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MAD Models

- Minimize the sum of absolute deviations
- The deviation is defined as ei = Yi - ∑XjiBj
- Similar to regression, except we minimize sum of the absolute values of the error term and not the sum of the squares of the error terms.

Problem Defined

- Minimize ∑ | ei |
- Subject to
- ei - ∑ Xji bj = Yi

- The bj are the variables to be determined in the model. They can be positive or negative, as can the ei.

Need Substitutions to solve

- Define ei = eipos – eineg
- Define bj = bjpos – bjneg
- If we replace these in the formulation, we can solve these. Remembering that the absolute value is always a positive number, we have an objective function that sums up over all eipos + eineg.

Orange Price Model

- Pricei = Bo + B1QOrangesi + B2QJuicei + e i
- The ei can be positive or negative, as can the Bj.
- Decision variables are the Bj and the ei which are determined by the choice of the Bj.
- Because I minimize the sum of absolute values, I must define two error terms for each observation, the positive and the negative.

The LP from this Problem

Here, I've let the Bj be unrestricted in sign,

because Solver will let me do that. I could

add more columns and define negative Bj.

Variation

- Minimize the biggest absolute deviation from the line.
- This can be accomplished by using inequality constraints and one e.
- Yi - ∑XjiBj ≤ e (if max e is positive)
- ∑XjiBj – Yi ≤ e (if max e is negative)
- Rewrite these so that Yi is on the RHS and e is on the left.
- Resolve our orange example, but restrict b1 to be non-positive and b2 to be non-negative.

Optimizing a Fraction

McCarl and Spreen Section 9.1.3

Type of Problem

Co + CjXj

MAX

do + djXj

S.T. aijXj le bi

Xj ge 0 and the denominator strictly positive

Do Some Math And . . .

Max CoYo + CjYj

s.t. -biYo + aijYj le 0

doYo + djYj = 1

Yo, Yj GE 0

Solve this and do the reverse

transformation to get the X values

Example Problem

1.8 X1 + 1.7 X2

MAX

10 + 4X1 + 4.1 X2

s.t. 1.5X1 + X2 le 6

3.0X1 + 4X2 le 20

X1, X2 ge 0

Transformed

Max 1.8 Y1 + 1.7Y2

s.t -6Yo + 1.5Y1 + Y2 le 0

-20Yo + 3Y1 + 4Y2 le 0

10Yo + 4Y1 + 4.1Y2 = 1

Yo, Y1, Y2 ge 0

Separable Programming

In some cases, the level of return to an

activity is not constant as the activity level

increases. Because of the law of diminishing

returns, returns to additional units of an

enterprise often decrease as more units

are added.

Modeling this Situation

Max Profit = R1X1 + R2X2 + R3X3

X1 LE A

X2 LE (B-A)

And X2 and X3 use the same (or higher) levels

of all resources. Because R1 > R2 > R3, X1

will enter solution to its full level A before X2 is

selected and so on.

Tableau

Totals: A-635; B-100, C-50

Alternative Method

This method of solving separable programming

is sometimes called the "delta method."

An alternative, presented in McCarl and

Spreen chapter 9, is sometimes called the

"lambda method" or "grid point approximation."

McCarl and Spreen

McCarl and Spreen present a different

application of the gridpoint method in

chapter 9, section 2.1. Also, they

show you how to extend this method

to functions of more than one variable.

The more complicated material won't

appear on the final but you may want

to read it on your own.

Using Integers to Approximate Nonlinear Functions

McCarl and Spreen, 15.1.6

We will come back to this

material. If you have already

seen integer programming, you

may want to look at it now.

Some Functions

Some functions cannot be approximated

using the delta or lambda methods

because the more "attractive" section

of the function takes place at larger

numbers. If, for example, unit

profits increase as output increases or

costs (in a cost min) decrease as a variable

gets larger, those techniques don't work.

An Example

Modification of our original 3 product

situation so that the profit from A

increases as output goes up.

Without Integers

We are skipping the intermediate steps.

Profit from 635 units of A is $6200, not

$6254.7 as it would appear from this answer.

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