MATLAB: Finding maximum value and corresponding variable

MATLABmatrix

Dear Matlab expert

P(m,d,L) is a matix and size of P(m,d,L) is [10 40 10]. Total element =10*40*10=4000;

Now I need to find maximum value(one maximum value) for P and corresponding m,d,L.

I am expecting your kind help.

Kind regards

Sohel

Best Answer

P = rand(10,40,10); % Some pretend data
[P_max,linear_index_max] = max(P(:)); % Find max and index, for linearized matrix
[mc,dc,Lc] = ind2sub([10 40 10],linear_index_max); % Convert the linear index to subscripts

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MATLAB: Max for intervals in along vector

M=max(reshape(L,[100,10]),[],2);

Best wishes

Torsten.

MATLAB: Can’t separate the two variables for which solution needs to be found

Well, you can reduce everything down to an expression in p and L, generating a 15th degree polynomial if solved for p, or a 12th degree polynomial if solved for L. There are, of course, no analytical roots for either of them.

To make the expression clearer, I will show it collected by p within L, and by L within p:

(-p^12 + (4*p^12-24*p^11+72*p^10-152*p^9+252*p^8-336*p^7+368*p^6-336*p^5+252*p^4-152*p^3+72*p^2-24*p+4)*L^11 + (-12*p^13+24*p^12+24*p^11-132*p^10+264*p^9-384*p^8+384*p^7-264*p^6+132*p^5-24*p^4-24*p^3+12*p^2)*L^10 + p^10 + (12*p^14+60*p^13-216*p^12+240*p^11-260*p^10+292*p^9-132*p^8+16*p^7+56*p^6-188*p^5+188*p^4-96*p^3+32*p^2-4*p)*L^9 + (-4*p^15-96*p^14-24*p^13+376*p^12-348*p^11+383*p^10-407*p^9+36*p^8+116*p^7-46*p^6-22*p^5+40*p^4-5*p^2+p)*L^8 + (36*p^15+252*p^14-240*p^13-120*p^12+39*p^11-138*p^10+276*p^9-312*p^8+194*p^7+48*p^6-52*p^5+16*p^4+3*p^3-2*p^2)*L^7 + (-108*p^15-204*p^14+332*p^13-175*p^12-105*p^11+150*p^10+98*p^9+22*p^8-62*p^7+63*p^6-p^5-12*p^4+2*p^3)*L^6 + (108*p^15-72*p^14-119*p^13+364*p^12-82*p^11-63*p^10+119*p^9+23*p^8-10*p^7-28*p^6+32*p^5-22*p^4+6*p^3)*L^5 + (108*p^14-54*p^13-37*p^12+166*p^11-43*p^10-29*p^9+18*p^8+26*p^7-45*p^6+16*p^5+3*p^4-p^3)*L^4 + (45*p^13-29*p^12-16*p^11+47*p^10-21*p^9-23*p^8+17*p^7+9*p^6-6*p^5+p^4)*L^3 + (p^13+17*p^12-20*p^11+3*p^10+11*p^9+5*p^8-9*p^7+3*p^6)*L^2 + (-3*p^13+3*p^12+3*p^11-2*p^10-4*p^9+3*p^8)*L) / ((p-1)*(p^4+(p^4-2*p^3+2*p^2-2*p+1)*L^3+p^3+(-p^5-2*p^4+2*p^3+p^2)*L^2+(3*p^5+p)*L)*(-p^2+(p-1)*L)^2*p^2)

Or alternately,

((-4*L^8+36*L^7-108*L^6+108*L^5)*p^15 + (12*L^9-96*L^8+252*L^7-204*L^6-72*L^5+108*L^4)*p^14 + (-12*L^10+60*L^9-24*L^8-240*L^7+332*L^6-119*L^5-54*L^4+45*L^3+L^2-3*L)*p^13 + (4*L^11+24*L^10-216*L^9+376*L^8-120*L^7-175*L^6+364*L^5-37*L^4-29*L^3+17*L^2+3*L-1)*p^12 + 4*L^11 + (-24*L^11+24*L^10+240*L^9-348*L^8+39*L^7-105*L^6-82*L^5+166*L^4-16*L^3-20*L^2+3*L)*p^11 + (72*L^11-132*L^10-260*L^9+383*L^8-138*L^7+150*L^6-63*L^5-43*L^4+47*L^3+3*L^2-2*L+1)*p^10 + (-152*L^11+264*L^10+292*L^9-407*L^8+276*L^7+98*L^6+119*L^5-29*L^4-21*L^3+11*L^2-4*L)*p^9 + (252*L^11-384*L^10-132*L^9+36*L^8-312*L^7+22*L^6+23*L^5+18*L^4-23*L^3+5*L^2+3*L)*p^8 + (-336*L^11+384*L^10+16*L^9+116*L^8+194*L^7-62*L^6-10*L^5+26*L^4+17*L^3-9*L^2)*p^7 + (368*L^11-264*L^10+56*L^9-46*L^8+48*L^7+63*L^6-28*L^5-45*L^4+9*L^3+3*L^2)*p^6 + (-336*L^11+132*L^10-188*L^9-22*L^8-52*L^7-L^6+32*L^5+16*L^4-6*L^3)*p^5 + (252*L^11-24*L^10+188*L^9+40*L^8+16*L^7-12*L^6-22*L^5+3*L^4+L^3)*p^4 + (-152*L^11-24*L^10-96*L^9+3*L^7+2*L^6+6*L^5-L^4)*p^3 + (72*L^11+12*L^10+32*L^9-5*L^8-2*L^7)*p^2 + (-24*L^11-4*L^9+L^8)*p) / ((p-1)*((-L^2+3*L)*p^5+(L^3-2*L^2+1)*p^4+L^3+(-2*L^3+2*L^2+1)*p^3+(2*L^3+L^2)*p^2+(-2*L^3+L)*p)*(L*p-p^2-L)^2*p^2)

Not so nice, either way you look at it.

If you examine the first variant, collected by L, and look at the denominator, you can see that the expression blows up for several cases -- p = 0, p = 1, p = L/2 +/- sqrt(L^2-4*L)/2, or p^4+(p^4-2*p^3+2*p^2-2*p+1)*L^3+p^3+(-p^5-2*p^4+2*p^3+p^2)*L^2+(3*p^5+p)*L = 0 which has 3 roots when solved for L.

The third of those conditions can be expressed as L = p^2/(p-1) which is real-valued for all p except p = 1, so no matter how the cubic behaves, there will be at least one non-solution for each p. In fact the cubic has a minimum of 1 real L root for each positive p (two real solutions beyond a point), and a minimum of two real L roots for each negative p. I believe that the cubic has three real roots for each negative p, but some fancy algebraic mangling would be necessary to prove that those vanishingly small imaginary components are necessarily zero.

Anyhow, for each p there are thus between 1 and 4 L for which x1 and x2 have no solution.

Best Answer

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