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Linear Programming Regression, Lecture Notes - Mathematics - Prof. J Vanderbei.pdf, Prof. J Vanderbei, Mathematics, Linear Programming, Regression, 1995 Adjusted Gross Incomes, Means and Medians, Regression Model for Algorithm Efficiency
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Robert J. Vanderbei
October 17, 2007
Operations Research and Financial Engineering Princeton University Princeton, NJ 08544 http://www.princeton.edu/∼rvdb
Median: x ˆ = b1+ 2 m ≈ $22,500.
Mean:
x ¯ =
m
∑^ m
i=
bi = $4,189,353,615,000/118,218,327 = $35,437.
x ˆ = argminx
∑m
i=
|x − bi|.
Proof:
f (x) =
∑^ m
i=
|x − bi|
f ′(x) =
∑^ m
i=
sgn (x − bi) where sgn(x) =
1 x > 0 0 x = 0 − 1 x < 0 = (# of bi’s smaller than x) − (# of bi’s larger than x).
If m is odd:
1
3
5
− − −
Observed Data:
t = # of iterations m = # of constraints n = # of variables
Model: t ≈ 2 α(m + n)β
Linearization: Take logs:
log t = α log 2 + β log(m + n) + ↑ error
Euclidean Distance: ‖x‖ 2 = (∑ i x^2 i^ )^1 /^2 Least Squares Regression: x¯ = argminx‖b − Ax‖^22 Calculus:
f (x) = ‖b − Ax‖^22 =
∑ i
bi − ∑ j
aij xj
2
∂f ∂xk^ (¯x) =^
∑ i
2
bi − ∑ j
aij ¯xj
(^) (−aik) = 0, k = 1, 2 ,... , n
Rearranging, (^) ∑
i
aikbi =
∑ i
∑ j
aikaij ¯xj , k = 1, 2 ,... , n
In matrix notation, AT^ b = AT^ Ax¯ Assuming AT^ A is invertible, ¯x =
( AT^ A
)− 1 AT^ b
Manhattan Distance: ‖x‖ 1 = ∑ i |xi| Least Absolute Deviation Regression: xˆ = argminx‖b − Ax‖ 1 Calculus:
f (x) = ‖b − Ax‖ 1 =
∑ i
∣∣ ∣∣ ∣∣bi^ −^
∑ j
aij xj
∣∣ ∣∣ ∣∣
∂f ∂xk^ (ˆx) =^
∑ i
∣^ bi^ −^ ∑^ j^ aij^ xˆj ∣∣bi − ∑ j aij^ xˆj
∣∣ ∣
(−aik) = 0, k = 1, 2 ,... , n
Rearranging, (^) ∑
i
aikbi i(ˆx) =^
∑ i
∑ j
aikaij xˆj i(ˆx) ,^ k^ = 1,^2 ,... , n
In matrix notation, AT^ E(ˆx)b = AT^ E(ˆx)Ax,ˆ where E(ˆx) = Diag((ˆx))−^1
Assuming AT^ E(ˆx)A is invertible, ˆx =
( AT^ E(ˆx)A
)− 1 AT^ E(ˆx)b
min
i
bi −
j
aij xj
Equivalent Linear Program:
min
i
ti
−ti ≤ bi −
j
aij xj ≤ ti i = 1, 2 ,... , m
param m; param n;
set I := {1..m}; set J := {1..n};
param A {I,J}; param b {I};
var x{J}; var t{I};
minimize sum_dev: sum {i in I} t[i];
subject to lower_bound {i in I}: -t[i] <= b[i] - sum {j in J} A[i,j]*x[j];
subject to upper_bound {i in I}: b[i] - sum {j in J} A[i,j]*x[j] <= t[i];