The document defines a set function v over subsets of a set N={1,2,3} that assigns values to singleton and composite sets. It then defines a vector x=(x1,x2,x3) and constrains the values of x to be consistent with v through linear inequalities and equations. It solves for the values of x1, x2, x3 that satisfy all the constraints, finding the solution x1=20, x2=60, x3=50. The document also discusses some properties and definitions related to set functions.

2. “
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4.
v({1,2,3}) = 100

5.
v({1,2}) = 50

6.
v({2,3}) = 80

7.
v({1,3}) = 40

8.
v({1}) = 0
v({2}) = 0
v({3}) = 0

9.
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
23
− 1 = 7
v
(N, v)
v

11.
S ⊃ T
(N, v)
v
v
v
v
v(S ∪ T) ≥ v(S) + v(T)
v(S) ≥ v(T)
v(N) >
n
∑
i=1
v(i)
v(N) = v(S) + v(NS)
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

12.
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
x = (x1, x2, x3)
xi ≥ v(i)
3
∑
i=1
xi = v({1,2,3})

13.
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
x = (x1, x2, x3)
xi ≥ v(i)
3
∑
i=1
xi = v({1,2,3})
v({1,2,3})
x
x3
x1
x2

14.
x
x3
x1
x2
(N, v) x, y
v(S) ≥
∑
i∈S
x(i)
xi > yi, ∀i ∈ S
x y
(N, v)

15.
x
x3
x1
x2
(N, v)
∑
i∈S
x(i) ≥ v(S), ∀S ⊂ N

16.
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
x = (x1, x2, x3)
x1 + x2 ≥ 50
x2 + x3 ≥ 80
x1 + x3 ≥ 40
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 + x3 = 100
x1 = 20
⇒ 20 ≥ x1

17.
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
x = (x1, x2, x3)
x1 + x2 ≥ 50
x2 + x3 ≥ 80
x1 + x3 ≥ 40
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 + x3 = 100
x1 = 20
⇒ 60 ≥ x2
x2 = 60

18.
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
x = (x1, x2, x3)
x3
x1 + x2 ≥ 50
x2 + x3 ≥ 80
x1 + x3 ≥ 40
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 + x3 = 100
x1 = 20
⇒ 50 ≥ x3
x2 = 60
x3 = 50

19.
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
x = (x1, x2, x3)
x3
x1 + x2 ≥ 50
x2 + x3 ≥ 80
x1 + x3 ≥ 40
x1 ≥ 0
x2 ≥ 0
x3 ≥ 0
x1 + x2 + x3 = 100
x1 = 20
x2 = 60
x3 = 50

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

21.
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

22.
(N, v) (N, v′)
βi(i = 1,2,⋯, n)
v′ = αv(S) +
∑
i∈S
βi
(N, v)
v′(i) = 0, i = 1,⋯, n
(N, v′)
v′(N) = 1

23.
(N, v)
v({1,2}) + v({2,3}) + v({1,3}) ≤ 2v({1,2,3})

24.
(N, v)
min
∑
i∈N
xi s . t .
∑
i∈S
xi ≥ v(S), ∀S ⊊ N
z * z* ≤ v(N)

27. See you next time