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hw9soln - 112:3 a A parking lot is a queueing system for...

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Unformatted text preview: 112:3 a) A parking lot is a queueing system for providing parking with cars as the customers, and parking spaces as the servers. The service time is the amount of time a car spends in a space. The queue capacity is O. b) L = 0(Po)+1(P,)+ 2(PZ) + 3(P3) = 0(0.2) + l(0.3) + 2(0.3) + 3(0.2) = 1.5 cars Lq = 0 cars W = (E) = (E) = 0.75 hours ft 2 L W] = [—f) = (-29) = 0 hours c) A car spends an average of 45 minutes in a parking space. 11‘4'5 85 mow/e59 7’»qu , U = min/7“,, 7;, 7;) T, ~exP(%.o), ,71~exf('/;) , T; ~exfl I//o) vweXf(-L-+,—’~+7g-)= «ff—6%) 3°. WM war/757 17kt = 7% = 4‘8 Inf/”0755 /3 b) w=wfi+72 , up.“ an/Mw Eu): 805+ E]? = 5m; =15 01;th = 2L 4,“, WW]: VMMfifi ”’7; =(7—li)z+ {ff—wi— =0.0547 0“ (,5: 51-005 (1:) KMWVI 2° ”I‘M. [MT Maloifizem wank/,2 cult); /¢«§ (9 =9 PL=(6-§P.~2p.)/ = E’P. as P5=5§W5Po '§Po=P~ @ =? P;=(2P°-2 -JP°)/ = 314% mu ML D: P Hurts” ’7“ LA; ___( P WWV" n kilw 14:6-5 3:3,»:11’ $;‘ ’ Pa3/l, The, sgstem wctkout the storage mstrccfion £5 a M/MH “Wen-IF ‘A square cast 0? Fkoor space we“2 QVacLablc fir wactcmfivxiproporh‘ox oF hime £th WOuLd be suFRc‘Ceht CS 33‘1"“ we, wamt to 43an ”At suck that :22- >/ %C For 8,4,3"), 'wkere $1:.5I%Afl.q’%§.qq‘ 4 4 . Now ER 2% (=> 3 (1-p)p‘ >, %¢¢‘>("P) (1.2%”); w an L:0 (l—p) ‘(1 'n .1 (=5) 1~P U 3 (k (a P'nvzs 4-%e<¢(7\t+a)tnp5£n(1_%cy aunt”); EMU—311$) ’"c 2 C (1-3 1 ”1 4 L: i=i=3 customers 11—). 40-30 W=—l—-=—1———=O.1hours 11—1 40—30 2. 30 W" = -————- = ————- = 0.075 hours ,u(,u — 1) 40(40 - 30) L4 = AW" = 30(0.075) = 2.25 customers P0 =l—p=1—O.75=O.25 P, = (1 — p)p = (1 - 0.75)0.75 = 0.188 P2 = (1 — p)p2 = (1 —0.75)0.752 = 0.141 There is a 42% chance of having more than 2 customers at the checkout stand. b) Data Results 30 (mean arrival rate) 40 (mean service rate) (# servers) PO+P1+P2= 0.578125 Igl-lf 116—9 6) 30 J_ .— ,u—x'L 60—30 W=;=_1__ 11—1 60—30 W 2:__4_______3_9_._ ' ”(11— A) 60(60—30) Ll] = qu = 30(0.075) = 0.5 customer P0 =1—p=1-0.5=O.5 1DI = (1 — p)p = (1 — 0.5)0.5 = 0.25 P2 = (1 -p)p2 = (1 —0.5)0.52 =0.125 L = = 1 customer = 0.033 hours = 0.017 hours There is a 12.5% chance of having more than 2 customers at the checkout stand. 30 (mean arrival rate) (mean service rate) (# servers) P0+P1+P2= 0.875 e) The manager should adopt the new approach of adding another person to bag the groceries. I746 Data Results 1 0 (mean arrival rate) 20 (mean service rate) (# servers) 0.0067379 0.5 P0+P1+P2+P3+P4+P5= 0.984375 All the criteria are currently being satisfied. b) Data Results (mean arrival rate) (mean service rate) (# servers) PO+P1+P2+P3+P4+P5= 0.822021484 None of the criteria are now satisfied. l7-l? (mean arrival rate) (mean service rate) (# servers) P0+P1+P2+P3+P4+P5= 0.926640437 In this case, the first and third criteria are satisfied but the second is not. Data 116-15 (mean arrival rate) (mean service rate) (4 servers) “ , 9493125., 2.061E-09 when l = 10 Prob( flu") =_ ' '«r-ul wwwuv. . °-§§§.3..333. 0 when t = 0.305555556 P0+P1= Data (mean arrival rate) (mean service rate) (# servers) f’.’°b‘.‘fa?.’)‘ 0245098 0 when t = 0897875817 P0+P1+P2= mean arrival rate) (mean service rate) (# servers) Pr(o)>t) = 8.047E-53 when t = 10 0.0577101 0 Prob( 0).») = when l = 0983969426 P0+P1+P2+P3= Data (mean arrival rate) (mean serwce rate) (# servers) . PM?!) -. 7.297553 when t = 10 Prob( n31): 6.0139241 .. ”ea-‘2 shy-www.- gm 0 when t = P0+P1+P2+P3+P4= 0997703314 17-20 I :°¥.1.388§§°9" ‘- -‘-"2’"...19:1j1.5740,74 ' ‘Pa "= 0.09645062- ._;,_:... 2259; r"— ~:m.x~&‘." Results Ls, 1100840336 ‘Laf' 9‘ 7597003 ' . P' " 6.1931572: P2 =. . 034297336 P= ;..9.0§9572.44 Results L = 0.85552951 0.02219618 008555295 000221962 . 027777778 .043213296 ' . 9.9601105 ' 045904617 (om/rib) 95 ‘~‘ 000271151 (couf'o) Data P0+P1+P2+P3+P4+P5= (mean arrival rate) (mean service rate) (# servers) 0.999708926 a) 2 servers b) 3 servers C) 2 servers d) Iserver e) 5 servers f) Iserver g) 3 servers ...
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