Economics 172A: Introduction to Operations Research
Winter 2008
Problem set 2
Due Thursday, March 13 at start of class (no late papers)
Instructions
Unless otherwise noted, you are required to supply complete answers and explain how
you got them. Simply stating a numerical answer is insufficient. For graphs, clearly label
the graph and identify what’s on it.
1. Consider the problem choose x (a scalar) to solve maximize 3x
subject to
x
≤
5
x
≤
b
x
≥
0
where b (also a scalar) > 0.
(a) Graphically or by educated guess, whichever you prefer, compute x*(b), the optimal
value of x, as a function of the parameter b. Your answer must tell what the optimal value
of x
is for any value of b > 0; that is, it must be a clearly specified function of b.
(Hint: Given that b > 0, could there ever be a solution with x* = 0? Which constraint
would you expect to determine the solution when b < 5? When b > 5?)
(a) Clearly x*(b) > 0 for all b > 0. For b > (<) 5, the second (first) constraint is
redundant. One of these constraints must always be binding, so x*(b) = b if b
≤
5,
and x*(b) = 5 if b > (or
≥
)
5.
(b) Write the dual, using y
i
to represent the dual control variable that is the shadow price
of the ith constraint in the primal.
(b) Choose y
1
, y
2
to solve
minimize 5y
1
+ by
2
subject to
y
1
+ y
2
≥
3
y
1
≥
0
y
2
≥
0.
(c) Graphically or by educated guess, whichever you prefer, compute y
1
*(b) and y
2
*(b),
the optimal values of y
1
and y
2
, as functions of the parameter b. Your answer must tell
what the optimal values of y
1
and y
2
are for any value of b. (When b = 5, identify all the
possible optimal values of y
1
*(b) and y
2
*(b).)
(c) If b < (or
≤
) 5, y
1
*(b) = 0 and y
2
*(b) = 3. If b > (or
≥
) 5, y
1
*(b) = 3 and y
2
*(b) = 0.
If b = 5, any y
1
and y
2
with y
1
+ y
2
= 3 and y
1
≥
0 and y
2
≥
0 is optimal, including but
not limited to the values specified above.