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| |
Hoylake Rescue Squad |
|
| Probability |
of
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| Time between calls |
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| Simulation |
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| P(x) |
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| Cumulative |
(lower bound)
Time between calls
simulation Number |
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| RN |
Time between calls
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| Cumulative clock |
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| 0 |
.1
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| 5 |
1 1
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| 0.1 |
2 2
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| 0.2 |
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| 3 |
3
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| 0.25 |
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| 4 |
4
0.2 5 5
0.1
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| 6 |
6
1
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| 7 |
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| 8 |
| EV = |
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| 9 |
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| Average |
Time =
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| 11 |
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| 12 |
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| 13 |
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| 20 |
P2
| Petroco service |
Simulation
Probability Cumulative
| Time between arrival (min) |
|
|
|
| 0.3 |
5
1
Counts |
RN Time between calls RN Time between calls Cumulative clock
0.25 2 1
0.2 3 2
0.2 4 3
1 4
5
| a. Avg Arrival time |
6
| b. Avg. arrival time |
7
8
| Compare a. and b. |
9
10
11
12
13
14
15
16
17
18
19
20
P3
|
| Dynaco Manufacturing |
| Probability breakdown per week |
Simulation
P(x) Cumulative
|
|
| Breakdown |
|
| Week |
RN
Breakdowns |
0.1 0 1
0.2 1 2
|
|
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| 0.15 |
2 3
0.3 3 4
0.15 4 5
0.1 5 6
1 7
8
9
| Simulated avg. breakdown |
10
| Average breakdowns = |
11
12
13
14
15
16
17
18
19
20
P4
|
| Sun Visor |
or
Umbrella |
?
Simulation
P(x) Cumulative Sun Visor Week RN
| SunVisor ($) |
RN
Umbrella ($) |
|
| 0.35 |
-400 |
1
0.25
| -200 |
2
|
| 0.4 |
1500 |
3
1 4
5
6
P(x) Cumulative Umbrella 7
0.35
| 2100 |
8
0.25 0 9
0.4
| -800 |
10
1 11
12
13
14
15
16
17
18
19
20
Average
P5
Dynaco Manufacturing
| Table from P3 |
| Repair Time |
Simulation Breakdown
P(x) Cumulative
| Repair (hrs) |
Week RN
Breakdown # |
RN
Repair time/breakdown |
Repair Time/week |
P(x) Cumulative Breakdown
0.2 1 1 0 0.1 0
| 0.5 |
2 2 0 0.2 1
0.3 3 3 0 0.15 2
1 4 0 0.3 3
5 0 0.15 4
6 0 0.1 5
| Simulated avg. repair time |
7 0
8 0
9 0
| Theoretically calculated |
10 0
11 0
12 0
13 0
14 0
15 0
16 0
17 0
18 0
19 0
20 0
| Average repair time |
MAT540 Homework
Week 3
Page 4 of 2
1. The Hoylake Rescue Squad receives an emergency call every 1, 2, 3, 4, 5, or 6 hours, according to the following probability distribution. The squad is on duty 24 hours per day, 7 days per week:
Time Between
Emergency Calls (hr.)
Probability
1
0.05
2
0.10
3
0.30
4
0.30
5
0.20
6
0.05
1.00
Solution
Time between emergency calls (hr)
probability
Cumulative probability
1
0.05
0.05
2
0.10
0.15
3
0.30
0.45
4
0.30
0.75
5
0.20
0.95
6
0.05
1.00
a. Simulate the emergency calls for 3 days (note that this will require a “running”, or cumulative, hourly clock), using the random number table.
Random number
Emergency call
10
2
88
6
29
3
b. Compute the average time between calls and compare this value with the expected value of the time between calls from the probability distribution. Why are the results different?
The average time between calls = (2 + 6 + 3)/3
= 11/3 = 3.667
The expected value of the time between calls from the probability distribution:
(1 x 0.05) + (2 x 0.10) +(3 x 0.30) +( 4 x 0.30) +(5 x 0.20) + 6 x 0.05 = 3.65
The results are different because only 3 values were simulated
2. The time between arrivals of cars at the Petroco Service Station is defined by the following probability distribution:
Time Between
Arrivals (min.)
Probability
1
0.25
2
0.30
3
0.35
4
0.10
1.00
Solution
Time between Arrivals (min)
probability
Cumulative probability
1
0.25
0.25
2
0.30
0.55
3
0.35
0.90
4
0.10
1.00
a. the arrival of cars at the service station for 20 arrivals and compute the average time between arrivals.
Solution
Random Number
Time between arrivals
52
2
88
3
50
2
70
3
40
2
77
3
39
2
17
1
9
1
40
2
38
2
45
2
90
3
14
1
49
2
18
1
88
3
87
3
10
1
44
2
Total
41
Average time between intervals = sum of time between arrivals divided by 20 arrivals i.e 41/20 = 2.05
b. Simulate the arrival of cars at the service station for 1 hour, using a different stream of random numbers from those used in (a) and compute the average time between arrivals.
Solution
Using stream of 20 different random numbers
Random Number
Time between arrivals
85
3
37
2
39
2
67
3
40
2
38
2
20
2
68
3
45
2
16
1
28
2
20
1
80
3
9
1
50
2
84
3
80
3
68
3
97
4
47
2
Total
46
Average time between intervals = sum of time between arrivals divided by sum of arrivals i.e 55/23 = 2.391
c. Compare the results obtained in (a) and (b).
The average obtained in (a) is less than that of obtained in (b). This is because the second generated random numbers are different from the first one and also the corresponding time between arrivals are different from the first one thus resulting to the difference in the two results.
3. The Dynaco Manufacturing Company produces a product in a process consisting of operations of five machines. The probability distribution of the number of machines that will break down in a week follows:
Machine Breakdowns
per Week
Probability
0
0.05
1
0.15
2
0.20
3
0.30
4
0.25
5
0.05
1.00
Solution
Machine Breakdowns per week
probability
Cumulative probability
0
0.05
0.05
1
0.15
0.20
2
0.20
0.40
3
0.30
0.70
4
0.25
0.95
5
0.05
1.00
a. Simulate the machine breakdowns per week for 20 weeks.
Solution
Random Number
Machine breakdowns per week
50
3
80
4
28
2
11
1
38
3
16
1
23
2
88
4
3
0
58
3
48
3
98
5
12
1
7
1
77
4
79
4
56
3
96
5
1
0
23
2
Total
51
b. Compute the average number of machines that will break down per week.
The average number of machines that will break down per week = Sum of machine breakdowns per week divided by 20 weeks i.e 51/20 = 2.55
4. Simulate the following decision situation for 20 weeks, and recommend the best decision.
A concessions manager at the Tech versus A&M football game must decide whether to have the vendors sell sun visors or umbrellas. There is a 30% chance of rain, a 15% chance of overcast skies, and a 55% chance of sunshine, according to the weather forecast in College Junction, where the game is to be held. The manager estimates that the following profits will result from each decision, given each set of weather conditions:
Weather Conditions
Decision
Rain
Overcast
Sunshine
.30
.15
.55
Sun visors
$-500
$-200
$1,500
Umbrellas
2,000
0
-900
Solution
Sun Visors
Umbrellas
Probability
Cumulative Probability
$–500
$2,000
0.30
0.30
$–300
$0
0.15
0.45
$1,500
$–900
0.55
1.00
Random Number
Sun Visors
Random Number
Umbrellas
78
$ 1,500.00
38
$ –
70
$ 1,500.00
98
$ (900.00)
22
$ (300.00)
90
$ (900.00)
15
$ (500.00)
14
$ 2,000.00
60
$ 1,500.00
20
$ 2,000.00
25
$ (500.00)
35
$ –
70
$ 1,500.00
50
$ (900.00)
89
$ 1,500.00
78
$ –
77
$ 1,500.00
62
$ (900.00)
70
$ 1,500.00
40
$ (900.00)
71
$ 1,500.00
20
$ 2,000.00
39
$ (500.00)
79
$ (900.00)
36
$ (500.00)
90
$ (900.00)
88
$ 1,500.00
58
$ (900.00)
80
$ 1,500.00
80
$ (900.00)
83
$ 1,500.00
42
$ –
90
$ 1,500.00
10
$ 2,000.00
30
$ (300.00)
26
$ –
88
$ 1,500.00
99
$ (900.00)
70
$ 1,500.00
80
$ (90.00)
Total
$ 18,400.00
Total
$ (1,090.00)
The average profit for Sun visors = $18,400/20 = $920.00
The average profit for Umbrellas = $1,090/20 = $54.50
Therefore, the decision of Sun visors is the best, since it has higher profit .
5. Every time a machine breaks down at the Dynaco Manufacturing Company (Problem 3), either 1, 2, or 3 hours are required to fix it, according to the following probability distribution:
Repair Time (hr.)
Probability
1
0.25
2
0.55
3
0.20
1.00
a. Simulate the repair time for 20 weeks and then compute the average weekly repair time.
Solution
Repair time (hr)
probability
Cumulative probability
1
0.25
0.30
2
0.55
0.50
3
0.20
1.00
Random Number
Weekly Repair Time
9
1
30
2
34
2
76
3
68
3
48
2
25
1
15
1
89
3
16
1
26
1
53
2
70
3
20
1
88
3
89
3
19
1
30
2
16
1
48
2
Total
38
Average repair time = Sum of weekly repair time divided by 20 weeks i.e 38/20 = 1.9 hrs.
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