t.homes.ieu.edu.tr/stunali/courses/chapter7_supp.pdf · 2014. 3. 25. · cl nd ro r proposal b....

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SUPPLEMENT 7 CAPACITY PLANNING 353 .. SOLVED PROBLEM S7.5 T. SmunL Manuracturing Corp. has th e process di sp layed below. Th e drilling operati on occu rs separ ately fr om an d simultaneously with the s (J wing and :o;a ndin g operatio ns. Th e producl on ly n eeds to go throu gh one of the thr ee assemb ly (the as sem bl y ope ration s are " parallel"). a) Which ope ra tio n is the bOl.tleneck? b) What. Ihe throug hpu t time fo r the overall sy s tem? Sawing 1-1 Sanding c) Jf th e firm ope rate s 8 ho urs per day, 22 d :'IY:> per monlh , whitl is the mo nthly capa c it y o rlhe manuracturing pro cess? d) S uppose th ,lt a seco nd drilljng Inuch ine is <tdded. and it wkes the same time <I S the origin al dri llin g machin e . W ha t is th e new bottleneck li me of the system? e) Suppose thaI a second d rilling ma c hin e j :-. Zl dd ed, tlnd il takes the Same li me as th e o ri ginal drilling mac:..: hin e. What. is lhe new th roug hput time? / l Assembly 78 min/unit '--------' 15 min!unit 15 min/unil Welding ·---1 Assembly 25 minlunit 78 min/unit Drill ing Asse mbly '''''1 27 min/unit 78 min/unit SOLUTION a) Th e time fo r {/.\".\·(·mhly is 78 m inu tes!3 ope r aLO rs = 26 minut es per lIn it, so the stali on that tak es (he longes t tim e, hence the bO ll ienee k, is dri!ling. at 27 minut es. b) System (hroughput time is the maximum o f(l 5 + 15 + 25 + 78), (27 + 25 + 78) = ma x. i mum o r (133, 130) = lJ3 minut es ; e) Mo nthl y capac it y = (60 minutes)( 8 ho ur s){22 da ys) j 27 m i nutes per unit = 10,560 minul es per m onl h/27 minutes pe r unit = 39 1.1Iu nits j m o nth. d) The bollienec k s hifts Lo A., ·semhly , w ith a ti me or26 minutes p er unit. e) Red undan cy does not affect th roug hput time. rl is sti ll 133 minute s. Problems Nole: Px means the problem may be solved wilh POM lor Wind ows and/or Excel OM. · · S7 .1 Southea ste rn O klahom a Sta te University 's bu si- ness program ha s th e facilities and raculty to handl e a n e nroll- ment of 2,000 new pe r se mes ter. However , in <In erro rt to limit cla ss sizes to a " re asonable" l eve l (under 200. genera ll y), So uthea s te rn 's dean, Holl y Lutze, pla ced a ce iling o n e nrollm ent of 1,500 n ew s tudent s. A lth ough there was a mpl e de mand for business courses last se me:'>ler. connicti ng sc hedules a ll owed o nl y 1,450 new student s to take busines s c o urses. What ar e the uti li za - ti on and effic ie ncy of this sys t em? 57.2 Amy Xia 's plant was designed to produ ce 7,000 i1all1- mers per day but is limit ed to making 6.000 ha mmers p er Ja y beca use o f the time need ed 10 c ha nge equ ipme nl be tween style s o f ha mmers. Wh a t is the u ti lizatio n? 57.3 If <J plant an efrecti vc capa city or 6.500 a nd an e fli ciellcy 0 1" 88'%. whilt the <letllal ( planned) ou tput? 57.4 A plant has all et Teclive c apacit y o f9 00 un its per d ay a nd proouces 800 lin its per da y with ir s producll11ix; Wh,l t is its efTi ciency? 57.5 M<lle ri< ll delays h ave ro utinel y limited pro du ctioll of hous ehold sinks to 40 0 nnil s pe r da y. If the pl<:ln t efficiency is 80%. what is th e efJ'l..:clive capacit y? 57.6 For lh e past month , the p] ':IJlI in Pr oble m S7.2. which has all e rr ect ive capacity or 6,500. h as ma de o nl y 4,5 00 hml1mers per da y because or ma te ri a l delay. emplo ye!.: a bse nces. and other prob lems . What is iLS e ffi cie ncy? 57.7 ThedTcctive capa c it y <.I nd e ffi cie ncy ror the next quar- ter at M MU Mfg. in Wac o. Tc x<Js. for each or three depa rtmen ts arc sh ow n: EffECTIVE 93.600 .95 Fabricat ion \56. 000 1. 03 F ini s hing 61.400 1.05 Compute the expected producLion for next qu a rt er fo r eac h depart ment. ·· 57.8 Unde r jdeal co nd it ions, a ser vice ba y at a Fast Lube ca n se r ve 6 cars pe r h ou r. The ca pacit y a nd efficien cy o r a F a st Lu be se rvice bay ar e known {O be 5.5 and 0.880,

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  • SUPPLEMENT 7 CAPACITY PLANNING 353 ..

    SOLVED PROBLEM S7.5 T. SmunL Manuracturing Corp. has th e process displayed be low. The drilling operation occu rs separa te ly fr om an d simultaneously with the s(J w ing and :o;a nding operations. The

    prod ucl only needs to go throu gh o ne o f t he three assembly openHion~ (the assem bly operations are " parallel").

    a) Which opera tio n is the bOl.tleneck?

    b) What. j~ Ihe throughpu t time fo r the overall sys tem?

    Sawing 1-1 Sanding

    c) Jf the firm operates 8 ho urs pe r day, 22 d :'IY:> per monlh , whitl is the m o nthly capac it y orlhe manuracturing process?

    d) Suppose th ,lt a second drilljng Inuch ine is ler. connicti ng schedules a ll owed o nl y 1,450 new students to take business co urses. What are the ut iliza tion and efficie ncy of thi s system?

    • 57.2 A m y X ia 's plant was designed to produce 7,000 i1all1mers pe r d a y but is limited to making 6.000 ha mmers per Ja y beca use o f the time needed 10 c ha nge equ ipme nl be tween styles o f ha mmers. Wha t is the u ti lization?

    • 57.3 If

  • 354 PART 2 DES IGNING OPERATION S

    respective ly Wh"ll is I he min imum number of service b

  • a) Which opera tio n is the bottleneck? b) Wh,H is the bo ttleneck time'! c) What is the throLlghpullime or the ove ra ll system'? d) If the Ii rlll opera tcs 8 hOlll's pe r da y. 20 Jays per month . wh

  • 356 PART 2 DESIG NING OPE RATI O NS

    price 0($6.25 wit h l:llvage val ue o f $45,000'l The anu ua l p rofit fro m the inves tment i:s $ 15,000 eaeh yea r fo r 5 ye