Learn R Programming

pid (version 0.50)

manufacture: Simulation of a manufacturing facility's profit when varying two factors

Description

The hourly profit made when selling the product at price P and the product is produced at a throughput rate T [parts per hour].

Usage

manufacture(P=0.75, T=325)

Arguments

P

the selling price of the product. The default amount, if unspecified, is $0.75.

T

the production rate (throughput), measured in parts per hour. The default value, if unspecified, is 325 parts per hour.

Value

Returns the hourly profit made. Random noise is added for some realism.

Details

This function simulates the hourly profit in a manufacturing facility. Two factors can be adjusted by the user to determine the optimum:

  • P: The selling price of the product (must be a positive value).

  • T: The production rate, also called throughput (must be a positive value).

Can you determine the best combination of conditions, using a systematic method (i.e. not by trial and error)? More defects are created when production rates are too high.

References

Please see Chapter 5 of the following book: Kevin Dunn, 2010 to 2019, Process Improvement using Data, https://learnche.org/pid

See Also

popcorn, grocery

Examples

Run this code
# NOT RUN {
# Producing at the default settings of price ($0.75)
# and throughput of 325 parts per hour:
manufacture()

# Let's try selling for a higher price, $1.05,
# and a slower throughput of 298 parts per hour:
manufacture(P=1.05, T=298)

# What happens if the product is sold too cheaply
# at high production rates?
manufacture(P=0.52, T=417)

# Can you find the optimum combination of settings to
# maximize the profit, but using only a few experiments?
# }

Run the code above in your browser using DataLab