ECON Answers Homework #2

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Silvester Rose
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1 ECON 33 - Answers Homework #2 Exercise : Denote by x the number of containers of tye H roduced, y the number of containers of tye T and z the number of containers of tye I. There are 3 inut equations that have to be satised, one for each tye of steel: i Steel s : 2x + 3y + 4z = 29; ii Steel s 2 : x + y + z = 3; iii Steel s 3 : 3x + 2y + z = 6. We end u with a system of 3 linear equations that deend on 3 unknowns, x, y and z. From here, you have 2 main strategies: Gaussian elimination or attemt Cramer's rule. Since the coecient matrix does not have many zeros, it will take some time to calculate its determinant and decide whether we can actually aly Cramer's rule or not. So in this case, I would go with Gaussian elimination. But again, if you refer to calculate the determinant of the coecient matrix it is non-zero here and aly Cramer's rule, this is erfectly valid. Gaussian elimination to solve the above system: x + y + z = 3 2x + 3y + 4z = 29 3x + 2y + z = 6 x + y + z = 3 y = 3 L 2 L 2 2L y 5z = 6 L 3 L 3 3L x + y + z = 3 y/5 + z = 23/5 L 2 L 3 /5 y = 3 L 3 L 2 x = = 2 z = 23/5 3/5 = 20/5 = 4 y = 3 To conclude, 2 containers of tye H, 3 containers of tye T and 4 containers of tye I

2 can be roduced given available steel. Exercise 2: a Denote by x the roduction level of Service, y the roduction level of Electricity and z the roduction level of Oil. There are 3 equations that need to be satised by x, y, z: they describe how each outut is used ithin the closed economy: i Outut Service: x = 0.3x + 0.3y + 0.3z ii Outut Electricity: y = 0.4x + 0.y + 0.5z iii Outut Oil: z = 0.3x + 0.6y + 0.2z We end u with a system of 3 linear equations that deend on 3 unknowns, x, y and z. Recall that this system is homogenous that is once we rewrite it in matrix form, the right-hand side coecient vector is zero. We also know that e.g. this system does not have a unique solution otherwise it would be the trivial one 0,0,0; this also means that the determinant of the coecient matrix is likely zero, and that Gaussian elimination is likely the only method to nd the solutions. Gaussian elimination to solve the above system: x = 0.3x + 0.3y + 0.3z y = 0.4x + 0.y + 0.5z z = 0.3x + 0.6y + 0.2z 0.7x 0.3y 0.3z = 0 0.4x + 0.9y 0.5z = 0 0.3x 0.6y+ 0.8z = 0 x 3/7y 3/7z = 0 L L / y 47z = 0 L 2 40L + 70L 2 5y + 47z = 0 L 3 30L + 70L 3 At this stage, it is imortant to realize that equations 2 and 3 are exactly identical! This is tyical from a system with degree of freedom. We can then eliminate either equation 2 or equation 3, without loosing any information. In other words, we have the 2

3 following system that is exactly equivalent to the revious ones: { x 3/7y 3/7z = 0 L L /0.7 { + 5y 47z = 0 L 2 40L + 70L 2 x = 3/7y + z = 3/747/5z + z = 4/4z y = 47/5z We have found an innite number of solutions that can be written as functions of z as follows: x = 4/7z, y = 47/5z, z is any real number. The system has one degree of freedom. In other words, every time I choose a dierent real number for z, I get a new solution by lugging it x and y. Note that if you exress your solutions as functions of y or z this is erfectly valid too. b If I want to x the roduction level of the industry service to 00, I have xed one degree of freedom of the system and I should nd the unique solution associated with it. Since I have exressed my solutions in a as functions of z, I need to reexress them as functions of x and lug in x = 00 to nd the associated solution: x = 4/7z z = 7/4x y = 47/5z y = 47/5 7/4x = 47/42x z real number x real number Now if I x x = 00, I get the associated unique solution: y = 39950/357.9 and z = 850/ Note that at this nal stage, it is aroriate to rovide an aroximate answer. Exercise 3: a - First, realize that the above model is a Markov chain model with 2 states: delay and no delay. So, in order to answer this question, we rst need to write the associated matrix transition. We call it T and we dene it as: T = Note that the convention I used in the notes is that the rows of T sum to. This is because each row, given the current state, describes the robability of being in either 3

4 state next eriod. More secically, in the matrix I dened above, the rst state is "no delay" and the second one is "delay". The rst row corresonds to what haens when the current state is "no delay": that is, next eriod there is a robability 0.85 of being in the state "no delay" and robability 0.5 of being in the state "delay". - Second, realize that the current state that is at time 0 right before you begin your commute to DT Vancouver is "delay", that is state 2 in my modelization. The associated state vector is then: 0 x 0 = - Third, realize that the eriod of reference is 5 minutes. Hence, in order to answer the question, we have to calculate x 2. We know that: x 2 = x 0T 2 = x 0T T x 2 = [0 ] x 2 = [ ] x 2 = [ ] Hence, the robability that for the next 30 minutes the trac will be delayed is equal to b We are now looking for the steady state of the system. That is, we are looking for the vector π = π N π D such that 2 conditions are satised: i π T = π ii π { 0.85π N π D = π N 0.5π N π D = π D = π N + π D = There are many ways to solve this system. I suggest using π D = π N since we are looking for π N and using one equation of the system to nd π N. The solution is: 4

5 π N = 5/8. To conclude, the robability that the trac will not be delayed in the long run is 5/ c The key assumtion of Markov chains models considered in class is that the transition robabilities are constant. This assumtion is very unlikely to be valid in this roblem. This is related to the fact that trac atterns are not constant during the day. For instance, there is a lot more trac delays between 7-9am than between -3m. Exercise 4: a The exercise rovides you with a candidate for the inverse matrix of I + qj, say M to avoid confusion. So, the only thing you have to do is verify that this candidate does the job: in other words, check 2 requirements, i MI + qj = I and ii I + qjm = I. Then use the fact that the inverse is unique to conclude that this candidate is the actual inverse. We start with requirement i: i MI + qj = I q + nq J I + qj = I 2 + qij + nq JI = I + qj + nq J = I + nq 2 + nq J q 2 = I + nj J 2 + nq Realize that we cannot go further unless we nd J 2 rst. Actually, we would like to nd that J 2 = nj in order to conclude that the candidate satises condition i. By denition, J is a matrix of size n with all its elements equal to. We are looking for the elements i,j for i and j from to n of J 2 : call them b ij to avoid confusion. We will use the formula of the roduct of 2 matrices. We call for now the elements of J a ij and we kee in mind a ij = for i,j from to n. b ij = n a ik a kj = n = n = n k= k= k= 5

6 The above calculation is valid for any i,j from to n, so we conclude that all the elements of J 2 are equal to n; in other words, J 2 = nj. Now back to condition i: it is now easy to conclude that condition i holds, that is:mi + qj = I. For comleteness, we need to check condition ii: ii I + qjm = I + qj q + nq J I = I 2 + qji + nq IJ = I + qj + nq J n + nq J = I And we conclude that the candidate is indeed the inverse of I qj. b Here we have to rove that an equation holds. This equation involves the inverse of C + aj. We do not have much information about C, so we start from the denition of the inverse: C + ajc + aj = I CC + aj + ajc + aj = I C + aj C + aj = I C + aj C + ac + aj J = I Somehow, we also need to use the fact that: JC = CJ = O n. If we remultily by J, we get: And i is roved. JCC + aj + ajjc + aj = JI 0 n + aj 2 C + aj = J anjc + aj = J JC + aj = an J 6

7 If we now ostmultily by J, we get: and ii is roved. C + aj CJ + ac + aj JJ = IJ 0 + ac + aj J 2 = J anc + aj J = J C + aj J = an J 7

ECON 331 Homework #2 - Solution. In a closed model the vector of external demand is zero, so the matrix equation writes:

ECON 331 Homework #2 - Solution. In a closed model the vector of external demand is zero, so the matrix equation writes: ECON 33 Homework #2 - Solution. (Leontief model) (a) (i) The matrix of input-output A and the vector of level of production X are, respectively:.2.3.2 x A =.5.2.3 and X = y.3.5.5 z In a closed model the