翻译:where ri = j∈B xij cij is the receiving rate of UE i, and Ui(ri) reflects the degree of satisfaction for UE i. Different degrees of satisfaction are obtained for different receiving rates. Therefore, the utility function is a monotonically increasing function of the receiving rate. This function can be modeled as either a logarithmic function or a sigmoid function. Notably, different types of UE devices may have different levels of satisfaction, even for the same receiving rate, and one can reasonably assume that the utility functions are different for different types of UE devices. In the simulation section, a specific utility function, such as a logarithmic function, is used. In addition, the proposed algorithm can be easily implemented when UEs have different utility functions.
时间: 2024-04-22 12:26:45 浏览: 12
其中ri = j∈B xij cij 是用户设备 i 的接收速率,而Ui(ri) 反映了用户设备 i 的满意程度。不同的接收速率会获得不同程度的满意度。因此,效用函数是接收速率的单调递增函数。这个函数可以建模为对数函数或S型函数。值得注意的是,不同类型的用户设备可能对相同的接收速率具有不同的满意度,可以合理地假设不同类型的用户设备具有不同的效用函数。在模拟部分,使用了特定的效用函数,比如对数函数。此外,当用户设备具有不同的效用函数时,所提出的算法可以很容易地实现。
相关问题
lingo表示约束条件xij <=xii,i=1,…,6,j=1,…,6
在 Lingo 中,可以使用如下语句表示该约束条件:
```
forall(i in 1..6, j in 1..6: i != j) xij(i,j) <= xii(i,i);
```
其中,`forall` 表示对所有的 `i` 和 `j` 进行约束,`xij(i,j)` 表示决策变量 `x_ij`,`xii(i,i)` 表示决策变量 `x_ii`。`i != j` 表示 `i` 和 `j` 不能相等,因为这个约束条件不适用于 `i=j` 的情况。
投掷一枚均匀硬币 n 次,如果第 i 次投掷和第 j 次投掷出现同一面,则令 Xij=1, 否则令 Xij=0。证明:Xij(i<j)两两独立但不相互独立。
好的,我明白了,您的第一个问题是有关硬币投掷的概率论问题。现在,我将针对这个问题作出解答。
首先,显然,Xij只能取0或1。设X =(Xij)为n×n矩阵,它的对角线上元素都是0。那么X的非对角线上的元素为随机变量,我们需要证明这些随机变量是两两独立的。
我们假设i < p < q < j,并分别考虑Xi p, Xpq和Xqj的相关性。显然,如果Xi p = 0,则Xpq和Xqj必须都是0;如果Xi p = 1,则Xpq和Xqj必须不相等(即一个为0,一个为1)。因此,我们可以得到以下三个条件概率:
P(Xpq = 0 | Xi p = 0) = 1
P(Xqj = 0 | Xpq = 0, Xi p = 0) = 1
P(Xqj ≠ Xpq | Xi p = 1) = 1
由此可以得到:
P(Xpq = 0, Xqj = 0 | Xi p = 0) = P(Xpq = 0 | Xi p = 0) × P(Xqj = 0 | Xpq = 0, Xi p = 0) = 1
P(Xpq = 1, Xqj = 0 | Xi p = 0) = P(Xpq = 1 | Xi p = 0) × P(Xqj = 0 | Xpq = 1, Xi p = 0) = 0
P(Xpq = 0, Xqj = 1 | Xi p = 0) = P(Xpq = 0 | Xi p = 0) × P(Xqj = 1 | Xpq=0, Xi p = 0) = 0
P(Xpq = 1, Xqj = 1 | Xi p = 0) = P(Xpq = 1 | Xi p = 0) × P(Xqj ≠ Xpq | Xi p = 0) = 0
P(Xpq = 0, Xqj = 0 | Xi p = 1) = P(Xpq = 0 | Xi p = 1) × P(Xqj = 0 | Xpq = 0, Xi p = 1) = 1/4
P(Xpq = 1, Xqj = 0 | Xi p = 1) = P(Xpq = 1 | Xi p = 1) × P(Xqj = 0 | Xpq = 1, Xi p = 1) = 1/4
P(Xpq = 0, Xqj = 1 | Xi p = 1) = P(Xpq = 0 | Xi p = 1) × P(Xqj = 1 | Xpq = 0, Xi p = 1) = 1/4
P(Xpq = 1, Xqj = 1 | Xi p = 1) = P(Xpq = 1 | Xi p = 1) × P(Xqj ≠ Xpq | Xi p = 1) = 1/4
从上面的概率可以看出,Xpq和Xqj在已知Xi p的条件下是不相互独立的。但是,当我们去掉Xi p的条件时,Xpq和Xqj是独立的。因此,Xij(i<j)两两独立但不相互独立。
这就是该问题的证明过程。希望我的解答能够满足您的需要。如果您有任何其他问题,欢迎随时提出,我将尽我所能给出解答。