From Proposition 1, we plug ri,O = li(μ)τi into (39) and rewrite problem (38) as maximize ri,O 􏰗ai − μ li (μ) − Yi(t)g [li(μ)]􏰘 ri,O li (μ)hi (41a) March 2, 2021 DRAFT maximize ˆr O subject to 0 ≤ ri,O ≤ Qi(t), (41b) 0, ifa − μ −Y(t)g[li(μ)] <0, subject to where the optimal solution is r∗ i,O Accordingly, we have τ∗ = r∗ ii,Oi i1 of μ in (32) as 1−􏰀i∈M1 τi∗. Then, we obtain the optimal dual variable μ∗ through the ellipsoid method (bi-section search in this case) over the range [0,∆], where ∆ is a sufficiently large value, until a prescribed precision requirement is met. Given the optimal μ∗, we denote the optimal ratio obtained from (40) as li (μ∗) 􏰝 r∗ /τ∗, i,O i ∀i ∈ M1. Notice that the optimal solution 􏰕τi∗, r∗ , ∀i ∈ M1􏰖 of the dual problem may not be i,O primal feasible. Therefore, to find a primal optimal solution to (31), we substitute τi = ri,O/li (μ∗) into (31) and simplify the problem as = i li(μ) i li(μ)hi (42) otherwise. Qi (t), /l (μ). After obtaining τ∗, ∀i ∈ M , we calculate the subgradient 􏰁 􏰗ai − Yi(t)g [li(μ∗)]􏰘 ri,O (43a) i ∈ M 1 h i l i ( μ ∗ ) 􏰁 ri,O ≤ 1, ri,O ≤ Qi(t), ∀i ∈ M1. (43b) i∈M1 li(μ∗) The above problem is a simple linear programming (LP) that can be easily solved. With a bit abuse of notation, we denote the optimal solution of (43) as ˆr∗ = 􏰕r∗ , ∀i ∈ M 􏰖 and retrieve 20 the optimal solution to (31) as τ∗=r∗ /l(μ∗),e∗ =τi∗g[li(μ∗)],∀i∈M. (44) i i,O i i,O hili(μ∗) 1 Denote τˆ∗ = {τi∗,∀i ∈ M1} and ˆe∗O = 􏰕e∗i,O,∀i ∈ M1􏰖. As {τˆ∗,ˆe∗O,ˆr∗O,μ∗} satisfies the KKT conditions, {τˆ∗,ˆe∗O,ˆr∗O} is an optimal solution to (31). By combining the optimal solutions in (30) and (44), we obtain an optimal solution of (P4). We summarize the pseudo-code of the O i,O 1 algorithm to solve (P4) in Algorithm 2.,翻译并解释li和hi是什么

在这段文字中，li和hi分别代表优化问题中的变量。li是一个非负数，它表示第i个用户在当前时刻能够分配的最大带宽。hi是一个非负数，它表示第i个用户的利润，即用户在当前时刻通过分配带宽所能获得的收益。

在算法中，我们需要通过求解一个线性规划问题来确定每个用户在当前时刻的最优带宽分配。通过计算每个用户的利润和最大可分配带宽，我们可以获得线性规划问题的约束条件。然后，我们采用一种叫做椭球法的优化算法来解决对偶问题，并计算出对应的最优解。最后，通过将最优解代入原问题中，我们可以得到每个用户在当前时刻的最优带宽分配方案。

相关问题

Recall that a clause is of the form (H1 ∨ H2 ∨ · · · ∨ Hk) ← (B1 ∧ B2 ∧ · · · ∧ B` ) for literals B1, . . . , B` , H1, . . . , Hk, k ≥ 1, and ` ≥ 0. Are these propositions clauses? If not, convert them into equivalent clause form, i.e., for proposition p, construct a set S of clauses such that any interpretation π satisfies p if and only if π satisfies S: (a) A ∧ B (b) A ∨ B (c) (A ∧ ¬B) ∨ (¬A ∧ B) (d) ¬((A → B) ∧ (C → ¬B)) Answer. (a) A ∧ B is not a clause. S = {A, B} (b) A ∨ B is a clause (c) (A ∧ ¬B) ∨ (¬A ∧ B) is not a clause. S = {A ← ¬B, ¬B ← A} (d) ¬((A → B) ∧ (C → ¬B)) is not a clause. S = {A ← ¬C, C ← B, A ← ¬B}中文解释

这道题目要求我们判断一些命题是否为子句（clause）形式，如果不是，则需要将其转化为等价的子句形式。其中，子句形式的定义为 (H1 ∨ H2 ∨ · · · ∨ Hk) ← (B1 ∧ B2 ∧ · · · ∧ B` )，其中 B1, . . . , B` , H1, . . . , Hk 为文字（literal），且 k ≥ 1，` ≥ 0。

以下是每个命题的答案及解释：

(a) A ∧ B 不是子句。为了将其转化为等价的子句形式，我们可以创建两个子句如下：S = {A, B}。

(b) A ∨ B 是子句。

(c) (A ∧ ¬B) ∨ (¬A ∧ B) 不是子句。为了将其转化为等价的子句形式，我们可以创建两个子句如下：S = {A ← ¬B, ¬B ← A}。

(d) ¬((A → B) ∧ (C → ¬B)) 不是子句。为了将其转化为等价的子句形式，我们可以创建三个子句如下：S = {A ← ¬C, C ← B, A ← ¬B}。

一般来说，我们可以使用上述转换方法将任何逻辑命题转化为等价的子句形式。在许多逻辑应用中，如自动推理和定理证明，这种转换方法非常有用。

value proposition

A value proposition is a statement that explains what makes a product or service unique and desirable to potential customers. It highlights the benefits and value that the product or service provides to customers and sets it apart from its competitors. A good value proposition should be clear, concise, and compelling, and it should address the customer's pain points and needs. It is an important element of a company's marketing strategy as it helps attract and retain customers.