I need a code to solve this problem

The budget to be shared is 40.5; The participant initially gets the following amount

v1= 7;
v2=10.5
v3= 13
v(1,2)= 28
v(1,3)=28
v(2,3)=28
v(1,2,3)=40.5

v1,v2 and v3 should get an amount greater 7, 10.5,13 respectively and the total should be equal to 40.5

At the same time the this conditions should be satisfied as well v1+v2 >=28 ; v1+v3 >=28 ; v2+v3 >=28

[Azzi Abdelmalek]

What is v(1,2)?

[Jeffrey Eiyike]

v (1,2) is the minimum that v1+v2 should get and similar thing goes for v (1,3) and v (2,3). But they should meet the conditions should be satisfied as well v1+v2 >=28 ; v1+v3 >=28 ; v2+v3 >=28

[Shameer Parmar]

@Jeffrey,

AS you mentioned V1=7, v2=10.5, v3=13, so how it will be v1+v2>=28 v1+v2 is coming as 17.5 only..

Please elaborate in details what you want?

[Jeffrey Eiyike]

I want 40.5 to be share among three persons x1, x2 and x3 such that x1 get a value larger than 7, x2 get a value larger than 10.5 and x3 gets a value larger than 13. While sharing the 40.5 among x1, x2 and x3, the values should also satisfy x1+x2 >= 28 , x1+x3>=28 and x2+x3>=28

[Walter Roberson]

That is not possible.

40.5 - 7 - 10.5 - 13 = 10, so you have 10 units beyond the minimums to distribute among the three items. The first two items have minimums of 7 and 10.5 so their minimum sum is 17.5. If you were to distribute all 10 of the available units between the two of them, then the total would reach 27.5. But you require that the total of the two be at least 28. You would need another 0.5 available units to make that possible, even without considering any increase in value that might be needed for x3 to satisfy the other conditions.

[Shameer Parmar]

Yes, Its seems difficult, because as per your inputs, if we provide x1=8 (because xx need larger than 7), then x2 will be 20 (because x2 is larger than 10.5, also x1+x2 >= 28), so now we left with 40.5-28=12.5 for x3, but see here, it is not satisfying the last condition i.e. x3 larger than 13.

It is not possible..

[Stefan Reich]

Take a look at

fmincon

http://de.mathworks.com/help/optim/ug/fmincon.html

Answer 1

[Walter Roberson]

By inspection we can see that under the pairwise constraints on v1, v2, v3, that the smallest values that would satisfy those would be v1 = 14, v2 = 14, v3 = 14, in which case the total would be 42. There is no possible solution to those pairwise constraints when the total must be 42.

If the total must be 40.5 then the least reduction in the pairwise constraints that could be satisfied would be v1 = 13.5, v2 = 13.5, v3 = 13.5, total 40.5, and the pairwise constraints would be 27 (instead of 28)