(1) What is the current basic feasible solution? Is the current solution optimal, why? (10 points) (2) If the current solution is not optimal, identify the pivot element and complete Gaussian Elimination for the pivot row and the first row of the tableau in the next iteration. (5 points) Then, answer the following questions: which variables are the basic variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5 points) Iteration 1: Solution to problem 2 元 x1 x2 x3 s1 s2 s3 s4 s5 1 0 -200 -100 0 300 0 0 constant 15000 20 0 0 (10) 2 1 0 -16 0 0 200 200/10 = 2 0 0 4 2 0 -8 0 100 100/4 = 25 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 80/1=80 0 1 0 0 0 0 1 150 Iteration 2: П x1 x2 x3 s1 s2 $3 s4 $5 constant 1 0 0 -60 00 1 2/10 20 0 -20 1/100 0 0 19000 -1.6 0 0 20 New basic Variables are : (X₂, S2, X₁, S4, S5) Solution from Iteration 2 is not optimal, because there are still negative values in the first row. This means there is further improvement to be made,

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.6: Applications Of Differential Equations
Problem 1YT: YOUR TURN 1 Suppose that an epidemic in a community of 50,000 starts with 80 people infected, and...
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how did they get to this solution?

(1) What is the current basic feasible solution? Is the current solution optimal, why? (10
points)
(2) If the current solution is not optimal, identify the pivot element and complete
Gaussian Elimination for the pivot row and the first row of the tableau in the next
iteration. (5 points) Then, answer the following questions: which variables are the basic
variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5
points)
Transcribed Image Text:(1) What is the current basic feasible solution? Is the current solution optimal, why? (10 points) (2) If the current solution is not optimal, identify the pivot element and complete Gaussian Elimination for the pivot row and the first row of the tableau in the next iteration. (5 points) Then, answer the following questions: which variables are the basic variables at iteration 2? Is the solution from Iteration 2 likely to be optimal, and why? (5 points)
Iteration 1:
Solution to problem 2
元
x1
x2
x3
s1
s2
s3
s4
s5
1
0
-200
-100
0
300
0
0
constant
15000
20
0
0
(10)
2
1
0
-16
0
0
200
200/10 = 2
0
0
4
2
0
-8
0
100
100/4 = 25
0
1
0
0
0
0
1
0
0
50
0
0
1
0
0
0
0
1
0
80
80/1=80
0
1
0
0
0
0
1
150
Iteration 2:
П
x1
x2
x3
s1
s2
$3
s4
$5
constant
1 0
0
-60
00
1
2/10
20 0 -20
1/100
0
0
19000
-1.6
0
0
20
New basic Variables are : (X₂, S2, X₁, S4, S5)
Solution from Iteration 2 is not optimal, because
there are still negative values in the first row.
This means there is further improvement to be
made,
Transcribed Image Text:Iteration 1: Solution to problem 2 元 x1 x2 x3 s1 s2 s3 s4 s5 1 0 -200 -100 0 300 0 0 constant 15000 20 0 0 (10) 2 1 0 -16 0 0 200 200/10 = 2 0 0 4 2 0 -8 0 100 100/4 = 25 0 1 0 0 0 0 1 0 0 50 0 0 1 0 0 0 0 1 0 80 80/1=80 0 1 0 0 0 0 1 150 Iteration 2: П x1 x2 x3 s1 s2 $3 s4 $5 constant 1 0 0 -60 00 1 2/10 20 0 -20 1/100 0 0 19000 -1.6 0 0 20 New basic Variables are : (X₂, S2, X₁, S4, S5) Solution from Iteration 2 is not optimal, because there are still negative values in the first row. This means there is further improvement to be made,
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