OR 508 Exercises Solution.pdf

Prepared by: Youssry Hamdy- CS Diploma

OR 508 (Modeling and Simulaon)

CS Diploma

1 Prepared by: Youssry Hamdy-CS Diploma

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Solu!on of Midcourse Exam Dec. 2010 Members of the Willow Creek Emergency Emergency Rescue Squad know from past experience that they will receive between zero and six emergency calls each night, according to the following discrete probability distribution:

Calls Probability

0 0.05

1 0.12

2 0.15

3 0.25

4 0.22 0. 22

5 0.15

6 0.6

The rescue squad classifies each emergency emergenc y call into one of o f three categories: minor, regular, or major emergency. The probability that a particular call will be each type of emergency e mergency is as follows: Emergency Probability

Minor 0.30

Regular 0.56

Major 0.14

The type of emergency call ca ll determines the size of the crew sent in response. A minor emergency requires a two-person crew, a regular call requires a three-person crew, and a major emergency requ ires a five person crew. Simulate the emergency calls received by the rescue squad for 10 nights, compute the average number of each type of emergency call each night, and determine the maximum number of crew members that might be needed on any given night. Solution Step 1. Develop Random Number Ranges for the Probability Distributions Calls Probability Cumulative Probability Random Number Range, r 1

0

.05

.05

1-5

1

.12

.17

6-17

2

.15

.32

18-32

3

.25

.57

33-57

4

.22

.79

58-79

5

.15

.94

80-94

6

.06

1.00

95-99

1.00 Emergency Type

Minor

Probability

Cumulative Probability

Random Number Range, r2

.30

.30

1-30




Solu!on of Midcourse Exam Dec. 2010 Members of the Willow Creek Emergency Emergency Rescue Squad know from past experience that they will receive between zero and six emergency calls each night, according to the following discrete probability distribution:

Calls Probability

0 0.05

1 0.12

2 0.15

3 0.25

4 0.22 0. 22

5 0.15

6 0.6

The rescue squad classifies each emergency emergenc y call into one of o f three categories: minor, regular, or major emergency. The probability that a particular call will be each type of emergency e mergency is as follows: Emergency Probability

Minor 0.30

Regular 0.56

Major 0.14

The type of emergency call ca ll determines the size of the crew sent in response. A minor emergency requires a two-person crew, a regular call requires a three-person crew, and a major emergency requ ires a five person crew. Simulate the emergency calls received by the rescue squad for 10 nights, compute the average number of each type of emergency call each night, and determine the maximum number of crew members that might be needed on any given night. Solution Step 1. Develop Random Number Ranges for the Probability Distributions Calls Probability Cumulative Probability Random Number Range, r 1

0

.05

.05

1-5

1

.12

.17

6-17

2

.15

.32

18-32

3

.25

.57

33-57

4

.22

.79

58-79

5

.15

.94

80-94

6

.06

1.00

95-99

1.00 Emergency Type

Minor

Probability


Random Number Range, r2

.30

.30

1-30




Regular

.56

.86

31-86

Major

.14

1.00

87-99, 00

1.00 Step 2. Set Up a Tabular Simulation

Use the second column of o f random numbers in Night r1 Number of Calls r2 Emergency Type Crew Size Total per Night

1

65

4

2 48

3

3

08

1

4

05

0

5

89

6

7

8

9

06

62

17

77

5

1

4

1

4

71

Regular

3

18

Minor

2

12

Minor

2

17

Minor

2

89

Major

5

18

Minor

2

83

Regular

3

90

Major

5

5

0

0

18

Minor

2

08

Minor

2

26

Minor

2

47

Regular

3

94

Major

5

72

Regular

3

47

Regular

3

68

Regular

3

60

Regular

3

88

Major

5

36

Regular

3

43

Regular

3

28

Minor

2

31

Regular

3



9

10

14

3

14

3

10


10

68

4

06

Minor

2

39

Regular

3

71

Regular

3

22

Minor

2

76

Regular

3

11

Step 3. Compute Results

Average number of Minor emnergency calls per night =

Average number of Regurlar emgergency calls ca lls per night = Average number of Major emgergency calls per night =

If all the calls came in at the same time, the maximum number of squad members required during any 1 night would be 14. Question: 2 Burlingham Mills produces denim cloth that it sells to jeans manufacturers. It is negotiating a contract with Troy Clothing Company to provide denim cloth on a weekly basis. Burlingham has established its monthly available production capacity for this contract to be between 0 and 600 yards, according to the following probability distribution:

Troy Clothing's weekly demand for denim cloth varies according to the following following probability pro bability distribution: 0 100 200 300 400 500 Demand (yd.) 0.3 0.12 0.20 0.35 0.20 0.10 Probability Simulate Troy Clothing's cloth orders for 5 weeks and determine the average weekly capacity capac ity and demand. Also determine the probability pro bability that Burlingham will have sufficient capacity to meet demand. Use the following sets of random numbers: {0.59, 0.43, 0.43, 0.11, 0.11, 0.82, 0.35} 0.35} { 0.18, 0.86, 0.22, 0.62, 0.70} Solution: Step: 1 Develop Random Rando m Number Ranges for the Probability Distributions Demand 0 100 200 300 400 500

Probability 0.3 0.12 0.20 0.35 0.20 0.10

Cumulative Probability 0.3 0.15 0.35 0.70 0.90 1.00

Range 0-2 3-14 15-34 35-69 70-89 90-99




1.00

Σ

100

Production is uniformly distributed variable ranges from 0 to 600, such that it needs to be converted to discrete using the following formula: a+r (b-a) = a=0, b=600 ∴ 0+r(600-0)= 600r Step: 2 Tabular Simulation: weeks 1 2 3 4 5

Rn1 0.59 0.43 0.11 0.82 0.35

Σ

Production 300 300 100 400 300 1400

Rn 2 0.18 0.86 0.22 0.62 0.70

Demand [a+r(b-a)] 600 0.18 x 600= 108 516 132 372 420 1548

Step: 3 Computations: Average weekly capacity production: Average weekly Demand:

= 280 Yards per week.

= 309.6 ≈ 310 Yards per week.

Ex: ComputerWorld Company demand data for laptops selling for $4,300 per unit over a period of 15 weeks. Simulate the demand and compute the average demand and revenue for this company. Use the following data: PCs Demanded per week 0 1 2 3 4 Total

Frequency of Demand 20 40 20 10 10 100

Probability of Demand, p(x) 0.2 0.4 0.2 0.10 0.10 1.00

Step: 1 Develop Random number ranges for the probability distribution Probability of Demand, p(x) 0.2 0.4 0.2 0.1 0.1 1.00

PCs Demanded per week 0 1 2 3 4 Total


Ranges of Random number 0.0-19 20-59 60-79 80-89 90-99

Step: 2 Use tabular simula on: Weeks 1

Random number of demand Demand Revenue ($) 39 1 (1x4,300) = 4,300 5 Prepared by: Youssry Hamdy-CS Diploma


Prepared by: Youssry Hamdy- CS Diploma 2 3 4 5 6 7 8 9 10 11 12 13 14 15

73 72 75 37 02 87 98 10 47 93 21 95 97 69

2 2 2 1 0 3 4 0 1 4 0 4 4 2 31

8,600 8,600 8,600 4,300 0 12,900 17,200 0 4,300 17,200 0 17,200 17,200 8,600 $ 133,300

Step: 3 Computa !ons: Average Demand= Average Revenue=

=2.07 laptop PCs per week. =$ 8,920.

Simulation of a queuing system of Burlingham Mills Given the following data simulate the system for 10 days and compute average waiting time and average time in the system. Distribution of Arrival Intervals Arrival interval (Days), x Probability 1 20 2 40 3 30 4 10

Distribution of Service Times Service Time (Days), y Probability 5 20 1 50 2 30

Step: 1 Develop Random number ranges for the probability distribution for arrival intervals: Arrival interval (Days), x 1 2 3 4

Probability 20 40 30 10

Cumulative Probability 20 60 90 100

Range 0-19 20-59 60-89 90-99

Develop Random number ranges for the probability distribution of service times: Service Time (Days), y 5 1 2

Probability 20 50 30

Cumulative Probability 20 70 100



Range 0-19 20-69 70-99


Step: 2 Tabular Simula "on: Arrival Arrival Interval, x Clock 1 0.0 2 71 3 3.0 3 12 1 4.0 4 48 2 6.0 5 18 1 7.0 6 08 1 8.0 7 05 1 9.0 8 18 1 10.0 9 26 2 12.0 10 94 4 16.0 Total Step: 3 Computa #ons: Batch

Enter Facility clock 0.0 3.0 4.0 6.0 8.0 10.0 12.0 14.0 14.5 16.0

Rn1

Average Waiting time= Average time in system=

Waiting time 0.0 0.0 0.0 0.0 1.0 2.0 3.0 4.0 2.5 0.0 12.5

Rn2 65 18 17 89 83 90 89 08 47 06

Service Time, y 1.0 0.5 0.5 2.0 2.0 2.0 2.0 5.0 1.0 0.5

Departure Clock 1.0 3.5 4.5 8.0 10.0 12.0 14.0 14.5 15.5 16.5

Time in system 1.0 0.5 0.5 2.0 3.0 4.0 5.0 4.5 3.5 0.5 24.5

= 1.25 days / 10 batches = 2.45 days / 10 batches

Ex: Harry’s Company is producing vehicles’ tires according to the following probability distribution. Simulate the demand for 10 days and compute average demand. Demand for Tires 0 1 2 3 4 5 Total

Frequency 10 20 40 60 40 30 200

Probability 5 10 20 30 20 15 100.0

Step: 1 Develop Random number ranges for the probability distribution of tires’ demand: Demand for Tires

Probability


Range

0 1 2 3 4

5 10 20 30 20

5 15 35 65 85

0-4 5-14 15-34 35-64 65-84



Prepared by: Youssry Hamdy- CS Diploma 5

15 100.0

100

85-99

Step: 2 Tabular Simula $on Days

Rn

Demand

1 2 3 4 5 6 7 8 9 10

52 37 82 69 98 96 33 50 88 90

3 3 4 4 5 5 2 3 5 5 39

Σ

Step: 3 Computa %ons: Demand per day=

= 3.9 tires / 10 days

Lecture Example: Compcomm, Inc. is an international communications and information technology company. A stock analyst would like to use simulation to predict the stock prices of Compcomm for an extended period. Based on historical data, the analyst has developed the following probability distribution for the movement of Compcomm stock prices per day: Stock Price Movement

Probability

Increase

0.45

Same

0.3 0.25 1

Decrease

The analyst has also developed the following probability distributions for the percentage of the increases or decreases in the stock price per day: Stock Price Change 25%

Increase 0.4 0.17

37.50%

0.12

12.50%

Decrease

0.12 0.15 0.18




0.1 0.08 62.50% 0.07 75% 0.04 87.50% 0.02 100% 1 The price of the stock is currently 62. 50%

0.21 0.14 0.1 0.05 0.05 1

Develop a simula&on model to track the stock price of Compcomm stock and simulate for 10 days. Indicate the new stock price at the end of the 10 days Use the following sets of random numbers: 17, 95, 21, 78, 58,24, 33, 45, 77, 48 37, 79, 49, 12, 38,48, 13, 93, 55, 96 Step: 1 Develop Random number ranges for the probability distribution for price movement: Stock Price Movement Increase Same Decrease

Probability 45 30 25 100.0

Cumulative Probability 45 75 100

Range 0-44 45-74 75-99

Develop Random number ranges for the probability distribution for the percentage of increases and decreases in stock price: Increase: Stock Price Change 12.50% 25% 37.50% 50% 62.50% 75% 87.50%

Probability


Range

40 17 12 10 8 7 6

40 57 69 79 87 94 100

0-39 40-56 57-68 69-78 79-86 87-93 94-99

Probability


Range

12 15 18 21

12 27 45 66

0-11 12-26 27-44 45-65

Decrease: Stock Price Change 12.50% 25% 37.50% 50%




14 10 10

62.50% 75% 87.50%

80 90 100

66-79 80-89 90-99

Step: 2 Tabular Simula 'on Days

Rn1

Stock Price Movement

Rn2

Stock Price Change

1

17

Increase

37

12.5%

=7.75+62=≈ 70

2 3 4 5 6 7 8 9 10

95 21 78 58 24 33 45 77 48

Decrease Increase Decrease Same Increase Increase Same Decrease Same

79 49 12 38 48 13 93 55 96

62.5% 25% 25% 0 25% 12.5% 0 50% 0

23.25 77.5 46.5 62 77.5 70 62 31 62

Price

Ex: 14 page 62 A bank has one drive-in teller and room for one additional customer to wait. Customers arriving when the queue is full park and go inside the bank to transact business. The times between arrivals and the service-time distribution follow: Time between Arrivals (Minutes) Probability Service Time (Minutes) Probability 0 9 1 20 1 17 2 40 2 27 3 28 3 20 4 12 5 12 Simulate the operation of the drive-in teller for 10 new customers. The (rst of the 10 new customers arrives at a )me determined at random. Start the simula *on with customer being served, leaving at +me 3, and on in the

queue. How many customers went into the bank to transact business?

Solution: Step: 1 Develop Random number ranges for the probability distribution for time between arrivals: Time between Arrivals

Probability




Range


0 1 2 3 4 5

9 17 27 20 15 12

9 26 53 73 88 100

0-8 9-25 26-52 53-72 73-87 88-99

Service Time

Probability


Range

1 2 3 4

20 40 28 12

20 60 88 100

0-19 20-59 60-87 88-99

Customer

Rn for Arrival

Interarrival Time

Clock Time

1 2 3 4 5 6 7 8 9 10

30 46 39 86 63 83 07 37 69 78

2 2 2 4 3 4 0 2 3 4

2 4 6 10 13 17 17 19 22 26

Rn for Service Time 27 26 99 72 12 17 78 91 82 62

Service Time

No. in Queue

2 2 4 3 1 1 3 4 3 3

1 0 0 0 0 0 1 1 0 0

Time Service Begin 4 6 10 13 17 18 22 26

Time Service Ends 6 10s 13 14 18 21 25 29

Go into the Bank? Yes No √ √ √ √ √ √ √ √ √ √

Two customers only will go into the bank to transact business. Lecture Example: PortaCom manufacturers’ personal computers and related equipments. PortaCom’s product design group developed a prototype for a new high-quality portable printer. The new printer features an innovative design and has the potential to capture a significant share of the portable printer market. Preliminary marketing and financial analyses provided the following selling price, first year administrative cost, and first-year advertising cost: Selling price=$ 249 per unit Administra ,ve cost= $ 400,000 Adver-sing cost= $ 600,00 in the simulation model for the PortaCom problem the proceeding values are constants and are referred to as parameters of model. The cost of parts and the first-year demand for the printer are not known with certainty and are considered probabilistic inputs. At this stage of planning process, PortaCom’s best estimates of these inputs are $ 45 per unit for the direct labor cost $ 90 per unit for parts cost, and 15,000 units for (rst-year 11 Prepared by: Youssry Hamdy-CS Diploma


Prepared by: Youssry Hamdy- CS Diploma demand. PortaCom would like an analysis of first-year profit potential for printer. Because of PortaCom’s tight cash flow situation, management is particularly concerned about the potential for a loss. where Direct labor cost and parts cost Direct labor cost per unit Probability 43 0.1 44 0.2 45 0.4 46 0.2 47 0.1 Parts cost: depends on the general economy, the overall demand for parts, and pricing policy of PoraCom’s parts suppliers. portCom believes that the parts cost will range from $80 to $100 per unit is described by uniform probability distribution. Solution: Profit = (selling price – Direct labor cost per unit – parts cost per unit)(demand)-(Admin cost +ad Cost) Letting c1 = direct labor cost per unit C2 = parts cost per unit X=first-year demand The profit model for the first year can be as follows: Pro(t=(249-c1-c2)x-1,000,000 Profit=(249-45-90)(15,000)-1,000,000=114 x 15,000=1,710,000-1,000,000=710,000 Direct labor cost per unit Probability Cumulative Range 43 0.1 0.1 0.0-0.1 44 0.2 0.3 0.1-0.3 45 0.4 0.7 0.3-0.7 46 0.2 0.9 0.7-0.9 47 0.1 1.0 0.9-1.0 Parts cost = a+r(b-a) = 80+r(100-80)=80+r20 we will use set of 10 random number to calculate the parts cost {0.6953, 0.0082, 0.1514, 0.4013, 0.3125, 0.8166,0.9439,0.5448,0.3813,0.5389} Parts cost Trials 1 2 3 4 5 6 7 8

Rn 0.6953 0.0082 0.1514 0.4013 0.3125 0.8166 0.9439 0.5448

Parts cost $ [a+r(b-a)] 80+0.6953(20)=94 80 83 88 86 96 99 91



Prepared by: Youssry Hamdy- CS Diploma 9 10

0.3813 0.5389

87 90

Estimate first-year demand: Trials 1 2 3 4 5 6 7 8 9 10

Rn 0.7005 0.3204 0.9868 0.1804 0.4346 0.9605 0.5647 0.7334 0.0216 0.3218

Demand 17,366 12,900 20,686 10,888 14,259 22,904 15,732 17,804 5,902 12,918

Tabular Simulation Trials 1 2 3 4 5 6 7 8 9 10 Total Average

Direct Labor Cost per unit ($) 47 44 45 43 45 44 45 45 45 46 449 44,9

Parts Cost per unit ($) 94 80 83 88 86 96 99 91 87 90 912.64 91.26

Units Sold 17,366 12,900 20,686 10,888 14,259 22,904 15,732 17,804 5,902 12,918 151,359 15,359

Profits $ 1,023,570 461,828 1,288,906 169,807 648,911 1,526,769 814,686 1,165,501 -350,131 385,585 7,137,432 713,743.2

---------------------------------------------------------------------------------------------------------------------------------------------Lecture Example:

25- Tracy McCoy is shopping for a new car. She has identified a particular sports utility vehicle she likes but has heard that it has high maintenance costs. Tracy has decided to develop a simulation model to help her estimate maintenance costs for the life of the car. Tracy estimates that the projected life of the car with the first owner (before it is sold) is uniformly distributed with a minimum of 2.0 years and a maximum of 8.0 years. Furthermore, she believes that the miles she will drive the car each year can be defined by a triangular distribution with a minimum value of 3,700 miles, a maximum value of 14,500 miles, and a most likely value of 9,000 miles. She has determined from Automobile Association data that the maintenance cost per mile driven for the vehicle she is interested in is normally distributed, with a mean of $0.08 per mile and a standard deviation of $0.02 per mile. Using Crystal Ball, develop a simulation model (using 10 trials) and determine the average maintenance cost for the life of the car with Tracy and the probability that the cost will be less than $3,000. Solution: 13 Prepared by: Youssry Hamdy-CS Diploma



1. The car life cost is uniformly distributed [a=2.0, b =8.0] 2. The Mile drive is triangular distributed [a=3,700, b=14,500, c=9,000] 3. Maintenance cost is normally distributed [. = 0.8, / = 0.02] Convert uniform distribution using the formula a+r(b-a) 2.0+r(8.0-2.0) = r 6.0 Convert Triangular distribution Find h value =

Convert Maintenance = Step: 1 Develop probability distribution ranges: (Car life cost) Trials 1 2 3 4 5 6 7 8 9 10

Rn 0.405 0.173 0.011 0.529 0.004 0.514 0.691 0.654 0.709 0.412 Σ

Car life cost ($) [a+r(b-a)] r6 2.43 1.038 0.066 3.174 0.024 3.084 4.146 3.924 4.254 2.472 24.612

Develop probability distribution ranges: (Mile Drive) Trials 1 2 3 4 5 6 7 8 9 10

Rn 0.726 0.009 0.369 0.571 0.828 0.237 0.605 0.275 0.949 0.865 Σ

Mile Drive (0.926) 0.672 0.008 0.342 0.529 0.767 0.219 0.560 0.255 0.879 0.672 4.903



Prepared by: Youssry Hamdy- CS Diploma Develop probability distribution ranges: (Maintenance cost) Z x / = x –., ∴ X = z/ + ., X= .-Z/, if r < 0.5, X= . if r =0.5, X= .+z/ if >0.5, if r > 0.5 (r-0.5) Trials 1 2 3 4 5 6 7 8 9 10

Rn 0.930 0.014 0.995 0.796 0.588 0.758 0.040 0.607 0.286 0.351

Maintenance cost (0.4177) 0.033 0.117 0.280 0.103 0.197 0.017 0.150 0.099 0.137 0.033 2.272

Σ

Trials

Car life cost

Mile driven/Year

Maintenance cost/mile

1 2 3 4 5 6 7 8 9 10

2.43 1.038 0.066 3.174 0.024 3.084 4.146 3.924 4.254 2.472

Σ

24.612 0.4475

0.672 0.008 0.342 0.529 0.767 0.219 0.560 0.255 0.879 0.672 4.903 0.0891

0.033 0.117 0.280 0.103 0.197 0.017 0.150 0.099 0.137 0.033 2.272 0.0313

Average

Total cost=car life cost Mile+ Maintenance 1.135 1.163 0.688 3.806 0.988 3.32 4.856 4.278 5.27 3.177 28.681 2.68

The average cost of car life = $2,680 The probability that the cost will be less than $3,000. P (x≤ 3,000) Z= ___________________________________________________________________________________ Ex: A city is served by two newspapers the Tribune and the Daily News. Each Sunday readers purchase one of the newspapers at a stand. The following matrix contains the probabilities of a custo mer's buying a particular newspaper in a week, given the newspaper purchased the previous Sunday: This Sunday Tribune

Next Sunday Daily News 15 Prepared by: Youssry Hamdy-CS Diploma



Tribune Daily News Simulate a customer's purchase of newspapers for 10 weeks to determine the steady-state probabilities of a customer buying each newspaper in the long run.

Solution: Step: 1 Develop distribution ranges for this week: Newspaper Tribune Daily News

Probability 0.65 0.35.

Cumulative Probability 0.65 1.00 1.00

Range 0-64 65-99


Range 0-44 45-99

Develop distribution ranges for next week: Newspaper Tribune Daily News

Probability 0.45 0.55

Step: 2 Tabular Simulation Trials

Rn

1 2 3 4 5 6 7 8 9 10

66 09 39 60 66 20 19 71 86 82

Total

Newspaper Daily News 1 0 0 0 1 0 0 1 1 1

Tribune 0 1 1 1 0 1 1 0 0 0

5

5




Probabilities of a customer buying each newspaper in the long run Daily News = Tribune =

Book Example: 2.5 page: 46 A milling machine has different bearings that fail in service. The distribution of the life of each bearing is identical, as it indicated below. When a bearing fails, the mill stops, a repairperson is called, and a new bearing is installed. The delay time of repairperson’s arriving at the milling is also a random variable having the distribution in the given table. Down0me for the es1mated at $10 per minute. The direct on-site cost of the repairperson is $30 per hour. It takes 20 minutes to change one bearing, 30 minutes to change two bearings, and 40 minutes to change three bearings. A proposal has been made to replace all three bearings whenever a bearing fails. Management needs an evaluation of proposal. The total cost per 1, 0000 bearings-hours will be used as the measure of performance. The below table represent a simula 2on of 15 bearings changes under the current method of operation. Note that there are instances where more than one bearing fails at the same time. This is unlikely to occur in practice and is due to using a rather coarse grid of 100 hours for bearing life. It will be assumed in this example that the times are never exactly the same and thus nor more than one bearing is changed at any breakdown. Bearing life hours 1000 Probability 0.10 Delay Time Distribution

1100 0.13

Delay Time (Minute) Probability

1200 0.25

1300 0.13

5 0.6

1400 0.0

1500 0.12

10 0.3

1600 0.02

1700 0.06

15 1.0

Step: 1 develop probability distribu3on range for Bearing-life Distribution. Bearing life hours Probability Cumulative probability 1000 0.10 0.10 1100 0.13 0.23 1200 0.25 0.48 1300 0.13 0.61 1400 0.09 0.70 1500 0.12 0.82 1600 0.02 0.84 1700 0.06 0.90 1800 0.0 5 0.95 1900 0.0 5 1.00 Develop probability distribution range for Bearing-life Distribution. Delay Time (Minutes) 5 10

Probability 0.6 0.3

Cumulative probability 0.6 0.9



Range 0-9 10-22 23-47 48-60 61-69 70-81 82-85 84-89 90-94 95-99

Range 0-5 6-8

1800 0.05

1900 0.05


0.1

1.0

9-00

2- Step: Tabular Simulation Bearing replacement under current method: Bearing 1 Life Rd2 Hours 1400 7 1300 3 1900 1 1500 6 1300 4 1400 8 1600 5 1800 7 1200 0 1000 5 1900 9 1400 4 1300 7 1700 1 1800 2

Bearing 2

Delay Life Delay Rd3 Rd4 Rd5 Time Hours Time 1 67 10 71 1500 8 10 18 2 55 5 21 1100 3 5 17 3 98 5 79 1600 3 5 65 4 76 10 88 1700 1 5 03 5 53 5 93 1800 0 5 54 6 69 10 77 1500 6 10 17 7 80 5 08 1000 9 15 19 8 93 10 21 1100 8 10 09 9 35 5 13 1100 3 5 61 10 02 5 03 1000 2 5 84 11 99 15 14 1100 1 5 11 12 65 5 05 1000 0 5 25 13 53 19 29 1200 2 5 86 14 87 5 07 1000 4 5 65 15 90 5 20 1100 3 5 44 Total 119 100 Bearing changed at any breakdown. The cost of the current is estimated as follows: cost of bearing = 47 bearings X $32/bearing = $ 1,504 Cost of delay 4me = (110=110+105) minutes x $10.minute = $ 3,250

Trials

Rd1

Bearing3 Life Hours

Rd6

1100 1100 1400 1000 1300 1100 1100 1000 1400 1700 1100 1200 1700 1400 1200

6 2 2 9 8 3 6 7 1 0 5 2 8 3 4

Delay Time 10 5 5 15 10 5 10 10 5 15 5 5 10 5 5 120

cost of down5me during repair = 45 bearings x 20 minutes/bearing x $10 minute = $9,000 Cost of repairperson = $45 bearings x 20 minutes/bearing x $30/60 minutes = $450 Total Cost = $ 1,440+$3,250+$9,000+$450=$ 14,140 Total life of bearings = (22,300+18,700+18,600) 59,600 hours. Therefore, the total cost per 10,000 bearings-hours is ($14,140/5.96) = $2,372. Proposed Method: Trails 1 2 3 4 5 6 7 8 9 10 11 12

Bearing1 Life(hours) 1700 1000 1500 1300 1200 1000 1500 1300 1800 1300 1400 1500

Bearing2 Life (hours) 1100 1800 1700 1100 1100 1200 1700 1700 1200 1300 1300 1300

Bearing3Life First Failure (hours) (hours) 1000 1000 1200 1000 1300 1300 1800 1100 1300 1100 1000 1100 1200 1200 1000 1000 1100 1100 1100 1100 1900 1300 1400 1300 18 Prepared by: Youssry Hamdy-CS Diploma


Delay (minute) 10 5 5 5 5 10 5 10 15 5 10 5

Prepared by: Youssry Hamdy- CS Diploma 13 1500 1800 14 1000 1900 15 1300 1700 Total Cost of bearings = 45 bearings x $ 32/bearing =$ 1,44,

1200 1400 1700

1200 1000 1300

10 5 5 110

Cost of delay 6me = 110 minutes x $ 10/minute = $ 1.100 Cost of down7me during repairs = 15 sets x 40 minutes/ set x $10/ minute = $6,000 Cost of repairpersons = 15 sets x 40 minutes/set x $30/60 minutes = $300. Total cost = $ 1,440 +$1,100+$6,000+$300=$8,840 The total life of bearings = (17,000x3) 51,000 hours. Therefore, the total cost per 10,000 bearings-hours is ($8,840/5.1)=$1.733 The new policy generates a savings of $ 634 per 10,000 hours bearing-life. Ex. Page 40: News Dealer's Problem The new stand buys the papers for 33 cents each and sells them for 50 cents each. Newspapers not sold at the end of the day are sold as scrap for 5 cents each. Newspapers can be purchased in bundles of 10 thus; the newsstand can buy 50, 60, and so on. There're three types of newsdays: 'Good", "Fair" and "Poor"; they have the probabili8es 0.35, 0.45 and 0.20, respec 9vely. The distribu:on of newspapers demanded on each of these days us given below. The problem is to compute the optimal number of papers the newsstand should purchase. This will be accomplished by simula;ng demands for 10 days and recording pro(ts from sales each day. Profit = (Revenue from sales)-(Cost of newspapers)-(lost profit from excess demand)+(salvage from sale of scrap papers). Demand

Good 0.03 0.05 0.15 0.20 0.35 0.15 0.07

40 50 60 70 80 90 100

Demand Probability Distribution Fair 0.10 0.18 0.40 0.20 0.08 0.04 0.00

poor 0.44 0.22 0.16 0.12 0.06 0.00 0.00

Type of News probability Distribution Type of News Good Fair Poor

Probability 0.35 0.45 0.20

Solution: Step: 1 Develop probability distribu
Probability 0.35

Cumulative Distribution 0.35



Range 0-34

Prepared by: Youssry Hamdy- CS Diploma Fair 0.45 0.75 Poor 0.20 1.0 Step: 2 Develop probability distribu=on range for newspaper type . Demand 40 50 60 70 80 90 100

Good 0.03 0.08 0.23 0.43 0.78 0.93 1.0

Cumulative Distribution Fair 0.10 0.28 0.68 0.88 0.96 0.1 0.1

Poor 0.44 0.66 0.82 0.94 0.1 0.1 0.1

35-74 75-99

Good 0-2 3-7 8-22 23-42 43-77 78-92 93-99

Ranges Fair 0-9 10-27 28-67 68-87 88-95 96-97 98-99

Poor 0-43 44-65 66-81 82-93 94-95 96-97 98-99

Step: 3 Tabular Simula "on:

Day 1 2 3 4 5 6 7 8 9 10

Rn1 for newspaper type 58 17 21 45 43 36 27 73 86 19

Type of newspaper

Rn2 for Demand

Demand

Fair Good Good Fair Fair Fair Good Fair Poor Good Total

93 63 31 19 75 84 37 23 02 53

80 80 70 50 70 80 70 50 40 70

Revenue from sales $35.00 35.00 35.00 25.00 35.00 35.00 35.00 30.00 20.00 20.00 $ 300.00

Lost profit from excess Demand $1.70 1.7 1.70 3.40 $ 17.00

Salvage from sale of scrap 1.00 0.50 1.50 1.50 $10.00

Daily profit $ 10.2 10.20 11.90 2.90 10.20 11.90 8.50 7.40 -1.60 -1.60 $161.00

The op>mal number the newsstand (10 days x $0.33 x 70) = $ 231

Ex: The time between arrivals of cars at the Petroco Service Station is defined by the following probability distribution:

1 2 3 4 Time Between Arrivals (min.) 0.15 0.30 0.40 0.15 Probability Simulate the arrival of cars at the service station for 10 arrivals and compute the average time between arrivals. 20 Prepared by: Youssry Hamdy-CS Diploma



Solution : Step:1 Develop probability distribution range for Time between Arrivals . Time Between Arrivals (min.)

Probability

Cumulative Distribution

Range

1 2 3 4

0.15 0.30 0.40 0.15

0.15 0.45 0.85 1.00

0-14 15-44 45-84 85-99

Step: 2 Tabular Simulation Trials 1 2 3 4 5 6 7 8 9 10

Rn 71 18 12 17 89 18 83 90 18 08

Time Between Arrivals (min.) 3 2 1 2 4 2 3 4 2 1 24

Σ

The Average Time between Arrivals =

24 =2.4 Minute/Car 10

Ex: The Dynaco Manufacturing Company produces a product in a process consisting of operations of five machines. The probability distribution of the nu mber of machines that will break down in a week follows: Machine Breakdowns per Week Probability

0 0.10

1 0.10

2 0.20

3 0.25

4 0.30

5 0.5

a. Simulate the machine breakdowns per week for 10 weeks. b. Compute the average number of machines that will break down per week. Solution: Step: 1 Develop probability distribution range for Time between Arrivals . Machine Breakdown Per week 0 1 2 3

Probability 0.10 0.10 0.20 0.25

Cumulative Distribution 0.10 0.20 0.40 0.65



Range 0-9 10-19 20-39 40-64

Prepared by: Youssry Hamdy- CS Diploma 4 5

0.30 0.5

0.95 1.00

65-94 95-99

Step: 2 Tabular Simulation Weeks

Rn1

Machine 1 Breakdown

Rn2

Machine 2 Breakdown

Rn3

Machine 3 Breakdown

Rn4

Machine 4 Breakdown

Rn5

Machine Breakdown 5

1 2 3 4 5 6 7 8 9 10

65 48 08 05 89 06 62 17 77 68

4 3 0 0 4 0 3 1 4 4

71 18 12 17 89 18 83 90 18 08

4 1 1 1 4 1 4 4 1 0

26 47 94 72 47 68 60 88 36 43

2 3 4 4 3 4 3 4 2 3

28 31 06 39 71 22 76 80 95 11

2 2 0 2 4 2 4 4 5 1

77 35 0 90 69 55 10 72 06 40

4 2 0 4 4 3 1 4 0 3

Σ

23

21

32

Average number of machines that will break down per week=

26

25

23 + 21 + 32 + 26 + 25 = 10.4 10

Ex: Every time a machine breaks down at the Dynaco Manufacturing Company , either 1, 2, or 3 hours are required to fix it, according to the following probability distribution:

1 0.30

Repair Time (hr.) Probability

2 0.50

3 0.20

a. Simulate the repair time for 10 weeks and t hen compute the average weekly repair time . b. If the random numbers that are used to simulate breakdowns per week are also used to simulate repair time per breakdown, will the results be affected in any way? Explain . c. If it costs $50 per hour to repair a machine when it breaks down (including lost productivity), determine the average weekly breakdown cost . d. The Dynaco Company is considering a preventive maintenance program that would alter the probabilities of machine breakdowns per week as shown in the following table: Machine Breakdowns per Week

Probability

0 0.20

1 0.30

2 0.20

3 0.15

4 0.10

5 0.5

The weekly cost of the preventive maintenance program is $150. Using simula ?on, determine whether the company should institute the preventive maintenance program. Solution:

Step: 1 Develop probability distribution machine repair time . Repair Time (hr.) 1 2


Cumulative Distribution 0.30 0.80



Range 0-29 30-79


0.20

1.00

80-99

a) Repair Time Simulation for 10 days: Weeks

1 2 3 4 5 6 7 8 9 10

Rn1

Machine 1 Repair Time

Rn2


Rn3


Rn4


Rn5


2 2 3 2 1 3 3 3 3 1

77 35 18 79 75 92 48 13 56 85

2 2 1 2 2 3 2 1 2 3

30 71 18 01 85 83 67 53 88 04

2 2 1 1 3 3 2 2 3 1

62 27 73 84 81 26 38 0 64 96

2 1 2 3 3 1 2 1 2 3

16 52 98 79 82 34 51 65 16 09

1 2 3 2 3 2 2 2 1 1

62 54 82 48 10 82 82 81 95 05 23

20

20

20

19

23 + 20 + 20 + 20 + 19 = 10.3 Hour per machine/Week 10 b) If the same random numbers that are used to simulate breakdowns per week are also used to simulate repair time per breakdown: Average weekly Repair Time =

Weeks

Rn1


Rn2


Rn3


Rn4


Rn5


1 2 3 4 5 6 7 8 9 10

65 48 08 05 89 06 62 17 77 68

2 2 1 1 3 1 2 1 2 2

71 18 12 17 89 18 83 90 18 08

2 1 1 1 3 1 3 3 1 1

26 47 94 72 47 68 60 88 36 43

1 2 3 2 2 2 2 3 2 2

28 31 06 39 71 22 76 80 95 11

1 2 1 2 2 1 2 3 3 1

77 35 0 90 69 55 10 72 06 40

2 2 1 3 2 2 1 2 1 2

Σ

17

17

21

18

18

if we used the random number used in problem # 3, the result of average repair time will be affected as follows: 17 + 17 + 21 + 18 + 18 = 9.1 Hour per machine/Week Average weekly Repair Time = 10 The difference between using di @ erent random numbers and the same random numbers are 10.3 – 9.1=1.2 hours per machine/week c) The Average weekly breakdown cost

Machine Breakdown Per week

Probability

Cumulative Distribution



Range

Prepared by: Youssry Hamdy- CS Diploma 0 1 2 3 4 5

Weeks

1 2 3 4 5 6 7 8 9 10

Rn1

62 54 82 48 10 82 82 81 95 05 22

0.30 0.20 0.20 0.15 0.10 0.5

0.30 0.50 0.70 0.85 0.95 1.00

0-29 30-49 50-69 70-83 84-94 95-99

Machine 1 Breakdown

Rn2

Machine 2 Breakdown

Rn3

Machine 3 Breakdown

Rn4

Machine 4 Breakdown

Rn5

Machine Breakdown 5

2 2 3 1 0 3 3 3 5 0

77 35 18 79 75 92 48 13 56 85

3 1 0 3 3 4 1 0 2 4

30 71 18 01 85 83 67 53 88 04

1 3 0 0 4 4 2 0 4 0

62 27 73 84 81 26 38 0 64 96

2 0 2 4 4 0 1 0 2 5

16 52 98 79 82 34 51 65 16 09

0 2 5 3 3 1 2 2 0 0

21

18

20

18

Weekly breakdown Cost per machine Machine 1 = 22 hrs x ($50) =$ 1,100 Machine2 =21 hrs x ($50) = $ 1,050 Machine3 = 18 hrs x ($50) = $ 900 Machine4 = 20 hrs x ($50) = $ 1,000 Machine5 = 18 hrs x ($50) = $ 900 Total weekly breakdown cost = $1,100+$1,050+$900+$1,000+$900=$ 4,950 $4,950 =$ 495 per machine/weekly Average weekly breakdown cost = 10 d) Preventive maintenance program weekly preventive cost program = $ 150 Weekly maintenance cost = Machine breakdown hours x weekly preventive cost Machine1 = 22 x $ 150 = $ 3,300 Machine2 = 21 x $ 150 = $ 3,150 Machine3 = 18 x $ 150 = $ 2,700 Machine4 = 20 x $ 150 = $ 3,000 Machine5 = 18 x $ 150 = $ 2,700 Total weekly cost of new Preventive maintenance program = 3,300+3,150+2,700+3,000+2,700= $ 14,850 $14,850 =$ 1,485 per machine/weekly Average cost of new Preventive maintenance program = 10 Difference between the old system and new system = $ 1,485 - $ 495 = $ 990 Decision: the old system is more feasible than the new Preventive maintenance program because the old system will save $ 4,950 ($ 990 x 5) weekly to the company. So it is recommended to keep the old system than adopting the new one. 24 Prepared by: Youssry Hamdy-CS Diploma



Ex: Sound Warehouse in Georgetown sells CD players (with speakers), which it orders from Fuji Electronics in Japan. Because of shipping and handling costs, each order must be for 5 CD players. Because of the time it takes to receive an order, the warehouse outlet places an order every time the present stock drops to 5 CD players. It costs $100 to place an order. It costs the warehouse $400 in lost sales when a customer asks for a CD player and the warehouse is out of stock. It costs $40 to keep each CD player stored in the warehouse. If a customer cannot purchase a CD player when it is requested, the customer will not wait until one comes in but will go to a competitor. The following probability distribution for demand for CD players has been determined:

0 0.4

Demand per Month Probability

1 0.8

2 0.28

3 0.40

4 0.16

5 0.2

6 0.2

The time required to receive an order once it is placed has the following probability distribution: 1 0.60

Time to Receive an Order (mo.) Probability

2 0.30

3 0.10

The warehouse has five CD players in stock. Orders are always received at the beginning of the week. Simulate Sound Warehouse's ordering and sales policy for 10 months and Compute the average monthly cost. Solution: Step: 1 Develop probability distribuAon for Monthly demand: Probability 0 0.4 1 0.8 2 0.28 3 0.40 4 0.16 5 0.2 6 0.2 Develop probability distribution for Time receives order:

Cumulative Distribution 0.8 0.12 0.40 0.80 0.96 0.98 1.00

Demand Per Month

Time to Receive an Order (mo.) 1 2 3


Range 0-7 8-11 12-39 40-79 80-95 96-97 98-99

Cumulative Distribution 0.60 0.90 1.00

Range 0-59 60-89 90-99

Step: 2 Tabular SimulaBon:

Months

Rn1 for Demand

Demand

Rn2 for Time to receive an order

1 2 3 4

39 65 76 45

2 3 3 3

19 90 69 64

Time to receive and order

Sales

stock

Cost (Order place lost sales +order holding)

1 3 2 2

3 2 0 3

2 0 0 2

40x2=$80 1x100x400=$40,000 3x100x400=$120,000 40x2=$80




5 6 7 8 9 10

96 73 71 00 70 99

Average Monthly Cost =

5 3 3 0 3 6 31

61 20 31 99 00 85

2 1 1 3 1 2 18

2 0 3 0 0 3

0 3 0 0 3 0

3x100x400=$120,000 40x3=$120 0 0 3x100x400=$120,000 2x100x400=$80,000 $ 480,280

$480,280 = $ 48,028 per month 10

Ex: The emergency room of the community hospital in Farmburg has one receptionist, one doctor, and one nurse. The emergency room opens at time zero, and patients begin to arrive some time later. Patients arrive at the emergency room according to the following probability distribution: Time Between Arrivals (min.) Probability

Patient Needs to See Probability

5 0.06

10 0.10

Doctor alone 0.50

15 0.23

20 0.29

25 0.18

Nurse alone 0.20

30 0.14

Both 0.30

The attention needed by a patient who comes to the emergency room is defined by the following probability distribution: If a patient needs to see both the doctor and the nurse, he or she cannot see one before the other that is, the patient must wait to see both together. The length of the patient's visit (in minutes) is defined by the following probability distributions: Doctor 10 15 20 25 30

Probability 0.22 0.31 0.25 0.12 0.10

Nurse 5 10 15 20

Probability .08 .24 .51 .17

Both 15 20 25 30 35 40

Probability 0.07 0.16 0.21 0.28 0.17 0.11

Simulate the arrival of 10 patients to t he emergency room and compute the probability that a patient must wait and the average waiting time. Based on this one simulation, does it appear that this system provides adequate patient care? Solution : Step: 1 Develop Probability Distribution ranges for Time Between Arrivals: Time Between Arrivals 5

Probability 0.06

Cumulative Probability 0.06



Range 0-5


10 15 20 25 30

0.16 0.23 0.29 0.18 0.14

0.16 0.39 0.68 0.86 1.00

6-15 16-38 39-67 68-85 86-99

Patient needs to see Patient needs to see Doctor only Nurse only Both



Range 0-49 50-69 70-99

Length of Visit (Doctor only)

Doctor only 10 15 20 25 30

Probability 0.22 0.31 0.25 0.12 0.10

Cumulative Probability 0.22 0.53 0.78 0.90 1.00

Range 0-21 22-52 53-77 78-89 90-99

Length of Visit (Nurse only)

Nurse only 5 10 15 20

Probability 0.08 0.24 0.51 0.17

Cumulative Probability 0.08 0.32 0.83 1.00

Range 0-7 8-31 32-82 83-99

Length of Visit (Both)

Both (Doctor & Nurse) 15 20 25 30 35 40

Probability 0.07 0.16 0.21 0.28 0.17 0.11


Range 0-6 7-22 23-43 44-71 72-88 89-99

Step: 2 Tabular Simulation

Patient

Rn1

1 2 3

37 90

Time between Arrivals 15 25

Arrival Clock

0.0 15.0 25.0

Enter Room Clock 0.0 15.0 30.0

Waiting Time

Rn2

Patient need to

Rn3

Length of Visit

Time in system

0.0 0.0 5.0

65 88 30

Nurse Both Doctor

52 60 77

15 30 20

15 30 25




4 5 6 7 8 9 10

40 10 53 08 93 60 96

20 10 20 10 25 20 30

45.0 55.0 75.0 85.0 110.0 130.0 160.0

50.0 65.0 95.0 135.0 145.0 220.0 325.0

5.0 10.0 20. 50.0 35.0 90.0 165.0 380

16 27 99 54 79 14 50

Doctor Doctor Both Nurse Both Doctor Nurse

44 67 08 15 90 35 24

15 20 20 10 40 15 10

20 30 40 60 75 105 175.0 575

Step: 3 Computation:

Average waiting time =

380 =38 Minutes per patient 10

Probability that the patient must wait: 5 5 10 20 =0.0131, P (5) = =0.0131, P (10) = =0.0263, =0.0526 P (5) = P (20) = 380 380 380 380 50 35 90 165 =0.0131, P (5) = =0.0921, P (5) = =0.2368, P (5) = =0.4342 P (5) = 380 380 380 380 ∴ Probability that the patient must wait=0.0131+0.0131+0.0263+0.0526+0.0131+0.0921+0.2368+0.4342=0.8813 The system doesn’t appear to provide adequate care to patient because the waiting time ranges between 0 and 165 minutes. Thus waiting time is longer in some cases which affect patient’s health condition.

Ex. 8 Page 60 Book An elevator in a manufacturing plant carries exactly 400 kilograms of materials. There’re three kinds of material, and these are in boxes, that arrive for a ride on the elevator. These materials and their distributions of time between arrivals are as follows:

Material A B C

Weight (Kilograms) 200 100 50

Interarrival Time (Minutes) 5±2 (uniform) 6 (constant) P(2)=0.33 P(3)=0.67

It takes 1 minute to go up to the second Coor, 2 minutes to unload, and 1 minute to return to the (rst Coor. The elevator doesn’t leave the first floor unless it has a full load. Simulate 1 hour of opera#on of the system. What’s the average transit time for a box of material A (Time from its arrival until it is unloaded)? What’s the average waiting time for a box of material B? How many boxes of material C made the trip 1 hour? Solution: Material A has a uniform distribuDon 5±2, such that b=(5+2) = 7, a=(5-2) = 3 a+r(b-a)=3+r(7-3) = 3r+4=r7

Interarrival Time

Probability




Range


3 4 5 6 7

0.2 0.2 0.2 0.2 0.2

0.2 0.4 0.6 0.8 1.0

0-19 20-39 40-59 60-79 80-99

Material A (200kg/box)

Box 1 2 3 4 5 6 7 8 9 10

Rn 55 89 17 38 69 71 90 18 94 47

Interarrival Time 5 7 4 4 6 6 7 3 7 5

Clock Time 5 12 16 20 26 32 39 42 49 54

Material B (100kg/box)

Box 1 2 3 4 5 6 7 8 9 10

Clock Time 6 12 18 24 30 36 42 48 54 60

Material C (50kg/box)

Interarrival Time 2 3

Box 1


Rn 39

Cumulative Probability 0.33 1.00

Interarrival Time 3 29 Prepared by: Youssry Hamdy-CS Diploma


Range 0-32 33-99

Clock Time 3


2 3 4 5 6 7 8 9 10

Clock Time 3 6 7 9 11 12 15 18 19 21

73 72 75 02 87 98 10 47 21

3 3 3 2 3 3 2 3 2

Material( A) Arrival 1 2 -

6 9 12 14 17 20 22 25 27

Material (B) Arrival 1 2 3 -

Material (C) Arrival 1 2 3 4 -

Ex: 9 The time between arrivals of oil tankers at a loading dock at Prudhoe Bay is given by the following probability distribution: Time Between Ship Arrivals (days) Probability

1 0.5

2 0.10

3 0.20

4 0.30

5 0.20

6 0.10

7 0.5

The time required to fill a tanker with oil and prepare it for sea is given by the following probability distribution: Time to Fill and Prepare (days) Probability

3 0.10

4 0.20

5 0.40

6 0.30

Simulate the movement of tankers to and from the single loading dock for the first 10 arrivals. Compute the average time between arrivals, average waiting time to load, and average number of tankers waiting to be loaded. Solution: Step:1 Develop Random Number Ranges for Time between Ship Arrivals Time Between Ship Arrivals (days) 3 4


Cumulative Probability 0.5 0.15



Range 0-4 5-14


5 6 5 6 7

0.20 0.30 0.20 0.10 0.5

0.35 0.65 0.85 0.95 1.00

15-34 35-64 65-84 85-94 95-99


Range 0-9 10-29 30-69 70-99

Develop Random Number Ranges for Time to Fill and prepare (days) Time to Fill and Prepare (days) 3 4 5 6

Probability 0.10 0.20 0.40 0.30

Step: 2 Tabular Simulation: Tanker Movement

1 2 3 4 5 6 7 8 9 10 Σ

Time Rn1 between Arrivals

17 06 47 80 23 25 77 32 96 71

3 2 4 5 3 3 5 3 6 5 39

Arrival Clock

Enter Clock

Waiting Time

6 7 8 9 10 11 12 13 14

6 12 17 22 27 33 36 41 44

4 5 9 13 17 22 24 28 30 152

Rn2

95 62 31 36 63 76 01 66 03 55

Time to Fill and Prepare (days) 6 5 5 5 5 6 3 5 3 5

No. in Queue

1 1 1 2 1 1 1 1 1 10

Time Service begin

Time Service End

Time in system

3 6 12 17 22 27 33 36 41 44

9 11 17 22 27 33 36 41 44 49

6 11 10 14 18 23 25 29 31 35 202

Step: 3 Computations

39 =3.9 Days per Ship 10 152 =15.2 Days per Ship. Average Waiting Time to load = 10 10 =1 Tanker per Day Average number of Tankers waiting to be loaded = 10 Average Time between Arrivals =

Ex: 10 The Saki automobile dealers in the Minneapolis St. Paul area orders the Saki sport compact, which gets 50 miles per gallon of gasoline, from the manufacturer in Japan. However, the dealer never knows for sure how many months it will take to receive an order once it is placed. It can take 1, 2, or 3 months, with the following probabilities: Months to Receive an Order

1

2

3




0.50

Probability

0.30

0.20

The demand per month is given by the following distribution: 1 0.10

Demand per Month (cars) Probability

2 0.30

3 0.40

4 0.20

The dealer orders when the number of cars on the lot gets down to a certain level. To determine the appropriate level of cars to use as an indicator of when to order, the dealer needs to know how many cars will be demanded during the time required to receive an order. Simulate the demand for 10 orders and compute the average number of cars demanded during the time required to receive an order. At what level of cars in stock should the dealer place an order? Solution: Step: 1 Develop Random Number Ranges for Months to Receive an Order Months to Receive an Order 1 2 3



Range 0-49 50-79 80-99

Develop Random Number Ranges for Demand per Month (cars) Demand per Month (cars) 1 2 3 4

Probability 0.10 0.30 0.40 0.20


Range 0-9 10-39 40-79 80-99

Step: 2 Tabular Simulation: Orders 1 2 3 4 5 6 7 8 9 10

Rn1 71 12 48 18 08 94 26 83 63 05

Months to Receive an Order 2 1 1 1 1 3 1 3 2 1

Rn 2 65 18 19 89 83 90 89 08 74 69

Σ



Demand 3 2 2 4 4 4 4 1 3 3 30


The Average number of Cars demanded during the time required to receive an order =

30 =3 Cars 10

The dealer should place an order when the demand is 4 cars per month.

Ex: 21 The Western Outfitters Store specializes in denim jeans. The variable cost of the jeans varies according to several factors, including the co st of the jeans from the distributor, labor costs, handling, packaging, and so on. Price also is a random variable that varies according to competitors' prices. Sales volume also varies each month. The probability distributions for volume, price, and variable co sts each month are as follows: Sales Volume Probability

Price $ Probability

300 0.12

400 0.18

22 0.07

23 0.16

8 0.17

Variable Cost $ Probability

500 0.20

600 0.23

24 0.24

25 0.25

9 0.32

10 0.29

700 0.17

26 0.18

11 0.14

800 0.10

27 0.10

12 0.08

Fixed costs are $9,000 per month for the store. Simulate 10 months of store sales and co mpute the probability that the store will at least break even a nd the average profit (or loss).

Solution :

Step: 1 Develop Random Number Ranges for Sales Volume Sales Volume 300 400 500 600 700 800

Probability 0.12 0.18 0.20 0.23 0.17 0.10


Range 0-11 12-29 30-49 50-72 73-89 90-99


Range 0-6

Develop Random Number Ranges for Price Price $ 22

Probability 0.07




23 24 25 26 27

0.16 0.24 0.25 0.18 0.10

0.23 0.47 0.72 0.90 1.00

7-22 23-46 47-71 72-89 90-99


Range 0-16 17-48 49-77 78-91 92-99

Develop Random Number Ranges for Variable Cost $ Variable Cost $ 8 9 10 11 12

Probability 0.17 0.32 0.29 0.14 0.08

Step: 2 Tabular Simulation: Month Rn1 1 2 3 4 5 6 7 8 9 10

71 12 48 18 08 94 26 83 63 05 Σ

Sales Volume 600 400 500 400 300 800 400 700 600 300 50,000

Rn2

65 18 19 89 83 90 89 08 74 69

Price $ Rn3 25 23 23 26 26 27 26 22 26 25

17 06 47 80 23 25 77 32 96 71

Variable Cost $ 9 8 9 11 9 9 10 9 12 10

Profits $

Loss $

$600 5,400 100 $ 6,100

-3,000 -2,900 -3,000 -3,900 -2,600 -600 -4,500 $ 20,500

Step: 3 Computations:

Profit = Sales volume x price per unit $ - (variable cost + Fixed Cost) Month 1 = (600 x $25) =$15,000 – (600 x 9 +$9,000) = (5,400+9,000) =14,400 Profit = $ 15,000 – $14,400= $ 600 Month 2 = (400 x $ 23) = $9,200 – (400 x $8 + $9,000) = $12,200 Loss = $9,200 - $12,200 = - $3,000 Month 3 = (500 x $23) = $11,500 – (500 x $9 +$9,000) = $ 13,500 Loss = $11,500 –$ 13,500 = $ -2,000 Month 4 = (400 x $ 26) = $10,400 – (400 x $ 11 + $9,000) = $ 13,400 Loss = $10,400 – $ 13,400 = $ -3,000 Month 5: (300x$26)=$7,800 – (300x9+$9,000) = $11,700 Loss = $7,800-$11,700= $ - 3,900 Month 6: (800x$27)=$21,600 – (800x$9+$9,000) = $16,200 Profit= $21,600 –$16,200= $5,400 34 Prepared by: Youssry Hamdy-CS Diploma



Month 7: (400x$26)=$10,400 – (400x$10+$9,000) =$13,000 Loss=$10,400 –$13,000= $ -2,600 Month 8: (700x$22)=$15,400 – (700x$9+$9,000) = $15,300 Profit= $15,400 –$15,300=$100 Month 9: (600x$26)=$15,600 – (600x$12+$9,000) = $16,200 Loss= $15,600 –$16,200 = $ -400 Month 10: (300x$25)=$7,500 –(300x$10+$9,000)=$12,000 Loss = $7,500 –$12,000= $ -4,500 $6,100 = $ 610 Per month Average profits = 10 $20,500 = $ 2,050 Per month Average Loss = 10 Ex: 21 Randolph College and Salem College are within 20 miles of each other, and the students at each college frequently date each other. The students at Randolph College are debating how good their dates are at Salem College. The Randolph students have sampled several hundred of their fellow students and asked them to rate their dates from 1 to 5 (in which 1 is excellent and 5 is poor) according to physical attractiveness, intelligence, and personality. Following are t he resulting probability distributions for these three traits for students at Salem College: Physical Attractiveness Probability

1 0.27

2 0.35

3 0.14

4 0.9

5 0.15

Intelligence Probability

1 0.10

2 0.16

3 0.45

4 0.17

5 0.12

Personality Probability

1 0.15

2 0.30

3 0.33

4 0.07

5 0.15

Simulate 10 dates and compute an average overall rating of the Salem students. Solution: Step: 1 Develop Random Number Ranges for Physical Attractiveness

Physical Attractiveness 1 2 3 4 5

Probability 0.27 0.35 0.14 0.09 0.15


Range 0-26 27-61 62-75 76-84 85-99


Range 0-9

Develop Random Number Ranges for Intelligence

Intelligence 1

Probability 0.10




2 3 4 5

0.16 0.45 0.17 0.12

0.26 0.71 0.88 1.00

10-25 26-70 71-87 88-99


Range 0-14 15-34 35-77 78-84 85-99

Develop Random Number Ranges for Personality

Personality 1 2 3 4 5

Probability 0.15 0.30 0.33 0.07 0.15

Step: 2 Tabular Simula $on

Date

Rn1

2 2 3 4 5 6 7 8 9 10

71 12 48 18 08 94 26 83 63 05

Physical Attractiveness

Σ

3 1 2 1 1 5 1 4 3 1 22

Rn2

Intelligence

Rn3

Personality

65 18 19 89 83 90 89 08 74 69

3 2 2 5 4 5 4 1 4 3 33

55 89 17 38 69 71 90 18 94 47

3 5 2 3 3 3 5 2 5 2 33

Step: 3 Computa %on Average Physical Attractiveness =

22 =2.2 10

33 =3.3 10 33 Average Personality = =3.3 10

Average Intelligence =

∴ The Overall RaEng of Salem Collage = 2.2+3.3+3.3=8.8

Random Number Generation using Linear congruential Method:

Formula: X +1= ax +c mod m I = 0 1 2 3 …. Where: X0 is the seed, a is mul Fplier, c is increment, m is modulus. 1.

, i =1, 2, 3, …. Example: 7.1 Page: 254 36 Prepared by: Youssry Hamdy-CS Diploma

Formula : Xi +1=(axi+c)mod m, I = 0,1,2,3,….


Prepared by: Youssry Hamdy- CS Diploma Use Linear congruenGal Method to generate a sequence of 10 random numbers with

X0=27, a=17, c=43, mod 100 Solution: Xi +1 =(17X 27+43) mod 100= 502 mod 100 = 2 =

= 0.02

X2 =(17X 2+43) mod 100= 77 mod 100 = 77 = 0.77 X3 =(17X 77+43) mod 100= 1352 mod 100 = 52 = 0.52 X4 =(17X 52+43) mod 100= 927 mod 100 = 27 = 0.27 X5 =(17X 27+43) mod 100= 502 mod 100 = 2 = 0.02 X6 =(17X 2+43) mod 100= 43.34 mod 100 = 77 = 0.77 X7 =(17X 77+43) mod 100= 1352 mod 100 = 52 = 0.52 X7=(17X 52+43) mod 100= 927 mod 100 = 27 = 0.27 X8=(17X 27+43) mod 100= 502 mod 100 = 2 = 0.02 X9=(17X 2+43) mod 100= 77 mod 100 =77 = 0.77 X10=(17X 77+43) mod 100= 1352 mod 100 =52 = 0.52 ∴ The

10 random number {0.02, 0.77, 0.52, 0,27, 0.02, 0.77, 0.52,0.27,0.02,0.77,0.52}

Ex: 4 Page 269 Use Linear congruenHal Method to generate a sequence of 3 two-digit random integers Let X0=27, a=8, c=47, mod= 100 Solution: Xi +1 =(8x 27+47) mod 100= 263 mod 100 = 63 37 Prepared by: Youssry Hamdy-CS Diploma



R1 =

63 =0.63 100

x2 =(8x63+47) = mod 100 = 551 mod 100 = 51

R2 =

51 =0.51 100

x3 =(8x51+47) = mod 100 = 455 mod 100 = 55

R3 =

55 =0.55 100

x4 =(8x55+47) = mod 100 = 487 mod 100 = 87

R4 =

87 =0.87 100

x5 =(8x87+47) = mod 100 = 743 mod 100 = 43

R5 =

43 =0.43 100

The Sequence of Random integer numbers are {63, 51, 55, 87, 43}

Ex: 6 Page: 269 Use the multiplicative congruential method to generate a sequence of four-digit random integers. Let X 0 =117, a=43, and m=1000. ax mod 100 X1=[43(117)]=mod 1000 = (5031) mod 1000 = 31 X2=[43(31)]=mod 1000 = (1333) mod 1000 = 333 X3=[43(333)]=mod 1000 = (14319) mod 1000 = 319 X4=[43(319)]=mod 1000 = (13717) mod 1000 = 717 The four-digit random integers are {31, 333, 319, 717}

Tests of autocorrelation: Ex: 7.8 Page: 266 Given the following probability distribution and intervals, compute the largest M that satisfting the inequality M=4 Random Probability

3 0.23

8 0.28

13 0.33

18 0.27

23 0.05

28 0.36

Formula: Where: i is starIng point, m is step size, M is the largest integer 4 such that i+(M+1)m<30




Solution: R3 +5k+1-0.25

0.25 = (0.23 x 0+02.8x 0.33+0.33x(0.27)+0.27x0.05+0.05x0.36-0.25=- 0.1945

Q:1 Define the simulation and draw the steps of simulation modeling and process. A simulation is the Imitation of the operation of real-world process or system. Simulation involves the generation of an artificial history of a system, and observation of that artificial history to draw inferences concerning the operating characteristics of the real system. Steps of simulation: 1- Problem formulation, 2- Se Jng of objecKves and overall project plan, 3- Model conceptualization 4- Data collecLon, 5- Model translaMon, 6-TesNng veri(caOon, 7- Testing validation 8- Experimental design, 9- Running and analysis, 10- More runs, 11- Repor Png, 12- Setting implementation




Q 2: De&ne the model, its importance and the di ' erent types of models. Model is a representation of a system for the purpose of studying the system. The importance of model is that it provides a structural approach to run the system and study i ts beaver without running the system itself to determine the problems of the system under investigation that may appear through implementation and solve them. Types of models: 1- Mathematical

2- simulation [static, Dynamic, Deterministic, Stochastic, Discrete, Continues]




Q 3: When and when not a Simula(on is an appropriate tool?

When simulation is a appropriate tool: 1. Simulation enables the study of and experimentation with, internal interaction of a complex system 2. Changes can be simulated, and the effect of these alternation can be observed 3. The knowledge gained in designing a simulation model may help in suggesting improvement in the system under investigation 4. Simulation can be used to test new designs or policies, so as to prepare for what may happen 5. Simulation can be used to verify the analytical solution 6. Simulation models designed for training allow learning without the cost of on-the-job learning 7. The modern systems (factory, service organization, etc) are so complex that the interactions can be treated only through simulation. When simulation is not a appropriate tool: If the problem can be solved using common sense 1. If the problem can be solved analytically 2. If it is easier to perform direct experiments 3. If the costs exceed the savings 4. If the resources or time are not available 5. If the system behavior is too complex or can’t be defined Q : 4 What’re the advantages and disadvantages of simula )on and areas of application? Advantages of Simulation: 1- Simulation appealing to a client because it mimics what happens in a real system in the design stage 2- It is possible to develop a simulation model without dubious mathematical assumptions 3- In contrast to optimization models, simulation models are run rather than solved Disadvantages of Simulation: 1. Model building requires special training 2. Simulation result may be difficult to interpret 3. Simulation modeling can be time consuming and expensive 41 Prepared by: Youssry Hamdy-CS Diploma


OR 508 Exercises Solution.pdf

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