Tamil Nadu Board of Secondary EducationHSC Commerce Class 11th

# Suppose the inter-industry flow of the product of two industries are given as under. - Business Mathematics and Statistics

Sum

Suppose the inter-industry flow of the product of two industries are given as under.

 Production sector Consumption sector Domestic demand Total output X Y X 30 40 50 120 Y 20 10 30 60

Determine the technology matrix and test Hawkin’s -Simon conditions for the viability of the system. If the domestic demand changes to 80 and 40 units respectively, what should be the gross output of each sector in order to meet the new demands.

#### Solution

a11 = 30, a12 = 40, x1 = 120

a21 = 20, a22 =10, x2 = 60

"b"_11 = "a"_11/"x"_1 = 1/4, "b"_12 = "a"_12/"x"_2 = 2/3

"b"_21 = "a"_21/"x"_1 = 20/120 = 1/6, "b"_22 = "a"_22/"x"_2 = 10/60 = 1/6

The technology matrix B = [(1/4,2/3),(1/6,1/6)]

I - B = [(1,0),(0,1)] - [(1/4,2/3),(1/6,1/6)]

= [(3/4,-2/3),(-1/6,5/6)], elements of main diagonal are positive.

|I - B| = [(3/4,-2/3),(-1/6,5/6)]

= 3/4 xx 5/6 - (- 1/6) xx ((-2)/3)

= 1/4 xx 5/2 - (1/3) xx (1/3)

= 5/8 - 1/9 = (5 xx 9 - 1 xx 8)/(8 xx 9) = (45 - 8)/72 = 37/72

The main diagonal elements of I – B are positive and |I – B| is positive. Therefore the system is viable.

adj (I - B) = [(5/6,2/3),(1/6,3/4)]

(I - B)-1 = 1/|"I - B"| adj(I - B)

= 1/((37/72)) [(5/6,2/3),(1/6,3/4)] = 72/37 [(5/6,2/3),(1/6,3/4)]

X = (I - B)-1D, where D = [(80),(40)]

= 72/37[(5/6,2/3),(1/6,3/4)][(80),(40)]

=> 1/37[(72xx5/6,72xx2/3),(72xx1/6,72xx3/4)][(80),(40)]

= 1/37[(60,48),(12,54)][(80),(40)]

=> 1/37[(60xx80+48xx40),(12xx80+54xx40)]

= 1/37[(6720),(3120)]

= [(181.62),(84.32)]

The output of industry X should be 181.62 and Y should be 84.32.

Concept: Input–Output Analysis
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