Consider the 0-1 knapsack problem: max {£₁ Σ cjxj : Σ ajx; ≤ b, x € j=1 jxj ≤ b,x € (0.1)"} with aj, c;> 0 for j = 1,..., n. (a) Show that if > ... > cn > 0, Σ;a; ≤ b and Σ';-1 a; > b, the solution of the LP relaxation is x; = 1 for j = 1,..., r − 1, xp = a₁ an - (b - Σ';= a;) /a,, and x; = 0 for j > r.
Consider the 0-1 knapsack problem: max {£₁ Σ cjxj : Σ ajx; ≤ b, x € j=1 jxj ≤ b,x € (0.1)"} with aj, c;> 0 for j = 1,..., n. (a) Show that if > ... > cn > 0, Σ;a; ≤ b and Σ';-1 a; > b, the solution of the LP relaxation is x; = 1 for j = 1,..., r − 1, xp = a₁ an - (b - Σ';= a;) /a,, and x; = 0 for j > r.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 78E
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![Consider the 0-1 knapsack problem:
{cjxj : Ëajxj;
ajxj
j=1
with aj, c; > 0 for j = 1,..., n.
(a) Show that if > ... > cn > 0, Σ' a; ≤ b and Σ';=1 a; > b, the
solution of the LP relaxation is xj = 1 for j = 1,...,r - 1,xp =
a₁
an
(b- Σ;a;)/a,, and x; = 0 for j > r.
(b) Solve the instance
max
xj ≤ b,x € (0,1)"}
max 17x1 + 10x2 + 25x3 + 17x4
5x1 + 3x2 + 8x3 + 7x4 ≤ 12
x = {0,1}4
by branch-and-bound. Always branch first on the node with the best
bound. You must number the nodes as you explore them. Name
the first node as "Root". Create a table with one row for each node.
At each row, indicate the value of the variables in the solution of
the node, the capacity used, the profit of the solution (or indicate
its infeasibility) and the status associated with the node (branched,
pruned by infeasibility, pruned by integrality or pruned by quality).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3b2b977-49e2-428f-b276-1fc0022629e3%2F2280467a-683a-4969-9baf-eb2cfa7cc8dd%2Fnx6qfmb_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the 0-1 knapsack problem:
{cjxj : Ëajxj;
ajxj
j=1
with aj, c; > 0 for j = 1,..., n.
(a) Show that if > ... > cn > 0, Σ' a; ≤ b and Σ';=1 a; > b, the
solution of the LP relaxation is xj = 1 for j = 1,...,r - 1,xp =
a₁
an
(b- Σ;a;)/a,, and x; = 0 for j > r.
(b) Solve the instance
max
xj ≤ b,x € (0,1)"}
max 17x1 + 10x2 + 25x3 + 17x4
5x1 + 3x2 + 8x3 + 7x4 ≤ 12
x = {0,1}4
by branch-and-bound. Always branch first on the node with the best
bound. You must number the nodes as you explore them. Name
the first node as "Root". Create a table with one row for each node.
At each row, indicate the value of the variables in the solution of
the node, the capacity used, the profit of the solution (or indicate
its infeasibility) and the status associated with the node (branched,
pruned by infeasibility, pruned by integrality or pruned by quality).
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