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I have

lensp=21; rho=0.975; e1=0; pid= 0.5212; qid=0.4788; tid=10;

based on my own calculations I have given vectors of dimension lenspx1

prf1[stsp]={2021.8180858663386, 2011.9930631686764, 1999.9243976405476, 
1985.0997308573862, 1966.88970219104, 1944.5212301788522, 
1917.0446924244275, 1883.2936106983718, 1841.8351297259853, 
1790.9091873076468, 1728.3537933238854, 1651.5132454491252, 
1557.125384999776, 1441.1831065203673, 1298.7642416896465, 
1123.8225955042703, 908.9312634716356, 644.9673326999568, 
320.7245813085048, -77.56226615034835, -566.8019243322452}

enpvc={22721.10462499856, 22610.691442815092, 22475.064299074696, 
22308.46533185524, 22103.821813506762, 21852.44588782723, 
21543.665739200012, 21164.372535491235, 20698.463910797407, 
20126.160361876526, 19423.1655368233, 18559.6347672108, 
17498.90805416407, 16195.953719050109, 14595.456646136767, 
12629.469956219407, 10214.53041628219, 7248.115123919593, 
3604.289039979013, -871.6414085290527, -6369.695629112519};

tenpvc = TableForm[enpvc];

enpva={1729.1679511432503, 1632.8636451934199, 1514.5672815927992, 
1369.2567526075109, 1190.763101650733, 971.5086288031089, 
702.1851900925085, 371.3590331761461, -35.01460672697431, 
-534.1877578084313, -1147.3521225541263, -1900.5387420256466, 
-2825.7231075362233, -3962.1826357017644, -5358.164136584017, 
-7072.932064968695, -9179.284510440477, -11766.643739117186, 
-14944.852491698064, -18848.837204384356, -23644.336123236586};

I am then creating a matrix of zeros and replacing the elements of the matrix as

newp = Table[e1, {i, lensp}, {j, lensp}]; 

Table[newp[[i, Min[i + 1, lensp]]] = pid, {i, lensp}]; 
Table[newp[[i, Max[i - 1, 1]]] = qid, {i, lensp}];

I am then creating another matrix "val" and replacing its last column with the vector enpvc

val = Table[e1, {i, lensp}, {j, tid + 1}]; 

val[[All, tid + 1]] = tenpvc; 

The goal of the problem is, starting backwards, you take the Max of (enpva, prf1[stsp]+ rho* newp.val(last column)). You are comparing element for element between the 2 vectors and taking the max to get the second-to-last column of "val". Then you do Max of (enpva, prf1[stsp]+ rho* newp.val(second-to-last column)) and get the previous column and so on. You keep replacing until you get the first column of val.

I have tried both the following options

Do[Table[val[[All, i]] = Table[Max[enpva[[j]], (prf1[stsp] + rho*newp.val[[All, i + 1]])[[j]]], {j, 1, lensp}], {i, 1, tid}], {tid}];

Do[Table[val[[All, i]] = Max /@ Transpose[{enpva, (prf1[stsp] +    rho*newp.val[[All, i + 1]])}], {i, 1, tid}], {tid}];

Both take forever to run..and so I cannot proceed with the problem. I have purposely put in the ugly looking vectors of prf1[stsp], enpvc and enpva as these are my real calculations and I am not sure if these long numbers are causing the problem.

If you see lensp=2*tid+1 and with greater tid's (20 or 30) and with even uglier numbers, what is the recourse??

Any help would be greatly appreciated. Thank you.

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This is rather an extended comment than an actual answer. I find your code very hard to read, in particular because you use very complicated syntax to achieve trivial goals. Here some examples:

Creating a matrix with a given constant value in all entries. Your code:

e1=0;
newp = Table[e1, {i, lensp}, {j, lensp}];

Alternative:

newp = ConstantArray[0, {lensp,lensp}];

Creating matrices with given entries on the kth diagonal. Your code:

e1=0;
newp = Table[e1, {i, lensp}, {j, lensp}];
Table[newp[[i, Min[i + 1, lensp]]] = pid, {i, lensp}]; 

Alternative:

newp=DiagonalMatrix[ConstantArray[pid, lensp], 1];
newp[lensp,lensp]=pid; (* are you sure you want this? *)

I couldn't quite follow as to what you want to achieve in the end. However, I suggest to apply the same concept of eliminating the Table and Do loops in the rest of the code.

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  • $\begingroup$ I can change newp your way with the command "Constant Array". Then "pid" and "qid" are probabilities and given the ith row of newp, I am changing the Min of (i+1th,last) column and the Max of (i-1th, first) column. So please keep that part unchanged. Now val=ConstantArray[0,{lensp,tid+1}], with the last column equal to the given vector enpvc. The goal of the problem is, starting backwards, you take the Max of (enpva, rho*newp.val(last column)). You are comparing element for element and taking the max to get the previous column of "val". U keep replacing until you get the first column of val. $\endgroup$ – Supratim Das Gupta Feb 4 '17 at 23:06
  • $\begingroup$ I think it would really help if you illustrate your process step by step with the expected outcome on a toy data set of maybe 3 dimensions instead of 21 and integer values. $\endgroup$ – Felix Feb 4 '17 at 23:16
  • $\begingroup$ Thanks..I had found my mistake. The probblem was with the command val[[All, tid + 1]] = tenpvc; It SHOULD HAVE BEEN enpvc (NOT the Table Form). Now the commands work. $\endgroup$ – Supratim Das Gupta Feb 11 '17 at 17:41

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