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Hoylake Rescue Squad
Probability of Time between calls
P(x)
0.15
0.1
0.2
0.25
0.2
0.1
1
Cumulative
(lower
Time
bound)
between calls
1
2
3
4
5
6
EV =
Average Time =
Simulation
simulation
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
RN
Time
between
calls
Cumulative
clock
Petroco service
Simulation
Probability
0.35
0.25
0.2
0.2
1
a. Avg Arrival time
b. Avg. arrival time
Compare a. and b.
Cumulative
Time
between
arrival (min)
1
2
3
4
Counts
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
RN
Time between calls
RN
Time between calls
Cumulative clock
Dynaco Manufacturing
Probability breakdown per week
P(x)
0.1
0.2
0.15
0.3
0.15
0.1
1
Cumulative
Simulated avg. breakdown
Average breakdowns =
Simulation
Breakdown
0
1
2
3
4
5
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
RN
Breakdowns
Sun Visor or Umbrella?
Simulation
P(x)
0.35
Cumulative
Sun Visor
-400
Week
1
0.25
-200
2
0.4
1500
3
1
P(x)
0.35
0.25
0.4
1
4
5
Cumulative
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Umbrella
2100
0
-800
Average
RN
SunVisor ($)
RN
Umbrella ($)
Dynaco Manufacturing
Table from P3
Repair Time
P(x)
0.2
0.5
0.3
1
Simulation
Cumulative
Simulated avg. repair time
Theoretically calculated
Repair (hrs)
1
2
3
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Breakdown
RN
Breakdown #
RN
Repair time/breakdown
Repair Time/week
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
P(x)
0.1
0.2
0.15
0.3
0.15
0.1
Average repair time
Cumulative
Breakdown
0
1
2
3
4
5
MAT540 Homework
Week 3
Page 1 of 3
MAT540
Week 3 Homework
Chapter 14
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
0.15
2
0.10
3
0.20
4
0.25
5
0.20
6
0.10
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.
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 result different?
2.
The time between arrivals of cars at the Petroco Services Station is defined by the following
probability distribution:
Time Between
Emergency Calls (hr.)
Probability
1
0.35
2
0.25
3
0.20
4
0.20
1.00
MAT540 Homework
Week 3
Page 2 of 3
a. Simulate the arrival of cars at the service station for 20 arrivals and compute the average time
between arrivals.
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.
c. Compare the results obtained in (a) and (b).
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
0.10
1
0.20
2
0.15
3
0.30
4
0.15
5
0.10
1.00
a. Simulate the machine breakdowns per week for 20 weeks.
b. Compute the average number of machines that will break down per week.
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:
MAT540 Homework
Week 3
Page 3 of 3
Decision
5.
Weather Conditions
Rain
Overcast
Sunshine
0.35
0.25
0.40
Sun visors
$-400
$-200
$1,500
Umbrellas
2,100
0
-800
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
0.20
2
0.50
3
0.30
1.00
Simulate the repair time for 20 weeks and then compute the average weekly repair time.

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