Question

The time between arrivals of parts in a single machine queuing system is uniformly distributed from 1 to 20 minutes (for simplicity round off all times to the nearest whole minute.) The part's processing time is either 8 minutes or 14 minutes. Consider the following case of probability mass function for service times: Prob. of processing (8 min.) = .5, Prob. of processing (14 min.) = .5 Simulate the case, you need to estimate average waiting time in system. Start the system out empty and generate the first arrival time, etc. You need to submit random numbers used (Use the third block of the Random number table to generate interarrival time and the fourth block to generate service time), interarrival time and service time for each part, discuss how you compute (10%) and (you need to generate 20 parts):

(1) clock arrival time for each part (10%),

(2) clock time starts to process each part (10%),

(3) clock departure time for each part (10%),

(4) total waiting time in system for each part (20%),

(5) plot average total waiting time in system versus part number (20%),

(6) plot average idle time of the machine versus part number (compute at the end of each processing) (20%)

Answer 1

The time between arrivals of parts in a single machine queuing system is uniformly distributed from 1 to 20 minutes. Therefore, the probabilities of a part arriving after 1 or 2 or 3 or so on till 20 minutes are equal.

Inter-arrival time

Inter-arrival time (minutes)	Probability	Cumulative Probability	Random Nos. assigned
1		0.05	00-04
2		0.1	05-09
3		0.15	10-14
4		0.2	15-19
5		0.25	20-24
6		0.3	25-29
7		0.35	30-34
8		0.4	35-39
9		0.45	40-44
10		0.5	45-49
11		0.55	50-54
12		0.6	55-59
13		0.65	60-64
14		0.7	65-69
15		0.75	70-74
16		0.8	75-79
17		0.85	80-84
18		0.9	85-89
19		0.95	90-94
20		1	95-99
Total	1

The processing time is either 8 minutes or 14 minutes with equal probability of 0.5

Processing or Service time:

Processing time (minutes)	Probability	Cumulative Probability	Random nos. assigned
8	0.5	0.5	00-49
14	0.5	1	50-99
Total	1

The question asks for 20 simulations. So there should be a set of 20 random numbers for inter-arrival time and another set of 20 numbers for processing/servicing time.

Random numbers for inter-arrival time: 38 84 74 30 34 96 09 46 61 41 02 93 94 90 00 71 84 98 30 82

Random numbers for service time: 22 04 86 13 33 00 99 74 75 27 43 68 22 59 20 66 00 24 01 96

Let the System start at 10:00 am

Run	Arrivals			Service
	Random no.	Inter-arrival time	Time as per clock	Random no.	Service time	Service starts	Service ends	Idle time	Parts Waiting time in Q
1	38	8	10:08	22	8	10:08	10:16	8
2	84	17	10:25	04	8	10:25	10:33	9
3	74	15	10:40	86	14	10:40	10:54	7
4	30	7	10:47	13	8	10:54	11:02		7
5	34	7	10:54	33	8	11:02	11:10		8
6	96	20	11:14	00	8	11:14	11:22	4
7	09	2	11:16	99	14	11:22	11:36		6
8	46	10	11:26	74	14	11:36	11:50		10
9	61	13	11:39	75	14	11:50	12:04		11
10	41	9	11:48	27	8	12:04	12:12		16
11	02	1	11:49	43	8	12:12	12:20		23
12	93	19	12:08	68	14	12:20	12:34		12
13	94	19	12:27	22	8	12:34	12:42		7
14	90	19	12:46	59	14	12:46	13:00	4
15	00	1	12:47	20	8	13:00	13:08		13
16	71	15	13:02	66	14	13:08	13:22		6
17	84	17	13:19	00	8	13:22	13:30		3
18	98	20	13:39	24	8	13:39	13:47	9
19	30	7	13:46	01	8	13:47	13:55		1
20	82	17	14:03	96	14	14:03	14:17	8
								Total idle time = 49 minutes	Total Waiting time = 123 minutes