a brokerage house offers three stock portfolios. portfolio a consists of 2 blocks of common stock and 1 municipal bond. portfolio ii consists of 4 blocks of common stock, 2 municipal bonds, and 3 blocks of preferred stock. portfolio iii consists of 7 blocks of common stock, 3 municipal bonds, and 3 blocks of preferred stock. a customer wants 21 blocks of common stock, 10 municipal bonds, and 9 blocks of preferred. stock. how many units of each portfolio should be offered?
x of portfolio 1
y of portfolio 2
z of portfolio 3
portfolio 1: 2 common stock, 1 muni bond
portfolio 2: 4 common stock, 2 muni bond, 3 preferred stock
portfolio 3: 7 common stock, 3 muni bond, 3 preferred stock
so
common 21 = 2 x + 4 y + 7 z
muni 10 = 1 x + 2 y + 3 z
preferred 9 = 3 y + 3 z or 3 = y+z
so
z = (3-y)
21 = 2x + 4 y + 7(3-y)
10 = 1x + 2 y + 3(3-y)
0 = 2 x - 3 y so x = 1.5 y
10 = 1.5 y + 2 y + 9 - 3 y
1 = .5 y
y = 2
go back and get x and z
4,6
To determine the number of units of each portfolio that should be offered, we can set up a system of linear equations.
Let's denote the number of units of portfolio A as x, the number of units of portfolio II as y, and the number of units of portfolio III as z.
Based on the given information, we can set up the following equations:
1. For the common stock: 2x + 4y + 7z = 21
2. For the municipal bonds: 1x + 2y + 3z = 10
3. For the preferred stock: 1y + 1z = 9
Now we can solve this system of equations using any method like substitution or elimination. Since this is a small system, let's solve it using substitution.
From equation 3, we can solve for y:
y = 9 - z
Substituting this value of y into equations 1 and 2, we have:
2x + 4(9 - z) + 7z = 21 --> 2x + 36 - 4z + 7z = 21 --> 2x + 3z = -15 (equation 4)
1x + 2(9 - z) + 3z = 10 --> x + 18 - 2z + 3z = 10 --> x + z = -8 (equation 5)
Now, we have two equations (4 and 5) with two variables (x and z). We can solve this system:
Multiply equation 5 by 2: 2(x + z) = 2(-8)
2x + 2z = -16
Subtract equation 4 from this new equation:
(2x + 2z) - (2x + 3z) = -16 - (-15)
-z = -1
z = 1
Substitute this value of z back into equation 5:
x + 1 = -8
x = -9
Therefore, the customer should be offered 9 units of Portfolio A (common stock), 1 unit of Portfolio II (municipal bond), and 1 unit of Portfolio III (preferred stock).
To determine how many units of each portfolio should be offered to meet the customer's desired allocation, we can use a system of equations.
Let's assign variables to represent the number of units for each portfolio:
- Let 'x' represent the number of units of Portfolio A.
- Let 'y' represent the number of units of Portfolio II.
- Let 'z' represent the number of units of Portfolio III.
We can now set up the equation based on the information given:
For common stock: 2x + 4y + 7z = 21 blocks
For municipal bonds: 1x + 2y + 3z = 10 bonds
For preferred stock: 0x + 3y + 3z = 9 blocks
Now we have a system of linear equations:
2x + 4y + 7z = 21 (Equation 1)
1x + 2y + 3z = 10 (Equation 2)
0x + 3y + 3z = 9 (Equation 3)
We can solve this system to find the values of x, y, and z.
One approach to solve this system is to use the method of elimination or substitution. However, since the coefficients are relatively simple in this case, we can use a method called "Gaussian elimination."
First, let's simplify Equation 3 by dividing all terms by 3:
0x + y + z = 3 (Equation 4)
Now, we can set up an augmented matrix to represent the system of equations:
[ 2 4 7 | 21 ]
[ 1 2 3 | 10 ]
[ 0 1 1 | 3 ]
Let's perform row operations to simplify the matrix:
- Replace R2 with R2 - R1:
[ 2 4 7 | 21 ]
[ -1 -2 -4 | -11 ]
[ 0 1 1 | 3 ]
- Replace R3 with R3 - (1/2)R2 (division by 2 is used to eliminate the -1 value):
[ 2 4 7 | 21 ]
[ -1 -2 -4 | -11 ]
[ 0 1 1 | 5 ]
- Replace R1 with R1 - 2R2:
[ 4 8 15 | 43 ]
[ -1 -2 -4 | -11 ]
[ 0 1 1 | 5 ]
- Replace R1 with R1/4 to simplify the coefficients:
[ 1 2 15/4 | 43/4 ]
[ -1 -2 -4 | -11 ]
[ 0 1 1 | 5 ]
We have achieved a triangular form that is easier to solve using back substitution.
Now, let's solve for z:
From Equation 3 (now simplified): z = 5
Next, let's solve for y:
From Equation 2: y + z = 3
Substitute the value of z: y + 5 = 3, which gives y = -2
Finally, let's solve for x:
From Equation 1: 2x + 4y + 7z = 21
Substitute the values of y and z: 2x + 4(-2) + 7(5) = 21
Solving this equation gives x = 4
Therefore, the solution to the system of equations is:
x = 4 (units of Portfolio A)
y = -2 (units of Portfolio II)
z = 5 (units of Portfolio III)
Since the customer cannot have a negative number of units, we can conclude that the brokerage house should offer 4 units of Portfolio A, 0 units of Portfolio II, and 5 units of Portfolio III in order to meet the customer's desired allocation.