Tag: Allocation

Banker’s algorithm – additional allocation

I have a problem with the banker’s algorithm. There are 6 processes and one type of resources. The allocation is (0, 29, 35, 10, 25, 35) and the MAX need is (50, 80, 50, 25, 40, 90). The available resources are equal to D. The first question is what is the minimum value for D as the state to be safe. I have managed to solve this and the value is 15. But the next question is: To how many processes can be allocated additional 5 resources to keep the state safe (multiple situations)? I am not sure I understand this. It means that I need to add 5 to the allocation array or both allocation array and MAX need?

Let $V_1$ be the variance of the estimated mean from a stratified random sample of size $n$ with proportional allocation.

Let $ V_1$ be the variance of the estimated mean from a stratified random sample of size $ n$ with proportional allocation. Assume that the strata sizes are such that the allocations are all integers.Let $ V_2$ be the variance of the estimated mean from a simple random sample of size $ n$ . Show that the ratio $ \frac{V1}{V2}$ is independent of $ n$ .

We define the estimator $ \bar y_{st}=\sum_{h}w_h \bar{y}_h$ , where $ w_h=\frac{N_h}{N}$ . We know this is an unbiased estimator of the population mean. For proportional allocation, $ n_h=nw_h $

$ V(\bar y_{st})=V_1=\frac{1}{n}\sum w_h \sigma^2_h$ (If we assume SRSWR is applied to sample from each strata)

Again, $ V_2=V(\bar y)=\frac{\sigma^2}{n}$

Now, $ \frac{V_1}{V_2}=\sum w_h \frac{\sigma^2_h}{\sigma^2}=\frac{1}{N}-\sum \frac{w_h(\bar Y_h- \bar Y)^2}{\sigma^2}$ Now, how can I proceed from here?