This preview shows pages 1–2. Sign up to view the full content.
Fall 2010
Optimization I (ORIE 3300/5300)
Recitation 4 Solution
Winery Problem
First, we solve the winery problem again using the model and data ﬁles provided (
prod.mod
,
plonk.dat
):
ampl: model prod.mod; data plonk.dat; solve;
MINOS 5.5: optimal solution found.
3 iterations, objective 2447.5
ampl: display X;
X [*] :=
ADWhite
200
BRed
300
lred
82.5
;
Following the instruction,
ampl: display Limit.body, Limit;
:
Limit.body
Limit
:=
ADWhite
200
0.5
BRed
300
0.625
lred
82.5
0
;
As we can read on page 35,
Limit.body
displays the “body” in the
Limit
constraints. In this
case, it displays
x[j]
for each
j in P
. Meanwhile, as we can read on page 17,
Limit
displays the
“dual values” or the marginal values associated with the
Limit
constraints. These dual values
describe how the objective function (in our case, total proﬁt) would change if the right hand side
of the corresponding constraint is increased, given the current optimal solution. For instance,
Limit[’ADWhite’] = 0.5
means that if the demand for AD White wine had increased (so we can
sell more each day), then we would gain an additional proﬁt of $0.50 for each additional unit up to
some level.
Limit[’BRed’] = 0.625
means that if the demand for Big Red wine had increased (so
we can sell more each day), then we would gain an additional proﬁt of $0.625 for each additional
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 TODD

Click to edit the document details