Simulation of Demand Forecasting Problem - Quantitati - Quantitati ques Management eee | | Simulation Problems and Solutions in Operations Research The simulation problems and solutions in operations research are mentioned below Example: An ice-cream parlor's record of previous month's sale of a particular variety of ice cream as follows (see Table) Simulation of Demand Problem Demand (No. of Ice-creams) No. of days 4 5 5 10 = or s 1 ‘Simulate the demand for first 10 days of the month Solution: Find the probability distribution of demand by expressing the frequencies in terms of proportion. Divide each value by 30. The demand per day has the following distribution as shown in, table Probability Distribution of Demand Demand 0.20 0.27 0.03 bility and assign a set of random number intervals to various demand levels | a}oren| & Find the cumulative probal ‘The probability figures are in two digits, hence we use two digitrandom numbers taken from a random. muumber table, The random numbers are selected from the table from any row oF column, but in a onsecutive manner and random intervals are set using the cumulative probability distribution as c shown in Table,Cee Patan | Coma Probability _[ Random } Sane Interval 4 O17 0.17 5 0.33 0.50 6 0.20 0.70 7 0.27 Od 8 0.03 1.00 To simulate the demand for ten days, select ten random numbers from random number tables. The random numbers selected are, 17, 46, 85, 09, 50, 58, 04, 77, 69 and 74 The first random number selected,|7 lies between the random number interval 17-49 corresponding to a demand of 5 ice-creams per day. Hence, the demand for day one is 5. Similarly, the demand for the remaining days is simulated as shown in Table. Demand Simulation Day 1 roo Te s [9] 20] Random Number 17 46 8s | 09 50 ss 04 77 | 69 74 Demand = So [070 4m) 6) irene eae ae ese l | ‘ Example : A dealer sells a particular model of washing machine for which the probability distribution of daily demand is as given in Table. Probability Distribution of Daily Demand [_ Pemandraay] 0 T oe 00s | 028 | 020 | 025 | 010) ons 2 3 + 5 { Demand Simulation Problems using Random numbers The simulation problems using random numbers are given below Find the average demand of washing machines per day. Solution: Assign sets of two digit random numbers to demand levels as shown in Table.Re 1 andom Numbers Assigned to Demand Demand : Probability 0 bab ‘Cumulative Prol ti z 05 a bability | Random Number Intervals, z 0.20 O80) 05-26 os pag 0:50 a 4 = 0.75 0.10 _ E vi "S a = aa Ten rand rs lom numbers that have been selected from random number tables are 68, 47, 92, 76, 86, 46, 16, 2! 5 - 28, 35, 54. To find the demand for ten days see the Table below. Table 15.7: Ten Random Numbers Selected Tratss Random Number Desa tay : = 5 = al oF s = 76 z 5 36 3 7 “SS = z x z = 35 = 77 3a $ Toral Demand a ‘Average demand =28/10 =2.8 washing machines per day. The expected demand /day can be computed as, Expected demand per day Expected demand per day = UN sey where, p = probability and x.= demand = (0.05 % 0) + (0.25 * 1) + (0.20 * 2) +(0.25 3)+(0.1 « 4) + (0.15 5) = 2,55 washing machines. ‘the average demand of 2.8 washing machines using ten-day simulation differs significantly when compared to the expected daily demand. If the simulation is repeated number of times, the answer would get closer tothe expected daily demand.Example: The Lajwaab Bakery Shop Problem The Lajwaab Bakery Shop keeps stock of a popular brand of cake. Previous experience indicates the daily demand as given below: Daily demand Probability | o 0.01 15 01s 25 0.20 35 0.50 <5 0.12 50 0.02 Consider the following sequence of random numbers: 21, 27, 47, 54, 60, 39, 43, 91, 25, 20 Using this sequence, simulate the demand for the next 10 days. Find out the stock situation, if the owner of the bakery shop decides to make 30 cakes ever demand for the cakes on the basis of simulated data. Solution. ry day. Also estimate the daily average Using the daily demand distribution, we obtain a probability distribution as shown in the following table. Table 1 Daily demand | Probability Cumulative probability Random Numbers ee 0.01 0.01 0 15 0.15 0.16 115 IE28 Sr 35 Keiteac 45 oa so 0.02 At the start of simulation, the first random number 21 ‘able 2. The demand is determined from the cumulat ‘generates a demand of 25 cakes as shown in 've probability values in table I. At the end of first day, the closing quantity is 5 (30-25) cakes, Similarly, we can calculate the next demand for others. Table 2 Demand | Random Numbers | Next demand _| Daily production = 30 cakes Leftout | shortage A 2 25 5 3 B 27 25 10 a 7 35 5 a 54 35 0 5 60 35 4 6 39. 35 Bo) 3 4B 35 8 91 459 25 2 10 20 25 Total 320 Total demand = 320 Average demand = Total demandino. of days The daily average demand for the cakes = 320/10 = 32 cakes Example: A farmer has 10 acres of agricultural land and is cultivating tomatoes on the entire land. Due to fluctuation in water availabilty, the yield per are differs. The probability distribution yields are given below: a. The farmer is interested to know the yield for the next 12 months if the same water availability exssts. Simulate the average yield using the following random numbers 50, 28, 68, 36, 90, 62, 27, 50, 18, 36, 61 and 21, given in table. ‘Vicia of tomatoes Probability 200 ous 20 02s 240 0.35 260 0.3 280 ou2 b. Due to fluctuating market price, the price per kg of tomatoes varies from Rs. 5,00 to Rs, 10.00 per kg, The probability of price variations is given in the Table below, Simulate the price for next 12 to determine the revenue per acre, Also find the average revenue per ace, Use the following 74, 05, 71, 06, 49, 11, 13, 62, 69, 85 and 69. months random numbers 53,Price per kg (Rp a ot Probability Ges 5 0.05 ee 35-— O15 | 3.00 02 [>a 025 O15 Solution: Table for Random Number Interval for Yield ‘Yield of tomatoes Probability ‘Cumulative Probability [Random Number per acre a 200 ous ous an 220 0.28 0.40 15-39 240 035 os 40-74 260 043 0.88 75-387 280 O12 1.00 8-99 Table for Random Number Interval for Price Price Per Kg | Probability ‘Cumulative Probabili ‘Random Number Interval 500 005 008 00-08 6.50 os 0.20 0s 19 7.50 0.50 20-49 8.00 025 075 50-74 10.00 02s 1,00 73-99 ‘Simulation for 12 months period23 (Rs) & Month Yield (2) 220 26 6 240° 7 220. s 240 9 220 10 220 ul 240 10.00 12 220 8.00 pane ZEOOY (21298) | AW revenue = 243642 = per acre 2290] \2_ tc precage. Je + 2e2 =Rs. 1373.50 e SARAAACC CE = 233-33 Example: 1.M Bakers has to supply only 200 pzzas every day to their outlet situated in eity bazar. ‘The production of pzzas varies due to the availability of raw materials and labor for which the probability distribution of production by observation made is as follows: ‘Simulation Problem 197 | 198 | 199 | 200 | 201 | 202 | 203 | 204 Production per day | 196 0.10 | 0.16 0.20 | 0.21 0.08 | 0.07 | 0.03 Probability 0.06 | 0.09 Simulate and find the average number of pzzas produced more than the requirement and the average number of shortage of p2zas supplied to the outlet. “Solution: Assign two digit random numbers to the demand levels as shown in table Random Numbers Assigned to the Demand LevelsDemand | Probability | | Cumulative Probability No of Pizzas shortage 196 0.06 0.06 00-05 197 0.09 O18 06-14 198 0.10 0.25 15-24 ies | es 0.41 25-40 200 0.20 0.61 41-60 201 0.21 0.82 61-81 202 0.08 0.90 82-89, 203 0.07 0.97 90-96 L 204 0.03 1.00 97-99 Selecting 15 random numbers from random numbers table and simulate the production per day as shown in table below Simulation of Production Per Day [Trial Number | Random Number | Production Per | No of Pizzas over | No of pizzas day produced shortage 1 199 5 1 2 | __200 : : 3 201 1 a d 201 1 i as T 74 201 1 causalThe average number of pizzas produced more than requirement = 12/15 = 0.8 per day The average number of shortage of pzzas s =4ns = 0.26 per day Simulation of Queuing Problems - Quantitative Techniques for Management Example : Mr. Srinivasan, owner of Citizens restaurant is thinking of introducing separate coffee shop facility in his restaurant. The manager plans for one service counter for the coffee shop customers. A market study has projected the inter-arrival times at the restaurant as given in the table. The counter can service the customers at the following rate: Simulation Problem with Solution Simulation of Queuing Problem 10SL] Random | Inver] Arrival ] Service | Rondom [Service ] Service [Waiting Time No. | Number | arvival | Timeat | Startsat | Number | Time Ends at [wate Gnvivay | Tine Gervieg | (in) Customer atin nee © | 705 | 706 | 36 7] 10 2) 7 Em 3s 1 a] 2 Ca 3 [720 | a + [7 [m2 [os 2 22 s[ a an |e osd| erase |S + | 7a7 - 6 | os |r| | 3 [730 2 7] ns 4 | 729 [730 | © | 136 1 +] 7 s_ [7 | 736 | 6 [72 2 9 | 70 739_|_7a_ | _15 3 [as A 5 10 | 07 7a_|_145_|_33 + | 749 4 un] 7a7_ [749 ot 2 731 2 = 2] a i 4 [756 1 Bf 3 35_ | 756 | 03 2 [758 1 = “| 7 s [so [so | ss [4 | a0 1 2 5 37 «| so | a0] s6_| 4 | sos 5 = 6 | 8 © | s10 | 81 | 0s om an 5 2 7 | 38 3] a9 | TN OE a 1 | 4 i] | Ai | E 2 | is 3 320 | _s21_| 3 [a 1 5 20 | 38 3 | 823) 28 | 76 3 [829 T 5 a ae 2 | Mean waiting time of eustomer before service = 20/20 = 1 minute ‘Average service idle time = 17/20 = 0.85 minutes ii Time spent by the customer inthe system = 3.6+ 1 = 46 minutes 2Example : Dr. Strong a dentist sch Pte ae ules al his patients for 30 minute sppointment, Some ofthe less than 30 minut : Rica 'es depending on the type of dental le shows the sum ae te nmary of the various cate : *pories of work, th i actually needed to complete the wor ae ik. Simulation Problem Category gory Time required (minutes) Probability of category Filling 45 0.40 Crown 6 1s Cleaning is 0.15 Extraction 45 0.10 Checkup 15 0.20 Simulate the dentist's clinic for four hours and determine the average waiting time for the patients as well as the idleness of the doctor. Assume that all the patients show up at the clinic exactly at their scheduled arrival time, starting at 8.00 am. Use the following random numbers for handling the above problem: 40, 82, 11, 34, 25, 66, 17, 79 Solution: Assign the random number intervals to the various categories of work as shown in table, Random Number Intervals Assigned to the Various Categories Category of work | Probability | Cumulative probability | Random Number Interval Filling 0.40 0.40 00-39 Crown 01s 40-54 Cleaning 01s 0.70 55-69 Extraction 0.10 0.80 70-19 1.00 Check-upAssuming the dent in table, Arrival Pattern of the Patients Scheduled Arrival Patient Number ist clinic starts at 8,00 am, the arrival pattern and the service category are shown Cleaning 15 2 3 [o Check-up 3 9.00) Filling 4 Fillin 5 5 Filling 45 6 g Filling 45 Extraction 45 The arrival, departure patterns and patients’ waiting time are tabulated. Time | Event (Patient Number) | Patient Number (Time to go) | Waiting (Patient Number) 8.00 Larrives 1(60) E 830 2 amives 130) 2 9.00 [1 departure, 3 arrives 2(15) 3 ois 2 depart 345) : 9.30 4 amive 330) 4 10.00] 3 depart. 5 arrive 4(45) 5 10.30 6 anive 4(15) 56 10.45 4 depart 5(45) 6 11.00 T arrive 5 (30) 67 11.30] 5 depart, 8 arrive 6(15) 78 1145 6 depart 745) 8 12.00 End 730) 8 ‘The dentist was not idle during the simulation period, The waiting times for the patients are as given in table below, 4Patient's Waiting Time Paties * nt Arrival Time Servi a oh x Starts Waiting time | 2 8.30 9. S z 3 9.00 a i : 10.00 10.45 e € 10.30 2 é 3 11.30 f 11.00 1145 i i130 12.30 a Total a The average waiting time of patients = 285/8, S minutes. Simulation of Inventory Problems - Quantitative Techniques for Management A dealer of electrical appliances has a certain product for which the probability distribution of demand per day and the probability distribution of the lead-time, developed by past records are as shown in the following tables. Probability distribution of lead demand ‘Demand (Units) 2 3 4 5 aia 10 Probability 0.05 | 0.07 | 0.09 oxo [007 | 0.06 Probability distribution of lead time Lead Time (Days, The various costs involved are, Ordering Cost = Rs. 50 per order Holding Cost = Rs.1 per unit per day Shortage Cost = Rs. 20 per unit per dayGa ‘| search Simulation Problems and Solutions in Operations Reseat fi given below The simulation problems and solutions in operational research are giv ete ‘der point and the The dealer is interested in having an inventory policy with two parameters, the reorder p a ber of order quantity, ie., at what level of existing inventory should an order be placed and the number 35 units to be ordered. Evaluate a simulation plan for 35 days, which calls for a reorder quantity of units and a re-order level of 20 units, with a beginning inventory balance of 45 units. ‘Solution: Assigning of random number intervals for the demand distribution and lead time distribution is shown in Tables 15.25 and 15.26 respectively. Random Numbers Assigned for Demand Per Day Random Nuinber Interval Demand perday | Provabittty | Cumulative probability 2 0.05 005 00-08 0.07 012 05-11 4 0.09 oat 12-20 5 015; 036 6 0.20 056 7 O21 O77 3 0.10 037 9 0.07 094 10 0.06 100 Random Numbers Assigned for Lead-time Lead Time (Days) Probability | Cumulative probability Random Number Interval 1 0.20 0.20 00-19 0.50 20-49 25 3 0.85 50-84 4 1.00 85-99 16ion Work-shee et for Inventory Problem (Case = 1) pales a Short- i van Tea | karte RO Ore aig | Sm Nabe Nabe Tt Wot | Recel-| ing | Meat? | ag Day | Nusher | Demand Lead | Tite | stent | Re |” ing ad ~ | Demand) Gead*| (ays))| ayaa a Ie Time)Total Reorder Quantity = 35 units, Reorder Level = 20 units, Beginning Inventory = 45 units 18Simulation Work-: sheet for Inventory Problem (Case = 1) Random Random | vi | Lead Inventory) Day| Number |Demana|\"€"] rie | at endo | Qt [Ordering Hotang [shortage (Demand) Gead |. Received] Cost | Cost | Cost Time {293} day oe : es : : - ire | a8 i ail ea i : 38 i 2) | seeds 6 ela L ges y 3 [eas 6 1 lee : ; 26 : ine nels o—|—3 Seien ; so | 20 - Si |e 46 6 : : 4 S z as 3 6 | 46 6 - |: 8 i = S é alent F 5 ale 30 A 31 : 3 5 : 29 5 29 5s = 3 8 = oy eae re ees : = 10] 40 6 ¢ 3 ea =a 2 - uf st 6 aie ae aa 2 : ie = nl 9 7S ea a 7 i i 32 : ae Z ila aS 30 : seule 5 3 a E fib IT 21 - 15 68 ol z = a = a re : 3 Fag ya as : 5 : 16 4 - 2 . 5 i - 7] 96 10 | le : : = [asa eee | ace ibe - melee : len | > 9] 83 . i m 19300 683 20 EZ | E Ti 20 units, Beginning Inventory = 45 units Reorder Quantity ~ 30 units, Reorder Level = ‘The simulation of 35 days with an inventory policy of reordering quantity of 35 units atthe time of as worked out in table, The table explains the demand inventory level at the end of day is 20 units, ‘ost, holding cost and shortage cost for each day. inventory level, quantity received, ordering ¢' Completing a 35 day period, the costs are Total ordering cost = (6 x 50) = Rs 300.00 Total holding cost = Rs. 768.00 Since the demand for each day is satisfied, there is no shortage cost. 20Therefore, Total cost = 300 + 768 =Rs. 1068.00 Fora different set of, A Parameters, with a re-order quanti b quantity of 30, units, ifthe 35-day simulation i t tion is performed, table. —— a Total ordering cost = 6 x 50=Rs, 300.00 Total holding cost = Rs. 683.0 Total shortage cost = Rs. 20.00 Therefore, Total cost = 300 + 683 +20 =Rs. 1003.00 which may lead to some cost. In this type of problems, the approach with various combinations of two pa a large number of times to find the total cost ‘ofeach experiment, compare the to optimum alternative, i.., that one which incurs the lowest cost. Simulation Sample Problems (with answers): Problem A) Given the following probabilities assign ‘the correctly simulate the probabilities. [Event Probability 7 [Random Number a 0.032 A B (0.504 (0.348D ‘0.16 000 to the events to B) Given the following probabilities assign the random numbers 0001 to 10 correctly simulate the probabilities. Event [Probability [Random Number i lV —— (oua3s if nee Re x fo.o02s | oa | fF 03450 iz (0.5290 7 | Problem Two A machine that we are concemed with has a breakdown rate as follows. On the first day of operation after repair it has an 8 % chance of breakdown. On the second day and each following day, the probability of breakdown increases by 3 %, ie., P(Failure) = .05 + .03X, where X is the # days of operation since repair. Assuming that the machine was repaired yesterday, and that the ‘machine is repaired on the same day that it breaks down, Also assume that the machine will break down if the random number given is less than the P(Failure) for that day. Simulate 10 days of operation and determine the following: A) What is_—the_~—total_~—snumber_~—of_—_—breakdowns. ithout a _—_ breakdown? B) What was the — most _—_days C) Did the machine ever break two days in a row? [Day |x |p) [RN# [Break? (Break? i El ih a ce are 3 (i | x [ ya [61 [29 2Problem A local company stocks Product AM22 each period. Tide unit and it sells the product for $ 200,00 Per unit. Units sold the next period at charges itself $ 10.00 ie a nae Sa own bel lost sales, The distribution of demand is shown bel 20 units in beginning inventory at the start of period one, Cost of Goods is based « Period. Simulate 10 periods of demand with an order policy of ordering 250 units per ordered in “"a period? are!) availAblel fOrmnNeal oman ‘not sold in one For your simulation determine the A) Average per period total stock B) Average per period sales in C) Average per period beginning —_ inventory and _— inventory —_ charges. D) Average per period lost sales in units and lost sales cost. E) Average per period total cost. F) What was the maximum ending inventory. G) Does it appear that we should change our order policy? Why? [Demand Random Number /[ Random ‘Number String Roo (01-05 ‘| 0 230 (06-15 [4 240 16-35 [ B 250 ibs [ pr 290 ee ce ae 300 [Pp - f oe [| ip 35Four a 1 and service rates do not fit any Key ‘A local company has a queuing system where the arival an he aon distabutionsom ae Below: 418 naa tibiae ul OCEL ON 4 Pi cane an Simulate the system for 10 on ve Picea A)imortthen 5 saverngolmeamtnbersjmgitOCA, gullet) xirvi a; ' ine . Beith wverngey nites, cok, tenting, the, diag served per, ‘hour. ©) the — average ~—snumbers._ of, items S Ca jueue). D) the maximum length of the ending line a E) the number of hours where there was zero items in the ending queue. Number Arriving lex ee Random Number pets Served per Server per hour ps p20 6 oe {Ie aids i 7 (46-80 8 a fis fio 8=—SC«*D 3 | Alll items wait in a single line, Items are served on a first come, first served basis. The maximum queue length is infinite as is the population for the system. THERE ARE TWO (2) SERVERS. The beginning queue is three @). RANDOM NUMBER STRINGS Arrival faa [Server One ~ ies or Solutions Problem ‘One A) 24Probability 5 =! 0.1235, 0001-1235, xX (0.0025, 1236-1260 f ly 0.3450 1261-4710 IZ 0.5290 ‘4711-10000 Problem Two ‘A) What is the total number of breakdowns. 2. B) What was the most days without a breakdown? _4 C) Did the machine ever break two days in a row? _No___ Day x Pe a Break? [Day XP) fi T [08 (57 No Cra lel 2 ma eial |” No 7 4 [a7 B 3 [a [a a 4 1 [os [6 (29 NolProblem Three Pr lora [Stock RN|Dem [Sale [EI {Ls |coG fic |cLs|rc jTR Pp [1 Po 50 [270 jg0 (290 270 (0 [20 [27000 |400 200 27600 [54000 | 26400 230 230 20 0 [23000/0 0 [23000 [46000 | 23000 240 240 Bo [0 [24000400 fo [24400 _ [48000 23600 240 [240 [40 (0 [24000/600 (0 ~ faaeo0 [48000 23400 320 [290 [0 [30 [29000 [800 “300 [30100 [58000 | 27900 ~ 20000 [40000 | 20000 200 (200 [50 o [20000 86 [300 300 jo {0 [30000 29000 31000 (60000 [25000 [50000 [25000 9 [25400 [50000 24600 fio fo 250 250 55 [250 [250 [0 (0 [2500/0 0 25000 [50000 25000 B 0 [250 [250 [41 [250 [250 Jo [o {25000 pe ps0 250 |79 290 (250 [0 [40 [25000 1 I 160, 2660 [2610 2520 [90 | [3200 [p00 [256100 [504000 [247900 [Ave 16| 266 ~ [261 [252 fo { [320 [90 | 25610 | so4o0| 24790 Based on this simulation they may want’ to increase order’ amount Note Revenue and Profit not requested.Solutions of Assignment.8 Q1.Monte carlo simulation j nis (Stochastic problems whe eterministic probi (©) Stochastic Te Wher PassaGe oft probl ge of time ive role, for solving Q2. Monte Cari (a) Staten Suton is generally i Dynamic c) Static or dynamic (a) None of these Solutions of questions fr om, Refer the following for a3-06, wa Astore has one counter. TI ival tir Arson he probability of inter-arrival time and service time of customers are given in Inter- service | onal Probability| time Probability ax teri) (olay 1 0.2 3 0.2 2 03 ra 05 3 0.357 | nara LOTS 4 0.1 5 04 Random numbers used for prediction of inter-arrival time and service time are as per the table given below: [Customer 1] 2[3]4/5|6\7 B | 9 | 10] 11] 12 Rn. for | | | Arrival 61 | 55| 4 | 33 | 19 | 25 | 79 | 93 | 18 | 49 | 92 | Ran. for | service | 28| 1 | 61 | 85 | 67 | 531 62) 79 | 66 | 63 | 33 | 7 It is assumed that first customer comes at 0 time. Random numbers used are from the set of 100 ease numbers from 00 to 99. Simulation is to be caried out to find the performance measures ofa queueing system. @3. The service start time for 10 customer will be at (a) 87 min. (b) 47 min. (0) 42 min (d) 52 min. Q4. Waiting time in queue by 6 customer wil be (in minutes) (a) 11 (b) 15 (c)6 (a) 18Q5. The arrival time of 7 customer will be at (a) 12 min. (b) 15 min. (c) 21 min. (4) 20 min, Qs. ihe time spend by 5" customer in the system will be (in minutes) (a) 5 (b)7 (c) 16 (d) 23 Refer the following for Q7 to Q10. A store has one counter. The probablity of inter-arrival time (in min) and service time (in min.) of customers are given in the following table. "nterarival | probability Stime | Probability 1 02 1 we 2 03 3 cm [arma 03 5 Ws 4 0.14 E 5 0.4 Random numbers used for prediction of interarrival ime and service time as per the table given below: Customer _[1 [2 [3 [4 [5 [6 [7 [a [9 [40 ]44 | 42] Ran. for faa | arrival 61 |55 |14_| 33 | 19 |25 | 79 | 93 | 48 | 49 | 92 Rn. for | service 28/1 | 61 | 85 | 67 | 53 | 62 | 79 | 66 | 63 | 33 | 77 It is assumed that first customer comes at 0 time. Random numbers used are from the set of 100 random numbers from 00 to 99. Simulation is to be carried out to find the performance measures of a queueing system, Q7. The service start time for 11" customer will be at (a) 3 min. (b) 5 min, (c) 31 min. (d) 34 min. Q8. Waiting time in queue by 5" customer will be (in minutes) (ao (b) 5 (c)7 (d)8 Q9. The arrival time of 4" customer will be at (a) 5 min. (b) 6 min, (c) 7 min (d) 9 min.gio. The time spend by gth 8 i YS" customer inthe syste mM will be (ir (13 be (in minutes) (a) 14 For Q 11-13: For a particular sh Op, the dail ily demand of an item with associated probabilities is given below: foal demand | 0 10 [20 [a )_ i) y 0.01 [0.20 [0.15 [0.50 12 0.02 If random number stream (X; nae + e)mod mi With Racer, ne tags = 0) esa using linear congruential generator (Xi = a"Xes ait, aces dally demand for first four days will be (b) 30 (c) 35 (d) 27 @12.Average daily demand for first ten days will be (a) 25 (b) 30 (c) 35 (d) 27 Q13.Demand on 5" day is expected to be (a) 10 (b) 20 (c) 30 (d) 40 For Q 14-15: For a particular shop, the daily demand of an item with associated probabilities is given below: 10 20 30 40 50. 0.50 [0.12 | 0.02 [Daily demand | 0 [Probability [0.01 [0.20 0.15 0 random numbers generated between 00-99) ce of random numbers (out of 10 ily demand will be a14.For the sequen saad ara a5.26, 39/65, 76) 12) 0573) Seyi a nGla alae da (a) 25 (b) 30 ,(c) 35 (d) 24 Q15.For the sequence of random numbers (out of 100 random numbers generated between 00-89) used are as 40,19, 87,83,73,84,29,09,02,20,the average daily demand will be (a) 25 (b) 30 (c) 22 (d) 27 Solutions For Q3-6 [Rn. Ty Time | Wait Time | Time Custome | (arrival |Intarr |R.n.for Arr | Serv |ser |time- serv _| cust in r ) time _| serv time |beg | queue _|ends _| syst 1 5 0 0 5 5 2 64 3 3 5 2 8 5 3 55 3 5 8 2 13 7 4 if 4 7 13 6 20 413, 5 33 2 5 20 MW 25 16. 6 19 1 STIG | B25 45 30 20 7 25 2 5 30 18 35 23 8 3 7 35. 20 42 27 9] 5| 5 42 22 47 27 10 1 5 47, 26, 52 34 Siri [eau 9) 2 5| 52 29 57 34 12 92 5 2 a 57, 29 64 36 Fal |nammres 4 7 64 32 74 39 14 61) 3 7 71 36 78 43, 7 oe ee 7 78 a 85 48 (a 16 78, 3 Z 85 45, 92 52. rd] 2 5 92, 50 97, 55 18 32 2 3 7 53] 100 56Rn. Int arr : leima” |for [Arr | Serv | tim Wait | Time | Tim |time serv [time | time | tim ‘ont, | case 2 aa 28 0 : serbeg | queue | ends _| syst. 3 8 1 3 o 0 3 3 55 3 i 3 0 4 ‘il ; 61 6 3 7 4 1 Ae | 85/_7 5 9 2 a 7 2 ea 3 2 14 7 6 19| 1 53 ra 3 14 a 7 a 7 25 2 3 ar 7 20 40 =n 7a 2 62] 12 3 20 8 23. 4 = 79 15 5 23 9 03 = a 8 28 13 70 18 ji a 20 3 28 8 31 14 1 49 21 3 34 40 34 13 Wf 2 8 23 3 34 14 37 14 28 5 37. 9 42 14 13 a ee | a a 32 5 42 10. 47 15 Te 4 | 3 86 35 5 47 | 12 52 17 28 2 79 37 5 52| 15 57. 20 16, iS a8] 40 5 57. 17 62 22 17 24 2 43| 42 3) 62 20 65 23 l 18 32 2 7441 1 65 24 66 22 For Q11-13 Daily Demand Probability ] Cumulative Random number a probability interval 0.01 0.01 00 10 0.20 0.21 [01-20 20 0.45 0.36 21-35 30 0.50 0.86 36-85 40 0.12 0.98 86-97 [50 0.02 1.00 98-99 Rand no LC generator 7 LeG(a=17,0=4, X0=27, m=100) S.No. Xin | Xi 1 27| 63 2 63 75 3 75 79 4 (79) amd 5 47 Bi 6 3 55 7 55 39 8 39 67 9 67 43 10] 43 35Jo] a]sJola]a]o]ry a ie S| 35| 20 ‘Avg demand for first four days= 30 ‘Avg. daily demand for first 10 days= 270/10=27 Demand on 5" day= 10 jae Fn a | | 14 DallyDemand Probability ‘Cumulative Random number probability interval o 0.01 0.01 00 10 0.20 (07 01-20 20 0.15 0.36 21-35 30. 0.50 Hi 0.86 36-85, 40 0.12 0.98 86-97 [50 [0.02 [4.00 [98-99 [Days |Randomnumbers [Demand | 1 25 re 2 2 200 iE 65" 4 76 5 12 6 05 7 73 8 89 9 19 10 [49 ‘Avg, daily demand 240/11Random numbers = 40 19 87 83 73 84 29 09 02 10 20 Avg. daily demand 220/10=22 ZJoloo| Jojo) s]o!