Question

The supervisor of a packaging unit of a milk plant is being pressurised to finish the job closer to the distribution time, thus giving the production staff more leeway to cater to last minute demand. He has the option of running the unit at normal speed or at 110% of normal - “fast speed”. He estimates that he will be able to run at the higher speed for 60% of the time. The packet is twice as likely to be damaged at the higher speed which would mean temporarily stopping the process. If a packet

Accepted Answer

B. Let $m$ be rate at normal speed. Output in time $t$ at normal speed = $0.40 	imes m \cdot t = 0.40 mt$. At fast speed (110%): $0.60 	imes 1.10 m \cdot t = 0.66 mt$. Proportion of output at normal speed = $\frac{0.34 mt}{mt} = 0.34$. Proportion at fast speed = $0.66$. Let $p$ be probability of damage at normal speed. At fast speed, damage probability = $2p$. Total damage probability = $0.34p + 0.66(2p) = 1.66p = 0.112 \implies p = \frac{0.112}{1.66} = 0.067$. Probability of NOT damaged at normal speed = $1 - 0.067 = 0.933 \approx 0.93$.

Packaging Unit Speed and Damage Probability

Explanation

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