1 | function test_gp_lakshminarayanan |
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2 | |
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3 | % Architectural parameters |
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4 | beta = 4.03 * 10 ^ -4; |
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5 | alpha = 2.6 * 10 ^ -7; |
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6 | tau = 2.02 * 10 ^ -7; |
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7 | |
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8 | % Number of processors |
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9 | Procs = 5; |
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10 | |
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11 | % ------------- End of Parameters (constants) ----------------------- |
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12 | |
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13 | obj3d = 1:5; |
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14 | obj2d = 1:5; |
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15 | objstripMP = 1:5; |
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16 | objstrip1P = 1:5; |
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17 | N = 1:5; |
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18 | |
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19 | MP_tis = 1:5; |
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20 | |
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21 | v2d_tis = 1:5; |
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22 | v2d_tjs = 1:5; |
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23 | |
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24 | v3d_tis = 1:5; |
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25 | v3d_tjs = 1:5; |
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26 | v3d_tks = 1:5; |
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27 | |
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28 | N_i = 1000; |
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29 | N_j = 1000; |
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30 | j = 1; |
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31 | for n = N_i/10: N_i/10 : N_i |
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32 | Nk(j) = n; |
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33 | j = j + 1; |
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34 | end; |
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35 | for n = 2*N_i : N_i : 10*N_i |
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36 | Nk(j) = n; |
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37 | j = j + 1; |
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38 | end; |
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39 | |
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40 | fprintf('Iteration '); |
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41 | for i = 1:5 |
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42 | |
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43 | N_k = Nk(i); |
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44 | |
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45 | % Call the 3D model |
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46 | [obj3d(i), v3d_tis(i), v3d_tjs(i), v3d_tks(i)] = f3D(alpha,beta,tau,Procs,N_i,N_j,N_k); |
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47 | |
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48 | % Call the 2D model |
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49 | [obj2d(i), v2d_tis(i), v2d_tjs(i)] = f2D_SemiOblique_PerPlane(alpha,beta,tau,Procs,N_i,N_j,N_k); |
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50 | |
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51 | % Call the Strip Baseline Multi Pass model |
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52 | [objstripMP(i), MP_tis(i)] = fStrip_MP(alpha,beta,tau,Procs,N_i,N_j,N_k); |
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53 | |
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54 | % Call the Strip Baseline One Pass model |
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55 | objstrip1P(i) = fStrip_1P(alpha,beta,tau,Procs,N_i,N_j,N_k); |
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56 | |
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57 | % Save N for plots |
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58 | N(i) = Nk(i); |
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59 | % print current iteration. |
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60 | fprintf('%g ',i); |
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61 | if mod(i,25) == 0 |
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62 | fprintf('\n'); |
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63 | end; |
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64 | end |
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65 | mbg_asserttolequal(norm(obj3d) + norm(obj2d) + norm(objstripMP) + norm(objstrip1P),1.742627952792674e+002,1e-5); |
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66 | |
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67 | function [obj_val, ti_val, tj_val, tk_val] = f3D( alpha, beta, tau, Procs, Ni, Nj, Nk) |
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68 | % Computes the optimal tile sizes given architectural parameters and domain sizes. |
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69 | % |
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70 | % INPUT ARGUMENTS: (alpha, beta, tau, Procs, Ni, Nj, Nk) |
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71 | % alpha -- time to compute an iteration |
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72 | % beta -- time to transfer a word |
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73 | % tau -- startup cost |
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74 | % Procs -- number of processors |
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75 | % Ni -- size of domain along dimension i |
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76 | % Nj -- size of domain along dimension j |
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77 | % Nk -- size of domain along dimension k |
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78 | % |
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79 | % RETURN VALUES: [obj_val, ti_val, tj_val, tk_val] |
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80 | % obj_val -- Value of objective function at the optimizer |
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81 | % ti_val, tj_val, tk_val -- the optimal tile sizes. |
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82 | |
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83 | |
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84 | % Notes |
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85 | % ----- |
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86 | % Linear processor array allocation |
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87 | % Tile and skew along k |
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88 | % projection along j for every plane |
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89 | % do the planes sequentially |
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90 | |
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91 | |
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92 | % ------------- Tile Variables ---------------------------- |
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93 | |
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94 | % ti, tj, and tk are tile sizes |
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95 | ti = sdpvar(1,1); |
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96 | tj = sdpvar(1,1); |
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97 | tk = sdpvar(1,1); |
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98 | |
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99 | % ------------- Execution Time Model ---------------------------- |
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100 | |
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101 | |
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102 | % Latency count |
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103 | Lambda = (1+ ti/tj) * (Procs-1); |
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104 | |
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105 | % Time for computing one tile |
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106 | TilePeriod = (alpha * ti * tj * tk) + (2 * tau * tj * tk) + ( 2 * beta) ; |
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107 | |
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108 | % Number of planes |
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109 | no_planes = Nk / tk ; |
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110 | |
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111 | % Number of passes per plane |
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112 | no_passes_per_plane = (1/Procs) * ( (Ni + tk) / ti ); |
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113 | |
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114 | % Number of tiles per pass per plane -- macro column |
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115 | no_tiles_per_pass_per_plane = (Nj + ti + 2 * tk) / tj; |
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116 | |
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117 | % Total running time |
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118 | T = ( Lambda + ( no_planes * no_passes_per_plane * no_tiles_per_pass_per_plane ) ) * TilePeriod ; |
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119 | |
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120 | % ------------- End of Execution Time Model ---------------------------- |
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121 | |
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122 | |
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123 | % ------------- Constraints ---------------------------- |
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124 | |
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125 | % Lower bounds on the tile vars |
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126 | F = set(ti > 1) ; |
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127 | F = F + set(tj > 1) ; |
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128 | F = F + set(tk > 1) ; |
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129 | |
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130 | % Upper bounds on the tile vars |
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131 | F = F + set(ti < ( Ni / Procs)) ; |
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132 | F = F + set(tj < Nj) ; |
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133 | % Number of tiles per macro column is atleast 2 |
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134 | % F = F + set( (Nj+ti+2tk) < 2*tj); |
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135 | F = F + set(tk < Nk) ; |
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136 | |
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137 | % No idle time between passes |
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138 | %F = F + set ( ((Procs-1)/Nj) * (ti+tj) <= 1) |
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139 | %F = F + set ( ((Procs-1)/ (2*Nj)) * (ti+tj) < 1) |
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140 | %F = F + set ( (Procs-1) * ( ti + tj ) * (1/(2*Nj)) < 1 ); |
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141 | F = F + set( (1/(Nj + 2*Nk) ) * ( (Procs-2) * ti + (Procs-1) * tj ) <= 1); |
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142 | % Not sure this is correct |
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143 | %F = F + set( tj <= ( Nj / (Procs-1) ) ) ; |
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144 | |
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145 | % solve an integer version of the problem |
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146 | intConstraints = set (integer(ti)) + set(integer(tj)) + set(integer(tk)); |
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147 | F = F + intConstraints; |
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148 | |
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149 | % + set( (Procs-1) * tj <= Nj + 2*tk) ; |
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150 | % No idle time between passes |
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151 | |
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152 | % ------------- End of Constraints ------------------------- |
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153 | |
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154 | %%%%%%% Obective function |
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155 | obj = T; |
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156 | |
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157 | %%%%% Solve the optimization problem |
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158 | |
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159 | solvesdp(F,obj, sdpsettings('verbose',0,'solver','bnb','debug',1)); |
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160 | |
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161 | %%%%% Print the solutions |
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162 | |
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163 | |
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164 | % Return values |
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165 | obj_val = double(obj); |
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166 | ti_val = double(ti); |
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167 | tj_val = double(tj); |
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168 | tk_val = double(tk); |
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169 | |
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170 | |
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171 | function [obj_val, ti_val, tj_val] = f2D_SemiOblique_PerPlane( alpha, beta, tau, Procs, Ni, Nj, Nk) |
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172 | % Computes the optimal tile sizes given architectural parameters and domain sizes. |
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173 | % |
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174 | % INPUT ARGUMENTS: (alpha, beta, tau, Procs, Ni, Nj, Nk) |
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175 | % alpha -- time to compute an iteration |
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176 | % beta -- time to transfer a word |
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177 | % tau -- startup cost |
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178 | % Procs -- number of processors |
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179 | % Ni -- size of domain along dimension i |
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180 | % Nj -- size of domain along dimension j |
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181 | % Nk -- size of domain along dimension k |
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182 | % |
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183 | % RETURN VALUES: [obj_val, ti_val, tj_val] |
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184 | % obj_val -- Value of objective function at the optimizer |
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185 | % ti_val, tj_val -- the optimal tile sizes. |
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186 | |
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187 | |
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188 | % Notes |
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189 | % ----- |
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190 | % In this model we skew the ij plane and do not skew or tile the |
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191 | % k dimension. We solve for ti and tj. |
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192 | |
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193 | % ------------- Tile Variables ---------------------------- |
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194 | |
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195 | % ti and tj are tile sizes |
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196 | ti = sdpvar(1,1); |
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197 | tj = sdpvar(1,1); |
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198 | |
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199 | % ------------- Execution Time Model ---------------------------- |
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200 | |
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201 | |
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202 | % Latency count |
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203 | Lambda = (1 + (ti/tj)) * (Procs-1); |
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204 | |
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205 | % Time for computing one tile |
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206 | TilePeriod = (alpha * ti * tj) + (2 * tau * tj) + (2 * beta) ; |
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207 | |
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208 | % Number of planes |
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209 | no_planes = Nk; |
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210 | |
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211 | % Number of passes per plane |
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212 | no_passes_per_plane = (1/Procs) * ( Ni / ti ); |
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213 | |
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214 | % Number of tiles per pass per plane -- macro column |
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215 | no_tiles_per_pass_per_plane = (Nj + ti) / tj; |
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216 | |
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217 | % Total running time |
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218 | T = ( Lambda + ( no_planes * no_passes_per_plane * no_tiles_per_pass_per_plane ) ) * TilePeriod ; |
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219 | |
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220 | % ------------- End of Execution Time Model ---------------------------- |
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221 | |
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222 | |
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223 | % ------------- Constraints ---------------------------- |
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224 | |
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225 | % Lower bounds on the tile vars |
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226 | F = set(ti > 1) ; |
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227 | |
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228 | % To avoid idle time between execution of successive planes |
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229 | % we need tj >= 2. |
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230 | F = F + set(tj >= 2 ) ; |
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231 | |
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232 | % Upper bounds on the tile vars |
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233 | F = F + set(ti < ( Ni / Procs)) ; |
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234 | F = F + set(tj < Nj) ; |
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235 | |
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236 | % No idle time between passes |
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237 | %F = F + set( ( (1/Nj) * tj * (Procs-1) * (ti+tj) ) <= 1 ); |
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238 | %F = F + set( ( (1/Nj) * (Procs-1) * (ti+tj) ) <= 1 ); |
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239 | F = F + set( (1/Nj) * ( (Procs-2) * ti + (Procs-1) * tj ) <= 1); |
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240 | |
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241 | % solve an integer version of the problem |
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242 | intConstraints = set (integer(ti)) + set(integer(tj)); |
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243 | F = F + intConstraints; |
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244 | |
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245 | % ------------- End of Constraints ------------------------- |
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246 | |
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247 | %%%%%%% Obective function |
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248 | obj = T; |
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249 | |
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250 | %%%%% Solve the optimization problem |
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251 | |
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252 | solvesdp(F,obj, sdpsettings('verbose',0,'solver','bnb')); |
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253 | |
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254 | %%%%% Print the solutions |
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255 | |
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256 | |
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257 | % Return values |
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258 | obj_val = double(obj); |
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259 | ti_val = double(ti); |
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260 | tj_val = double(tj); |
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261 | |
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262 | |
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263 | function [obj_val, ti_val] = fStrip_MP(alpha, beta, tau, Procs, Ni, Nj, Nk) |
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264 | % Computes the optimal tile sizes given architectural parameters and domain sizes. |
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265 | % |
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266 | % INPUT ARGUMENTS: (alpha, beta, tau, Procs, Ni, Nj, Nk) |
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267 | % alpha -- time to compute an iteration |
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268 | % beta -- time to transfer a word |
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269 | % tau -- startup cost |
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270 | % Procs -- number of processors |
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271 | % Ni -- size of domain along dimension i |
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272 | % Nj -- size of domain along dimension j |
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273 | % Nk -- size of domain along dimension k |
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274 | % |
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275 | % RETURN VALUES: [obj_val, ti_val, tj_val] |
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276 | % obj_val -- Value of objective function at the optimizer |
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277 | % ti_val -- the optimal tile size. |
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278 | |
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279 | |
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280 | % Notes |
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281 | % ----- |
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282 | % Divide the ij plane into strips. |
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283 | % In this model we allow the strip width to vary, so that |
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284 | % there may be multiple passes. |
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285 | % We solve to ti. |
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286 | |
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287 | % ------------- Tile Variables ---------------------------- |
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288 | |
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289 | % ti and tj are tile sizes |
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290 | ti = sdpvar(1,1); |
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291 | |
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292 | % ------------- Execution Time Model ---------------------------- |
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293 | |
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294 | |
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295 | % Latency count |
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296 | Lambda = (Procs-1); |
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297 | |
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298 | % Time for computing one tile |
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299 | TilePeriod = (alpha * ti * Nj) + (2 * tau * Nj) + ( 2 * beta) ; |
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300 | |
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301 | % Number of planes |
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302 | no_planes = Nk; |
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303 | |
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304 | % Number of passes per plane |
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305 | no_passes_per_plane = (1/Procs) * ( Ni / ti ); |
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306 | |
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307 | % Total running time |
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308 | T = ( Lambda + ( no_planes * no_passes_per_plane ) ) * TilePeriod ; |
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309 | |
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310 | % ------------- End of Execution Time Model ---------------------------- |
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311 | |
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312 | |
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313 | % ------------- Constraints ---------------------------- |
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314 | |
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315 | % Lower bounds on the tile vars |
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316 | %F = set(ti > 1) ; |
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317 | % To avoid idle time between execution of successive planes |
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318 | % we need ti >= 2. |
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319 | F = set(ti >= 2) ; |
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320 | |
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321 | % Upper bounds on the tile vars |
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322 | F = F + set(ti < ( Ni / Procs)) ; |
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323 | |
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324 | % No idle time between passes |
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325 | %F = F + set( (Procs-1) <= Nk ); |
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326 | |
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327 | % solve an integer version of the problem |
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328 | intConstraints = set (integer(ti)) ; |
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329 | F = F + intConstraints; |
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330 | |
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331 | % ------------- End of Constraints ------------------------- |
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332 | |
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333 | %%%%%%% Obective function |
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334 | obj = T; |
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335 | |
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336 | %%%%% Solve the optimization problem |
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337 | |
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338 | solvesdp(F,obj, sdpsettings('verbose',0,'solver','bnb')); |
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339 | |
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340 | %%%%% Print the solutions |
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341 | |
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342 | %fprintf('Objective : %g \n', double(obj)); |
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343 | %fprintf('ti : %g \n', double(ti)); |
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344 | |
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345 | % Return values |
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346 | obj_val = double(obj); |
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347 | ti_val = double(ti); |
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348 | |
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349 | |
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350 | |
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351 | function [obj_val] = fStrip_1P(alpha, beta, tau, Procs, Ni, Nj, Nk) |
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352 | % Computes the running time given architectural parameters and domain sizes. |
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353 | % |
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354 | % INPUT ARGUMENTS: (alpha, beta, tau, Procs, Ni, Nj, Nk) |
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355 | % alpha -- time to compute an iteration |
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356 | % beta -- time to transfer a word |
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357 | % tau -- startup cost |
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358 | % Procs -- number of processors |
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359 | % Ni -- size of domain along dimension i |
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360 | % Nj -- size of domain along dimension j |
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361 | % Nk -- size of domain along dimension k |
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362 | % |
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363 | % RETURN VALUES: [obj_val, ti_val, tj_val] |
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364 | % obj_val -- the running time. |
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365 | |
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366 | |
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367 | % Notes |
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368 | % ----- |
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369 | % Divide the ij plane into strips and do the plane in one pass. |
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370 | % In this model we DO NOT allow the strip width to vary, so that |
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371 | % there is only one pass. |
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372 | |
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373 | |
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374 | % ------------- Execution Time Model ---------------------------- |
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375 | |
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376 | |
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377 | % Latency count |
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378 | Lambda = (Procs-1); |
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379 | |
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380 | % Time for computing one tile |
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381 | TilePeriod = (alpha * Ni * Nj / Procs ) + (2* tau * Nj) + (2*beta) ; |
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382 | |
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383 | % Number of planes |
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384 | no_planes = Nk; |
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385 | |
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386 | % Total running time |
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387 | T = ( Lambda + no_planes ) * TilePeriod ; |
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388 | |
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389 | % ------------- End of Execution Time Model ---------------------------- |
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390 | |
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391 | if (Ni / Procs) < 2 |
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392 | fprintf('>>>>> fStrip_1P: Constraint Ni/Procs >= 2 violated\n'); |
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393 | end; |
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394 | |
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395 | % Return values |
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396 | obj_val = T; |
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397 | |
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398 | |
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