14

It holds $$ \begin{array}{rcl} \operatorname V(x) &= &\dfrac1N\left\| x-\dfrac{e^\top x}{N} e \right\|^2 \\ & = & \dfrac1N\left(x^\top x+\dfrac{(e^\top x)^2 e^\top e}{N^2}-2\dfrac{(e^\top x)^2}N\right) \\ & = & \dfrac{x^\top x}{N} - \dfrac{(e^\top x)^2}{N^2}. \end{array} $$ So you are minimizing the $\ell^2$-norm of an ...


12

For strictly increasing CDFs, you can invert: $$x \le \Phi^{-1}(b)$$


12

I had the same doubt, and I arrived at the conclusion that the formula given in the textbooks is, at best, a practical approximation. The lead-time demand, in fact, is not normally distributed. Let $L$ denote the lead-time and $d$ denote the demand per unit of time. Working under the assumption that both of them are normally distributed, then the random ...


12

In reference to the first question, I think it often comes down to the information you have about the underlying uncertainty. If you only have intervals or ranges, robust is the way to go. If you have all of the distributional information (or assume it), stochastic programming is an option. As @TheSimpliFire mentioned, you can include risk measures in ...


10

There is indeed a paper titled Loss Distributions that provides the limited expected value functions $L(x)$ for several probability distributions (on page 15). It is directly related to the first-order loss function $n(x)$ through $$n(x)=\Bbb E(X)-L(x)\tag1$$ and notice that the loss function can also be written as $$n(x)=\int_x^\infty yf(y)\,dy-x(1-F(x))\...


10

The following papers discuss this extensively with numerical experiments, but they tackle specific examples. Emphasis is mine. Kazamzadeh et al. (2017) This is a comparison of the two techniques using the example of unit commitment, answering your first question. A popular impression has arisen that the robust approach, with its focus on the worst case, is ...


9

Let $a = \hat a$ and $b = \hat b$. Denote $r = \frac{\hat b - \hat m}{\hat m - \hat a}$. Then choose the shape parameters to be $\alpha_1 = \frac{4 + 3r + r^2}{1 + r^2}$ and $\alpha_2 = \frac{1 + 3r + 4r^2}{1 + r^2}$. This will produce a Generalized Beta distribution with mode $m = \hat m$ and a variance $\text{Var}(X) = \left(\frac{\hat b - \hat a}{...


9

Regarding your first question, I think other answers have summed it up pretty good. Two things I would add are as follows: Stochastic programming models (besides chance constraint/probabilistic programming ones) allow you to correct your decision using the concept of recourse. In this idea, you have to make some decisions before the realization of uncertain ...


6

Not directly answering your question of how to code it manually but for discrete simulation of queues in R I would strongly recommend the simmer package. The minimal code for your example would look like this (adapted from the tutorial). library(simmer) library(simmer.plot) lambda <- 2 queue <- trajectory() %>% seize("server", amount=1) %>% ...


6

To calculate the base-stock to meet a 99% type-1 service level, we need the 0.99 fractile of the demand distribution. The safety stock level is the base-stock level minus the mean demand. For the lognormal case, the author has fit a lognormal distribution and found that the parameters are $\mu=2.645$ and $\sigma=0.83255$. (Note that for a lognormal ...


5

In order to find the best upper bound for variance, for given input values of $u_i$ and $\sigma_i^2$, you should globally maximize variance with respect to the $w_i$, subject to the constraints $w_i \ge 0, \Sigma w_i = 1$. This can be formulated as a convex QP (Quadratic Programming problem), i.e., maximizing a concave quadratic subject to linear constraints....


4

After having read Chapter 5.3 of Decision Making Under Uncertainty by Mykel J. Kochenderfer, I have come to some conclusions. We are dealing with model uncertainty, in which case we can formulate a Bayes Adaptive Model. In the book that I read, the term model uncertainty refers more to not knowing what the transition probabilities nor the structure of the ...


4

I tried simulating lots of normally distributed lead times and the normally distributed demand in each. The lead time demand sure looks normal: But a normality test gives $p = 0$ to at least 9 decimal places. Edit: Since the null hypothesis for this test is “the data comes from a normal distribution,” this means we can conclude the lead-time demands are not ...


3

For a Poisson process the rate of events is constant. The distribution of time between events in the Poisson process is exponential with $F(v)=1-e^{-\lambda v}$ for $v\ge 0$ which gives the hazard rate $\lambda$. So a non-constant hazard rate can be seen as a way of comparing with a Poisson process, an increasing hazard rate means the events come faster and ...


3

Cycle stock is used to meet the mean demand. Safety stock is used to protect against randomness in demand. So, I would use your historical data to calculate the mean demand over time and the inventory level over time. The mean demand will equal the cycle stock and any excess is the safety stock. You’ll have to decide how to calculate the averages — moving ...


3

I've thought about this for a bit, and I now believe that leadtime demand in most common situations is not normally distributed, although it may be as usual a good approximation. Of course, we know that the normal distribution has infinite tails which means you could argue that it is not ever appropriate for non-negative random variables, like demand. This ...


3

This is a $M/M/1$ queue with Poisson arrival distribution with $\lambda =1/10$ and Exponential service distribution with $\mu=1/3$. The proportion of the time that the system is busy can be calculated by using the following equation: $$\rho=\frac{\lambda}{\mu} = \frac{1/10}{1/3}=0.3$$ which is exactly the proportion of time that a person arriving at the ...


2

Stochastic Optimization (SO) requires the probability distributions (PDF) of the uncertain variables which are usually hard to fit. Then, a large number of scenarios are required to be sampled from these PDFs with their probabilities. This makes some computational complexities and intractability so, scenario reductions are needed but some information will be ...


2

I have done extensive analysis of procurement lead time distribution across industries. In my experience, I found most of the distribution are heavily skewed towards the left. Yes, you are right! Most of the books talk about mathematical modeling by considering lead time as normally distributed. I have yet to see mathematical models that leverage actual ...


2

I'm still not sure I understand the question, but I'll suggest an answer to what I think is being asked. I'm going to assume that an a priori upper bound $O$ exists for $o_t$. To keep what follows somewhat compact, I'm going to assume that $\mathcal{M} = \lbrace 20, 21, 22, 23\rbrace$ and $O=6$. To start, for each $t\in T$ generate a random sample of $O$ ...


1

My professor helped me for solving this error like below: clc clear cov_x=[2.5 0.1 0.2;0.1 0.4 0.3;0.2 0.3 0.9]; E_x=[10 6 5]; [V, landa]=eig(cov_x); E_w=V'*E_x'; var_w= landa; W=repmat(E_w,1,1000)+sqrt(var_w)*(randn(3,1000)); X=V*W; A= X(1,:); B= X(2,:); C=X(3,:); %D= A'*B.^2*C.^3'; for k=1:1000 D(:,k)=A(:,k)'*B(:,k).^2*C(:,k).^3; end %now we use ...


1

So far this is what I have come up with lambda <- 2 interarrivals <- rexp(5000,lambda) ## (2 items per minute) Provided the $\mu$ we expect that the interarrivals is about half a minute mean(interarrivals) <- 0.516 service.times <- rnorm(5000, mean=8,sd=1) where the service distribution is $N(8,1)$ arrival.times <- cumsum(interarrivals) ...


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