A repairable machine operates 2400 hours in a year and breaks down eight times. Mean repair time including waiting is 20 hours. If it were reduced to ten hours, find availability improvement (% to two decimals).
Mean operating time between failures is 2400/8 = 300 hours. Initial availability is $A_1=300/(300+20)=0.9375$; improved availability is $A_2=300/(300+10)=0.967742$. Relative improvement is $100(A_2-A_1)/A_1=3.225806\%$, or **3.23%**. The absolute gain would instead be 3.02 percentage points; the official range uses relative improvement.
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