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2 years ago

6

What information you would gain by watching the video. Plz Help due today.

1 answer:

Archy [21]2 years ago

3 0

Answer: It doesnt let me watch the link im sorry :(

Explanation:

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2 years ago

An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t

IRINA_888 [86]

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

$= \frac{39.2}{n} + (n * 0.01 * 5.0)$

$= \frac{39.2}{n} + (n * 0.05)$

At minimum pt. = 0, we have:

dTp/dn = 0

$= \frac{-39.2}{n^2} + 0.05 = 0$

Solving for n²:

$n^2 = \frac{39.2}{0.05} = 784$

$n = \sqrt{784} = 28$

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) $Tp = \frac{39.2}{28} + (28 * 0.01 * 5)$

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

$\frac{60min}{2.8} = 21.43$

The proportion uptime,

$E = \frac{1.4}{2.8} = 0.5$

3 0

2 years ago