Absolute value to linear program

[gdespont]

Hi,

I have to convert the first to the second program, but i don't know how to do this:
MAD.png

ABStolinear.xlsx: the first sheet is the first program which is correct (i think), the second sheet is a failed attempt to write the second program
Can somebody please help me out?

Gilles

 

[sijpie]

Not that i exactly understand your problem. If ABS is correct why would you want to do it in another way?

In sheet linear, is it only about the values with an arrow next to them? Why not use the ABS() function? Else you need
=if(h12<0, -h12,h12) in I12

 

[gdespont]

That's the frustrating thing, it works with ABS but I can't use it because i need the program to be linear (comparison between a quadratic and a linear program for my thesis).
It's not clear in excel, but the arrows mean "smaller or equal". The cells H12:K16 represent the constraints:
Yt >= sum (Ait*x)
Yt >= -sum (Ait*x)

To simply explain what i try to do: I know that i can change a non-linear program with an absolute value to a linear program in the following way:

min abs (b)

to

min a

with constraints:
a >= b
a >= -b

where in my model a = Yt and b = Ait*x.
But i don't know how to apply this to my model in excel.

 

[Rick Rothstein]

min a

with constraints:
a >= b
a >= -b

Are you sure about your constraints? Are they meant to be applied at the same time? If so, then I would point out that if a >= b then by definition it is also automatically greater than (or equal to if b=0) -b. Did you perhaps switch the comparison operator symbol in that first one and actually meant these for your constraints?

a <= b
a >= -b

 

[sijpie]

Sorry, but this type of math was too long ago for me.

 

[gdespont]

Are you sure about your constraints? Are they meant to be applied at the same time? If so, then I would point out that if a >= b then by definition it is also automatically greater than (or equal to if b=0) -b. Did you perhaps switch the comparison operator symbol in that first one and actually meant these for your constraints?

a <= b
a >= -b

No, because b can also be negative.
The result of these constraints is that a >= abs (b), but since we minimize it, it is actually equivalent to min abs (b).

@sijpie: Thanks anyway.