Question

Students are going through a three-step process to obtain their ID cards. Each student will spend 2 minutes at the registration desk before going to one of three cashiers to pay a fee for the ID card. After that, he/she will visit one of four ID processing stations to have his/her picture taken and ID card printed. Visits to the cashier and ID processing station take 10 and 20 minutes respectively. If the demand rate is 0.125 student per minute, how long does it take to process 20 students assuming the system is full?
A. 100 minutes

B. 152 minutes

C. 160 minutes

D. 184 minutes

Answer 1

Answer:

C. 160 minutes

Step-by-step explanation:

The calculation to be made will be to process 20 students

We have, according to the exercise, the following data:

For process 1: Registration table, Number of servers = 1, Time spent = 2 minutes

For process 2: cashier, number of servers = 3, time spent = 10 minutes

For process 3: ID processing station, number of servers = 4, time spent = 20 minutes

Demand rate = 0.125

To solve it, we will look for the capacity that the server has of the previously mentioned processes, calculating the following:

Capacity = Number of students served per minute

We can say that at the registration table we observe:

Time needed to serve 1 student = 2 minutes

Capacity = 0.5 students per minute

With the cashier we analyze the following:

Time needed to serve 1 student = 10 minutes

Number of servers = 3

Capacity = 0.3 students per minute

ID processing station

Time needed to serve 1 student = 20 minutes

Number of servers = 4

Capacity = 0.2 students per minute

When comparing the processes, it is definitely found that the bottleneck is the ID processing station, where it takes more time to serve a student, which leads us to infer that the capacity of the process is comparable to the capacity of the process bottleneck

Process capacity = 0.2 students per minute

Given, demand rate = 0.125 students per minute

We observe that the demand rate is less than the capacity of the process, therefore we can infer that the number of students served during each minute is the same as the demand rate.

In this way we find that:

Number of students served per minute = 0.125

Time needed to serve 1 student = 1 / 0.125 = 8 minutes

Time needed to serve 20 students = 8 x 20 = 160 minutes.

We conclude that the answer is that it will take 160 minutes to serve 20 students.

.

Answer 2

Answer:

C) 160 minutes

Explanation:

Given:

Time spent at registration desk = 2mins

Time spent at the cashier = 10 mins

Time spent at the ID processing station = 20 mins

Number of registration desk = 1

Number of cashiers = 3

Number of ID processing stations = 4

Let's calculate the capacity of each process using the expression: number of agents / process time

Therefore,

Capacity of registration desk =

$\frac{1}{2} = 0.5 students per min$

Capacity of cashiers =

$\frac{3}{10} = 0.3 students per min$

Capacity of ID processing stations =

$\frac{4}{20} = 0.2 students per min$

The process capacity is equal to the bottleneck process. Here, the bottleneck process is the process with the longest time per minute which is the ID processing station.

Therefore, given a rate of 0.125 student per min and Process capacity of 0.2 student per min, we'll take 0.125 as the number of students per minute since it is lower than the process capacity.

Therefore, time taken to serve one student = $\frac{1}{0.125} = 8mins$

Time taken to serve 20 students would be = 20 * 8 = 160 minutes