Answer:
a) 28 stations
b) Rp = 21.43
E = 0.5
Explanation:
Given:
Average downtime per occurrence = 5.0 min
Probability that leads to downtime, d= 0.01
Total work time, Tc = 39.2 min
a) For the optimum number of stations on the line that will maximize production rate.
Maximizing Rp =minimizing Tp
Tp = Tc + Ftd
[tex] = \frac{39.2}{n} + (n * 0.01 * 5.0)[/tex]
[tex] = \frac{39.2}{n} + (n * 0.05)[/tex]
At minimum pt. = 0, we have:
dTp/dn = 0
[tex] = \frac{-39.2}{n^2} + 0.05 = 0 [/tex]
Solving for n²:
[tex] n^2 = \frac{39.2}{0.05} = 784[/tex]
[tex] n = \sqrt{784} = 28[/tex]
The optimum number of stations on the line that will maximize production rate is 28 stations.
b) [tex] Tp = \frac{39.2}{28} + (28 * 0.01 * 5) [/tex]
Tp = 1.4 +1.4 = 2.8
The production rate, Rp =
[tex] \frac{60min}{2.8} = 21.43 [/tex]
The proportion uptime,
[tex] E = \frac{1.4}{2.8} = 0.5 [/tex]
