Show that: sum x_ ie_i=0 and also show that sum y_i e_i=0. Now I do believe that being able to solve the first sum will make the solution to the second sum more clear. So far I have proved that sum e_i=0.

Mariah Sparks 2022-07-22 Answered
Show that: x i e i = 0 and also show that y ^ i e i = 0. Now I do believe that being able to solve the first sum will make the solution to the second sum more clear. So far I have proved that e i = 0.
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Answers (1)

Kitamiliseakekw
Answered 2022-07-23 Author has 23 answers
Here's one way of viewing it. We want to write
[ y 1 y n ] = α ^ [ 1 1 ] + β ^ [ x 1 x n ] + [ ε ^ 1 ε n ]
and choose the values of α ^ and β ^ that minimze ε ^ 1 2 + + ε ^ n 2 . The sum of the first two terms on the right is [ y ^ 1 , , y ^ n ] T .
That means the point [ y ^ 1 y n ] = α ^ [ 1 1 ] + β ^ [ x 1 x n ] is closer to [ y 1 y n ] in ordinary Euclidean distance than is any other point in the plane spanned by [ 1 1 ] and [ x 1 x n ] . The point in a plane that is closet to y is the point you get by dropping a perpendicular from y to the plane. That means ε ^ = y ^ y is perpendicular to the two columns that span the plane, and thus perpendicular to every linear combination of them, such as y ^ . "Perpendicular" means the dot-product is zero. Q.E.D.
The vector of fitted values y ^ = [ y ^ 1 y ^ n ] is the orthogonal projection of the vector y = [ y 1 y n ] onto the column space of the design matrix X = [ 1 x 1 1 x n ] .
The orthogonal projection is a linear transformation whose matrix is the "hat matrix" is H = X ( X T X ) 1 X T , an n × n matrix of rank 2. Observe that if w is orthogonal to that column space then X w = 0 so H w = 0, and if w is in the column space, then w = X u for some u R 2 , and so H w = w.
It follows that ε ^ = y y ^ = ( I H ) y is orthogonal to the column space. Since y ^ is in the column space, ε ^ is orthogonal to y ^ .
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The Porsche Club of America sponsors driver education events that provide high-performance driving instruction on actual racetracks. Because safety is a primary consideration at such events, many owners elect to install roll bars in their cars. Deegan Industries manufactures two types of roll bars for Porsches. Model DRB is bolted to the car using existing holes in the car's frame. Model DRW is a heavier roll bar that must be welded to the car's frame. Model DRB requires 20 pounds of a special high alloy steel, 40 minutes of manufacturing time, and 60 minutes of assembly time. Model DRW requires 25 pounds of the special high alloy steel, 100 minutes of manufacturing time, and 40 minutes of assembly time. Deegan's steel supplier indicated that at most 40,000 pounds of the high-alloy steel will be available next quarter. In addition, Deegan estimates that 2000 hours of manufacturing time and 1600 hours of assembly time will be available next quarter. The pro?t contributions are $200 per unit for model DRB and $280 per unit for model DRW. The linear programming model for this problem is as follows:
Max 200DRB + 280DRW
s.t.
20DRB + 25DRW 40,000 Steel Available
40DRB + 100DRW ? 120,000 Manufacturing minutes
60DRB + 40DRW ? 96,000 Assembly minutes
DRB, DRW ? 0
Optimal Objective Value = 424000.00000
Variable Value blackuced Cost
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
DRB 1000.00000 0.00000
DRW 800.00000 0.00000
Constraint Slack/ Surplus Dual Value
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1 0.00000 8.80000
2 0.00000 0.60000
3 4000.00000 0.00000
Objective Allowable Allowable
Variable Coef?cient Increase Decrease
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
DRB 200.00000 24.00000 88.00000
DRW 280.00000 220.00000 30.00000
RHS Allowable Allowable
Constraint Value Increase Decrease
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1 40000.00000 909.09091 10000.00000
2 120000.00000 40000.00000 5714.28571
3 96000.00000 Infnite 4000.00000
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e. If the available manufacturing time is increased by 500 hours, will the dual value for the manufacturing time constraint change? Explain.