ANSWERS set 7 WAIT LINE. L=24 30-24 (1)(a)L=4 people in drug store.

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Presentation transcript:

ANSWERS set 7 WAIT LINE

L= (1)(a)L=4 people in drug store

(1)(b)Lq Lq = 24*24 30(30-24) Lq= 3.2 people waiting for service

(1)(c) W= W=.167 hr Interpret: W = 10 minutes from time you enter drug store until time you leave

(1)(d) Wq Wq = 24 30(30-24) Wq=.13 hour Interpret: 8 minutes before service

(1)(e) U=24/30=.80 80% efficiency P(server busy)

(1)(f) Po=1-.8=.2 P(no waiting)

(2) Cost of Waiting Av 3 customers arrive per hour(given) 10 hours per day(given) You compute: 30 customers/day Cost of waiting $5 per hour(given)

WAIT COST (A)(B) #SERVED/HR60/5=1260/10=6 WAIT TIME 3 = (12-3) 3 =.167 6(6-3) DAILY WAIT TIME 30(.028)=.8330(.167)=5 WAIT COST.83(5)=$4.175(5)=$25

Labor Cost (a)(b) Hourly wage$16$8 Labor Cost10(16)=$16010(8)=$ 80

Total Cost (a)(b) Wait Labor16080 Total$164.17$105=MIN

(3) SIMULATION ARRIVALPROBCUMULARR RN to to to to to to 00

(3) MAXIMUM UNLOADED MAX UNLPROBCUMULRN UNL TO TO TO TO TO TO 00

FIRST 5 DAYS: RN ARR RNSIM ARRUNL RNSIM UNL

delay Case 1: Demand > Supply Delay = Demand – Supply Case 2: Demand < Supply Delay =0 Next slide: Demand = 2B UNL Supply = UNL

FIRST 5 DAYS DELAYARRIV2B UNLMAX UUNL

DAYS 6-10 DELAYARRIV2B UNLMAX UUNL

DAYS DELAYARRIV2B UNLMAX UUNL 4-3= = = = =12333

Av number delayed per day Total delayed= = 6 Number of days = 15 Av = 6/15= 0.4