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larisa [96]
3 years ago
11

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

he probability of a station failure that leads to a downtime occurrence is 0.01. The total work content time = 39.2 min and is to be divided evenly among the workstations, so the ideal cycle time for each station = 39.2/n, where n is the number of workstations.
Determine; (a) the optimum number of stations on the line that will maximize production rate and b) the production rate Rp and proportion uptime for answer to part (a).
Engineering
1 answer:
IRINA_888 [86]3 years ago
3 0

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

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Answer:

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#include <iostream>

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using namespace std;

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void circleType::setRadius(double r)

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}

double circleType::getRadius()

{

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