Exam 2

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A random number of 0.1835 falls within the range for demand of 1. A random number of 0.3094 falls within the range for demand of 2. Demand Probability Corresponding Random Numbers 1 0.20 0 ≤ x ≤ 0.20 2 0.20 0.20 < x ≤ 0.40 3 0.20 0.40 < x ≤ 0.60 4 0.20 0.60 < x ≤ 0.8050.200.80 < x ≤ 1.00

A manager is simulating the number of times a machine operator stops a machine to make adjustments. After careful study the manager found that the number of stops ranged from one to five per cycle and that each number of stops was equally likely. Using the random numbers 0.1835 and 0.3094 (in that order), the next two simulated cycles would respectively have stops for adjustment of:

A random number of 0.49 falls within the range for a total number of stops of 3. A random number of 0.14 falls within the range for a total number of stops of 1. Stops Probability Corresponding Random Numbers 1 0.20 0 ≤ x ≤ 0.20 2 0.20 0.20 < x ≤ 0.40 3 0.20 0.40 < x ≤ 0.60 4 0.20 0.60 < x ≤ 0.80 5 0.20 0.80 < x ≤ 1.00

A manager is simulating the number of times a machine operator stops a machine to make adjustments. After careful study the manager found that the number of stops ranged from one to five per cycle and that each number of stops was equally likely. Using the random numbers 0.49 and 0.14 (in that order), determine how many stops for adjustments each of the next two cycles will have. The first cycle will have ? selected answer correct stops and the second cycle will have ? selected answer correct stops.

With 4 supply nodes and 4 demand nodes, there will be 16 {4 × 4} arcs in the network.

A manufacturing firm has four plants and wants to find the most efficient means of meeting the requirements of its four customers. The relevant information for the plants and customers, along with shipping costs in dollars per unit, are shown in the table below: Customer (requirement)Factory (capacity)Customer 1 (125)Customer 2 (150)Customer 3 (175)Customer 4 (75)A (100)$15 $10 $20 $17 B (75)$20 $12 $19 $20 C (100)$22 $20 $25 $14 D (250)$21 $15 $28 $12 How many arcs will the network have?

Using Solver, the optimal quantity to ship from source B to destination 3 is 75 units. The problem formulation is Min∑4i=1∑Dj=Aci,jxi,jMin∑i=14∑j=ADci,jxi,j (where i represents the destinations and j represents the source) s.t.∑4i=1xi,j≤capacityj j∈{A,B,C,D}s.t.∑i=14xi,j≤capacityj j∈{A,B,C,D} (capacity constraint for each source) ∑Dj=Axi,j≥demandi i∈{1,2,3,4}∑j=ADxi,j≥demandi i∈{1,2,3,4} (demand constraint for each destination) xi,j≥0xi,j≥0 (quantity shipped must be non-negative)

Each of the four customers is a demand node.

A manufacturing firm has three plants and wants to find the most efficient means of meeting the requirements of its four customers. The relevant information for the plants and customers, along with shipping costs in dollars per unit, are shown in the table below: Customer (requirement)Factory (capacity)Customer 1 (25)Customer 2 (50)Customer 3 (125)Customer 4 (75)A (100)$15 $10 $20 $17 B (75)$20 $12 $19 $20 C (100)$22 $20 $25 $14 How many demand nodes are present in this problem?

Each of the three factories is a supply node.

Using Solver, minimum cost to meet all customer demand is $4,475. The problem formulation is Min∑4i=1∑cj=Aci,jxi,jMin∑i=14∑j=Acci,jxi,j (where i represents the destinations and j represents the source) s.t.∑4i=1xi,j≤capacityj j∈{A,B,C}s.t.∑i=14xi,j≤capacityj j∈{A,B,C} (capacity constraint for each source) ∑cj=Axi,j≥demandi i∈{1,2,3,4}∑j=Acxi,j≥demandi i∈{1,2,3,4} (demand constraint for each destination) xi,j≥0xi,j≥0 (quantity shipped must be non-negative)

Using Solver, the optimal quantity to ship from source A to destination 2 is 50 units. The problem formulation is Min ∑4i=1∑c j=A ci,jxi,j Min ∑i=14∑j=Acci,jxi,j (where i represents the destinations and j represents the source) s.t.∑4i=1 x i,j≤ capacity j j∈{A, B, C}s.t.∑i=14xi,j≤capacityj j∈{A, B, C} (capacity constraint for each source) ∑cj=Axi,j≥demandi i∈{1,2,3,4}∑j=Acxi,j≥demandi i∈{1,2,3,4} (demand constraint for each destination) xi,j≥0xi,j≥0 (quantity shipped must be non-negative)

Minimum-Cost Flow problem

First, convert service time to service rate (μ = 1/40 minutes = 0.025 customers per minute = 1.5 customers per hour). Next, set the utilization equal to 1 and solve for s. ρ = λ/sμ = 5 customers per hour/s*(1.5 customers per hour) = 1⇔s = λ/ρμ = 5/ 1.5 = 3.33 servers. Round up to 4 servers.

A multiple-server system has customers arriving at an average rate of five per hour and an average service time of forty minutes. The minimum number of servers for this system to have a utilization factor under 1 is:

True

A number is a random number between 0 and 1 if it is generated in such a way that every possible number within this interval has an equal chance of occurring. True or False

Each crew is considered a server because they work together.

A queueing system has four crews with three members each. The number of "servers" is:

The utilization factor can be calculated to be 0.4. First, convert inter-arrival time to arrival rate ( λ = 1/10 minutes = 0.1 customers per minute ) and convert service time to service rate ( μ= 1/4 minutes = 0.25 customers per minute ). ρ= λ/μ = 0.1 minutes / 0.25 minutes = 0.4

A single bay car wash with an exponential arrival rate and service time has cars arriving an average of 10 minutes apart, and an average service time of 4 minutes. The utilization factor is:

Expected interarrival time = 1/λ' rearranging to solve for λ (the arrival rate) we see that λ = 1 customer/10 minutes × 60 minutes/hour = 6 customers/hour

A single server queueing system has an average service time of 8 minutes and an average time between arrivals of 10 minutes. The arrival rate is:

Sales of 100 occurred in 12 out of 40 months. This is a probability of 0.3 (12 / 40 = 0.3). Therefore, if the lower end of the range of random numbers is 0, the upper end of the range will be 0.3. Sales (units) Frequency Corresponding Random Numbers 100 120 ≤ x < 0.3000 150 6 200 8 250 4 300 10

After reviewing past history, you have assembled the following table showing the frequency of certain levels of sales over the past 40 months. If the smallest random number used to represent sales of 100 units is 0, what will be the largest number used to represent sales of 100 units? Sales (units) Frequency Corresponding Random Numbers 100 120 ≤ x < ?? 150 6 200 8 250 4 300 10

Sales of 200 occurred in 8 out of 40 months. This is a probability of 0.2 (8 / 40 = 0.2). Therefore, if the lower end of the range of random numbers is 0.45, the upper end of the range will be 0.65 (0.45 + 0.2). Sales (units) Frequency Corresponding Random Numbers 100 12 150 6 200 8 0.45 ≤ x < 0.65 250 4 300 10

After reviewing past history, you have assembled the following table showing the frequency of certain levels of sales over the past 40 months. If the smallest random number used to represent sales of 200 units is 0.45, what will be the largest number used to represent sales of 200 units? Sales (units) Frequency Corresponding Random Numbers 100 12 150 6 200 8 0.45 ≤ x < ?? 250 4 300 10

True

Any node where the net amount of flow generated is fixed at zero is a transshipment node. True or False

Customers move through the system more slowly because utilization is increased.

As the ratio of arrival rate to service rate is increased, which of the following is likely?

To simulate an exponential distribution, use the natural logarithm formula in Excel. In this case, the correct formula is = −20 × LN(RAND()). Random Number Time Between Arrivals 0.49 14 0.15 38 0.28 25

Customers arrive at a carwash on average once every 20 minutes. It seems likely that customer arrivals follow an exponential distribution. For each of these random numbers, calculate the simulated time between arrivals at the carwash.

To simulate an exponential distribution, use the natural logarithm formula in Excel. In this case, the correct formula is = −20 × LN(RAND()). With a random number of 0.1398, this equation returns an arrival time of 39.35 minutes.

Customers arrive at a carwash with on average once every 20 minutes. It seems likely that customer arrivals follow an exponential distribution. In a simulation, the random number 0.1398 is generated. How long it will be until the next simulated arrival occurs?

To simulate an exponential distribution, use the natural logarithm formula in Excel. In this case, the correct formula is = −20 × LN(RAND()).

Customers arrive at a carwash with on average once every 20 minutes. It seems likely that customer arrivals follow an exponential distribution. In a simulation, what formula would you use to estimate how long it will be until the next arrival occurs?

First, convert service time to service rate (μ= 1/3 minutes = 0.333 customers per minute = 20 customers per hour) ρ =λ/μ= 14/20 = 0.7. Lq = λ^2/μ(μ−λ) = 196/120 = 1.633 customers

Customers arrive at a suburban ticket outlet at the rate of 14 per hour on Monday mornings (exponential inter-arrival times). Selling the tickets and providing general information takes an average of 3 minutes per customer, and varies exponentially. There is 1 ticket agent on duty on Mondays. What is the average number of customers waiting in line?

First, convert service time to service rate (μ = 13 minutes = 0.333 customers per minute = 20 customers per hour). W = 1/μ−λ = 1/6 = 0.1667 hour = 10 minutes

Customers arrive at a suburban ticket outlet at the rate of 14 per hour on Monday mornings (exponential inter-arrival times). Selling the tickets and providing general information takes an average of 3 minutes per customer, and varies exponentially. There is 1 ticket agent on duty on Mondays. How many minutes does the average customer spend in the system?

First, convert service time to service rate (μ = 13 minutes = 0.333 customers per minute = 20 customers per hour). Wq = λ/μ(μ−λ) = 14/20(20−14) = 0.1167 hour = 7 minutes

Customers arrive at a suburban ticket outlet at the rate of 14 per hour on Monday mornings (exponential inter-arrival times). Selling the tickets and providing general information takes an average of 3 minutes per customer, and varies exponentially. There is 1 ticket agent on duty on Mondays. How many minutes does the average customer wait in line?

First, convert service time to service rate μ = 13 minutes = 0.333 customers per minute = 20 customers per hour. ρ = 14/20 = 0.7.

λ = 1 customer per minute, μ = 0.4 customer per minute. Using the MMs textbook spreadsheet with 4 servers, the probability of three or fewer customers is the sum of the individual probabilities: P0 + P1 + P2 + P3 = 0.074 + 0.184 + 0.230 + 0.192 = 0.680

Customers arrive at a video rental desk at the rate of one per minute (exponential inter-arrival times). Each server can handle 0.4 customers per minute (exponential service times). If there are 4 servers, what is the probability of three or fewer customers in the system?

Using the MMs textbook spreadsheet, trial and error show that with 5 servers average time in the system is 2.63 minutes but with 4 servers average time in the system increases to 3.03 minutes.

Customers arrive at a video rental desk at the rate of one per minute (exponential inter-arrival times). Each server can handle 0.4 customers per minute (exponential service times). What is the minimum number of servers needed to achieve an average time in the system of less than three minutes?

First, convert inter-arrival time to arrival rate (λ = 11 minutes = 1 customers per minute = 60 customers per hour). Using the MMs textbook spreadsheet, this results in L = 4.059 ≅ 4

Customers filter into a record shop at an average of 1 per minute (exponential inter-arrivals) where the service rate is 15 per hour (exponential service times). What is the average number of customers in the system with 8 servers?

First, convert service time to service rate (μ = 14 minutes = 0.25 customers per minute = 15 customers per hour). Using the MMs textbook spreadsheet, this results in Lq = 0.099≅ 0.1

During the early morning hours, customers arrive at a branch post office at the average rate of 45 per hour (exponential inter-arrival times), while clerks can handle transactions on an average of 4 minutes each (exponential). What is the average number of customers waiting for service if 6 clerks are used?

False

Each node in a minimum cost flow problem where the net amount of flow generated is a fixed positive number is a demand node. True or False

There is an equal amount of supply and demand.

For a minimum cost flow problem to have a feasible solution, which of the following must be true?

The frequencies of each event can be used to calculate the probability of each event by dividing the frequency of each event by the total of all observations. (pi=Frequencyi∑nj=1Frequencyj)pi=Frequencyi∑j=1nFrequencyj . The results are then used to determine the corresponding random numbers to each event. A random number of 0.2258 falls within the range for demand of 0. Demand Frequency Probability Corresponding Random Numbers 0 38 0.38 0 ≤ x ≤ 0.38 1 22 0.22 0.38 < x ≤ 0.60 2 22 0.22 0.60 < x ≤ 0.82 3 18 0.18 0.82 < x ≤ 1.00

Given this frequency distribution, the random number 0.2258 would be interpreted as a demand of: Demand Frequency 0 38 1 22 2 22 3 18

The frequencies of each event can be used to calculate the probability of each event by dividing the frequency of each event by the total of all observations. (pi=Frequencyi∑nj=1Frequencyj)pi=Frequencyi∑j=1nFrequencyj . The results are then used to determine the corresponding random numbers to each event.A random number of 0. 5211 falls within the range for demand of 1. Demand Frequency Probability Corresponding Random Numbers 0 38 0.38 0 ≤ x ≤ 0.38 1 22 0.22 0.38 < x ≤ 0.60 2 22 0.22 0.60 < x ≤ 0.82 3 18 0.18 0.82 < x ≤ 1.00

Given this frequency distribution, the random number 0.5211 would be interpreted as a demand of: DemandFrequency 0 38 1 22 2 22 3 18

The frequencies of each event can be used to calculate the probability of each event by dividing the frequency of each event by the total of all observations. (pi=Frequencyi∑nj=1 Frequencyj)pi=Frequencyi∑j=1n Frequencyj . The results are then used to determine the corresponding random numbers to each event. Demand Frequency Probability Corresponding Random Numbers 0 23 0.23 0 ≤ x ≤ 0.23 1 16 0.16 0.23 < x ≤ 0.39 2 18 0.18 0.39 < x ≤ 0.57 3 43 0.43 0.57 < x ≤ 1.00 With the random numbers given, the simulated demand is Random Number Simulated Demand 0.1 0 0.6 3 0.4 2

Given this frequency distribution, what demand values would be associated with the following random numbers? (Do not round intermediate calculations.) Demand Frequency 0 23 1 16 2 18 3 43

It increases exponentially.

If a manager increases the system utilization standard (assuming no change in the customer arrival rate) what happens to the customer waiting time?

A random number of 0.6246 falls within the range for demand of 2. Demand Probability Corresponding Random Numbers 0 0.15 0 ≤ x ≤ 0.15 1 0.30 0.15 < x ≤ 0.45 2 0.25 0.45 < x ≤ 0.70 3 0.15 0.70 < x ≤ 0.85 4 0.15 0.85 < x ≤ 1.00

If a simulation begins with the first random number, the first simulated value would be:Random numbers: 0.6246, 0.2594, 0.4055 Demand Frequency 0 0.15 1 0.30 2 0.25 3 0.15 4 0.15

A random number of 0.2594 falls within the range for demand of 1. Demand Probability Corresponding Random Numbers 0 0.15 0 ≤ x ≤ 0.15 1 0.30 0.15 < x ≤ 0.45 2 0.25 0.45 < x ≤ 0.70 3 0.15 0.70 < x ≤ 0.85 4 0.15 0.85 < x ≤ 1.00

If a simulation begins with the first random number, the second simulated value would be:Random numbers: 0.6246, 0.2594, 0.4055 Demand Frequency 0 0.15 1 0.30 2 0.25 3 0.15 4 0.15

λ = 7 and μ = 12. Utilization is given by ρ = λ/μ = 7/12 =0.583. L = λ/μ−λ =7/(12−7) = 1.400 customers. W = 1/μ−λ = 1/(12−7) = 0.200 hours. Wq = λ/μ(μ−λ) =7/12(12−7) = 0.117 hours and Lq = λ^2/μ(μ−λ) = 7^2/12(12−7) = 0.817 customers.

In an M/M/1 queueing system, the arrival rate is 7 customers per hour and the service rate is 12 customers per hour. What is the utilization? What is the expected number of customers in the system (L)? What is the expected waiting time in the system (W)? What is the expected number of customers in the queue (Lq)? What is the expected waiting time in the queue (Wq)?

False

Managers can use simulation to obtain optimal answers for a wide range of problems. True or False

The Queueing Simulator returns a total time in the system (L) of about 4 customers. Your value may vary somewhat due to the use of the random number generator.

Note: This question requires the use of the Queueing Simulator spreadsheet in Excel. Use the queueing simulator to simulate a queueing system with 2 servers, exponential inter-arrival times with a mean of 20 seconds and constant service times of 35 seconds. Use a simulation run length of 10,000 arrivals. What is the point estimate for the average number of customers in the system?

The Queueing Simulator returns a total time in the system (W) of about 90 seconds. Your value may vary somewhat due to the use of the random number generator.

Note: This question requires the use of the Queueing Simulator spreadsheet in Excel. Use the queueing simulator to simulate a queueing system with 3 servers, uniform interarrival times between 20 and 50 seconds and exponential service times with an average of 75 seconds. Use a simulation run length of 10,000 arrivals. What is the point estimate for the total amount of time a customer will spend in the system?

Using a probability of less than 0.22 as indicating the car did not start, two values (0.0527 and 0.0131) indicate that David's car did not start.

On cold mornings, the probability that David's car won't start is 0.22. When it doesn't start, he takes the bus and is late for work. When it does start, he drives to work on the freeway. Sixty-five percent of the time the freeway is clear, and he gets to work on time. The rest of the time he is late. Simulate 10 consecutive days' worth of trips to work using the random numbers given (use the smaller numbers to represent "car won't start" and "freeway clear"). Random numbers: Car Traffic 0.7772 0.2282 0.2902 0.3674 0.8120 0.1654 0.2259 0.6909 0.0527 0.7010 0.2958 0.9802 0.2891 0.5615 0.0131 0.5604 0.5219 0.4361 0.9949 0.1405 How many times is David late for work because his car won't start?

Using the textbook spreadsheet Nonpreemptive Priorities, the average time in system (W ) for high priority items is 0.579 hour (34.74 minutes).

Priority Average Arrival Rate (exponential inter-arrival times) High 3 per hour Low 5 per hour Number of servers:5 Service rate: 2 per hour (exponential service times) What is the average time in the system (in minutes) for a high priority item?

ρ=λHigh+λLow/sμ = 8/5(2) = 0.8

Priority Average Arrival Rate (exponential inter-arrival times) High 3 per hour Low 5 per hour Number of servers: 5 Service rate: 2 per hour (exponential service times) What is the system utilization factor?

λTotal = λHigh + λLow = 5 + 3 = 8 customers per hour

PriorityAverage Arrival Rate (exponential inter-arrival times) High 3 per hour Low 5 per hour Number of servers: 5 Service rate: 2 per hour (exponential service times) What is the overall arrival rate per hour?

False

Random numbers can be generated in Excel by using the VLOOKUP function. True or False

To simulate an exponential distribution, use the natural logarithm formula in Excel. In this case, the correct formula is = −40 × LN(RAND()). With a random number of 0.7588, this equation returns an arrival time of 11.04 minutes.

Service times at a doctor's office take an average of 40 minutes. It seems likely that service times follow an exponential distribution. In a simulation, the random number 0.7588 is generated. How long it will be until the next simulated arrival occurs?

To simulate an exponential distribution, use the natural logarithm formula in Excel. In this case, the correct formula is = −40 × LN(RAND()).

Service times at a doctor's office take an average of 40 minutes. It seems likely that service times follow an exponential distribution. In a simulation, what formula would you use to estimate how long the next service will take?

True

Simulation is especially useful for situations too complex to be analyzed using analytical models. True or False

According to the Queueing Simulator results, only a total of 5 customers in the system (L = 5) falls within the 95% confidence intervals.

The Queueing Simulator returned the results shown below for a system with 2 servers. Which of the following is most likely to occur?

The Queueing Simulator returned a total time in the queue (Wq) range of about 15.3 to 17.3 seconds (95% confidence interval). The only choice that falls within this range is c, 15 seconds.

The Queueing Simulator returned the results shown below. Which of the following waiting times in the queue is most likely to occur?

False

The amount of flow that is eventually sent through an arc is called the capacity of that arc. True or False

Using Solver, the optimal quantity to ship from node A to node I is 33 TB/s. The problem formulation is Max xI,AMax xI,A s.t.xi,j≤capacityj i,j∈{A,B,C,D,E,F,G,H,I}s.t.xi,j≤capacityj i,j∈{A,B,C,D,E,F,G,H,I} (capacity constraint for each arc) xin−xout=0xin-xout=0 (flow into node equals flow out of node) xi,j≥0xi,j≥0 (quantity shipped must be non-negative)

The figure below shows the nodes (A-I) and capacities (labelled on arcs in TB/s) of a computer network. The firm would like to know how much information can flow from node A to node I. Note: This question requires Solver. Formulate the problem in Solver and find the optimal solution. What is the maximum amount of data that can be transmitted from node A to node I?

Node A is the source, Node I is the sink.

The figure below shows the nodes (A-I) and capacities (labelled on arcs in packages/day) of a shipping network. The firm would like to know how many packages per day can flow from node A to node I. How many transshipment nodes are present in this problem?

Using Solver, the shortest path is A-C-F-I-M. Therefore, node B is not visited. The problem formulation is: Min∑Mi=A∑Mj=Aci,jxi,jMin∑i=AM∑j=AMci,jxi,j (where i and j are the start and end of each arc) s.t.xi,j={1 if the arc is traveled 0 otherwises.t.xi,j=1 if the arc is traveled 0 otherwise xA,B+xA,C=1xA,B+xA,C=1 (traveler must leave node A exactly once) xI,M+xK,M+xL,M=1xI,M+xK,M+xL,M=1 (traveler must arrive at node M exactly once) xin−xout=0 ∀ {B,C,... ,K,L}xin-xout=0 ∀ {B,C,... ,K,L} (flow into node equals flow out of node for all nodes except nodes A and M) xi,j≥0xi,j≥0 (quantity shipped must be non-negative)

The figure below shows the possible routes from city A to city M as well as the cost (in dollars) of a trip between each pair of cities (note that if no arc joins two cities it is not possible to travel non-stop between those two cities). A traveler wishes to find the lowest cost option to travel from city A to city M. Note: This question requires Solver.Formulate the problem in Solver and find the optimal solution. Which of the following nodes are not visited?

Node A is the origin, Node M is the destination.

The figure below shows the possible routes from city A to city M as well as the cost (in dollars) of a trip between each pair of cities (note that if no arc joins two cities it is not possible to travel non-stop between those two cities). A traveler wishes to find the lowest cost option to travel from city A to city M. Which nodes are the origin and destination for this problem?

A-C-F-H-K-M

Shortest Path Problem

The figure below shows the possible routes from city A to city M as well as the cost (in dollars) of a trip between each pair of cities (note that if no arc joins two cities it is not possible to travel non-stop between those two cities). A traveler wishes to find the lowest cost option to travel from city A to city M. Which type of network optimization problem is used to solve this problem?

the sum of customer waiting costs and service costs.

The goal of queueing analysis is to minimize:

the order in which customers are processed.

The term queue discipline refers to:

The utilization factor can be calculated to be 0.4. First, convert service time to service rate (μ = 1/20 minutes = 0.05 customers per minute = 3 customers per hour). ρ=λ/sμ = 6 customers per hour / 5*(3 customers per hour) = 0.4

There are 5 servers in a system with an arrival rate of 6 per hour and a service time of 20 minutes. What is the system utilization?

The problem can be addressed as a maximal flow problem with the school as the source and the park as the sink. The maximum flow along this source to sink arc is 30.

WK Qu. 6-14 A city would like to know how many vehicles... A city would like to know how many vehicles it could move through its streets in an emergency. The most likely situation is an evacuation starting at the school and ending at the park. The connections between these two locations, plus their capacity and direction, is shown below. PathABCDEFGCapacity30102015301515

Maximize the amount flowing through a network.

What is the objective of a maximum flow problem?

All of the choices are correct. (Storage facilities, Processing facilities, Short-term investment options, & Warehouses)

Which of the following is an example of a transshipment node?

Flow can move toward the sink and away from the sink.

Which of the following is not an assumption of a maximum flow problem?

There is an equal number of supply and demand nodes.

Which of the following is not an assumption of a minimum cost flow problem?

The lines connecting certain pairs of nodes always allow travel in either direction.

Which of the following is not an assumption of a shortest path problem?

Service time

Which of the following is not generally considered as a measure of performance in queueing analysis?

Demand Nodes

Which of the following will have negative net flow in a minimum cost flow problem?

It generates an optimal solution.

Which of the following would not be considered a main advantage of simulation?

The analyst used a formula that generates a uniformly distributed cost between 15 and 45. Therefore, I and III are true.

You are reviewing a simulation model and find that the analyst who prepared the model used the formula = 15 + (45 − 15) × RAND() to generate the simulated cost of a product. Which of the following assumptions did the analyst make about the cost of the product? The minimum cost of the product is 15. The maximum cost of the product is 60. The cost of the product is uniformly distributed.

The formula to generate a uniformly generated variable that ranges between a and b is = a + (b − a) × RAND(). In this case, a = 5 and b = 12, so the formula is = 5 + (12 − 5 ) × RAND().

You have determined that waiting times at a restaurant are uniformly distributed over the interval 5 to 12 minutes. What formula would you use in Excel to generate random values in this range that follow the uniform distribution?

The formula to generate a uniformly generated variable that ranges between a and b is = a + (b − a) × RAND(). In this case, a = 5 and b = 12, so the formula is = 5 + (12 − 5 ) × RAND(). With a random number of 0.2154, this leads to a waiting time of 6.5 minutes (5 + (12 − 5 ) × 0.2154 = 6.5078).

You have determined that waiting times at a restaurant are uniformly distributed over the interval 5 to 12 minutes. The first random number your simulation returns is 0.2154. What is the waiting time that this random number generates?

The formula to generate a uniformly generated variable that ranges between a and b is = a + (b − a) × RAND(). In this case, a = 20 and b = 60, so the formula is= 20 + (60 − 20) × RAND().

You have determined that waiting times at a toll booth are uniformly distributed over the interval 20 to 60 seconds. What formula would you use in Excel to generate random values in this range that follow the uniform distribution?

The formula to generate a uniformly generated variable that ranges between a and b is = a + (b − a) * RAND(). In this case, a = 20 and b = 60, so the formula is= 20 + (60 − 20) * RAND(). With a random number of 0.4732, this leads to a waiting time of 38.928 seconds (20 + (60 − 20) * 0.4732 = 38.928).

You have determined that waiting times at a toll booth are uniformly distributed over the interval 20 to 60 seconds. The first random number your simulation returns is 0.4732. What is the waiting time that this random number generates?

Random Value 0.126 0.482 0.539 0.507 0.555 0.731 0.745 0.231 0.749 0.577 Outcome Heads Heads Tails Tails Tails Tails Tails Heads Tails Tails

You have prepared a coin flipping simulation. In your simulation, a value of less than 0.5 is designated as "Heads". Your simulation generates the following random numbers: 0.126, 0.482, 0.539, 0.507, 0.555, 0.731, 0.745, 0.231, 0.749, 0.577. How many trials ended in the result "Heads"? How many trials ended in the result "Tails"?

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