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Input-Output Analysis

An economy is an immensely complicated network of interdependence. Changes in one part can ripple out to affect other parts. Economists have struggled to be able to describe, and to make predictions about, such a complicated object. Mathematical models using systems of linear equations are a key tool. One example is Input-Output Analysis, pioneered by W. Leontief, who won the 1973 Nobel Prize in Economics.

Consider an economy with many parts, two of which are the steel industry and the auto industry. These two interact tightly as they work to meet the demand for their product from other parts of the economy, from users external to the steel and auto sectors. For instance, should the external demand for autos go up, that would increase in the auto industry’s usage of steel. Or should the external demand for steel fall, then it would lower steel’s purchase of trucks. The model that we consider here takes in the external demands and predicts how the two interact to meet those demands.

We start with production and consumption statistics. (These numbers, giving dollar values in millions, are from [Leontief 1965] describing the 1958 U.S. economy. Today’s statistics would be different because of inflation and because of technical changes in the industries.)

 
used by
steel
used by
auto
used by
others
total
value of
steel
5 395 2 664 25 448
value of
auto
48 9 030 30 346

For instance, the dollar value of steel used by the auto industry in this year is 2 , 664 million. Note that industries may consume some of their own output.

We can fill in the external demands. This year’s value of the steel used by others is 17 , 389 and the external value of autos is 21 , 268 . With that we have a complete description of how auto and steel interact to meet their demands.

Now imagine that the external demand for steel has recently been going up by 200 per year and so we estimate that next year it will be 17 , 589 . We also estimate that next year’s external demand for autos will be down 25 to 21 , 243 . We wish to predict next year’s total outputs.

That prediction isn’t as simple as adding 200 to this year’s steel total and subtracting 25 from this year’s auto total. For one thing, a rise in steel will cause that industry to have an increased demand for autos, which will mitigate the loss in external demand for autos. On the other hand, the drop in external demand for autos will cause the auto industry to use less steel and so lessen somewhat the upswing in steel’s business. In short, these two industries form a system. We must predict where the system as a whole will settle.

Here are the equations.

next year’s production of steel = next year’s use of steel by steel + next year’s use of steel by auto + next year’s use of steel by others next year’s production of autos = next year’s use of autos by steel + next year’s use of autos by auto + next year’s use of autos by others

For the left side let s be next years total production of steel and let a be next year’s total output of autos. For the right side, as discussed above, our external demand estimates are 17 , 589 and 21 , 243 .

For next year’s use of steel by steel, note that this year the steel industry used 5 , 395 units of steel input to produce 25 , 448 units of steel output. So next year, when the steel industry will produce s units out, we guess that doing so will take s ⋅ ( 5 395 ) / ( 25 448 ) units of steel input—this is simply the assumption that input is proportional to output. (We are assuming that the ratio of input to output remains constant over time; in practice, models may try to take account of trends of change in the ratios.)

Next year’s use of steel by the auto industry is similar. This year the auto industry uses 2 , 664 units of steel input to produce 30 , 346 units of auto output. So next year, when the auto industry’s total output is a , we expect it to consume a ⋅ ( 2 664 ) / ( 30 346 ) units of steel.

Filling in the other equation in the same way gives this system of linear equations.

( 5 395 / 25 448 ) ⋅ s + ( 2 664 / 30 346 ) ⋅ a + 17 589 = s ( 48 / 25 448 ) ⋅ s + ( 9 030 / 30 346 ) ⋅ a + 21 243 = a

Moving the variables to one side and the constants to the other

( 20 053 / 25 448 ) s − ( 2 664 / 30 346 ) a = 17 589 − ( 48 / 25 448 ) s + ( 21 316 / 30 346 ) a = 21 243

and applying Gauss’s Method or using Sage, as here

sage: var('a,s')
(a, s)
sage: eqns=[(20053/25448)*s - (2664/30346)*a == 17589,  
....:        (-48/25448)*s + (21316/30346)*a == 21243] 
sage: solve(eqns, s, a)
[[s == (2745320544312/106830469), a == (6476293881123/213660938)]]
sage: n(2745320544312/106830469)
25697.9171767186
sage: n(6476293881123/213660938)
30311.0804517904

gives our prediction: s = 25 , 698 and a = 30 , 311 .

Above, we discussed that the prediction of next year’s totals isn’t as simple as adding 200 to last year’s steel total and subtracting 25 from last year’s auto total. Comparing these predictions to the numbers for the current year shows that the total production of the steel industry should rise by  250 while auto’s total drops by 35 . The increase in external demand for steel causes an increase in internal demand by the steel industry, which is lessened somewhat by the drop in autos, but results in a total that is more than 200 higher. Similarly, auto’s total drops more than  25 despite the help that it gets from steel.

One of the advantages of having a mathematical model is that we can ask “What if …?” questions. For instance, we can ask, “What if our estimates for next year’s external demands are somewhat off?” To try to understand how much the model’s predictions change in reaction to changes in our estimates, we can revise the estimate of next year’s external steel demand from 17 , 589 down to 17 , 489 , while keeping the assumption of next year’s external demand for autos fixed at 21 , 243 . The resulting system

( 20 053 / 25 448 ) s − ( 2 664 / 30 346 ) a = 17 489 − ( 48 / 25 448 ) s + ( 21 316 / 30 346 ) a = 21 243

gives s = 25 , 571 and a = 30 , 311 . This is sensitivity analysis. We are seeing how sensitive the predictions of our model are to the accuracy of the assumptions.

Naturally, we can consider larger models that detail the interactions among more sectors of an economy; these models are typically solved on a computer. Naturally also, a single model does not suit every case and assuring that the assumptions underlying a model are reasonable for a particular prediction requires the judgments of experts. With those caveats however, this model has proven in practice to be a useful and accurate tool for economic analysis. For further reading, try [Leontief 1951] and [Leontief 1965].

Exercises

  1. Exercise 1 Worked answer

    With the steel-auto system given above, estimate next year’s total productions in these cases.

    1. Next year’s external demands are up 200 from this year for steel and are unchanged for autos.

    2. Next year’s external demands are up 100 for steel and are up 200 for autos.

    3. Next year’s external demands are up 200 for steel and are up 200 for autos.

    Back to Exercise 1

    Answer. These answers are from Octave.

    1. Sage gets s ≈ 25 952 and a ≈ 30 312 .

      sage: var('s,a')
(s, a)
sage: system = [(20053/25448)*s -  (2664/30346)*a == 17789,
....:             -(48/25448)*s + (21316/30346)*a == 21243]
sage: solve(system, s,a)
[[s == (2772443022712/106830469), a == (6476439541923/213660938)]]
sage: n(2772443022712/106830469), n(6476439541923/213660938)
(25951.8005365305, 30311.7621898814)

    2. Sage gets s ≈ 25 857 and a ≈ 30 596 .

      sage: system = [(20053/25448)*s -  (2664/30346)*a == 17689,
....:             -(48/25448)*s + (21316/30346)*a == 21443]
sage: solve(system, s,a)
[[s == (2762271457112/106830469), a == (6537219545323/213660938)]]
sage: n(2762271457112/106830469), n(6537219545323/213660938)
(25856.5883213711, 30596.2316112410)

    3. Sage gets s ≈ 25 984 and a ≈ 30 597 .

      sage: system = [(20053/25448)*s -  (2664/30346)*a == 17789,
....:             -(48/25448)*s + (21316/30346)*a == 21443]
sage: solve(system, s,a)
[[s == (2775832696312/106830469), a == (6537292375723/213660938)]]
sage: n(2775832696312/106830469), n(6537292375723/213660938)
(25983.5300012771, 30596.5724802865)

  2. Exercise 2 Worked answer

    For the steel-auto system, with the external demand estimates of 17 , 589 and 21 , 243 discussed above, what will be the value of steel used by steel, the value of steel used by auto, etc.?

    Back to Exercise 2

    Answer. This Sage session

    sage: var('a,s')
(a, s)
sage: eqns=[(20053/25448)*s - (2664/30346)*a == 17589,
....:       (-48/25448)*s + (21316/30346)*a == 21243] 
sage: solve(eqns, s, a)
[[s == (2745320544312/106830469), a == (6476293881123/213660938)]]
sage: s_by_s = (5395/25448)*2745320544312/106830469
sage: s_by_s
582010544505/106830469
sage: n(s_by_s)
5447.98267716114
sage: s_by_a = (2664/30346)*6476293881123/213660938  
sage: n(s_by_a)
2660.93449955743
sage: a_by_s = (48/25448)*2745320544312/106830469
sage: n(a_by_s)
48.4713936058822
sage: a_by_a = (9030/30346)*6476293881123/213660938
sage: n(a_by_a)
9019.60905818451

    gives this table.

     
    used by
    steel
    used by
    auto
    used by
    others
    		total
    value of
    steel
    		5 448 2 661 17 589 25 698
    value of
    auto
    		48 9 020 21 243 30 311

    For comparison here is the original table.

     
    used by
    steel
    used by
    auto
    used by
    others
    		total
    value of
    steel
    		5 395 2 664 17 389 25 448
    value of
    auto
    		48 9 030 21 268 30 346

  3. Exercise 3 Worked answer

    In the steel-auto system, the ratio for the use of steel by the auto industry is 2 , 664 / 30 , 346 , about 0.0878 . Imagine that a new process for making autos reduces this ratio to .0500 .

    1. How will the predictions for next year’s total productions change compared to the first example discussed above (i.e., taking next year’s external demands to be 17 , 589 for steel and 21 , 243 for autos)?

    2. Predict next year’s totals if, in addition, the external demand for autos rises to be 21 , 500 because the new cars are cheaper.

    Back to Exercise 3

    Answer.

    1. Sage gives s = 24 , 244 , a = 30 , 307 .

       sage: var('s,a')
(s, a)
sage: system = [(20053/25448)*s -  0.0500*a == 17589,
....:             -(48/25448)*s + (21316/30346)*a == 21243]
sage: solve(system, s,a)
[[s == (12951731858499/534221147), a == (32381469405615/1068442294)]]
sage: n(12951731858499/534221147), n(32381469405615/1068442294)
(24244.1392131918, 30307.1767071166)

    2. Sage gives s = 24 , 267 , a = 30 , 673 .

      sage: system = [(20053/25448)*s -  0.0500*a == 17589,
....:             -(48/25448)*s + (21316/30346)*a == 21500]
sage: solve(system, s,a)
[[s == (12964136043940/534221147), a == (16386224431390/534221147)]]
sage: n(12964136043940/534221147), n(16386224431390/534221147)
(24267.3584090448, 30673.1107957993)

  4. Exercise 4 Worked answer

    This table gives the numbers for the auto-steel system from a different year, 1947 (see [Leontief 1951]). The units here are billions of 1947 dollars.

     
    used by
    steel
    used by
    auto
    used by
    others
    		total
    value of
    steel
    		6.90 1.28 18.69
    value of
    autos
    		0 4.40 14.27

    1. Solve for total output if next year’s external demands are: steel’s demand up 10 % and auto’s demand up 15 %.

    2. How do the ratios compare to those given above in the discussion for the 1958 economy?

    3. Solve the 1947 equations with the 1958 external demands (note the difference in units; a 1947 dollar buys about what $ 1.30 in 1958 dollars buys). How far off are the predictions for total output?

    Back to Exercise 4

    Answer.

    1. These are the equations.

      ( 11.79 / 18.69 ) s − ( 1.28 / 14.27 ) a = 10.51 − ( 0 / 18.69 ) s + ( 9.87 / 14.27 ) a = 9.87

      Sage gives this result (including an inital check).

      sage: var('a,s')
(a, s)
sage: eqns=[(11.79/18.69)*s - (1.28/14.27)*a == 10.51,
....:       (-0)*s         + (9.87/14.27)*a == 9.87]
sage: solve(eqns,s,a)
[[s == (1869/100), a == (1427/100)]]
sage: n(1869/100)
18.6900000000000
sage: n(1427/100)
14.2700000000000
sage: eqns=[(11.79/18.69)*s - (1.28/14.27)*a == 11.561,
....:       (-0)*s         + (9.87/14.27)*a == 11.3505]
sage: solve(eqns,s,a)
[[s == (8119559/393000), a == (32821/2000)]]
sage: n(8119559/393000)
20.6604554707379
sage: n(32821/2000)
16.4105000000000

    2. These are the 1947 ratios, to three decimal places.

      1947 by steel by autos
      use of steel ( 6.90 / 18.69 ) = 0.369 ( 1.28 / 14.27 ) = 0.090
      use of autos ( 0 / 18.69 ) = 0.00 ( 4.40 / 14.27 ) = 0.308

      These are the 1958 ratios.

      1958 by steel by autos
      use of steel ( 5395 / 25448 ) = 0.212 ( 2664 / 30346 ) = 0.088
      use of autos ( 48 / 25448 ) = 0.002 ( 9030 / 30346 ) = 0.298

    3. The external demands in 1958 are 17.589 and  21.243 , in billions of 1958 dollars. In 1948 dollars those are ( 17.589 / 1.3 ) = 13.53 and ( 21.243 / 1.3 ) = 16.34 . Sage gives these numbers.

      sage: eqns=[(11.79/18.69)*s - (1.28/14.27)*a == 13.53,
....:       (-0)*s         + (9.87/14.27)*a == 16.34]
sage: solve(eqns,s,a)
[[s == (137466107/5541300), a == (1165859/49350)]]
sage: n(137466107/5541300)
24.8075554472777
sage: n(1165859/49350)
23.6242958459980

      Converting back to 1958 dollars by multiplying by 1.3 gives 32.25 and  30.71 .

  5. Exercise 5 No answer supplied in the original source.

    Predict next year’s total productions of each of the three sectors of the hypothetical economy shown below

     
    used by
    farm
    used by
    rail
    used by
    shipping
    used by
    others
    		total
    value of
    farm
    		25 50 100 800
    value of
    rail
    		25 50 50 300
    value of
    shipping
    		15 10 0 500

    if next year’s external demands are as stated.

    1. 625 for farm, 200 for rail, 475 for shipping

    2. 650 for farm, 150 for rail, 450 for shipping

  6. Exercise 6 No answer supplied in the original source.

    This table gives the interrelationships among three segments of an economy (see [Clark & Coupe]).

     
    used by
    food
    used by
    wholesale
    used by
    retail
    used by
    others
    		total
    value of
    food
    		0 2 318 4 679 11 869
    value of
    wholesale
    		393 1 089 22 459 122 242
    value of
    retail
    		3 53 75 116 041

    We will do an Input-Output analysis on this system.

    1. Fill in the numbers for this year’s external demands.

    2. Set up the linear system, leaving next year’s external demands blank.

    3. Solve the system where we get next year’s external demands by taking this year’s external demands and inflating them 10 %. Do all three sectors increase their total business by 10 %? Do they all even increase at the same rate?

    4. Solve the system where we get next year’s external demands by taking this year’s external demands and reducing them 7 %. (The study from which these numbers come concluded that because of the closing of a local military facility, overall personal income in the area would fall 7 %, so this might be a first guess at what would actually happen.)

References cited in this section

Leontief 1965

Wassily W. Leontief, The Structure of the U.S. Economy, Scientific American, volume 212 number 4 (Apr. 1965), p. 25.

Leontief 1951

Wassily W. Leontief, Input-Output Economics, Scientific American, volume 185 number 4 (Oct. 1951), p. 15.

Clark & Coupe

David H. Clark, John D. Coupe, The Bangor Area Economy Its Present and Future, report to the city of Bangor ME, Mar. 1967.