Transcript MA3264_HW1

MA3264
Homework 1 Due Friday 12 September
Answers
Page 50 Problem 6 a, b
Compute and plot the points
( Pn , Qn ), n  1,...,20
Answer
(a) The model makes sense intuitively.The constant 100 is the
equilibrium value of price, and 500 is the equilibrium value of
supply. The model asserts that
(i) prices decreases (represented by the sign of -0.1) if the supply
exceeds the equilibrium supply
(ii) the supply increases (represented by the sign of 0.2) if the
price exceeds the equilibrium price.
These facts are explained by the common sense “law of supply
and demand” in economics. Consumers differ in their utility
[desire] for goods and therefore differ in their the price that they
are willing to pay. Therefore lower prices are required to sell a
larger supply of goods. Producers differ in the price that they
can charge for a good and still remain profitable because more
efficient producers can produce at a lower cost. So producers
supply more of a good as its price increases.
Page 50 Problem 6 a, b
Compute and plot the points
( Pn , Qn ), n  1,...,20
Answer
(b) The following MATLAB commands compute and plot the
values of price and supply for n = 1:20 for cases A, B, C, D
Case A
P(1) = 100;
Q(1) = 500;
for n = 1:19
P(n+1) = P(n) -0.1*(Q(n)-500);
Q(n+1) = Q(n) + 0.2*(P(n)-100);
end
figure(1);
plot(P,Q,’b*’)
Case B, C, D similar
Page 50 Problem 6 a, b
Compute and plot the points
( Pn , Qn ), n  1,...,20
Q
A. Initial (100,500)
B. Initial (200,500)
C. Initial (100,600)
D. Initial (100,400)
P
Page 50 Problem 6 a, b
Compute and plot the points
Answer
Long-term Behavior
 Pn1 
Q   M
 n1 
 Pn   50 
 1  0.1
Q    20, M  0.2

1



 n 
 Pn1  100 

M

Qn1  500
>> eig(M)
ans =
1.0000 + 0.1414i
1.0000 - 0.1414i
 Pn  100 
n
Q  500    M
 n

Since M has
eigenvalues having
abosolute values > 1,
cases B, C, D
diverge to infinity,
case A is an unstable
equilibrium
 P1  100 
Q  500
 1

>> M^10000
ans =
1.0e+042 *
-8.2407 4.0303
-8.0606 -8.2407
Page 83 Problem 3 : Solution 1. S Constant
Fn  normal gravity force
Ft
Ff  friction force
Fd  drag force

m
Ff  Fd
Fn  mg cos 
Ft  tangential gravity force  mgsin 
At terminal velocity
Ft  Ff  Fd
mg sin   amg cos  bSv , a, b real
2
 v  c m , c real
Terminal velocity is proportional to the square root of mass.
Box with mass 800 lbs has terminal velocity 1.1547 times
the terminal velocity of the box with mass 600
2/3
S

k
m
Page 83 Problem 3 : Solution 2.
Fn  normal gravity force
Ft
Ff  friction force
Fd  drag force

m
Ff  Fd
Fn  mg cos 
Ft  tangential gravity force  mgsin 
At terminal velocity
Ft  Ff  Fd
mg sin   amg cos   bkm v , a, b , k real
2/3 2
 v  cm
1/ 6
, c real
Terminal velocity is proportional to the 6th root of mass.
Box with mass 800 lbs has terminal velocity 1.0491 times
the terminal velocity of the box with mass 600
Page 124 Problem 1
The proposed model is y  ax
It is NOT a linear function of the parameter b
Therefore we must TRANSFORM the data and then fit
the transformed data to the following model:
b
log y  log a  b log x
The following MATLAB comands input data,
transform the data, and compute a least squares
fit of the model and plots the data and the model.
x = [17 19 20 22 23 25 28 31 32 33 36 37 38 39 41];
y = [19 25 32 51 57 71 113 141 123 187 192 205 252 259 294];
logx = log(x); logy = log(y);
m = 15; xy = logx*logy’;
sx = sum(logx); sy = sum(logy);
xx = logx*logx’;b = (m*xy-sx*sy)/(m*xx – sx*sx);
loga = (xx*sy-xy*sx)/(m*xx-sx*sx); a = exp(loga);
xs = 16:.01:42; ys = a*xs.^b; plot(xs,ys); hold on; plot(x,y,’r*’);
dev = y – a*x.^b; maxdev = max(abs(dev)); S = dev*dev’;
Page 124 Problem 1
a = 0.0032, b = 3.0919, sqrt(S/m) = 13.7271 <= cmax <= maxdev = 28.1377