Homework Solutions #1:
1.1
1.4
1.7
1.11
1.15
1.26
2.12
2.29
2.54
1.1
Let Xij represent ingredients “i” in feed “j”, where i = 1, 2, 3, 4 represent ingredient Corn, Limestone, Soybeans,
Fish meal, and j = 1, 2, 3 represents feed Cattle, Sheep, and Chicken.
3
3
3
3
j =1
CORN
j =1
Lime
j =1
Soy
j =1
Fish
Z = 0.20 ∑ X 1 j +0.12 ∑ X 2 j + 0.24∑ X 3 j + 0.12∑ X 4 j
minimize
s.t.

X 1 j ≤ 6 ∑ X 2 j ≤ 10
∑
j =1
j =1

CORN
LIME
Supply
3
3
X
X
4
5
≤
≤
∑
∑ 4 j 
3j
j =1
j =1
SOY
FISH

3
3
4
∑X
i1
≤ 10
i =1
Cattle
4
∑X
i =1
Chicken
4
∑X
i =1
Sheep
i3
≤8
i2

≤ 6

demand



Vitamins8X11 + 6X21 + 10X31 + 4X41 ≥ 6(X11 + X21 + X31 + X41)
in Cattle Feed
8X12 + 6X22 + 10X32 + 4X42 ≥ 6(X12 + X22 + X32 + X42)
in Sheep Feed
4(X13 + X23 + X33 + X43)≥ 8X13 + 6X23 + 10X33 + 4X43 ≥ 6(X13 + X23 + X33 + X43) in Chicken Feed
Protein 10X11 + 5X21 + 12X31 + 8X41 ≥ 6(X11 + X21 + X31 + X41)
10X12 + 5X22 + 12X32 + 8X42 ≥ 6(X12 + X22 + X32 + X42)
10X13 + 5X23 + 12X33 + 8X43 ≥ 6(X13 + X23 + X33 + X43)
in Cattle Feed
in Sheep Feed
in Chicken Feed
Calcium 6X11 + 10X21 + 6X31 + 6X41 ≥ 7(X11 + X21 + X31 + X41)
6X12 + 10X22 + 6X32 + 6X42 ≥ 6(X12 + X22 + X32 + X42)
6X13 + 10X23 + 6X33 + 6X43 ≥ 6(X13 + X23 + X33 + X43)
in Cattle Feed
in Sheep Feed
in Chicken Feed
Crude Fat –
4(X11 + X21 + X31 + X41)≥ 8X11 + 6X21 + 6X31 + 9X41 ≥ 8(X11 + X21 + X31 + X41) in Cattle Feed
4(X12 + X22 + X32 + X42)≥ 8X12 + 6X22 + 6X32 + 9X42 ≥ 6(X12 + X22 + X32 + X42) in Sheep Feed
4(X13 + X23 + X33 + X43)≥ 8X13 + 6X23 + 6X33 + 9X43 ≥ 6(X13 + X23 + X33 + X43) in Chicken Feed
1.4 –
Launching a rocket to a fixed altitude b in a given time T let a(t) be the acceleration force exerted at time t, and
y(t) be the rocket altitude at time t. We have:
min
s.t.
∫
T
0
u (t ) dt
y(t ) = u (t ) − g
y (T ) = b
y (t ) ≥ 0
t ∈ [0, T ]
Discretize the above formulation.
let n be the # of subintervals, of width ∆ = T/n
let uj = u(tj) = u(j∆)
let yj = y(tj) = y(j∆)
let xj = |uj|
Notice that a finite difference approximation yields:
y (t j + ∆ ) =
y(t j + ∆ ) =
y (t j + ∆ )− y (t j )
∆
y (t j + ∆ )− y (t j )
∆
y (t j + ∆ ) − y (t j )
=
∆
−
∆
y (t j ) − y (t j − ∆ )
∆
=
y (t j + ∆ )− 2 y (t j )+ y (t j − ∆ )
∆2
New Formulation:
Minimize
n −1
∑x ∆
j
∆ = T/n
j =0
s.t.
y j +1 − 2 y j + y j −1 = ∆2 (u j − g )
yn = b
(yo = 0 )
yj ≥ 0
xj ≥ uj
xj ≥ -uj
for j = 1, ..., n-1
(I added this)
for j = 0,1, ..., n-1
This works for xj = |uj| because xj ≥ uj and xj ≥ -uj forces xj to be either one or the other.
Now the formulation has uj unrestricted in sign.
Another way is
let
uj = xj+ - xjthen |uj| = xj+ - xjand we need (xj+)(xj-) = 0
substituting throughout, we get,
Minimize
n −1
∑x
+
j
+ x −j
j =0
s.t.
(
y j +1 − 2 y j + y j −1 − ∆2 x +j − x −j − g
and
yn = b
(yo = 0 )
yj ≥ 0
xj ≥ uj
xj ≥ -uj
(xj+)(xj-) = 0
)
for j = 1, ..., n-1
for j = 0,1, ..., n-1
but this is satisfied by properties of BFS
1.7 –
Xij is the amount of money invested at the beginning of year i and a j year time deposit
year 1
X11
year 2
year 3
year 4
year 5
X12
X21
X22
X23
X31
X32
X33
X41
X42
X51
max
1.27X33 + 1.17X42 + 1.08X51
amount invested at begining
X11 + X12 ≤ $2,200
X21 + X22 + X23 ≤ 1.08X11 + (2,200 - X11 - X12) - amount at end of yr 2, beg yr if allowing not
to invest all (≤), need to include that in all constrains or because it is always profitable to invest, we could use
(=).
X11 + X12 = $2,200
X21 + X22 + X23 = 1.08X11
X31 + X32 + X33 = 1.17X12 + 1.08X21
X41 + X42 = 1.08X31 + 1.17X22
X51 = 1.08X41 + 1.17X32 + 1.27X23
s.t.
X11, X12, X21, X22, X23, X31, X32, X33, X41, X42, X51 ≥ 0
1.11 –
X1 = # of barrels of light crude oil
X2 = # of barrels of heavy crude oil
minimize
X2
11X1 + 9X2
s.t.
0.4X1 + 0.32X2 ≥ 1,000,000
0.2X1 + 0.4X2 ≥ 400,000
0.35X1 + 0.2X2 ≥ 250,000
X1 , X2 ≥ 0
X1
1.15 –
Let Xij be the number of product i where i = 1,2,3 produced on machine j where j = 1,2,3,4
minimize
s.t.
(4X11 + 4X12 + 5X13 + 7X14)
+(6X21 + 7X22 + 5X23 + 6X24)
+(4X31 + 4X32 + 8X33 + 11X34)
X 11 + X 12 + X 13 + X 14 ≥ 4000 

X 21 + X 22 + X 23 + X 24 ≥ 5000 units of product required
X 31 + X 32 + X 33 + X 34 ≥ 3000 
X 11 + X 21 + X 31 ≥ 1500 
X 12 + X 22 + X 32 ≥ 1200 
 available machine hours for production
X 13 + X 23 + X 33 ≥ 1500 
X 14 + X 24 + X 34 ≥ 2000
X11 X12 X13 X14 X21 X22 X23 X24 X31 X32 X33 X34 ≥ 0
l2
1.26 –
c
C4
b
C3
C1
1.
2.
3.
4.
5.
6.
7.
8.
a
C2
C4
C3
l1
If C is same direction as C1, then all points on halfline l1 are optimal.
If C lies between C1 and C2, then point a is optimal
If C is same direction as C2, then all points on line ab are optimal.
If C lies between C2 and C3, then point b is optimal.
If C is same direction as C3, then all points on line bc are optimal.
If C lies between C3 and C4, then point c is optimal.
If C is same direction as C4, then all points on halfline l2 are optimal.
If C lies between C4 and C1, then no optimal solution exists (unbounded).
2.12
Show that if A and B are n x n matrices that are both invertible, then (AB)-1 = B-1A-1.
If A and B are invertible than AA-1 = I and BB-1 = I.
Check if AB(AB)-1 = I;
AB(B-1A-1) = ABB-1A-1 = AA-1 = I;
Also
So
BA(A-1B-1) = BAA-1B-1 = BB-1 = I;
(AB)-1 = A-1B-1
2.29 –
a. {(X1, X2): X12 + X22 ≥ 1}
False,
|X| = 1,1
|X| = -1,-1
(0,0) = λ(-1,-1) + λ(1,1)
λ = 0.5
(0,0) is not in the set. Set is not convex.
b.
{(X1,X2,X3): X1 + 2X2 ≤ 1, X1 + X3 ≤ 2 }
True, convex - finite intersection of half spaces.
c.
{(X1, X2): X2 - X12 = 1}
False,
|X| = 1,1
|X| = -1,-1
(0,1) = λ(-1,-1) + λ(1,1)
λ = 0.5
(0,1) is not in the set. Set is not convex.
d.
{(X1,X2,X3): X2 ≥ X12 , X1 + X2 + X3 ≤ 6 }
True, convex – finite intersection of convex sets
e.
{(X1, X2): X1 = 1, |X2| ≤ 4}
True, convex – finite intersection of convex sets
f.
{(X1,X2,X3): X3 = |X2| , X1 ≤ 4 }
X3 = |X1|
X1 ≤ 4
not convext
convex
X3
|X| = 1,1 - (X2,X3)
|X| = -1,-1 - (X2,X3)
λ = 0.5
(0,1) = λ(-1,-1) + λ(1,1)
(X2,X3) not in set
X2
X1
2.54 –
a.
Is it possible for X = {x : Ax ≤ b, x ≥ 0} to be empty and D = {d : Ad ≤ 0,1d = 1, d ≥ 0} to be non-empty.
Yes consider
X = {(X1, X2): X1 – X2 = 0, X1 – X2 = 1, X1, X2 ≥ 0 }
X is empty
D = {(d1, d2): d1 – d2 = 0, d1 + d2 = 1, d1, d2 ≥ 0 }
= {1/2 , 1/2}
b.
Is there a relationship between redundancy and degeneracy of a polyhedral set?
A redundant constraint may cause degeneracy but only if it “supports the polyhedron X ”
Degeneracy may or may not be resolved by eliminating redundant constraints.
c.
Does degeneracy imply redundancy in two dimensions?
Yes
d.
If the intersection of a finite number of halfspaces is nonempty, then this set has at least one extreme
point?
False
e.
An unbounded n-dimensional polyhedral set can have at most n extreme directions?
False
f.
What is the maximum (actual) dimension of
X = {x : Ax ≤ b, x ≥ 0}, where A is m x n of rant t, t ≤ m ≤ N?
n–t
on page 64, “Now given any face F of X if r(f) is the maximum number of linearly indys. defining
hyperplanes binding at all points feasible to X then the dim of F is dim(F) = n – r(F)”
“Similarly, if r( X ) is defined w.r.t. X itself, then although X ≤ F, it is actually of dimension,
dim X = n – r( X )”