s := 1
This is required for
some plots below.
Supplement to 16.881 Homework#2
Exploration of the Quadratic Loss Function
Dan Frey, 6 JUN 98
The areas I expect you'll need to change to do homework#2 are highlighted.
ORIGIN := 1
Let 1 be the first index in any vector.
A o := 10
Cost to scrap the resistor [cents]
m := 100
Nominal value of the resistance (in ohms)
D o := 5%�m
Allowable variation in the diameter is +/- 5%
Ao
k1 :=
k2 :=
2
Do
L(y) :=
Ao
2
Do
k1�(y - m)
2
k2�(y - m)
2
if y > m
Define the quadratic loss function
if y £ m
y := m - 1.2�D o , m - 1.2�D o +
Do
10
.. m + 1.2�D o
Define a range over y for the purpose
of plotting
Quadratic Loss Function
m- D o
m+ D o
10
L(y)
5
0
94
96
98
100
102
y
1
104
106
Create a Monte Carlo simulation of the manufacture of the resistors.
n := 1000
s :=
Number of resistors to be manufactured
Do
6
m := m + 1.5� s
R := rnorm( n , m , s)
Create a vector with all of the resistance values of the
resistors we manufactured.
Set up the format for a histogram of the data.
number_of_bins := 20
width_of_bins :=
4�D o
number_of_bins
j := 1.. number_of_bins + 1
bin := m - 2�D o + width_of_bins� j
j
rel_freq :=
hist(bin , R)
n
Define a vector with the start and end points of
the bins.
Compute the relative frequency distribution
over interval.
bin_center := bin + 0.5� width_of_bins
2
Simulation Results for the Resistors
m- D o
L( y)
m+ D o
1
Ao
rel_freq
0.5
0
90
95
100
105
110
115
y , bin_center
Average_quality_loss :=
1
n
n
�
� L(Ri)
i=1
Average_quality_loss = 0.845
in cents
How does this compare to the theoretically derived figure?
m+ 2�D o
�
�
�
�
�
ım-2�D
2
Ø
- (y-m ) ø
Œ
œ
2
œ
1
2Œ
2�s
k1�(y - m) � Œ
�e
œ dy = 0.903
º s� 2�p
ß
o
3
in cents