manufacture {pid} | R Documentation |
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. |
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.
Value
Returns the hourly profit made. Random noise is added for some realism.
Author(s)
Kevin Dunn, <kgdunn@gmail.com>
References
Please see Chapter 5 of the following book: Kevin Dunn, 2010 to 2019, Process Improvement using Data, https://learnche.org/pid
See Also
Examples
# 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?