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General idea: Lets say there's a finite supply (S) of resources which should be allocated to two tasks based on their demands, D_{1}and D_{2}.If Supply is larger than total demand (D), each task simply gets whatever it demands for:

If S=1000 and D=400 (D

_{1}=100, D_{2}=300)Then 100 units are assigned to task one (let's call it utilization 1, so U

_{1}=100) and 300 units to task two (U_{2}=300) and the remaining 600 units are unused.However, S is often less than D. In that case, each task receives a proportion of S, such as:

S=500 and D=1000 (D

_{1}=200, D_{2}=800)Then: U

_{1}=D_{1}/D*S and U_{2}=D_{2}/D*STo make sure that the tasks do not receive more than what they demand for:

U

_{1}=min(D_{1}, D_{1}/D_{2}*S) and U_{2}=min(D_{2}, D_{2}/D*S)And let's call this as the

base case.Problem: I'd like to compare this simple allocation to a a more general allocation formulation that changes the priority of allocation based on α (a continuum that can vary between 0 and 1).α=0 represents full priority to task one (so only resources remaining after meeting the demand for task one is spent on task two) and α=1 represents full priority to task two, and the

base caseresults are obtained for α=0.5. To be more specific, here is an example:D

_{1}=100, D_{2}=900, and S=200Three scenarios:

α=0: Full priority to task one

So, U

_{1}=100, then S-U_{1}goes to task 2, so U_{2}=100α=0.5: Base case

So, U

_{1}=20 and U2=180α=1: Full priority to task two

So, U

_{2}=200 and nothing is left for task one, so U_{1}=0I'd like to formulate this (calculation of U

_{1}and U_{2}) without IF THEN ELSE and lots of Min and Max, just with mathematical equation(s). I already have something which may be too complex so I would like to see what others think of this formulation.Thanks!