quiz2-2009
-
Upload
welcometoankit -
Category
Documents
-
view
212 -
download
0
Transcript of quiz2-2009
![Page 1: quiz2-2009](https://reader031.fdocuments.us/reader031/viewer/2022021405/577cd57e1a28ab9e789aed23/html5/thumbnails/1.jpg)
8/13/2019 quiz2-2009
http://slidepdf.com/reader/full/quiz2-2009 1/2
Indian Institute of Technology, Bombay Chemical Engineering
CL603: Optimization Quiz 2, Spring 2009
Follow ALL instructions carefully: This is a CLOSED BOOK exam. Calculators are permitted. Make reasonable
assumptions and CLEARLY indicate them in your answer book. Where possible, box your final answers. Start each
problem on a new sheet.
1. A thin-walled cylindrical pressure vessel with spherical ends is to be designed to minimize the total volume
of the material used in its manufacture. The vessel must contain at least 25 m of gas at a pressure, , of
0.35 MPa. The hoop stress,
, in the cylinder walls must not exceed 200 MPa (
). A schematic
of the vessel is provided below.
(a) Graphically identify the feasible region on an
vs.
plot. [2]
(b) Using the Kuhn-Tucker conditions (i.e first order conditions), determine the values of radius
and
wall thickness
which minimize the volume of the material subject to the constraints. [6]
(c) Calculate the sensitivity of the optimal solution to small changes in the right hand sides of the
inequality constraints. [2]
Soln:
1. Let
. We want to minimize
where
m, subject to
MPa
(
Mpa) and
. Then
.
. The two constraints intersect at
m.
2. KKT is
x
T
g
x
0
Tg x
0
Which results in 4 cases
A:
and which implies or and or
and hence no feasible solution exists.
![Page 2: quiz2-2009](https://reader031.fdocuments.us/reader031/viewer/2022021405/577cd57e1a28ab9e789aed23/html5/thumbnails/2.jpg)
8/13/2019 quiz2-2009
http://slidepdf.com/reader/full/quiz2-2009 2/2
B:
and hence
and and
.
The last eq. gives
m while the second gives
or
and hence no feasible
solution again.
C:
and
and
. Taking ratio of
first two eqs,
which gives
which is not feasible.
D:
.
gives m.
gives
m.
Second eq gives
and first gives
and hence we have a global minimum.
3.
x
d
at d
is
. and hence the sensitivities are calculated already from above. The solutionis more sensitive to changes in the second constraint (volume) rather than the first (hoop stress).