Question

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Consider an economy consisting of 3 sectors: M ('manufacturing'), E ('energy') and T ('transportation'). The following is known about the required inputs in each of these sectors from the outputs of different sectors for production of yearly outputs x1, x2 and x3 in sectors M, E and T, respectively: (i) M requires 0.5 fraction of itself, 0.1 fraction of the output of E, and 0.2 fraction of T; (ii) E requires 0.2 fraction of itself, 0.3 fraction of M, and 0.3 fraction of T; (iii) T requires 0.4 fraction of itself, 0.2 fraction of M, and 0.4 units of E. (a). Write the input-output matrix for this economy, call this matrix A. Explain whether this economy corresponds to open or closed Leontieff model. Write the equation (in symbolic form) for this model using A for the matrix of coefficients of the system, and X for the production solution) vector. (b) Assume now some external demand vector b of your choice! (including the zero-demand vector if appropriate) that would make sense for this economy, and find the production vector that solves that model.

Answer 1

MET The corput output matrix can be written as: A = 0.5 0.3 0.2] on 0.2 0.4 input TL0.2 0.3 0.4 where the first row depicts that 0 0.5 of M is wed bym, o.3 of M is used by E and 0.2 of m is und by T. Similarly, now 2 and 3 can be interpreted. On the other hand, the firest column is cnterpreted as follows. To produce theory of $1 of product , M requires ois from itself, Ol from E and 0.2 from T. Similarly, column 2 and 3 represent the production preocess and their respective deput use. fan economy contains an open rector, which determines final demand. for the product of industries and the open sector supplies a prinarcy input not produced by the above 3 industries themselves. In such a case, the model is an open model. An open sector exists in an model it tu column sums of the atoon exput output matrix are less than 1. Each column sun represents partial aput cost incurereed en producing & worth of god. In the present case the first column sums to oise. 0.5 +0.1 +0.2), the second column sums to o scie. 0.3+0.2+0.3), but column 3 surns to 1 Cine 0.2 +0.4+0.4). Hence, except for transporciation (7) Industry, the other two industries cater to final demand of the open sector. Hence, the model is an open model. Let x, x₂ and 23 represent output produced by M, E and respectiven we may represent the model as follows: x = 0.5 x, +0.3x₂ + 0.2x₂ + b, where buba, b, are final demomds for 2, R2 + Rz x2 = 0.1 xito.2 x₂ + 0.403 + b2 z = 0.2 x+0.3 + 0.4x₂ tba respectively → (1-05) X, -0.3 2 -0.2*3 = b; -0.1%, + (1-0.21%2 -0.423 = b2 - 0.2x - 0.3x₂ + (-0.4) 3 = 63 sri 7 To.s 0.3 0.272727 o I 0 -101 0.2 -> (1-A) X = 6 0.4 o J 10.2 0.3 0.4) Premultiplying Q-A), we get (x = (1-4) 6] where x 'es the preoduction vector. (b) As mentioned, T has no final demand, hence we may take the final Troo demand vector as b=[bi]. Consider a specific rector as b=live 22 = J Lb 1b2

Let us ferest calculate (IA) = 10.5 -0.3 -0. 2 7-1 0 % -04 L-02-03 06 → Wrcite the augmented mateix ) A1 A2 A2B, 0.5 -03-2 1 2 -0.0.9 -04 13 1-0.2 -3 0.6 1. → Divide ist row by 0.5 =) A₂ A, B B. Bal 0 0 O -06-04 2 0 0.74 -0.44 0. 21 042 052 0.4 0 B3 - Divide second column by 0.74 TA A A Bi B2 -0.6 -0.4 2 o -0.594 0 2702 351 -0.42 0.52 0.4 0 Please note that some numbers quee expressed rep to 4 decimal places Byl - Eliminate second column A A A₂ B B. 110 -757 2:1620.8108 o -0-599 02703 1357 0 0 2703 0:514 0.568 0 in - dm -oo - → Divide third column by 0.2703 A A A Bi B2 B3] 0 -0.757 O S103 2.162 2 0 -095 2703 351 0 0 0 I 1.9 2.1 3.7 a Eliminate third column B, B2 B3 1 0 0 3.6 2.4 2.8 0 0 14 2.6 2.2 o o 1.9 2.1 3.7 a) (I-AJ-13.6 2.4 2.87 = 1.4 2.6 2.2 L 1.9 2.1 3.7 ) T 600 = x=(1-4) b = [3.6 2.4 2.81100] 2 X -"2 .4 2.6 2.2 yoo & the final 1.9 2.4 3:11° Production vector where x = 600, kg = x=400. L400