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?

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%.

Utilization = 63.88%

Efficiency = 95.83%

Explanation:

Given that:

The actual usage number of students served were 1150 students

The Design Capacity  of the system is 1,800 students per semester

The  highest number (effective capacity ) for the orientation session is 1,,200

The objective here is to determine the utilization and efficiency of the system.

Using the formula

Utilization of the system = actual usage number  / design capacity  of the system

Efficiency of the system = actual usage number  / effective capacity for the orientation session

Therefore; we have

Utilization = 1150/1800 = 0.6388

= 63.88%

Efficiency = 1150/1200 = 0.9583

= 95.83%

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