Example 1 is almost identical to the “closed exchange economy” example handed out in class during the first week of the course although there are a few more comments and updates in light of what we have learned since then.
Then, Example 2 illustrates (on a very small scale) some of the ideas behind making a more complicated model of an “open” economy.
Example 1 Closed Exchange Economy
Suppose an economy has only 4 sectors: agriculture (A), energy (E), manufacturing (M) and transportation (T).
In this simple “closed exchange economy,” we assume that all goods are manufactured and used among these four sectors.
The production/consumption among the sectors is summarized in an exchange matrix
Production of sector ÆÆÆÆ EIQX
Þ'& Þ$! Þ$! Þ#! E Þ"! Þ"! Þ"& Þ"! I Ä consumed by Þ#& Þ$& Þ"& Þ$! Q ! Þ#& Þ%! Þ%! X Going down a column, we can read off what part of a given sector's production goes to each of the other sectors: for example, the production of the manufacturing sector (Q ) is consumed as
30% to the agriculture sector, 15% to the energy sector, 15% to the manufacturing sector (it uses up some of its own goods), and 40% to the transportation sector.
Therefore, of course, the sum of each column is " (100%).
Going across a particular row, we can read off what part of each sector's production is consumed by one particular sector: for example, the energy sector consumes 10% of the production of the agriculture, energy and transportation sectors and 15% of the production of the manufacturing sector. Each row is a consumption vector for one of the sectors. Suppose :ß:ß:ß:E IQX represent the total production of goods in each sector (measured for convenience, say, in $). Is it possible to set production levels for the four sectors so that “everybody's happy” that is, every sector gets what it needs (using the value of its own goods to pay) and nothing is left over? If so, then the values :ß:ß:ß:E IQX give an equilibrium for this “closed exchange economy.”
The cost to sector E for what it needs is
Þ'&TE Þ$!TIQX Þ$!T Þ#!T
This expense must be paid for by the value of sector ETß 's goods, E so
Þ'&TEE Þ$!TIQX Þ$!T Þ#!T œ T , and similarly (*) Þ"!TE Þ"!TIQXI Þ"&T Þ"!T œ T Þ#&TE Þ$&TIQXQ Þ"&T Þ$!T œ T Þ#&TIQXX Þ%!T Þ%!T œ T
Rearranging these equations gives the linear system
Þ$&TE Þ$!TIQX Þ$!T Þ#!T œ ! (**) Þ"!TE Þ*!TIQX Þ"&T Þ"!T œ ! Þ#&TE Þ$&TIQX Þ)&T Þ$!T œ ! Þ#&TIQX Þ%!T Þ'!T œ !
whose augmented matrix is
Þ$& Þ$! Þ$! Þ#! ! Þ"! Þ*! Þ"& Þ"! ! . The row reduced echelon Þ#& Þ$& Þ)& Þ$! ! ! Þ#& Þ%! Þ'!!
form (with entries rounded to 2 decimal places for convenience) is
"!! #Þ!$! !"! !Þ&$! !!" "Þ"(! !!!!!
Writing ¸œ instead of (since I rounded decimals), we have
TE ¸ #Þ!$TX TIX ¸ !Þ&$T TQX ¸ "Þ"(T TX is free Therefore: If a total production TX ($) for the transportation sector is determined, then the productions of the other sectors can be adjusted to create an equilibrium. Of course, one would choose a value of T !X for the solutions to be economically feasible.
Some additional comments about Example 1
The system of equations (*) in Example " can be written in matrix form:
Iœ::
Þ'& Þ$! Þ$! Þ#! :E Þ"! Þ"! Þ"& Þ"! : where I: is the “exchange matrix” and œÞI Þ#& Þ$& Þ"& Þ$! :Q ! Þ#& Þ%! Þ%! :X
Each row is a consumption vector for one of the sectors. For example, the row ÒÞ"! Þ"! Þ"& Þ"!Ó indicates that the energy sector consumes 10% of the production of the agriculture and energy sectors, 15% of the production of the manufacturing sector, and 10% of the production of the transportation sector.
We could rewrite the matrix equation as follows:
IM:œ : (where M is the identity matrix M% ) IMœ::! ÐI MÑ:! œ (or equivalently , ÐM IÑ :! œ if we multiply both sides of the equation by " )
The last equation is just the matrix form of the system (**) in Example 1, because
Þ'&Þ$!Þ$!Þ#! "!!! Þ$&Þ$! Þ$! Þ#! Þ"!Þ"!Þ"&Þ"! !"!! Þ"!Þ*!Þ"& Þ"! IMœ œ Þ#&Þ$&Þ"&Þ$! !!"! Þ#& Þ$&Þ)& Þ$! ! Þ#&Þ%!Þ%! !!!" ! Þ#& Þ%! Þ'!
Þ$& Þ$! Þ$! Þ#! ! Þ"! Þ*! Þ"& Þ"! ! We could row reduce the augmented matrix to Þ#& Þ$& Þ)& Þ$! ! ! Þ#& Þ%! Þ'! !
find the general solution. It will turn out that there are nontrivial solutions, as the following more general discussion indicates. Here is the general version of this “closed exchange economy” model:
We have an economy divided into 8 sectors W"# ß W ß ÞÞÞß W 8 . The total production of each sector is consumed by (“exchanged with”) the other sectors. Each row of the exchange matrix is a consumption vector for one of the sectors: the entries //3" , 3# ,..., / 34 ,... / 38 in row 3 give the proportion of the outputs of W"# ß W ß ÞÞÞß W 8 consumed by W 3 Þ
Fraction of Production of
WâW"4 âW 8 Æ /â / â/ W "" "4 "8 " ã ã Iœ /â/3" 34 â/ 38 Ä Consumed by W3 ã ã /â8" / 84 â / 88 W8
Of course, each entry / !34 . Since, in this model, all the production of each sector is consumed by the other sectors, each column sum is "4 : that is, for each fixed , 8 /œ"Þ34 3œ" B" We are interested in finding a production vector BBBœIœã for which . Such B8 an B represents an equilibrium production from each sector in which all needs are met with no surplus.
We can rewrite this equation as an equivalent homogenous system:
ÐI MÑB! œ
Of course, we have an equilibrium if all the sectors produce nothing, that is, for the trivial solutionB!œÞ So we are really interested in the nontrivial solutions (if any) for IœBB. Moreover, in the economic situation, we are interested not just in a nontrivial mathematical solution, but an “economically feasible” nontrivial solution one where all B 3 0.
Can we always find such an B ? The next theorem offers a partial answer. Theorem Let I8‚8 be an matrix whose entries are all nonnegative and where each column sum is "ÐIMÑœÞ . Then there is a nontrivial solution for B!
Proof For each 4ß the 4>2 column of ÐI MÑ is the same as the 4 >2 column of I except that the diagonal entry /444 in column has been replaced by /" 44 . Since /"4 / #4 ÞÞÞ/ 44 ÞÞÞ/ 84 œ"ßthen / "4 / #4 ÞÞÞÐ/ 44 "ÑÞÞÞ/ 84 œ!À that is, each column in ÐI MÑ has sum œ !Þ Begin row reducing the augmented matrix ÒI M l! Ó by adding each row to the bottom row: first add row 1, then row 2, ..., etc. This means that, in each column, the bottom entry will end up being the sum of all numbers in that column! namely, . So, after these steps in a row reduction, we end up with a bottom row of ! 's. Therefore the matrix IM (which is 8‚8 ) cannot have 8 pivot positions. Therefore the system homogeneous system ÐI MÑB! œ has at least one free variable, so it must have nontrivial solutions (in fact, infinitely many of them). ñ
Note: It is possible to prove, in addition, that there must be a nontrivial solution B where all the entries B 3 0. But the proof is considerably harder and we can't do it here.
Examples Here are two very simple economies only two sectors; they are here to make a mathematical point, not because they're realistic.
"" ##!! a) If Iœ"" , then ÐIMÑBB! œ œ has the general ##"! solution
! B œ= "
Looking at the rows of I" (the consumption vectors) we see that, unlike Sector , Sector #" consumes all of its own product. Half the product of Sector is “out of balance”" in the sense that Sector doesn't use it and Sector # has nothing left with which to pay for it. An equilibrium is impossible unless Sector " produces nothing (“half of nothing out of balance” is not a problem).
And that's what the solution shows: Sector "# must produce nothing, and Sector can produce as much or little as it likes. "! !! b) If Iœ , then ÐIMÑBB! œ œ !" !!
Looking at the rows of I (the consumption vectors) we see that Sector 1 and Sector # each consume all of their own product and nothing from the other sector. The production of the sectors is “independent.” Therefore, each sector can produce as much it likes and the economy will be in equilibrium.
Mathematically, every B in ‘# is a solution: both variables in the system are ="! free, and we can write the set of solutions as BBœœ=> , or . >!" An economically feasible solution requires choosing any values = ! and > !Þ
In this example the possible production vectors cannot all be described as multiples of a single vector that is, there are production vectors where one is not just a rescaling of the other.
There's a deeper theorem that can be proven in a course on stochastic processes.
Theorem Let I8‚8 be an matrix whose entries are all nonnegative and for which all column sums are "5 . Suppose, in addition, that there is some positive integer for which the entries in the matrix IÐ!ß 5 are all strictly positive that is, all arenot merely 0). Then the solution set of ÐI MÑB! œ is spanned by a single vector so the general solution is B@œ> for some @ in ‘8 Þ The vector @ can be chosen so that all its entries are ! .
Therefore, if the exchange matrix satisfies the condition in the Theorem, outcomes like Examples a) and b) cannot happen. Example 2 Leontief Model of an Open Economy (Input/Output Model)
Assume again that an economy has 8 sectors W"8 ß ÞÞÞß W . As before, a production vector
B" B B œ # gives the production of each sector (measured in $). ã B8
We can look at the general “setup” for Leontief's model of an “open economy” a term which we will explain below. The setup has similarities to Example " , but it also has differences: for example, we look at not at the total consumption of Sector 4 from the other sectors, but the consumption by Sector 44 per unit of Sector product.
We can also solve the system we set up in the case of a simple example. However, proofs of the interesting theorems about “what happens in general” are beyond the scope of this course.
We begin with a consumption matrix G4 in which the >2 column lists the inputs ($) from W"# ß W ß ÞÞÞß W 8 consumed by sector W 4 to produce one unit ($) of output. Of course, each - !Þ34
WÄ -â - â- " "" "4 "8 ã ã Input from WÄ3 -â-3" 34 â- 38 œG ã ã WÄ8 -â8" - 84 â - 88 ÅÅ Å Consumption vectors for
WâW"84 âW
-"" ã For example, Column " œ"-3" lists the separate inputs ($) from Sectors ,..., ã -8"
Sector 8 consumed by Sector " to produce one unit ($) of its own product. These “inputs” are costs to Sector " .
The sum of the entries in Column " gives the total cost of products consumed by Sector to produce one unit of its product. The sum of the entries in Column 4 represents the total input ($) from W"# ß W ß ÞÞÞß W 8 consumed by WW44 in order for to produce one unit ($) of product. This is the cost to >2 Sector 44"W to produce one unit. If the column sum is , we say that sector 4 is profitable. -B-"4 " "4 ãã If we weight Column 4B Bœ with a scalar 44 0 , then the vector -B-34 " 34 lists ãã -B-84 " 84 the inputs ($) consumed from other sectors by sector WB44 to produce units ($) of its product.
If Sectors 1,2,...,8 produce B"# ß B ß ÞÞÞß B 8 units, then if we have a production vector
B" B# Bœ, where the B !3 , what does the vector G B mean? ã B8
--"" "# - "8 -"" B " ÞÞÞ- "8 B 8 ãã ãã GœBB "8--33#381 B2 ÞÞÞB -œ -3" B " ÞÞÞ -38 B 8 ãã ãã --88#881 --8" B " ÞÞÞ- 88 B 8
total ($) consumed by Sectors W"8 ß ÞÞÞß W from W " to produce B ã œ total ($) consumed by Sectors W"8 ß ÞÞÞß W from W 3 to produce B ã total ($) consumed by Sectors W"8 ß ÞÞÞß W from W 8 to produce B
GÐÑB lists the inputs demanded consumed from each sector if they are to deliver the production vector B Þ
If we examine the equation GœBB , then we are looking for an equilibrium value an B for which the total amount produced by each sector = the amount of its product consumed by all the other sectors. The production from each sector is not too large, not small but “just right.” This is a question about a “closed” economy. Now we add another feature to the model: an open sector . It is nonproductive ; it produces nothing that sectors W"8 ß ÞÞÞß W use; the open sector simply “demands” goods produced by sectors W"8 ß ÞÞÞß W Þ
." . A final demand vector . œ # tells how many units ($) the open sector demands from ã .8 sectors Wß ÞÞÞß W8 Þ For example, the nonproductive sector might be the government, or perhaps a sector consisting of charitable organizations.
We want to know if it is possible to set a level of production B so that both the productive and open sectors are satisfied, with nothing left over. That is, we want to find an B so that
Leontief Open Economy Production Model
BB. œG ÅÅÅ (***) Total productionœ Demand from + Demand from productive sectors open sector to produce B
We can rewrite this equation in the form ÐM GÑB. œ
To take a simple example, suppose the economy has 4 productive sectors W"% ß ÞÞÞW and the consumption matrix is
Þ"! Þ!& Þ$! Þ#! Þ"& Þ#& Þ!& Þ"! Gœ Þ$! Þ"! Þ"! Þ#& Þ"& Þ#! Þ"! Þ#! ÅÅ ÅÅ Consumption vectors for WW"# WW $%
ÐNote: The column sums in G" are all so each sector is profitable.)
Þ*! Þ!& Þ$! Þ#! Þ"& Þ(& Þ!& Þ"! Then ÐM GÑ œ Þ$! Þ"! Þ*! Þ#& Þ"& Þ#! Þ"! Þ)! #&!!! "!!!! If the demand from the open sector is . œ , then we want to try to solve $!!!! &!!!!
Þ*! Þ!& Þ$! Þ#! #&!!! Þ"& Þ(& Þ!& Þ"! "!!!! ÐM GÑB. œB œ œ Þ$! Þ"! Þ*! Þ#& $!!!! Þ"& Þ#! Þ"! Þ)! &!!!!
Row reducing the augmented matrix gives
Þ*! Þ!& Þ$! Þ#! #&!!! "!!! )&&)! Þ"& Þ(& Þ!& Þ"! "!!!! !"!! &!'#! µ Þ$! Þ"! Þ*! Þ#& $!!!! !!"! *'"'! Þ"& Þ#! Þ"! Þ)! &!!!! !!!""!$##!
(The calculations were carried out to many decimal places, but the final numbers displayed are rounded to the nearest unit ($) ).
Equivalently (but harder to compute by hand), we could try to find ÐM GÑ" and, if it exists, then compute BœÐM GÑ" .Þ (In this example, it turns out that MG is invertible so this approach would work.)
The equation (***) will be satisfied if WW"# produces 85580 units ($) of product, produces 50620 ($) , etc.
Here's a theorem that's too hard for us to prove:
Theorem (using the same notation as above ) If G and . have nonnegative entries and the column sums in G" are all (that is, if every sector in profitable ), then ÐM GÑ is invertible so there is a unique production vector satisfying the equation ÐM GÑB. œ Ànamely B œ ÐM GÑ" . . Moreover, this B is economically feasible in the sense that all its entries are !Þ Here is an informal, imprecise discussion that might lead you to believe the theorem is true.
1) Suppose G8‚8 is any matrix with entries all 0 and all column sums "Þ Then GÄ7 ! as 7Ä∞. Å (the 8‚8 zero matrix Ñ
This statement just means that every entry in G! will be made as close to as we like provided we choose a large enough value for 7 .
+, To keep things simple, we just look at the case of a #‚# matrix Gœ . -.
Let αα be the largest column sum in G!Ÿ"Þ . Our assumptions give that Therefore each column sum +-Ÿαα and ,.Ÿ .
BC Now an observation: suppose is any #‚# matrix with BßCßDßA ! . DA +, BC +B,D+C,A Then œ and this new matrix has -. DA -B.D-C.A
column sums
Column 1: Ð+ -ÑB Ð, .ÑD Ÿαα B D œ α ÐB DÑ Column 2: Ð+ -ÑC Ð, .ÑA Ÿαα C A œ α ÐC AÑ Å œ α times the old column BC sums of DA
BC +, If we apply this observation starting with œ we get that the DA -. +, +, column sums for œG# satisfy: -. -.
sum for column " of GŸ##ααααα (old column 1 sum) œÐ+-ÑŸ†œ sum for column 2 of GŸ##ααααα (old column 2 sum) œÐ,.ÑŸ†œ . BC Repeating this argument again with œG# , we see that in the matrix DA +, GœG†Gœ$# †G #, the sum of each column is Ÿα $ . -. Continuing in this way, we find that the sum of each column in GŸ77 is α , and (since all the elements in G 7 are nonnegative) this means that every entry in the matrix G77 is Ÿααα Þ Since ! Ÿ "ß 7 Ä ! as 7 Ä ∞Þ Therefore G 7 Ä ! as 7Ä∞Þ
2) We can write a matrix equation
MG7" œÐMGÑÐMGG # ÞÞÞG 7 Ñ
No hand-waving here; just multiply it out to verify:
ÐMGÑÐMGG#7 ÞÞÞG Ñ œ ÐM GÑM ÐMGÑG ÐMGÑG#7 ÞÞÞÐMGÑG œM###$77" GMMGG MG G ÞÞÞMG G œMGGG## G ÞÞ . ÞÞÞG 777" G G œMG7"
3) Since GÄ77"!! as 7Ä∞ , we have that G ¸ for large 7 , so for large 7 ,
M ¸ÐMGÑÐMGG#7 ÞÞÞG Ñ which means ÐMGG#7 ÞÞÞG Ñ is approximately an inverse for ÐMGÑÞ
This suggest= that MG is invertible. Moreover, ÐMGG#7 ÞÞÞG Ñ has nonnegative entries and ÐMGÑ" ¸ÐMGG # ÞÞÞG 7 Ñ ; this suggests that the elements of ÐM GÑ" are all nonnegative.
4) Of course, if ÐM GÑ is invertible and ÐM GÑ" has all nonnegative elements, then we have a unique solution B.œÐMGÑ" and ÐMGÑ " . has all nonnegative entries.
Having done all this, we can also notice that if ÐMGÑ" ¸ ÐMGG # ÞÞÞG 7 Ñ , then
BœÐMGÑ" .¸ ÐMGG # ÞÞÞG 7 ц . or
B..¸ G G#7 . ÞÞÞG .
(See text, p. 154-155 for additional comments about the economic significance of ÐM GÑ".