Expected loss of having fewer for the old medicine:
0.80 $1 + 0.20 ($1 + $10) = $3
Expected loss of having fewer for the new medicine:
0.95 $2 + 0.05 ($2 + $10) = $2.5
The patient is better off (moneywise) with the new medicine,
so the cost of $2 is justified.
Let P be the break even price (same expected loss for both). 3.0 = 0.95 P + 0.05*(P + 10) = P + 0.5 => P = $2.5 When the price for the new medicine is below (or at) $2.5, we will buy it, otherwise when the price is higher than $2.5, we will not buy it.
+--------+
| /20 |
+--------+
# Maple code, count number of successes
N := 10000; S := 0;
for k from 1 to N do
# one simulation with 8 new tires
tires := stats[random,normald[600,200]](8);
# we start race with first 4 tires
life := [seq(tires[i],i=1..4)];
# increase the life span of the lowest tires
for j from 1 to 4 do
life := sort(life);
life[1] := life[1] + tires[4+j];
end do;
# check if we are reaching the 1000 mile limit
reach := min(op(life));
if reach < 1000
then S := S+1;
end if;
end do;
S = 3729, so the probability is about .3729
+--------+
| /20 |
+--------+
B(n) = .8*B(n-1) + 400, B(0) = 0
B(n) = .8*B(n-1) + 400
= .8*(.8*B(n-2) + 400) + 400
...
n-1
---
= (.8^n)*B(0) + 400 > (.8)^k
---
k=0
= 400*(.8^n - 1)/(.8 - 1)
= 2000*(1 - .8^n)
We will never get more than $2000.
+--------+
| /20 |
+--------+
The transfer function of a digital filter
is the z-transform of the unit impulse response,
i.e.:
infty
-----
H(z) = > h(k) z^(-k)
-----
k=0
where { h(0), h(1), h(2), ... } is the response of the
filter to { 1, 0, 0, ...}.
Example: the averaging filter is defined by
y(k) = (u(k) + u(k-1)/2
and has z-transform H(z) = (1+z^(-1))/2.
We write the input signal u as a linear combination
of delayed impulses d and apply the linearity of the filter:
y = F u
= F( u(0)*d(0) + u(1)*d(1) + u(2)*d(2) + ... )
= u(0)*(F d(0)) + u(1)*(F d(1)) + u(2)*(F d(2)) + ... )
= u(0)*{ h(0), h(1), h(2), ...}
+ u(1)*{ 0, h(0), h(1), h(2), ...}
+ u(2)*{ 0, 0, h(0), h(1), h(2), ...}
= { u(0)*h(0), u(0)*h(1) + u(1)*h(0),
u(0)*h(2) + u(1)*h(1) + u(2)*h(0), ...}
= { y(0), y(1), y(2), ... }
So we see that once h(k) is known, we can compute y(k)
from u(k) via the convolution operator '*' : y = h'*'u.
+--------+
| /20 |
+--------+
+-----------+-------------+-----------+---------+-----------+ | boat type | storage | purchase | rental | need | | #seats | area | cost | price | for boats | +-----------+-------------+-----------+---------+-----------+ | 1 | f(1)=2 ft^2 | c(1)=$122 | p(1)=$5 | n(1)=20 | | 2 | f(2)=3 ft^2 | c(2)=$130 | p(2)=$7 | n(2)=15 | | 3 | f(3)=4 ft^2 | c(3)=$150 | p(3)=$9 | n(3)=10 | +-----------+-------------+-----------+---------+-----------+Boat i requires f(i) square feet to store, costs c(i) dollars to purchase and can be rented at p(i) dollars a trip. Our constraints are as follows. We need at least n(i) boats of type i, but have only 400 square feet to store the boats and our budget is limited to $10,000.
max 5*x(1) + 7*x(2) + 9*x(3)
subject to
2*x(1) + 3*x(2) + 4*x(3) <= 400
122*x(1) + 130*x(2) + 150*x(3) <= 10000
- x(1) <= -20
- x(2) <= -15
- x(3) <= -10
min 400*y(1) + 10000*y(2) - 20*y(3) - 15*y(3) - 10*y(5)
subject to
2*y(1) + 122*y(2) - y(3) >= 5
3*y(1) + 130*y(2) - y(4) >= 7
4*y(1) + 150*y(2) - y(5) >= 9
f = [400 10000 -20 -15 -10]';
A = [-2 -122 1 0 0;
-3 -130 0 1 0;
-4 -150 0 0 1;
-1 0 0 0 0;
0 -1 0 0 0;
0 0 -1 0 0;
0 0 0 -1 0;
0 0 0 0 -1];
b = [-5 -7 -9 0 0 0 0 0];
[y,lambda]=simlp(f,A,b)
y =
0
0.0600
2.3200
0.8000
-0.0000
lambda =
20.0000
15.0000
37.4000
165.4000
0
0
0
27.4000
The answer is x(1) = 20, x(2) = 15, x(3) = 37.4.
+--------+
| /20 |
+--------+
We take the equation y(k) = a*u(k) + b*u(k+1) + c*u(k+2)
and evaluate it at the 8 samples, for the signal u(t) and v(t).
This gives 16 linear equations in 3 unknowns:
% MATLAB CODE:
dt = 2*pi/8; % sampling interval
t0 = 0:dt:2*pi-dt; % range for u(k)
t1 = dt:dt:2*pi; % range for u(k+1)
t2 = 2*dt:dt:2*pi+dt; % range for u(k+2)
u0 = cos(t0); u1 = cos(t1); u2 = cos(t2); % sample u(t)
v0 = cos(3*t0); v1 = cos(3*t1); v2 = cos(3*t2); % sample v(t)
A1 = [u0' u1' u2']; % 1st conditions
b1 = u1';
A2 = [v0' v1' v2']; % 2nd conditions
b2 = zeros(8,1);
A = [A1; A2]; % A x = b
b = [b1; b2];
x = A\b
x =
0.3536
0.5000
0.3536
+--------+
| /20 |
+--------+
Use the Maple functions, for the options (A = 1st, B = 2nd):
A := n -> 100000*(1 + 0.07)^n
B := n -> 100000*(1 + 0.05)^n + 8000*n
Evaluate in 2nd year:
A(2) = $114,490 < $126,250 = B(2)
Evaluate in 20th year:
A(20) = $386,968 < $425,330 = B(20)
Option B, buying the house is better in the short run,
even at year 20 it is still the better option.
A plot of A and B reveals that the two curves intersect
between year 24 and year 25. Evaluation of A and B gives
A(24) = $507,236.70 < $514,509.99 = B(24)
A(25) = $542,743.26 > $538,635.49 = B(25)
We should sell the house in year 24.
+--------+
| /20 |
+--------+
D(p) = 5 + 10/p, S(p) = p^2 - 3
D(p) = S(p) gives the equilibrium price p = $3.32 (Maple)
Revenue = p*S(p) = 3.32*8.01 = 26.59
new supply nS := S(p+1) = (p+1)^2 - 3
new equilibrium price: $2.47
new supply: $9.05
new revenue: $22.35
With a subsidy, the producer will supply more goods,
which will decrease the price and the revenue.
The benefit of the consumer is the price decrease:
$3.32 - $2.47 = $0.85.
So the consumer gets 85 cents of the $1 subsidy per item,
the other 25 cents goes to the produces who loses revenue.
+--------+
| /20 |
+--------+