Chebyshev center Example: find the Chebyshev center of the polyhedron defined by the following inequalities: 2xy +2z ≤ 2, -x +2y + 4z ≤ 16, x + 2y – 2z ≤ 8, x ≥ 0, y≥ 0, z ≥ 0 0 0.0 2 0.5 1.0 1.5 X 2.0 4-19 Example: building a house A small sample: Let t, to, tm, tn, tt, to be start times of the associated tasks. Now use the graph to write the dependency constraints: 0,3 FINISH $,2 CARPENTRY Source: HBR 1963 43 FINISH FLOORING m,1 VARNISH FLOORS KITCHEN FIXTURES n,2 13 PAINT FINISH PLUMBING • Tasks o, m, and n can't start until task / is finished, and task / takes 3 days to finish. So the constraints are: ti +3≤ to, t₁ + 3 ≤ tm, ti +3 ≤ tn . Task t can't start until tasks m and n are finished. Therefore: tm +1≤tt, tn + 2 ≤ tt, • Task s can't start until tasks o and t are finished. Therefore: to + 3 ts, tt + 3 ≤ ts 4-24
Chebyshev center Example: find the Chebyshev center of the polyhedron defined by the following inequalities: 2xy +2z ≤ 2, -x +2y + 4z ≤ 16, x + 2y – 2z ≤ 8, x ≥ 0, y≥ 0, z ≥ 0 0 0.0 2 0.5 1.0 1.5 X 2.0 4-19 Example: building a house A small sample: Let t, to, tm, tn, tt, to be start times of the associated tasks. Now use the graph to write the dependency constraints: 0,3 FINISH $,2 CARPENTRY Source: HBR 1963 43 FINISH FLOORING m,1 VARNISH FLOORS KITCHEN FIXTURES n,2 13 PAINT FINISH PLUMBING • Tasks o, m, and n can't start until task / is finished, and task / takes 3 days to finish. So the constraints are: ti +3≤ to, t₁ + 3 ≤ tm, ti +3 ≤ tn . Task t can't start until tasks m and n are finished. Therefore: tm +1≤tt, tn + 2 ≤ tt, • Task s can't start until tasks o and t are finished. Therefore: to + 3 ts, tt + 3 ≤ ts 4-24
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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upyter nootebook python by julia with (JuMP, Clp): find output value: value(t1), value(t0),value(tm), value(tn),value(tt),value(ts),objective_value(model)?
Modify your models and coding of Chebyshev center and building a house to get the right anwsers.
# Given matrix A and
A = [2 -1 2; -1 2 4; 1 2 -2; -1 0 0; 0 -1 0; 0 0 -1]
b = [2; 16; 8; 0; 0; 0]

Transcribed Image Text:Chebyshev center
Example: find the Chebyshev center of the polyhedron
defined by the following inequalities:
2xy +2z ≤ 2, -x +2y + 4z ≤ 16, x + 2y – 2z ≤ 8,
x ≥ 0, y≥ 0, z ≥ 0
0
0.0
2
0.5 1.0 1.5
X
2.0
4-19

Transcribed Image Text:Example: building a house
A small sample:
Let t, to, tm, tn, tt, to be start
times of the associated tasks.
Now use the graph to write the
dependency constraints:
0,3 FINISH
$,2
CARPENTRY
Source: HBR 1963
43 FINISH FLOORING
m,1
VARNISH FLOORS
KITCHEN
FIXTURES
n,2
13 PAINT
FINISH
PLUMBING
• Tasks o, m, and n can't start until task / is finished, and task /
takes 3 days to finish. So the constraints are:
ti +3≤ to, t₁ + 3 ≤ tm, ti +3 ≤ tn
. Task t can't start until tasks m and n are finished. Therefore:
tm +1≤tt, tn + 2 ≤ tt,
• Task s can't start until tasks o and t are finished. Therefore:
to + 3 ts, tt + 3 ≤ ts
4-24
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