Contents

Aaron Klapheck

Assignment #3

clear, clc
date
ans =

19-Sep-2007

Problem 9b,d

A = [3, 7, -4, 12; -5, 9, 10, 2; 6, 13, 8, 11; 15, 5, 4, 1] % given

x = sort(A')
C = x'                  % b.

E = sum(A')             % d.
A =

     3     7    -4    12
    -5     9    10     2
     6    13     8    11
    15     5     4     1


x =

    -4    -5     6     1
     3     2     8     4
     7     9    11     5
    12    10    13    15


C =

    -4     3     7    12
    -5     2     9    10
     6     8    11    13
     1     4     5    15


E =

    18    16    38    25

Problem 10b,d

A = [1, 4, 2; 2, 4, 100; 7, 9, 7; 3, pi, 42]    % given
B = log(A)                                      % given

B_sum_second_row = sum(B(2, :))                 % b.

x = B(:, 2).*A(:, 1)
max(x)                                          % d.
A =

    1.0000    4.0000    2.0000
    2.0000    4.0000  100.0000
    7.0000    9.0000    7.0000
    3.0000    3.1416   42.0000


B =

         0    1.3863    0.6931
    0.6931    1.3863    4.6052
    1.9459    2.1972    1.9459
    1.0986    1.1447    3.7377


B_sum_second_row =

    6.6846


x =

    1.3863
    2.7726
   15.3806
    3.4342


ans =

   15.3806

Problem 12a,b,c

A = [-7, 16; 4, 9], B = [6, -5; 12, -2], ...
    C = [-3, -9; 6, 8]                          % Given information.

Sum = A + B + C                                 % a.

Sum_and_Difference = A - B + C                  % b.

x = (A + B) + C
y = A + (B + C)
y == x                                  % c. Returns matrix of 1's if true.
A =

    -7    16
     4     9


B =

     6    -5
    12    -2


C =

    -3    -9
     6     8


Sum =

    -4     2
    22    15


Sum_and_Difference =

   -16    12
    -2    19


x =

    -4     2
    22    15


y =

    -4     2
    22    15


ans =

     1     1
     1     1

Problem 13a,c

A = [64, 32; 24, -16], B = [16, -4; 6, -2]      % Given information

Array_Product = A.*B                            % a.

Element_Exponentiation = B.^3                   % c.
A =

    64    32
    24   -16


B =

    16    -4
     6    -2


Array_Product =

        1024        -128
         144          32


Element_Exponentiation =

        4096         -64
         216          -8

Problem 16

% For Five_Widget_Makers: row 1 contains the worker number,
% row 2 contains their hourly wages($),
% row 3 containes their number of hours worked (per week),
% row 4 containes their output(widgets per week).
Five_Widget_Makers = [1, 2, 3, 4, 5; 5, 5.50, 6.50, 6, 6.25; ...
    40, 43, 37, 50, 45; 1000, 1100, 1000, 1200, 1100]

weekly_Earnings = Five_Widget_Makers(2, :).*Five_Widget_Makers(3, :)  % a. (in $)

Total_Salary = sum(weekly_Earnings')                        % b.

Total_Widgets = sum(Five_Widget_Makers(4, :)')              % c.

Average_cost = Total_Salary/Total_Widgets                   % d. ($/widget)

% e. (hours/widget)
Hour_Widgets = sum(Five_Widget_Makers(3, :)')/sum(Five_Widget_Makers(4, :)')

% for part f: The first array product gives each workers sallary per week,
% deviding this by the number of widgets each worker makes per week
% gives the efficiency of each worker in $/widget.
f = (Five_Widget_Makers(2, :).*Five_Widget_Makers(3, :))./ ...
    Five_Widget_Makers(4, :)

% Least efficiant worker, where Lest_efficiant is the worker number.
[Efficiancy, Lest_efficiant] = min(f)

% Most efficiant worker, where Most_efficiant is the worker number.
[Efficiancy, Most_efficiant] = max(f)
Five_Widget_Makers =

  1.0e+003 *

    0.0010    0.0020    0.0030    0.0040    0.0050
    0.0050    0.0055    0.0065    0.0060    0.0063
    0.0400    0.0430    0.0370    0.0500    0.0450
    1.0000    1.1000    1.0000    1.2000    1.1000


weekly_Earnings =

  200.0000  236.5000  240.5000  300.0000  281.2500


Total_Salary =

  1.2583e+003


Total_Widgets =

        5400


Average_cost =

    0.2330


Hour_Widgets =

    0.0398


f =

    0.2000    0.2150    0.2405    0.2500    0.2557


Efficiancy =

    0.2000


Lest_efficiant =

     1


Efficiancy =

    0.2557


Most_efficiant =

     5

Problem 19

% For Material_Price: column 1 is the material number,
% column 2 is the price corresponding to each material ($/ton)
Material_Price = [1, 300; 2, 550; 3, 400; 4, 250; 5, 500]

% For Quantity_Purchased(tons): column 1, 2, and  3 is the ammount of
% material purchased in the month of May, June, and July respectively.
Quantity_Purchased = [5, 4, 6; 3, 2, 4; 6, 5, 3; 3, 5, 4; 2, 4, 3]

l = Material_Price(:, 2).*Quantity_Purchased(:, 1)
m = Material_Price(:, 2).*Quantity_Purchased(:, 2)
n = Material_Price(:, 2).*Quantity_Purchased(:, 3)

v = [l,m,n]                 % a.

% b.
May = sum(l)                % Total spent in May
June = sum(m)               % Total spent in June
July = sum(n)               % Total spent in July

% c.
Mat1 = sum(v(1, :))         % Total spent on Material 1
Mat2 = sum(v(2, :))         % Total spent on Material 2
Mat3 = sum(v(3, :))         % Total spent on Material 3
Mat4 = sum(v(4, :))         % Total spent on Material 4
Mat5 = sum(v(5, :))         % Total spent on Material 5

Total_Spent = sum([May,June,July])        % d.
Material_Price =

     1   300
     2   550
     3   400
     4   250
     5   500


Quantity_Purchased =

     5     4     6
     3     2     4
     6     5     3
     3     5     4
     2     4     3


l =

        1500
        1650
        2400
         750
        1000


m =

        1200
        1100
        2000
        1250
        2000


n =

        1800
        2200
        1200
        1000
        1500


v =

        1500        1200        1800
        1650        1100        2200
        2400        2000        1200
         750        1250        1000
        1000        2000        1500


May =

        7300


June =

        7550


July =

        7700


Mat1 =

        4500


Mat2 =

        4950


Mat3 =

        5600


Mat4 =

        3000


Mat5 =

        4500


Total_Spent =

       22550