Midterm IOE512 Solution
1. (50 points) Your state legislature has a total of R representatives, and
the state population is divided into N districts (e.g., towns, cities, or counties),
N < R. Let
Pi = population in district i, i = 1, . . . , N ;
N
P = total population in the state, i.e., P = i=1 Pi .
Clearly, if fractional solution is allowed then district i would get RPi /P
representatives. But, we want the number of representatives in each district to
be an integer.
Let xi be the number of representatives in district i. A state-wide debate
agreed that the best way is to allocate the number of representatives is to nd
xi such that the following is minimized:
N
xi R
min
i=1
where, of course,
N
i=1
Pi
,
P
xi = R .
(a) (25 points)Formulate a dynamic program for nding the number of representatives for each district.
(b) (25 points)Solve the problem numerically with data R = 4, N = 3,
P1 = 1.2, P2 = 6 and P3 = 2.8.
1
Solution:
(a) Let fn (y ) denote the optimal value of the following objective function:
n
|xi R
min
i=1
n
s.t.
Pi
|
P
xi = y
i=1
So for the problem, we want to nd fN (R). The dynamic program for nding
fN (R) should be as the following:
fn (y ) = min (|xn R
x n y
Pn
| + fn1 (y xn ))
P
when n = N, N 1, ..., 2. xn and y must be integers. The boundary condition
of the dynamic program should be f1 (y ), which is equivalent to the following
optimization problem:
P1
min |x1 R |
P
s.t. x1 = y
So the boundary condition should be f1 (y ) = |y R P1 |
P
(b) R = 4, N = 3, P1 = 1.2, P2 = 6 and P3 = 2.8
0.48
y=0
0.52
y=1
f1 (y ) = 1.52
y=2
2.52
y=3
3.52
y=4
Since f2 (y ) = minx2 y (|x2 R P2 | + f1 (y x2 )). We could get:
P
2.88
y=0
1.88
y=1
f2 (y ) = 0.88
y=2
0.92
y=3
1.12
y=4
Since f3 (y ) = minx3 y (|x3 R P3 | + f3 (y x3 )). We could get:
P
f3 (4) = min (2.88 + f2 (0), 1.88 + f2 (1), 0.88 + f2 (2), 0.12 + f2 (3), 1.12 + f2 (4))
= min (5.76, 3.76, 1.76, 1.04, 2.24)
= 1.04
So the optimal solution is x3 = 1, x2 = 2, x1 = 1. The optimal value is 1.04.
2
2. (50 points) You have D doctor-years (if one doctor works for one year, it
is one doctor year) available over the N years to help the public health of an
African country. The number doctor-years of committed to year i can be further
broken down to three activities: doctoral-year vaccination vi , education ei , and
working in clinic ci . vi , ei , ci must be nonnegative integers. Such a commitment
will yield Li (vi , ei , ci ) lives saved in year i. Note that Li (vi , ei , ci ) is increasing
in all arguments, and for (vi , ei , ci ), i = 1, . . . , N to be a feasible arrangement,
N
they must satisfy i=1 (vi + ei + ci ) = D.
(a) (25 points)Formulate a dynamic program to maximize total lives saved
over the N years.
(b) (25 points)Solve the problem with N = 2, D = 6, and
L1 (v1 , e1 , c1 )
=
2v1 + v1 e1 + +3e2 + 4 c1 ,
1
L2 (v2 , e2 , c2 )
=
2
v2 + 6e2 + 2c2 + e2 c2 .
3
Solution:
(a) The problem is:
N
max
Li (vi , ei , ci )
i=1
N
s.t.
(vi + ei + ci ) = D
i=1
vi , ei , ci are nonnegative integers
Let fn (y ) denote the optimal value of the following objective function:
n
max
Li (vi , ei , ci )
i=1
n
s.t.
( vi + e i + c i ) = y
i=1
vi , ei , ci are nonnegative integers
So for the problem, we want to nd fN (D).
Let gi (di ) denote the optimal objective function value of the following optimization problem:
max Li (vi , ei , ci )
s.t. vi + ei + ci = di
vi , ei , ci are nonnegative integers
Then the dynamic program of the original problem is as the following:
fn (y ) = max (gn (dn ) + fn1 (y dn ))
dn y
When n = N, N 1, ..., 2. The boundary condition should be:
f1 (y ) = g1 (y )
(b) we can get:
0
4
12
g1 (d1 ) = 27
48
75
108
d1
d1
d1
d1
d1
d1
d1
=0
=1
=2
=3
=4
=5
=6
4
(v1
(v1
(v1
(v1
(v1
(v1
(v1
= e1 = c1 = 0)
= e1 = 0, c1 = 1)
= c1 = 0, e1 = 2)
= c1 = 0, e1 = 3)
= c1 = 0, e1 = 4)
= c1 = 0, e1 = 5)
= c1 = 0, e1 = 6)
and
0
6
12
g2 (d2 ) = 18
24
30
37
d2
d2
d2
d2
d2
d2
d2
=0
=1
=2
=3
=4
=5
=6
(v2
(v2
(v2
(v2
(v2
(v2
(v2
= e2 = c2 = 0)
= c2 = 0, e2 = 1)
= c2 = 0, e2 = 2)
= c2 = 0, e2 = 3)
= c2 = 0, e2 = 4)
= c2 = 0, e2 = 5)
= 0, c2 = 1, e2 = 5)
So
f2 (6) = max (g2 (d2 ) + f1 (6 d2 ))
d 2 6
= max (g2 (d2 ) + g1 (6 d2 ))
d 2 6
= max(37 + 0, 30 + 4, 24 + 12, 18 + 27, 12 + 48, 6 + 75, 0 + 108)
= 108
So d1 = 6, d2 = 0
And
e1 = 6, v1 = 0, c1 = 0
e2 = 0, v2 = 0, c2 = 0
The optimal value is 108.
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