Test-Bank-for-Heizer-Operations-Management-9e

PROBLEMS

68.A waiting-line problem that cannot be modeled by standard distributions has been simulated. The table below shows the result of a Monte Carlo simulation. (Assume that the simulation began at 8:00 a.m. and there is only one server.)

Customer Number	Arrival Time	Service Time	Service Ends
1	8:06	2	8:08
2	8:07	10	8:18
3	8:12	10	8:28
4	8:24	11	8:39
5	8:30	5	8:44

a.What is the average waiting time in line?

b.What is the average time in the system?

(a) Waiting time is 0 + 1 + 6 + 4 + 9 = 20. Average waiting time is 20/5 = 4.0 min; (b) Total time in system is 2 + 11 + 16 + 15 + 14 = 58. Average time in system is 58/5 = 11.6 min. (Simulation of a queuing problem, moderate) {AACSB: Analytic Skills}

69.A distribution of service times at a waiting line shows that service takes 6 minutes 40 percent of the time, 7 minutes 30 percent of the time, 8 minutes 20 percent of the time, and 9 minutes 10 percent of the time. Prepare the probability distribution, the cumulative probability distribution, and the random number intervals for this problem.

Service time	Probability	Cumulative	Random number
		probability	intervals
6	.40	.40	01-40
7	.30	.70	41-70
8	.20	.90	71-90
9	.10	1.00	91-00

(Simulation of a queuing problem, easy) {AACSB: Analytic Skills}

591

70.A warehouse manager needs to simulate the demand placed on a product that does not fit standard models. The concept being measured is "demand during lead time," where both lead time and daily demand are variable. The historical record for this product suggests the following probability distribution. Convert this distribution into random number intervals.

Demand during lead time	Probability
100	.02
120	.15
140	.25
160	.15
180	.13
200	.30
Demand during lead time	Probability	Cumulative	Random number
		probability	intervals
100	.02	.02	01-02
120	.15	.17	03-17
140	.25	.42	18-42
160	.15	.57	43-57
180	.13	.70	58-70
200	.30	1.00	71-00

(Simulation and inventory analysis, moderate) {AACSB: Analytic Skills}

71.A distribution of service times at a waiting line shows that service takes 6 minutes 40 percent of the time, 7 minutes 30 percent of the time, 8 minutes 20 percent of the time, and 9 minutes 10 percent of the time. Prepare the probability distribution, the cumulative probability distribution, and the random number intervals for this problem. The first five random numbers are 37, 69, 53, 80, and 60. What is the average service time of this simulation run?

Service time	Probability	Cumulative	Random number		Simulation
		probability	intervals		frequency
6	.40	.40	01-40	1	(37)
7	.30	.70	41-70	3	(69, 53, 60)
8	.20	.90	71-90	1	(80)
9	.10	1.00	91-00	0

The average service time is 1*6 + 3*7 + 1*8 = 35 / 5 = 7 minutes. (Simulation of a queuing problem, moderate) {AACSB: Analytic Skills}

72.A distribution of service times at a waiting line indicates that service takes 12 minutes 30 percent of the time and 14 minutes 70 percent of the time. Prepare the probability distribution, the cumulative probability distribution, and the random number intervals for this problem.

Service time	Probability	Cumulative	Random number
		probability	intervals
12	.30	.30	01-30
14	.70	1.00	31-00

(Simulation of a queuing problem, easy) {AACSB: Analytic Skills}

592

73.	A distribution of service times at a waiting line indicates that service takes 12 minutes 30 percent
	of the time and 14 minutes 70 percent of the time. Prepare the probability distribution, the
	cumulative probability distribution, and the random number intervals for this problem. The first six
	random numbers were 99, 29, 27, 75, 89, and 78. What is the average service time for this
	simulation run?
	Service time	Probability			Cumulative				Random number			Simulation
					probability				intervals			frequency
	12		.30		.30			01-30				2	(29, 27)
	14		.70		1.00			31-00				4 (99, 75, 89, 78)
	The average	service time is			2*12 + 4*14 = 80				/ 6 = 13.33 minutes. (Simulation of a queuing
	problem, easy) {AACSB: Analytic Skills}
74.	Historical records on a certain product indicate the following behavior for demand. The data
	represent the 300 days that the business was open during 2000. Convert these data into random
	number intervals.
	Demand in cases		Number of occurrences
	7			52
	8			9
	9			14
	10			39
	11			72
	12			102
	Demand in cases		Number of				Probability				Cumulative			Random number
			occurrences								probability			intervals
	7			52			.18				.18			01-18
	8			9			.03				.21			19-21
	9			14			.05				.26			22-26
	10			39			.14				.40			27-40
	11			72			.25				.65			41-65
	12			102			.35				1.00			66-00

(Simulation and inventory analysis, moderate) {AACSB: Analytic Skills}

593

75.A small store is trying to determine if its current checkout system is adequate. Currently, there is only one cashier, so it is a single-channel, single-phase system. The store has collected information on the interarrival time, and service time distributions. They are represented in the tables below. Use the following two-digit random numbers given below to simulate 10 customers through the checkout system. What is the average time in line, and average time in system? (Set first arrival time to the interarrival time generated by first random number.)

Interarrival time	Probability		Service time	Probability
(minutes)			(minutes)
3	.25		1	.30
4	.25		2	.40
5	.30		3	.20
6	.20		4	.10

Random numbers for interarrival times: 07, 60, 77, 49, 76, 95, 51, 16, 14, 85
Random numbers of service times:					57, 17, 36, 72, 85, 31, 44, 30, 26, 09
Interarrival			Probability		RN			Service time					Probability					RN
time (minutes)					assignment				(minutes)								assignment
3		.25			01-25				1				.30					01-30
4		.25			26-50				2				.40					31-70
5		.30			51-80				3				.20					71-90
6		.20			81-00				4				.10					91-00
Customer	RN		Interarrival	Arrival		Service				RN	Service			Service		Time		Time
number			time		time		begins				time			ends		in		in
																line		System
1	07		3		3		3			57	2			5		0		2
2	60		5		5		5			17	1			6		0		1
3	77		5		10		10			36	2			12		0		2
4	49		4		14		14			72	3			17		0		3
5	76		5		19		19			85	3			22		0		3
6	95		6		25		25			31	2			27		0		2
7	51		5		30		30			44	2			32		0		2
8	16		1		31		32			30	1			33		1		2
9	14		1		32		33			26	1			34		1		2
10	85		6		38		38			09	1			39		0		1

Average time in line = 2/10 = 0.2 minutes; Average time in system = 20/10 = 2.0 minutes. (Simulation of a queuing problem, moderate) {AACSB: Analytic Skills}

594

76.Sam's hardware store has an order policy of ordering 12 gallons of a specific primer whenever 7 gallons are on hand. The store would like to see how well their policy works. Assume that beginning inventory in period 1 is 10 units, that orders are placed at the end of the week to be received one week later. (In other words, if an order is placed at the end of week one, it is available at the beginning of week 3.) Assume that if inventory is not on hand, it will result in a lost sale. The weekly demand distribution obtained from past sales is found in the table below. Also, use the random numbers that are provided and simulate 10 weeks worth of sales. How many sales are lost?

Weekly sales		Probability
3			.20
4			.30
5			.20
6			.20
7			.10
Random numbers for sales: 37, 60, 79, 21, 85, 71, 48, 39, 31, 35
Weekly sales		Probability				RN
					assignment
3			.20			01-20
4			.30			21-50
5			.20			51-70
6			.20			71-90
7			.10			91-00
Week	Order		Beginning			RN		Sales		Ending	Order?	Lost
	received		inventory							inventory		sales
1			10			37		4		6	Y
2			6			60		5		1
3	12		13			79		6		7	Y
4			7			21		4		3
5	12		15			85		6		9
6			9			71		6		3	Y
7			3			48		4		0		1
8	12		12			39		4		8
9			8			31		4		4	Y
10			4			35		4		0

Over the 10 weeks only 1 gallon of sales is lost.

(Simulation and inventory analysis, moderate) {AACSB: Analytic Skills}

595

77.The lunch counte at a small restaurant has difficulty handling the lunch business. Currently, there is only one cashier in a single-channel, single-phase system. The restaurant has collected information on the interarrival time, and service time distributions from past lunch hours. They are represented in the tables below. Use the following two-digit random numbers given below to simulate 10 customers through the checkout system. What is the average time in line, and average time in system? (Set first arrival time to the interarrival time generated by first random number.)

Interarrival time	Probability		Service time	Probability
(minutes)			(minutes)
1	.20		1	.20
2	.20		2	.30
3	.30		3	.30
4	.20		4	.20
5	.10

Random numbers for interarrival times:						32, 73, 41, 38, 73, 01, 09, 64, 34, 55
Random numbers of service times:						84, 55, 25, 71, 34, 57, 50, 44, 95, 64
Interarrival			Probability			RN			Service time					Probability					RN
time (minutes)					assignment					(minutes)								assignment
1		.20			01-20					1				.20					01-20
2		.20			21-40					2				.30					21-50
3		.30			41-70					3				.30					51-80
4		.20			71-90					4				.20					81-00
5		.10			91-00
Customer	RN		Interarrival	Arrival				Service			RN	Service			Service		Time		Time
number			time		time			begins				time			ends		in		in
																	line		System
1	32		2	2				2			84	4			6		0		4
2	73		4	6				6			55	3			9		0		3
3	41		3	9				9			25	2			11		0		2
4	38		2	11				11			71	3			14		0		3
5	73		4	15				15			34	2			17		0		2
6	01		1	16				17			57	3			20		1		4
7	09		1	17				20			50	2			22		3		5
8	64		3	20				22			44	2			24		2		4
9	34		2	22				24			95	4			28		2		6
10	44		3	25				28			64	3			31		3		6

Average time in line = 11/10 = 1.1 minutes; Average time in system = 39/10 = 3.9 minutes. (Simulation of a queuing problem, moderate) {AACSB: Analytic Skills}

596

78.Julie's Diamond Boutique is very concerned with its order policies related to one-carat diamond solitaires. Their current policy is to order 10 diamonds whenever their inventory reaches 6 diamonds. Currently there are 8 diamonds on hand. Orders are placed at the end of the month and take one month to arrive (e.g., if an order is placed at the end of month 1, it will be available at the beginning of month 3). The following distribution of monthly sales has been developed using historical sales. If Julie's does not have a diamond on hand, it will result in a lost sale. Use the following random numbers to determine the number of lost sales of one-carat solitaires at Julie's over 12 months.

Monthly sales			Probability
3				.20
4				.30
5				.20
6				.20
7				.10
Random numbers for sales: 10, 24, 03, 32, 23, 59, 95, 34, 34, 51, 08, 48
Monthly sales			Probability				RN
						assignment
3				.20			01-20
4				.30			21-50
5				.20			51-70
6				.20			71-90
7				.10			91-00
Month	Order			Beginning			RN		Sales		Ending	Order?	Lost
	received			inventory							inventory		sales
1				8			10		3		5	Y
2				5			24		4		1
3	10			11			03		3		8
4				8			32		4		4	Y
5				4			23		4		0
6	10			10			59		5		5	Y
7				5			95		7		0		2
8	10			10			34		4		6	Y
9				6			34		4		2
10	10			12			51		5		7
11				7			08		3		4	Y
12				4			48		4		0

Over the 12 months 2 sales are lost.

(Simulation and inventory analysis, moderate) {AACSB: Analytic Skills}

597