Question 1 of 11: Availability and Utilization

Question 1.1: Availability and Utilization (8 marks, compulsory)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

1.1.1 – Definitions. Availability is the fraction of the scheduled (reference) time that a machine is mechanically capable of operating – i.e. NOT under repair or unscheduled maintenance – regardless of whether it was actually put to work. Utilization is the fraction of the time the machine WAS available that it actually spent producing. The two answer different questions: availability asks “could the machine have run,” utilization asks “of the hours it could have run, how many did it.” Because the mine has ruled the 2 daily hours lost to shift change, lunch and coffee breaks structurally unavailable to any equipment, both ratios here are referenced to a 22-hour maximum, not the full 24-hour calendar day.

Given. Daily reference (maximum) time $=22$ h (24 h less 2 h unavoidable shift-change/lunch/coffee loss). Three independent daily operating patterns (1.1.2–1.1.4).

Find. Availability (%) and utilization (%) for each of the three patterns.

Approach. $$\text{Availability}=\dfrac{\text{Reference time}-\text{Repair/downtime}}{\text{Reference time}}\times100\%\qquad \text{Utilization}=\dfrac{\text{Operating time}}{\text{Available time}}\times100\%$$

1. 1.1.2 – Two 6-hour periods (12 h operating, no repair time stated). With no downtime recorded, every one of the 22 reference hours is mechanically available: $$\text{Availability}=\dfrac{22-0}{22}=\boxed{100\%}$$ Of those 22 available hours, only the 12 h actually worked count as production: $$\text{Utilization}=\dfrac{12}{22}=\boxed{54.5\%}$$ The remaining 10 h were available but unused (idle by choice/schedule, not breakdown) – that gap is exactly what utilization is designed to expose.
2. 1.1.3 – 4 h under repair, idle the remainder (0 h operating). $$\text{Availability}=\dfrac{22-4}{22}=\boxed{81.8\%}$$ $$\text{Utilization}=\dfrac{0}{22-4}=\boxed{0\%}$$ This is the diagnostic case: a respectable 81.8% availability (the machine was mechanically fit most of the day) coexists with 0% utilization (it was never actually put to work) – a pure dispatch/scheduling failure, not a maintenance one.
3. 1.1.4 – 20 h operating, 2 h under repair (accounts for the full 22 h). $$\text{Availability}=\dfrac{22-2}{22}=\boxed{90.9\%}$$ $$\text{Utilization}=\dfrac{20}{22-2}=\dfrac{20}{20}=\boxed{100\%}$$ Every available hour was used – utilization is capped at 100% by definition, and reaches it only when operating time exactly fills whatever time availability leaves on the table.