WAITING LINES AND SIMULATION • I. WAITING LINES (QUEUEING) : • II. SIMULATION WAITING LINES • • • • I. Length of line: number of people in queue II. Time waiting in line III. Efficiency: waiting vs idle server IV. Cost of waiting I. WAITING LINES • ASSUMTIONS • 1) FIRST COME FIRST SERVE • 2) ARRIVALS COME FROM VERY LARGE POPULATION • 3) NUMBER OF ARRIVALS IS POISSON • 4) SERVICE TIME IS EXPONENTIAL • 5) ARRIVALS INDEPENDENT APPLICATIONS • • • • • BANK TELLER LINE, CAR WASH INTERNET: CABLE VS PHONE LINE WAITING FOR CABLE GUY METERED FREEWAY ON RAMPS WAREHOUSE: ORDERS WAIT TO BE SHIPPED • AIRPLANES WAITING TO LAND av# arrivals av# served EXAMPLE: AUTO REPAIR • ONE MECHANIC • MAY NOT BE POISSON IF CUSTOMERS ARE CLUSTERED EARLY MORNING OR AFTER WORK • MAY NEED TO USE SIMULATION LATER 2cars / hr 3cars / hr L=Average Length • ALL customers in system • Waiting AND being served L 2 2 3 2 Lq=Average Length of queue • Customers waiting in line • Number waiting to be served Lq ( ) 2 2 2 Lq 1.3 3(3 2) W=Av Time customer in system • From arrival time to departure time • Time waiting and being served 1 W 1 W 1hr 3 2 Wq=Av time customer waits in queue • • • • • Waiting to be served Marketing, Service operations management Customers may go to competitor if Wq big Exception: lowest price(trade off) Car dealer: Wq=0 Wq ( ) 2 Wq .67hr 3(3 2) Interpret Wq • Wq=40 minutes waiting in line • W=60 minutes in system • 20 minutes being served U=Utilization • U=efficiency • Probability server is busy • Probability customer has to wait U U=2/3 67% efficiency Po=P(zero customers in system) • • • • Po=1-U P(server is idle) P(customer does not have to wait) Here: Po = .33 COST OF WAITING SUPPOSE EACH HOUR A CUSTOMER WAITS COSTS $10 INTANGIBLE COST • NOT ACCOUNTING COST • MARKETING ESTIMATE • USED FOR DECISION MAKING SUPPOSE MECHANIC RESIGNS • TWO ALTERNATIVE ACTIONS • ACT 1: MECHANIC #1, $17/HR LABOR COST, 3 CARS/HR • ACT 2: MECHANIC #2, $19/HR, 4 CARS/HR • 8 HRS/DAY MINIMIZE TOTAL COST • TOTAL COST = WAITING COST + LABOR COST • LABOR COST = (8)(COST/HR) • WAIT COST = (#HRS WAITING)($10) • AVERAGE #CARS ARRIVE/HR= 2 • TOTAL #CARS/DAY = 8(2)=16 Wq ( ) MECHANIC #1 3 CARS/HOUR 2 Wq .67hr 3(3 2) MECHANIC #2 • 4 CARS/HOUR 2 Wq .25 4(4 2) WAIT COST MECHANIC#1 MECHANIC#2 #SERVED/HR 3 4 WAIT TIME .67 HR .25 HR DAILY WAIT TIME WAIT COST .67(16)= .25(16)= 4HR 10.67HR 10.67(10)=$107 4(10)=$40 LABOR COST MECHANIC#1 MECHANIC#2 HOURLY WAGE $17/HR DAILY LABOR 8(17)=$136 COST $19/HR 8(19)=$152 Total cost MECHANIC#1 MECHANIC#2 WAIT COST $107 $40 LABOR COST $136 $152 TOTAL COST $243 $192=MIN HIRE SECOND MECHANIC? SIMILAR TABLE: SERVERS VS 1 SERVER 2 II. SIMULATION • DEFINE PROBLEM • DEFINE VARIABLES • BUILD MODEL: IMITATE BEHAVIOR OF REAL WORLD • LIST ALTERNATIVE ACTIONS • RANDOM NUMBERS • CHOOSE BEST ALTERNATIVE MONTE CARLO SIMULATION • • • • ADVANTAGES Flexibility Probabilities: Client understands model • Familiar simulations: dice, board games, video games, flight simulator • DISADVANTAGES • No mathematical optimization (LP guarantees optimum) • Trial and error • Might not try best action EXAMPLES • APOLLO 13 EMERGENCY RETURN • WEATHER FORECAST • SUGAR PLANTATION DECISION WHICH FIELD TO BURN EXAMPLE: WAIT LINE • PREVIOUS SECTION • RESTRICTIVE ASSUMPTIONS • EXACT FORMULAS • SIMULATION • NO RESTRICTIVE ASSUMPTIONS • ONLY APPROXIMATIONS EXAMPLE: WAIT LINE • • • • • • REFERENCE: RENDER, BARRY QUANTITATIVE ANALYSIS, P 708 BARGES ARRIVE AT PORT BARGES UNLOADED IN PORT OBJECTIVE: MINIMIZE DELAY FCFS:FIRST COME FIRST SERVED GIVEN: PROBABILITY DISTRIBUTIONS • X1= NUMBER OF BARGES ARRIVING AT PORT • X2= MAXIMUM NUMBER OF BARGES UNLOADED IN PORT ARRIVALS X1 P(X1) O .13 1 .17 2 .15 3 .25 4 .20 5 .10 STEP1:CUMULATIVE PROB X1 P(X1) P(X1<x) O .13 .13 P(X1<0) 1 .17 .30 P(X1<1) 2 .15 .45 P(X1<2) 3 .25 .70 4 .20 .90 5 .10 1 STEP 2: RANDOM NUMBER INTERVALS X1 P(X1) P(X<x) X1 RN O .13 .13 P(X1<0) 01 to 13 1 .17 .30 P(X1<1) 14 to 30 2 .15 .45 P(X2<2) 31 to 45 3 .25 .70 46 to 70 4 .20 .90 71 to 90 5 .10 1 91 to 00 STEP 3: SIMULATE ARRIVALS DAY 1 X1 RN (GIVEN) 06 SIMULATED ARRIVALS 0 2 50 3 3 88 4 4 53 3 MAX UNLOADED X2 1 P(X2); GIVEN .05 2 .15 3 .50 4 .20 5 .10 STEP 4: CUMULATIVE PROB X2 P(X2) P(X2<x) 1 .05 .05 2 .15 .20 3 .50 .70 4 .20 .90 5 .10 1 STEP 5: RANDOM NUMBER INTERVALS X2 P(X2) P(X2<x) X2 RN 1 .05 .05 01 to 05 2 .15 .20 06 to 20 3 .50 .70 21 to 70 4 .20 .90 71 to 90 5 .10 1 91 to 00 STEP 6: SIMULATE UNLOADING DAY X2 RN (GIVEN) 1 63 SIMULATED MAXIMUM UNLOADED 3 2 28 3 3 02 1 4 74 4 UNLOADED=MIN(3),(4) (1)#DELAYED 0 (2) ARRIV 0 (3) TOTAL 0 0 3 3 0 4 4 4-1=3 3 3+3=6 (4)MAX UNLOA UNL DED 3 MIN(0,3 =0 3 MIN(3,3 =3 1 MIN(4,1 =1 4 MIN(6,4 =4 AVERAGE NUMBER DELAYED • AV = TOTAL DELAYED = TOTAL NUMBER DAYS = ¾ = 0.75 • REAL-WORLD: WOULD RE-DO SIMULATION WITH MORE WORKERS TO UNLOAD BARGES TO RECALCULATE AV