The document describes an optimization problem to allocate resources among a set of items in a way that minimizes excess amounts above predefined valuations for different combinations of items. The problem is formulated as linear constraints with a variable M representing the maximum excess. Solutions are obtained by solving the linear program for different valuations scenarios. The optimal allocation balances minimizing M while satisfying all constraints.

2. “
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3. Google

4.
(N, v)

5.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
(N, v)

6.
(N, v)
∑
i∈S
xi ≥ v(S), ∀S ⊂ N
e(S, x) = v(S) −
∑
i∈S
xi
e(S, x) ≤ 0

7.
N
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
2N
− 2 θ(x)
x1 = 30, x2 = 40, x3 = 30
e({1,2}, x) = v({1,2}) − (x1 + x2) = 50 − (30 + 40) = − 20
e({2,3}, x) = v({2,3}) − (x2 + x3) = 80 − (40 + 30) = 10
e({1,3}, x) = v({1,3}) − (x1 + x3) = 40 − (30 + 30) = − 20
e({1}, x) = v({1}) − x1 = 0 − 30 = − 30
e({2}, x) = v({2}) − x2 = 0 − 40 = − 40
e({3}, x) = v({3}) − x3 = 0 − 30 = − 30
θ(x) = (10, − 20, − 20, − 30, − 30, − 40)

8.
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
x′
1 = 30, x′
2 = 50, x′
3 = 20
e({1,2}, x′
) = v({1,2}) − (x′
1 + x′
2) = 50 − (30 + 50) = − 30
e({2,3}, x′
) = v({2,3}) − (x′
2 + x′
3) = 80 − (50 + 20) = 10
e({1,3}, x′
) = v({1,3}) − (x′
1 + x′
3) = 40 − (30 + 20) = − 10
e({1}, x′
) = v({1}) − x′
1 = 0 − 30 = − 30
e({2}, x′
) = v({2}) − x′
2 = 0 − 50 = − 50
e({3}, x′
) = v({3}) − x′
3 = 0 − 20 = − 20
θ(x′
) = (10, − 10, − 20, − 30, − 30, − 50)

9.
x, x′
θ(x′
) = (10, − 10, − 20, − 30, − 30, − 50)
θ(x) = (10, − 20, − 20, − 30, − 30, − 40)
x >L x′

10.
x, x′
k, 1 ≤ k ≤ 2n
− 4
θl(x) = θl(x′) ∀l = 1,2,⋯k − 1, θk(x) < θk(x′)
x x′
x >L x′
x x
x
𝒩(v) = {x ∈ 𝒜(v)|x′ >L x x′ ∈ 𝒜(v)

11.
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 40
v({1}) = 0
v({2}) = 0
v({3}) = 0
x = (x1, x2, x3) ∈ 𝒜(v)
e({1,2}, x) = v({1,2}) − (x1 + x2) = 50 − (x1 + x2)
e({2,3}, x) = v({2,3}) − (x2 + x3) = 80 − (x2 + x3)
e({1,3}, x) = v({1,3}) − (x1 + x3) = 40 − (x1 + x3)
e({1}, x) = v({1}) − x1 = 0 − x1
e({2}, x) = v({2}) − x2 = 0 − x2
e({3}, x) = v({3}) − x3 = 0 − x3

12.
min M
e({1,2}, x) = v({1,2}) − (x1 + x2) = 50 − (x1 + x2) ≤ M
e({2,3}, x) = v({2,3}) − (x2 + x3) = 80 − (x2 + x3) ≤ M
e({1,3}, x) = v({1,3}) − (x1 + x3) = 40 − (x1 + x3) ≤ M
e({1}, x) = v({1}) − x1 = 0 − x1 ≤ M
e({2}, x) = v({2}) − x2 = 0 − x2 ≤ M
e({3}, x) = v({3}) − x3 = 0 − x3 ≤ M
s . t .
x1 + x2 + x3 = 100
x1, x2, x3 ≥ 0
(1)
(2)
(3)
(4)
(5)
(6)

13.
min M
0 ≤ x3 ≤ M + 50
0 ≤ x1 ≤ M + 20
0 ≤ x2 ≤ M + 60
s . t .
0 ≤ (x1 + x2 + x3) = 100 ≤ 3M + 130
⇔ − 10 ≤ M
(x1, x2, x3) = (10,50,40)
⇔ − 50 ≤ M
⇔ − 20 ≤ M
⇔ − 60 ≤ M
M

14.
min M
e({1,2}, x) = 50 − (x1 + x2) ≤ M
e({2,3}, x) = 80 − (x2 + x3) ≤ M
e({1,3}, x) = 40 − (x1 + x3) ≤ M
e({1}, x) = 0 − x1 ≤ M
e({2}, x) = 0 − x2 ≤ M
e({3}, x) = 0 − x3 ≤ M
s . t .
x1 + x2 + x3 = 100
x1, x2, x3 ≥ 0
(1)
(2)
(3)
(4)
(5)
(6)

15.
x3
x1 = 20
x2 = 60
x3 = 50

16.
v({1,2,3}) = 100
v({1,2}) = 50
v({2,3}) = 80
v({1,3}) = 80
v({1}) = 0
v({2}) = 0
v({3}) = 0
x3
x1 + x2 ≥ 50
x2 + x3 ≥ 80
x1 + x3 ≥ 80
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 + x3 = 100
x1 = 20
x2 = 20
x3 = 50

17.
min M
e({1,2}, x) = v({1,2}) − (x1 + x2) = 50 − (x1 + x2) ≤ M
e({2,3}, x) = v({2,3}) − (x2 + x3) = 80 − (x2 + x3) ≤ M
e({1,3}, x) = v({1,3}) − (x1 + x3) = 80 − (x1 + x3) ≤ M
e({1}, x) = v({1}) − x1 = 0 − x1 ≤ M
e({2}, x) = v({2}) − x2 = 0 − x2 ≤ M
e({3}, x) = v({3}) − x3 = 0 − x3 ≤ M
s . t .
x1 + x2 + x3 = 100
x1, x2, x3 ≥ 0
(1)
(2)
(3)
(4)
(5)
(6)

18.
min M
0 ≤ x3 ≤ M + 50
0 ≤ x1 ≤ M + 20
0 ≤ x2 ≤ M + 20
s . t .
0 ≤ (x1 + x2 + x3) = 100 ≤ 3M + 90
⇔
10
3
≤ M
⇔ − 50 ≤ M
⇔ − 20 ≤ M
⇔ − 20 ≤ M
(x1, x2, x3) = (20 +
10
3
,20 +
10
3
,50 +
10
3
)

19.
v({1,2,3}) = 100
v({1,2}) = 30
v({2,3}) = 60
v({1,3}) = 100
v({1}) = 0
v({2}) = 0
v({3}) = 0
x1 + x2 ≥ 30
x2 + x3 ≥ 60
x1 + x3 ≥ 100
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 + x3 = 100
x3 = 70
x2 = 0
x1 = 40

20.
min M
e({1,2}, x) = v({1,2}) − (x1 + x2) = 30 − (x1 + x2) ≤ M
e({2,3}, x) = v({2,3}) − (x2 + x3) = 60 − (x2 + x3) ≤ M
e({1,3}, x) = v({1,3}) − (x1 + x3) = 100 − (x1 + x3) ≤ M
e({1}, x) = v({1}) − x1 = 0 − x1 ≤ M
e({2}, x) = v({2}) − x2 = 0 − x2 ≤ M
e({3}, x) = v({3}) − x3 = 0 − x3 ≤ M
s . t .
x1 + x2 + x3 = 100
x1, x2, x3 ≥ 0
(1)
(2)
(3)
(4)
(5)
(6)

21.
min M
0 ≤ x3 ≤ M + 70
0 ≤ x1 ≤ M + 40
0 ≤ x2 ≤ M + 0
s . t .
0 ≤ (x1 + x2 + x3) = 100 ≤ 3M + 110
⇔ −
10
3
≤ M
⇔ − 70 ≤ M
⇔ − 40 ≤ M
⇔ 0 ≤ M
x2 = 0 30 ≤ x1 ≤ 40 60 ≤ x3 ≤ 70 x1 + x3 = 100
30 − (x1 + x2) ≤ 0 ⇔ 30 ≤ x1

22.
min M′
e({1,2}, x) = v({1,2}) − (x1 + x2) = 30 − x1 ≤ M′
e({2,3}, x) = v({2,3}) − (x2 + x3) = 60 − x3 ≤ M′
e({1,3}, x) = v({1,3}) − (x1 + x3) = 0
e({1}, x) = v({1}) − x1 = 0 − x1 ≤ M′
e({2}, x) = v({2}) − x2 = 0 − x2 = 0
e({3}, x) = v({3}) − x3 = 0 − x3 ≤ M′
s . t .
x1 + x3 = 100
x1, x3 ≥ 0
(1)′
(2)′
(3)′
(4)′
(5)′
(6)′
x2 = 0
30 ≤ x1 ≤ 40, 60 ≤ x3 ≤ 70

23.
min M′
s . t .
(x1, x2, x3) = (35, 0, 65)
30 − x1 ≤ M′
60 − x3 ≤ M′ ⇔ x1 − 40 ≤ M′ x1 + x3 = 100
−x1 ≤ M′
−x3 ≤ M′ ⇔ x1 − 100 ≤ M′ x1 + x3 = 100
30 ≤ x1 ≤ 40
30 − M′ ≤ x1 ≤ M′+ 40 ⇔ − 5 ≤ M′
30 − M′ M′+ 400 M′+ 100
4030
x1
−M′

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