A computer glitch (of all things) was discovered at The Baruch Academic Computing Center, and it was determined that only 1,150 students were actually served. If the design capacity of the system is 1,800 students per semester and the highest number of students who can actually go to their 93. 18 orientation session is 1,200, what is the utilization and efficiency of the system?

Question

Manuel Koch

Business

28 December, 04:37

A computer glitch (of all things) was discovered at The Baruch Academic Computing Center, and it was determined that only 1,150 students were actually served. If the design capacity of the system is 1,800 students per semester and the highest number of students who can actually go to their 93. 18 orientation session is 1,200, what is the utilization and efficiency of the system?

+4

Answers (1)

Know the Answer?

Jaslene Casey · Answer 1

Utilization = 0.6388 = 63.88%

Efficiency = 0.9583 = 95.83%.

Explanation:

So, in this question we are given the following parameters or information or data that is going to assist us in solving this question/problem and they are;

=> "only 1,150 students were actually served. If the design capacity of the system is 1,800 students per semester''

=> " the highest number of students who can actually go to their orientation session is 1,200".

(A). Thus, the Utilization of the system = real capacity / designed capacity.

=> 1150/1800 = 0.6388.

Hence, 0.6388 * 100% = 63.88%

(B). Efficiency of the system = real capacity / designed capacity.

=> 1150/1200 = 0.9583.

Therefore, 0.9583 * 100% = 95.83%.