 gigs -.-—_ (1—575 9Q???) +6.79%; = /. We) Mir '- §(7,):/5 CSCwa) = %C2.5él)+2-’2, = .2 (7! A34: :25 55555 ,qu 4(.2=—) = 0-98 4&5” CE!” = 60584—15) ﬁjtzz + .22. R .12 : 3.. 8’?7’AMI wrm): 98.297): 193m 7,33% ’H\5(¢/7: 5/ W \NIP —: \NJ:@03+ 4— maﬁa» ...
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Homework Sol Due 10-4 Nonserial Set 2 - 5M Homework Due...

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