l Question about costs - Linear algebra/matrices -

Njords

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An engineering department charges 25% of its total monthly costs to the computer department. The computer department charges 20% of its costs to the engineering department.

Total monthly cost of Eng dept. is $x and comp dept is $y

At any given month the eng dept cost, not including what it pays to comp dept is $95k. And the comp dept cost not including what is paid to eng dept is $85500.

1) write equation for x and y
2) determine matrix A in the equation x=d + Ax
3) calculate the total monthly cost for each dept.

_____________________________________________________________________________________________________________________________________________

1) equations i've come up with for total cost to each department are: a) x = 0.2y + 95000 ; b) y = 0.25x + 85500

2) matrix A I am saying is ...
00.2
0.250

3) I believe this to be a open Leontief economic modeling question based on the x=Ax+d part in 2)

or x = (I-A)-1x * d

(I-A) is Identity matrix minus A which gives
1
-0.2
-0.25
1

det of (I-A) is 1/0.95

(I-A)-1 . is

0.95
0.2375
0.19
0.95

multiplying this by the 2x1 matrix of

95000
85500

gives x = 110556.25 or just $110,556
and y = $99275

________________________________

would appreciate if anyone could let me know am I on the right track, if not, where I'm going wrong. Cheers.
 
would appreciate if anyone could let me know am I on the right track, if not, where I'm going wrong.
You can check the solution to any "solving" exercise by plugging it back in to the original question. If you plug your values in, what do you get?

. . .Engineering:
. . . . .total: 110,556 (proposed value)
. . . . .25%: 27,639 (computed)
. . . . .remainder: 95,000 (given)
. . . . .sum: 122,639

Since this does not match your proposed value, then we needn't go any further, I think.

An engineering department charges 25% of its total monthly costs to the computer department. The computer department charges 20% of its costs to the engineering department.

The total monthly cost for the Engineering department is "x" dollars, and the total monthly cost for the computer department is "y" dollars.

In any given month, the cost for the Engineering department, not including what it pays to the computer department, is $95,000. Similarly, the cost for the computer department, not including what is paid to the engineering department, is $85,500.

1) Write two equations in x and y
2) Determine the matrix A in the equation x = d + Ax
I'm assuming that "x" here stands for "a column vector" containing as follows:

. . . . .\(\displaystyle \left[\begin{array}{c}x\\y\end{array}\right]\)

But what is "d" in this context?

3) Calculate the total monthly cost for each department.

_____________________________________________________________________________________________________________________________________________

1) The two equations I've come up with for total cost to each department are: a) x = 0.2y + 95000 ; b) y = 0.25x + 85500
These look good.

2) matrix A I am saying is:

. . . . .\(\displaystyle \left[\begin{array}{rr}0&0.2\\0.25&0\end{array}\right]\)
How did you get to this matrix? Why not instead do the following?

. . . . .x = 0.2y + 95,000
. . . . .y = 0.25x + 85,500

. . . . .x - 0.2y = 95,000
. . . . .-0.25x + y = 85,500

. . . . .\(\displaystyle Ax\, =\, b\)

. . . . .\(\displaystyle \left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\, =\, \left[\begin{array}{c}95,000\\85,500\end{array}\right]\)

Then multiply both sides, on the left, by the inverse of A.

3) I believe this to be a open Leontief economic modeling question based on the x=Ax+d part in 2)

or x = (I-A)-1x * d
Does this mean that "d" is the "demand vector" (as shown, for instance, here and here)? Also, how are you getting "x = (I - A)-1x * d" to be equivalent to "x = d + Ax"? The latter rearranges as:

. . . . .x = d + Ax
. . . . .x - Ax = d
. . . . .Ix - Ax = d
. . . . .(I - A)x = d
. . . . .(I - A)-1 (I - A)x = (I - A)-1 d
. . . . .x = (I - A)-1 d

How are you getting that extra "x" on the right-hand side?

(I-A) is Identity matrix minus A which gives

. . . . .\(\displaystyle \left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\)

The determinant of (I-A) is 1/0.95
This 1/0.95 equals 1/(95/100) = 1/(19/20) = 20/19.

The matrix (I-A)-1 is

. . . . .\(\displaystyle \left[\begin{array}{rr}0.95&0.2375\\0.19&0.95\end{array}\right]\)
If you multiply this and your matrix for I - A, do you get the identity matrix? I don't. You might want to check this.

Multiplying this by the 2x1 matrix of

. . . . .\(\displaystyle \left[\begin{array}{c}95,000\\85,500\end{array}\right]\)

gives x = 110556.25 or just $110,556
and y = $99275
Are you supposed to be rounding?

P.S. You can check your answer by solving the system that I provided above; namely,

. . . . .\(\displaystyle \left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\, =\, \left[\begin{array}{c}95,000\\85,500\end{array}\right]\)

. . . . .\(\displaystyle \left[\begin{array}{rr}\frac{20}{19}&\frac{4}{19}\\\frac{5}{19}&\frac{20}{19}\end{array}\right]\left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\, =\, \left[\begin{array}{rr}\frac{20}{19}&\frac{4}{19}\\\frac{5}{19}&\frac{20}{19}\end{array}\right]\left[\begin{array}{c}95,000\\85,500\end{array}\right]\)

...and so forth. ;)
 
You can check the solution to any "solving" exercise by plugging it back in to the original question. If you plug your values in, what do you get?

. . .Engineering:
. . . . .total: 110,556 (proposed value)
. . . . .25%: 27,639 (computed)
. . . . .remainder: 95,000 (given)
. . . . .sum: 122,639

Since this does not match your proposed value, then we needn't go any further, I think.


I'm assuming that "x" here stands for "a column vector" containing as follows:

. . . . .\(\displaystyle \left[\begin{array}{c}x\\y\end{array}\right]\)

But what is "d" in this context?


These look good.


How did you get to this matrix? Why not instead do the following?

. . . . .x = 0.2y + 95,000
. . . . .y = 0.25x + 85,500

. . . . .x - 0.2y = 95,000
. . . . .-0.25x + y = 85,500

. . . . .\(\displaystyle Ax\, =\, b\)

. . . . .\(\displaystyle \left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\, =\, \left[\begin{array}{c}95,000\\85,500\end{array}\right]\)

Then multiply both sides, on the left, by the inverse of A.


Does this mean that "d" is the "demand vector" (as shown, for instance, here and here)? Also, how are you getting "x = (I - A)-1x * d" to be equivalent to "x = d + Ax"? The latter rearranges as:

. . . . .x = d + Ax
. . . . .x - Ax = d
. . . . .Ix - Ax = d
. . . . .(I - A)x = d
. . . . .(I - A)-1 (I - A)x = (I - A)-1 d
. . . . .x = (I - A)-1 d :) Thanks.

How are you getting that extra "x" on the right-hand side? -That's me being careless. Nice pick up, thanks.


This 1/0.95 equals 1/(95/100) = 1/(19/20) = 20/19.


If you multiply this and your matrix for I - A, do you get the identity matrix? I don't. You might want to check this.


Are you supposed to be rounding? I'm guessing no, I'll go with the fractions. :)

P.S. You can check your answer by solving the system that I provided above; namely,

. . . . .\(\displaystyle \left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\, =\, \left[\begin{array}{c}95,000\\85,500\end{array}\right]\)

. . . . .\(\displaystyle \left[\begin{array}{rr}\frac{20}{19}&\frac{4}{19}\\\frac{5}{19}&\frac{20}{19}\end{array}\right]\left[\begin{array}{rr}1&-0.2\\-0.25&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\, =\, \left[\begin{array}{rr}\frac{20}{19}&\frac{4}{19}\\\frac{5}{19}&\frac{20}{19}\end{array}\right]\left[\begin{array}{c}95,000\\85,500\end{array}\right]\) Awesome!!!

...and so forth. ;)

I made a mess of the inverse so no surprise the first part didn't check out when putting figures back in. Thank you very much on picking up some silly errors there, on the right track now thanks again for looking into this.
 
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