Fall 2010
Optimization I (ORIE 3300/5300)
Assignment 6 Solution
Problem 1
In this problem, we have ﬁve people (supply) and only four stroke styles (demand). Since the
supply quantity is greater than the demand quantity, we need to add a “dummy” demand: we add
a new stroke style called “bench” that everyone can swim in 0 seconds. The person swimming this
style will stay at the bench.
After this, we have ﬁve styles, ﬁve swimmers, and the problem is clearly an assignment problem
of assigning swimmers to diﬀerent strokes. To ﬁt it into a transportation model ﬁle, we have to
make sure that all demands and supplies are equal to one, while the costs are clearly the times of
swimmers in diﬀerent strokes.
We use the model ﬁle
transp.mod
from the amplcml’s MODELS folder. The data ﬁle is (there
are more than one way to do this):
set ORIG:= Barbara Carol Dianne Elizabeth Olympia;
set DEST:= breaststroke butterfly backstroke freestyle bench;
param supply default 1;
param demand default 1;
param cost (tr):
Barbara
Carol
Dianne
Elizabeth
Olympia:=
breaststroke
45.4
42.8
36.1
43.2
36.7
butterfly
34.3
36.6
29.5
39.9
33.4
backstroke
39.7
37.4
33.9
36.8
39.0
freestyle
31.2
33.1
29.4
31.6
32.5
bench
0.00
0
0
0
0;
The output is:
ampl: reset; model transp.mod; data transpdata.dat;
ampl: solve;
MINOS 5.5: optimal solution found.
14 iterations, objective 134.2
ampl: display Trans;
Trans [*,*]
:
backstroke
bench
breaststroke
butterfly
freestyle
:=
Barbara
0
0
0
7.91731e17
1
Carol
1.80525e17
1
0
0
0
Dianne
0
0
7.91731e17
1
0
Elizabeth
1
0
0
0
9.02627e18
Olympia
0
0
1
0
0
;
To get a nicer output, we can type:
1
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View Full DocumentFall 2010
Optimization I (ORIE 3300/5300)
ampl: option omit_zero_rows 1;
ampl: display Trans;
Trans :=
Barbara
butterfly
7.91731e17
Barbara
freestyle
1
Carol
backstroke
1.80525e17
Carol
bench
1
Dianne
breaststroke
7.91731e17
Dianne
butterfly
1
Elizabeth backstroke
1
Elizabeth freestyle
9.02627e18
Olympia
breaststroke
1
;
(Note: here, 7.91731e17 is the same as zero. It is not displayed as zero due to roundoﬀ error.)
Grading scheme:
2 points for adding a dummy demand (the new stroke called “bench”) such
that the supply quantity is equal to the demand quantity; 2 points for using the
transp.mod
ﬁle
provided; 6 points for correct data ﬁle and output.
Problem 2
(a) Clearly, the problem is feasible. The point (
x
1
,x
2
) = (0
,
0) is feasible:
x
1

2
x
2
= 0, so
the ﬁrst and second inequality constraints are both satisﬁed. It is also easy to see that
the problem is bounded, since the ﬁrst constraint provides an upper bound and the second
constraint provides a lower bound to the objective value of any feasible solution.
(b) The feasible region:
Now, we prove that this region does not have any extreme point. In general, to show that a
feasible region does not have any extreme point, we do the following. Consider an arbitrary
point in the region (that is, an arbitrary feasible solution), call it
x
, and show that we can
ﬁnd two other feasible solutions, call them
y
and
z
such that
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 Fall '08
 TODD
 Optimization, CPLEX, Barbara Carol Dianne Elizabeth Olympia

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