Question

please write matlab code it should be around 20 lines

Consider an open economy with three industries: coal-mining operation, electricity- generating plant and an auto-manufacturing plant. To produce $1 of coal, the mining operation must purchase $0.1 of its own production, $0.30 of electricity and $0.1 worth of automobile for its transportation. To produce $1 of electricity, it takes $0.25 of coal, $0.4 of electricity and $0.15 of automobile. Finally, to produce $1 worth of automobile, the auto-manufacturing plant must purchase $0.2 of coal, $0.5 of electricity and consume $0.1 of automobile. Assume also that during a period of one week, the economy has an exterior demand of $50,000 worth of coal, $75,000 worth of electricity, and $125,000 worth of autos. Find the production level of each of the three industries in that period of one week in order to exactly satisfy both the internal and the external demands. To help guide you, imagine the system to be this one: AP = d In this system, the vector d is the external demand. The vector P is the production level of each industry, and A is the coefficient matrix that describes how much each industry contributes, or consumes, to the overall external demand of the industry. For example: The coal industry contributes $0.9 to the external coal demand, the electricity industry contributes -$0.25 to the external coal demand, and the automobile industry contributes -$0.20 to the external coal demand. Use the LU decomposition function to solve for the production levels to meet the external demand. Output your results to the command window for each industry. To check your answer, it must satisfy the original system, no sample output will be given.

Answer 1

clc; clear; format shortg a=[0.1 0.30 0.1;0.25 0.4 0.15;0.2 0.5 0.1]; % given coefficient * given external demand d=[50000;75000;125000]; ab=[a, d]; ** LU decomposition n=length (a); 1=eye (n); % indentity matrix ** Row change operation in the form of lower matrix and upper matrix form for i=2:3 alpha=ab (i,1)/ab (1,1); 1(i,1)=alpha; ab (i,:)=ab (i,:) -alpha* ab (1, :): end 88 a (3,2) value change i=3; alpha=ab (3, 2)/ab (2,2); 1(i,2)=alpha; ab (3, :)=ab (3,:) -alpha*ab (2,:); fprintf ('lower triangle\n'); disp (1) fprintf ('upper triangle\n'); disp (ab)

8* find the production level * pl is the coal mining operatipon ;p2 is the electricity generating plant Sp3 is the auto-manufacturing plant p=zeros (3,1); ]for i=n:-1:1 p(i)= (ab (i, end) -ab (i,i+l:n) *p(i+l:n))/ab (i,i); Sassume production level of each industry end fprintf ('production level of each industry in the form of column matrix\n') disp (p) ; ** check the answer * product of lower triangle and upper triangle is given matrix fprintf ('check the answer\n'); disp (1*ab)

lower triangle 2.5 0.28571 upper triangle 0.1 0.3 0.1 50000 -50000 -0.35 -0.1 -1.3878e-17 -0.071429 39286 production level of each industry in the form of column matrix 1.5e+05 3e+05 -5.5e+05 check the answer 0.1 50000 0.3 0.1 0.25 0.4 0.15 75000 0.2 0.5 0.1 1.25e+05 *>>
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