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HW3 - Problem 5.1 Determine time when gueue clears m;= 5.2...

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Unformatted text preview: Problem 5.1? Determine time when gueue clears. m) ;= 5.2 — {LC-1t p“) ;= 3.3 + 2.44 (given) Integrate to obtain it of an'ivals at speciﬁc time f Arrivalslt) := J m) dt —; 5.20.1 — 500—15342 lp—t Integrate to obtain it of departures at speciﬁc time Departuresﬁ) :=J Mt) dt a 3.301 + 1.2m2 11‘ Since vehicle service begins (i.e._ toll booth opens} 10 minutes alter vehicles begin to arrive, Departuresﬂ) := 3.3(t — 10) + 1.2-(t — 10)2 Time to queue clearance: At what time does arrival rate equal constant 10 vehfrnin? MI}! 10 t=2.792 lat toon ;= t + 1n tcon = 12792 time at which service becomes constant at 10 vehfmin Departures{t = 13.555 con] 5.2-: - ODDS-t2: 18.568 + wit — 12.792} 2.01 P r:- .... 1 = 22255 queue clears 22 minutes and 15.9 seconds after ?:50 AM Check that total arrivals equals total departures Arrivals{22.265) = 113.3 Departures 19(22255 _ 12192) = 9473 arrivals after arrival rate becomes 1C! vehi'mil 94_73 + 13553 = 1133 add arrivals before constant arrival rate Problem 5.15 Determine the delay; and gueue information. M1} ;= 52 _ [12m vehfmin “ 3:3 vehfmin {given} Integrate to obtain # of arrivals and departures at speciﬁc time 1 t Arnvals-{t} := Jr Mt) dt —) 5.261- .16612 i 1 t Departuresﬂ) := Jr p. dt —> 3-t ‘1 Time to queue clearance Q(t} := Arrivals“) — Departures(t) —) 2.261 — _1CI-t2 or f f 2 lit Om :=J Mt) dt —J p dt —> 2.261 — .1DU-t When Qtt}=6, queue is cleared 52t—ﬂ1t2—3t-D _ . — 1 pt I = 22 Total delaglr is area between arrival and departure curve: 22 Delay :='[ Qtt] dt —) “146? veh— min lit (1 Maximum queue length 2.2 d—Q{t)=2_2—_2-t=ﬂ ;=— t=11 1J—t dt 0.2 Q(11}=12.1 veh m Wait time for 26th vehicle 5.2t-0_1t2-2(]=|] t = 4.133 time when 20th vehicle arrives 1.5! 3t — 26 = l] t = 5557 time when 26th vehicle departs 0-5Et wait := 6.66? — 4.1 83 wait = 2.48 min Determine the average length of gueueI average time sgent in the gigs-tumI and average waiting time in the Queue. Problem 5.22 MIDH p. = 5 J. ;= 4 (given) in. p = — p = u 8 [Eq— 5.27) it p2 1 pt — u —1 e veh (31.5.28) bar 2_(1_p} bar 2 — p _ 1 _ t = 0 5 mlnl‘veh [Eq. 5.30) lat bar 2P'(1—P) bar p _ w _ w _ 0 4 mInNeh [Eq_ 5.29) 1 pt bar 2_”(1_p) bar Determine the average length of gueue, PrOblem 5'23 average time sgent in the system, and average waiting time sgent in the Queue. MIMH p. := 5 9‘. := 4 {given} in. p =— p =08 (Eq. 5.2?) u 2 Qbar:= 1"— ﬁber = 3.2 veh (Eq. 5.31} 1F“ - P 1 - 1 t tbar __ tbar _ 1 mlnlveh (Eq. 5.33} D p, — i. J. _ wbar _ “bar = [13 mlnfveh (Eq. 5.32} 1 pt Problem 5.27 Determine the arrival rate. 1-1 ;= 4 vehi'min -: 30 min (time to queue clearance) ta. at queue clearance 1 t ML. = “(ta —tp) _p_ Where tp is the time until processing begins (i.e._ time at which queuing lane becomes full) tc lat IP = I substituting expression for In and solving for arrival rate gives, tc‘1 l-tczp- tic—I) 1 = 2 vehfmin P tn Determine how many vehJ'min should be Erocessed. length of queue at ﬁmet is ‘ f 0m =I 3.3 —ﬂ.1-tdt— p. dt 0 J Q(t} = 3.31— 0.05.12 — In for maximum \$013) = 0 = 3.3 — 0.1-t— 1,; _ 3.3 — u. [1.1 ﬁrst substitute for t, then ﬁnd departure rate, 3.3 — M 005(33 — p12 Qtt}_3'3'[ 0.1 ) [1.1 ) I-L =2_41 vehfmin 4 3.3—;1} [1.1 J 4 1pt 1pt Problem 5.31 1pt Determine the total time sEent in system by all vehicles in a one hour geriod. 1 _ 1. := 4311— :1. = 7.167 vehfmln so 1 _ ;=—.6 = 5 vehfmln u m H J. p :=— p =1.194 u. N :=2 3 =0.597 N 1 PD I=—2 P0 = 0.252 1 + E + p 1 1 21. — 3‘1 N) O PUPNH 1 Q n 662 h bar'=W' —2 W - “e _ El N) p "' Qbar _ tbar3=T tbar= {1259 mlnr'veh 50 total time spent in one hour: 43n.tbar= 111.4 "mm Problem 5.34 {given} (Eq. 5.27) (Eq. 5.34) (Eq. 5.33) (Eq. 5.40) 1pt 1pt 1pt 0.5pt Problem 5.35 Determine the minimum number of booths needed. 1 . -= ._ = vehimln . 1. 5m] 60 a. 8.333 (given) - 1 50 4 vehlmin " '_ 15' *1 _ l p := — p = 2.083 (ECI- 5-27} p. . 0.5pt N must be greater than 2 for the equations to apply, try 3 booﬂts N := 3 1 P0 := ﬁ P0 = 0.098 ('31- 5-34} 1 + — + p_ + p 11 1 31—[1 — 31 J N+1 PM 1 Qbari= W - —2 Qbar= 1.101 veh (Eq 53B) 91 1 _ _ E N) P + oh 1 . E .5.39 29" “ban: —ar _ _ “bar: [1132 mlnfveh ( q } 3'» 1L whar-BD = 7.924 sedveh try 4 booths N :=4 1 PD:— 2 3 4 P0 = 0.119 (E9. 5—34) 1 P P P P + — + — — 11 21 31 [ p1 4!— 1 — — NJ Pﬂ-pNH 1 0212 h (E9 533) Qbar-= W —12 Qbar= - V6 ' ' 1 _ 3 M N) J P + 0 “bar; Thar _ 1 WW: {1925 mirWeh (Eq. 5.39} ZDt P “bare” = 1.526 sedveh so a minimum 0f4 booths must be open 0-501: ...
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HW3 - Problem 5.1 Determine time when gueue clears m;= 5.2...

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