Homework 7

CHE 116: Numerical Methods and Statistics

Prof. Andrew White

Version 1.1 (3/6/2015)

0. Warmup (1 Point)

We saw many optimization algorithms. Collect all the cells which gave a summary of the different techniques and paste them into your homework below. An example is given below:

Newton's Method

Type: Root Finding

Discrete/Continuous: Continuous

Dimensions: 1

Derivative: optional

Convex: yes

Python: newton

1. Exercises (5 Points)

Minimize the following expression $$\frac{(x - 4)^2}{2} + \frac{(x - 2)^2}{4}$$
Minimize the following expression: $$4 \left[ r^{-12} - r^{-6}\right]$$
Solve the following system of equations: $$ 3 x^2 - 2 y^2 = 4$$ $$ x ^ 3 - 4 y = -2$$
$\xi$ is a fraction between 0 and 1. Maximize the following expression $$ -\xi \ln \xi$$

2. Box Folding (5 Points)

Your primary grade teacher gives you a piece of cardboard that has dimensions: 4 cm by 3 cm. You are supposed to cut the corners as shown below and fold up the corners to make an open box. Determine the height of the box that gives a maximum volume. See the picture below of the box



In [1]:

    
from IPython.display import Image
Image(url="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAQIAAADDCAMAAABeUu/HAAAAA3NCSVQICAjb4U/gAAAAflBMVEX/\
//9sbGylpaXW1tbZ2dmhoaEAAAD5+fn19fXk5OT8/PzQ0NDy8vLc3NzKysrs7Oy0tLS9vb2KiorF\
xcW4uLg9PT14eHiZmZmurq5UVFQ2NjYiIiJxcXEpKSlZWVmSkpJ/f39kZGQwMDAcHBxYWFhISEhL\
S0t8fHxBQUETExNOLLFOAAAACXBIWXMAAA+IAAAPiAEWyKWGAAAF80lEQVR4nO2di3ajKhSG/0SD\
4o0Y02jubabttO//ggd6mZ5LdTwwIOr+V1aWWUZLv2xgw4YtQCJ5qOT+aegiDK2QzR1Bsru+DF2G\
gbXY1ndDl2FYBQzXmSO4r2aP4CWaO4L9Ig93hzAauhwDas/elA9djgEVRRE2864ISjNvC5Tq/dAl\
IJFIJBKJRCKRSCRSp6Jg0uozOxNelhNWc+iDYGlqR17r1OM74Urv3pXbOVCuOd+46PEdXQQ80btO\
U1mmd51NBCORTQSrWO86TaWF3nVkBYSAKgLICmAXgWYnpatI0w2xiWDt1jXKA73rqCJYRRC4tQLB\
9a6zisCtgyyE3nVUEawiaNxWhNDD5tCtZ+RlpzgSkYNMVkAIYBfB8qMiVIU00bJK9e7SpbyqMgRV\
+eaABJp/wIUV1KwCNiw0u8t34myRyLed0U2cTJ9ezzHsrCfNWInF9ePYQ++w+Oyoc7ZsNP2W3yhZ\
MFF/HHMPxwhf4vLHsqR9n0hIp2wi+AqlrB7PdqwA/Hz6ZOBjKEV8tgXNAQ922oKYFSn7bAs8DqUk\
e8nisjG7yfday4agOpjN0LlwkJ+YtNUDq7u/raMte+VYsh9vNY4cZG0RAhopkhVgWqEUTaOziWBL\
oZSRyCaC1PFaIw+HSSGFUkaiCYVSfJw4c7z1OtGsd1QRyEEmK4BdBG77RG25CKU4ko89wkhkE0Hu\
eCW6h95hQWOEkcgmgtLxfIFm3NZJKMWNYo9DKZ6rJ4IyxxXYIF4pP7QR8lC++BpViDqRh1GNoMKa\
v58QDYqUHOTRiBBMCUFaRAIiRi47ByRcuUo8ev8UyxPZdycijvixx71HgkB2QQECgVJ2x4gK5SoV\
8funLEAoUKnDpFILEdLs40QK3ifLaCuChEu2kXobs4w27GYvLIU42FhL5lBmbUFyWCJkmsMTX2TY\
HNYsX2mOTryRaY9wfRq5DZgj2Oyfx56m1BDB9Qls7PkZzRBEzyqDue+NQRzmAlketvTdRgiyO1ap\
TtHSsso/JVlCDn48tvxURggSoVwjIRxvQvr/Yg2qY9vJCY0ROrRhom51YeeBQHZc7U78TBAsHi6t\
s3jzQPBwilnrmHgWCMRFPXahzY2dAwL+ygqJoG07wBwQJCrVvHxraQ3mgOA3Wqhcfsfu6bPpI9jd\
1+yxa+5r8ghUKr+w082fPII8BaofXRMfk0cARaBzoDMDBPzSvet3BgjUTq8uBrNAgIZ1fGceCNaz\
RpCopnDVta168ghSVqGZd6cYLdil6oyHTB7B70UICAEhACEAIcD4EYjVMkC+bFsIMQcEyekcI757\
aFkFMAcEAFugubSdnAeCLeMbO9EkwdgaOVMpzDxXw9onhcysQNwK8FcLKez+sKLb6mgrsrxjeeO/\
DeC8xY/W6QDTtuC0898GUOzVUpO2n8oUQcOsZK36o8oPhxLF3a0lUGCIYH1MDBMuDi9DBPuoe9px\
DDJCEKn8dfHe+1V33TJDUBU5RFXMGME0RAj6IhARpLVzJELtiPz8FAlk8fuhfP3nhOOtmlmm/mLP\
gv3tRE8EqcBWdoGy9qudPVUmD+UrKxBwlIk8TErwAMXXiZAP8Bi5z4KJ9B8FC/9VsDJSn7ZqL5P8\
x6gi0Ob9aT1M0MMUDiMRpXMhKyAEoARfICsAPTcJk0r56OPTs0YiepIePU+RKgLo2aqYVlp4DzvF\
kYgcZLICQgCqCCArwLSmTz3Meuk+lKIlFxVhf/uJ5Pls4QFi69sTx+p2Z8TahRXEL1tg96B3k27V\
lwTBx/oGH63gV1tQszJqXfxppIw94+FjkOxjW/ClzclCLXhTeGxMF3i48QsiZvz4zzZdfiXo8Nov\
EKx4blsHbqj7S2q61MkJgusalZ0sYNke2YthQ+uiIpQqM2jNLEwfxOd7ZWPv5fO5IqyWDZJ6uTW7\
y3cqViuBcll76xdQKMX13CGFUrQ1pVCKh3EEQjAWUXNInSJVBNAMMsgKQAhAFQFkBaBQCiiUAqoI\
ICsAtQXoiWCpd++RqE+IIz+cFtPV6fYXvytASAA4qaEAAAAASUVORK5CYII=")









    Out[1]:

3. Fitting a Line (3 Points)

Find a best fit line to the data below and plot the data along with the best fit line.



In [9]:

    
data_3_x = [0.0, 0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9,  1.0,  1.1,  1.2,  1.3,  1.4,  1.5,  1.6,  1.7,  1.8,  1.9,  2.0,  2.1,  2.2,  2.3,  2.4,  2.5,  2.6,  2.7,  2.8,  2.9,  3.0,  3.1,  3.2,  3.3,  3.4,  3.5,  3.6,  3.7,  3.8,  3.9,  4.0,  4.1,  4.2,  4.3,  4.4,  4.5,  4.6,  4.7,  4.8,  4.9,  5.0,  5.1,  5.2,  5.3,  5.4,  5.5,  5.6,  5.7,  5.8,  5.9,  6.0,  6.1,  6.2,  6.3,  6.4,  6.5,  6.6,  6.7,  6.8,  6.9,  7.0,  7.1,  7.2,  7.3,  7.4,  7.5,  7.6,  7.7,  7.8,  7.9,  8.0,  8.1,  8.2,  8.3,  8.4,  8.5,  8.6,  8.7,  8.8,  8.9,  9.0,  9.1,  9.2,  9.3,  9.4,  9.5,  9.6,  9.7,  9.8,  9.9]
data_3_y = [0.05, 0.45,  0.18,  -0.15,  -0.14,  -0.38,  -0.88,  -0.42,  -0.99,  -0.92,  -1.53,  -1.94,  -1.75,  -2.37,  -2.09,  -2.99,  -3.04,  -2.85,  -2.66,  -3.03,  -3.39,  -3.36,  -3.7,  -4.41,  -4.7,  -4.51,  -4.29,  -5.39,  -4.81,  -5.63,  -5.01,  -5.79,  -6.03,  -6.04,  -6.35,  -6.82,  -6.48,  -6.6,  -6.69,  -7.05,  -7.4,  -8.07,  -7.81,  -7.97,  -8.08,  -8.29,  -8.99,  -9.17,  -9.38,  -9.1,  -9.62,  -9.85,  -9.99,  -9.64,  -10.78,  -10.76,  -10.84,  -11.1,  -11.03,  -11.48,  -11.47,  -11.4,  -11.58,  -11.77,  -11.97,  -12.1,  -12.65,  -12.52,  -12.79,  -13.21,  -13.24,  -13.85,  -13.5,  -13.9,  -14.66,  -14.44,  -14.65,  -14.72,  -14.7,  -14.87,  -15.47,  -15.21,  -15.82,  -16.37,  -16.42,  -16.67,  -16.52,  -16.62,  -17.39,  -16.94,  -17.48,  -18.17,  -18.31,  -17.75,  -17.86,  -18.6,  -18.43,  -19.04,  -19.1,  -18.83]

4. Fitting a Binomial Distribution to Data (2 Points)

In the following cell there is a function which takes in data, a known trial number, and a proposed success probability. It then returns how well the binomial distribution fits the data. Complete the program so that it finds the optimal success probability



In [1]:

    
from scipy.special import comb
import numpy as np

bin_data = [4, 3,  6,  1,  2,  1,  1,  1,  3,  0,  4,  2,  4,  2,  2,  3,  2,  4,  4,  3]
N = 20

def bin_fit(p, data, N):
    fit = 0
    for di in data:
        fit += np.log( comb(N, di) * (1 - p)**(N - di) * p**di )
    return fit

#Example of how to use
print bin_fit(0.8, bin_data, N)









    



-452.321017887

5. Solution Reaction (5 Points)

The follow reaction occurs in water and is highly exothermic ($\Delta$H = $863.9$ BTU / lb-mol):

$$\textrm{AB}\rightarrow \textrm{A}^+ + \textrm{B}^-$$

Its equilibrium constant has been shown to fit the following empirical relationship from 400 $^\circ{}$ R to 800 $^\circ{}$ R:

$$ k = Ae^{\frac{-B}{RT}} + C\left(\frac{B}{RT} - 0.5\right)^2 $$

where $A = 10^{1} $ lb-mol / gal, $B = 500$ BTU, $C = 10^{-3}$ lb-mol / gal, $T$ is temperature and $R$ is the universal gas constant.

I add 1.5 lb-mol of AB to a 25.0 gal tank of water which is at 510.0 $^\circ{}$R. Assume that the heat capacity of the tank is well-apprximated by that of water, 17.89 BTU / (lb-mol $^\circ{}$ R) and that all reaction enthalpy goes to heating the solution. How much AB remains in lb-mol and what is the final temperature of the solution?

6. 2D Minimization (5 Points)

Subject to the constraint that $x$ and $y$ have opposite signs, find the $x$ and $y$ that minimize the following expression:

$$f(x,y) = \cos(x + 1)\sin y + (x + 2)^2$$

where $x$ and $y$ are both between $-2\pi$ and $2\pi$. Hint: The minimum function value is about -0.7

7. Sample Covariance (3 Points)

Use the data from Question 3. Compute its sample covariance and sample correlation using numpy. Compare its covariance and correlation with your answer from problem 3.

8. Functions of Distributions (1 Point)

The time between eruptions at a volcano is exponentially distributed with a rate of 1 eruption every 10 years. There is a nearby city. Each time the volcano erupts, everyone dies in the city except for 1 person. People gradually come back. If a volcano erupted at $t=0$, the population can be modeled by: $$N(t) = e^{0.9t}$$

Write down $P(N)$ in markdown using equiations from Unit 8. You must only set-up the integral that finds $P(N)$. Do not evaluate the integral.



In [ ]: