#!/usr/bin/env python
# coding: utf-8
#
# *This notebook contains course material from [CBE20255](https://jckantor.github.io/CBE20255)
# by Jeffrey Kantor (jeff at nd.edu); the content is available [on Github](https://github.com/jckantor/CBE20255.git).
# The text is released under the [CC-BY-NC-ND-4.0 license](https://creativecommons.org/licenses/by-nc-nd/4.0/legalcode),
# and code is released under the [MIT license](https://opensource.org/licenses/MIT).*
#
# < [CO2 Production by Automobiles](http://nbviewer.jupyter.org/github/jckantor/CBE20255/blob/master/notebooks/03.02-CO2-Production-by-Automobiles.ipynb) | [Contents](toc.ipynb) | [Adipic Acid Flowsheet](http://nbviewer.jupyter.org/github/jckantor/CBE20255/blob/master/notebooks/03.05-Adipic-Acid-Flowsheet.ipynb) >
# # Separating Milk
# ## Summary
#
# This example describes a situation where the material balances and problem specifications do not result in a unique solution. Instead, there is at least one degree of freedom that must be resolved by further considerations. In this case, those considerations include the testing for physically meaningful answers, and to maximize a business objective. What you should learn from this example -
#
# * Degrees of freedom may remain after all of the material balance and process specifications have are satisfied.
# * The importance of testing for feasible solutions.
# * How to use a degree of freedom to maximize a process objective.
# ## Problem Statement
# A dairy processor can separate raw milk into a number of commodity products, each with a price determined by the spot market. An example of commodity prices and product specifications is given in this table.
#
# 
#
# A commercial milk separator manufactured by Gruppo Pieralsi.
#
# | Component | Fat (wt%) | Non-fat Solids (wt%) |Price |
# | :--- | :--- | :--- | :--- |
# | Raw Milk | 3.85 | 8.85 | |
# | Regular Milk | 3.25 | 8.25 | 21.48 \$/cwt |
# | Skim Milk | 0.10 | 8.25 |16.01 \$/cwt |
# | Nonfat Dry Milk | 0.0 | 100.0 | 1.95 \$/lb |
# | Butterfat | 100 | 0.0 |1.77 \$/lb |
#
# Suppose a source can provide 5000 lb/hour of raw whole milk containing 3.85 wt% fat, and 8.85 wt% non-fat solids. The processor can produce a slate of products consisting of regular milk, skim milk, non-fat dry milk, and butterfat in whatever quantities are needed. Specify the product flows that maximize the revenue returned by this operation. What is the highest price the processor can pay for raw milk and still generate a positive cash flow?
#
# (See [The Daily Dairy Report](http://www.dailydairyreport.com/) for more background on the diary industry.)
#
# ## Solution
# In[1]:
get_ipython().run_line_magic('matplotlib', 'inline')
import sympy as sym
sym.init_printing()
# ### Define Streams
#
# The first step is to label the streams, and create mass flow variables corresponding to each stream.
# In[2]:
sym.var(['M1','M2','M3','M4','M5'])
# In[3]:
prods = {};
prods[M1] = "Raw Milk"
prods[M2] = "Regular Milk Product"
prods[M3] = "Skim Milk Product"
prods[M4] = "Nonfat Dry Milk Product"
prods[M5] = "Butterfat Product"
prods
# ### Material Balances
# In[4]:
mbal = [
sym.Eq(M1, M2 + M3 + M4 + M5),
sym.Eq(0.0385*M1, 0.0325*M2 + 0.001*M3 + 0*M4 + 1.00*M5),
sym.Eq(0.0885*M1, 0.0825*M2 + 0.0825*M3 + 1.00*M4 + 0*M5)
]
for m in mbal:
print(m)
# ### Specifications
# In[5]:
specs = [
sym.Eq(M1, 5000)
]
# ### Finding Feasible Solutions to the Material Balances
# In[6]:
soln = sym.solve(specs + mbal)
for m in soln.keys():
print(m, ":", soln[m])
# Let's repeat this step, this time specifying precisely which variables to solve for. The solutions will be found in terms of the remaining variables of the problem.
# In[7]:
soln = sym.solve(specs + mbal,[M1,M2,M3,M4])
for m in soln.keys():
print(m, ":", soln[m])
# What are the feasible values of $M_5$?
# In[8]:
sym.plot(M2.subs(soln),(M5,0,200))
print(sym.solve(M2.subs(soln),M5))
# In[9]:
sym.plot(M3.subs(soln),(M5,0,200))
print(sym.solve(M3.subs(soln),M5))
# In[10]:
sym.plot(M4.subs(soln),(M5,0,200))
print(sym.solve(M4.subs(soln),M5))
# In[11]:
M5_max = sym.solve(M2.subs(soln))[0]
M5_min = sym.solve(M3.subs(soln))[0]
# ### Finding a Solution to Maximize Revenue
# In[12]:
Revenue = 21.48*M2.subs(soln)/100.0 \
+ 16.01*M3.subs(soln)/100.0 \
+ 1.95*M4.subs(soln) \
+ 1.77*M5
print(Revenue)
# In[13]:
sym.plot(Revenue,(M5,M5_min,M5_max))
# In[14]:
opt = [sym.Eq(M5_min,M5)]
soln = sym.solve(mbal + specs + opt)
for m in soln.keys():
print("{0:2s} : {1:8.1f} lb/hr {2}".format(str(m), soln[m], str(prods[m])))
# Revenue and Milk Price Calculations
# In[15]:
print(" Maximum Revenue = ${0:8.2f}/hr".format(Revenue.subs(soln)))
print("Max Raw Milk Price = ${0:8.2f}/lb".format(Revenue.subs(soln)/5000.0))
# In[ ]:
#
# < [CO2 Production by Automobiles](http://nbviewer.jupyter.org/github/jckantor/CBE20255/blob/master/notebooks/03.02-CO2-Production-by-Automobiles.ipynb) | [Contents](toc.ipynb) | [Adipic Acid Flowsheet](http://nbviewer.jupyter.org/github/jckantor/CBE20255/blob/master/notebooks/03.05-Adipic-Acid-Flowsheet.ipynb) >