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Solutions to selected exercises from Jehle and Reny (2001): Advanced Microeconomic Theory Thomas Herzfeld September 2010 Contents 1 Mathematical Appendix 1.1 Chapter A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Chapter A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Consumer Theory 2.1 Preferences and Utility . . . . . . 2.2 The Consumer’s Problem . . . . . 2.3 Indirect Utility and Expenditure . 2.4 Properties of Consumer Demand 2.5 Equilibrium and Welfare . . . . . 3 Producer Theory 3.1 Production . . . . . 3.2 Cost . . . . . . . . . 3.3 Duality in production 3.4 The competitive firm

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1 Mathematical Appendix

1 Mathematical Appendix 1.1 Chapter A1 A1.7 Graph each of the following sets. If the set is convex, give a proof. If it is not convex, give a counterexample. Answer (a) (x, y)|y = ex This set is not convex. Any combination of points would be outside the set. For example, (0, 1) and (1, e) ∈ (x, y)|y = ex , but combination of the two vectors with t = 21 not: ( 12 , e+1 )∈ / 2 (x, y)|y = ex . (b) (x, y)|y ≥ ex This set is convex. Proof: Let (x1 , y1 ), (x2 , y2 ) ∈ S = (x, y)|y ≥ ex . Since y = ex is a continuous function, it is sufficient to show that (tx1 + (1 − t)x2 , ty1 + (1 − t)y2 ) ∈ S for any particular t ∈ (0, 1). Set t = 21. Our task is to show that 21 (x1 + x2 ), 12 (y1 + y2 ) ∈ S. 12 (y1 + y2 ) ≥ 21 (ex1 + ex2 ), since yi ≥ ex1 for i = 1, 2. Also,

x1

x1 x2 1 1 x1 (e + ex2 ) ≥ e2 (x1 +x2 = e 2 · e 2 2 x2 x1 ⇔ ex1 + ex2 ≥ 2e 2 · e 2

⇔ ex1 − 2e 2 · e

x2 2

+ ex2 ≥ 0 ⇔ (ex1 − ex2 )2 ≥ 0.

(c) (x, y)|y ≥ 2x − x2 ; x > 0, y > 0 This set is notconvex.   9 1 1 1 2 For ,  1 example,  1  1 1 10 2 1 , 1910 ,1 2 ∈ S = (x, y)|y ≥ 2x − x ; x > 0, y > 0. However, 1, 2 = 2 10 , 2 + 2 1 10 , 2 ∈ /S

(d) (x, y)|xy ≥ 1; x > 0, y > 0 This set is convex. Proof: Consider any (x1 , y1 ), (x2 , y2 ) ∈ S = (x, y)|xy ≥ 1; x > 0, y > 0. For any t ∈ [0, 1], (tx1 + (1 − t)x2 )(ty1 + (1 − t)y2 ) = t2 x1 y1 + t(1 − t)(x1 y2 + x2 y1 ) + (1 − t)2 x2 y2

> t2 + (1 − t)2 + t(1 − t)(x1 y2 + x2 y1 ), since xi yi > 1. = 1 + 2t2 − 2t + t(1 − t)(x1 y2 + x2 y1 )

= 1 + 2t(t − 1) + t(1 − t)(x1 y2 + x2 y1 ) = 1 + t(1 − t)(x1 y2 + x2 y1 − 2) ≥ 1 iff x1 y2 + x2 y1 ≥ 0. y2 y1 y2 y1 −2≥0 x1 y2 + x2 y1 = x1 y1 + x2 y2 − 2 ≥ + y2 y1 y1 y2 y − 1 − 2y1 y2 + y2 ≥ 0 (y1 − y2 )2 ≥ 0, 2

1 Mathematical Appendix which is always true and therefore, (tx1 + (1 − t)x2 , ty1 + (1 − t)y2 ) ∈ S which is convex. (e) (x, y)|y ≤ ln(x) This set is convex. Proof. Let (x1 , y1 ) + (x2 , y2 ) ∈ S. Then 12 (y1 + y2 ) ≤ (ln(x1 ) + ln(x2 )). S is convex   1 1 if 1 ⇒ ln(x1 ) + ln(x2 ) ≤ ln( x1 + x2 ) 2 2 2 1 1 1 ⇔ ln(x1 x2 ) ≤ ln( x1 + x2 ) 2 2 2 1 1 ⇔ (x1 x2 )1/2 ≤ ( x1 + x2 ) 2 2 1/2 ⇔ x1 − 2(x1 x2 ) + x2 ≥ 0 2  1/2 1/2 ≥0 ⇔ x 1 + x2 which is always true.

A1.40 Sketch a few level sets for the following functions: y = x1 x2 , y = x1 + x2 and y = min[x1 , x2 ]. Answer x2

x2

✻

✲ (a) y = x1 x2

x1

x2

✻ ❅

❅ ❅ ❅ ❅❅❅❅ ❅ ❅ ✲

✻

✲

x1

(b) y = x1 + x2

x1

(c) y = min(x1 , x2 )

Figure 1: Sets to Exercise A1.40 A1.42 Let D = [−2, 2] and f : D → R be y = 4 − x2 . Carefully sketch this function. Using the definition of a concave function, prove that f is concave. Demonstrate that the set A is a convex set. Answer Proof of concavity: Derive the first and second order partial derivative: ∂ 2y = −2 ∂x2

∂y = −2x ∂x

The first derivative is strictly positive for values x < 0 and negative for values x > 0. The second order partial derivative is always less than zero. Therefore, the function is concave. Proof of convexity: The area below a concave function forms a convex set (Theorem

3

1 Mathematical Appendix A1.13). Alternatively, from the definition of convexity the following inequality should hold 4 − (tx1 + (1 − t)x2 )2 ≥ t(4 − (x1 )2 ) + (1 − t)(4 − (x2 )2 ). Multiply out to get 4 − (tx1 + x2 − tx2 )2 ≥ 4 − x22 + t[(x1 )2 − (x2 )2 ]. Again, the area below the function forms a convex set. y ✻

✲x

Figure 2: Graph to Exercise A1.42 A1.46 Consider any linear function f (x) = a · x + b for a ∈ Rn and b ∈ R. (a) Show that every linear function is both concave and convex, though neither is strictly concave nor strictly convex. Answer The statement is true iff, for any x1 , x2 ∈ Rn , t ∈ [0, 1], it is true that f (tx1 + (1 − t)x2 ) = tf (x1 ) + (1 − t)f (x2 ). Substituting any linear equation in this statement gives f (tx1 +(1−t)x2 ) = a[tx1 +(1−t)x2 ]+b = tax1 +(1−t)ax2 +tb+(1−t)b = tf (x1 )+(1−t)f (x2 ) for all x1 , x2 ∈ Rn , t ∈ [0, 1]. (b) Show that every linear function is both quasiconcave and quasiconvex and, for n > 1, neither strictly so. (There is a slight inaccuracy in the book.) Answer As it is shown in (a) that a linear function is concave and convex, it must also be quasiconcave and quasiconvex (Theorem A1.19). More formally, the statement is true iff, for any x1 , x2 ∈ Rn (x1 6= x2 ) and t ∈ [0, 1], we have f (tx1 + (1 − t)x2 ) ≥ min[f (x1 ), f (x2 )](quasiconcavity)

f (tx1 + (1 − t)x2 ) ≤ max[f (x1 ), f (x2 )](quasiconvexity) Again by substituting the equation into the definition, we get tf (x1 ) + (1 − t)f (x2 ) ≥ min[f (x1 ), f (x2 )]

tf (x1 ) + (1 − t)f (x2 ) ≤ max[f (x1 ), f (x2 )]

∀t ∈ [0, 1]

A1.47 Let f (x) be a concave (convex) real-valued function. Let g(t) be an increasing concave (convex) function of a single variable. Show that the composite function, h(x) = g(f (x)) is a concave (convex) function. Answer The composition with an affine function preserves concavity (convexity). Assume that both functions are twice differentiable. Then the second order partial derivative of the composite function, applying chain rule and product rule, is defined as h′′ (x) = g ′′ (f (x)) f ′ (x)2 + g ′ (f (x)) f ′′(x)2

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1 Mathematical Appendix For any concave function, ∇2 f (x) ≤ 0, ∇2 g(x) ≤ 0, it should hold ∇2 h(x) ≤ 0. In the case the two functions are convex: ∇2 f (x) ≥ 0 and ∇2 g(x) ≥ 0, it should hold ∇2 h(x) ≥ 0. A1.48 Let f (x1 , x2 ) = −(x1 − 5)2 − (x2 − 5)2 . Prove that f is quasiconcave. Answer Proof: f is concave iff H(x) is negative semidefinite and it is strictly concave if the Hessian is negative definite.   −2 0 H= 0 −2 T z H(x)z = −2z12 − 2z22 < 0, for z = (z1 , z2 ) 6= 0 Alternatively, we can check the leading principal minors of H: H1 (x) = −2 < 0 and H2 (x) = 4 > 0. The determinants of the Hessian alternate in sign beginning with a negative value. Therefore, the function is even strictly concave. Since f is concave, it is also quasiconcave. A1.49 Answer each of the following questions “yes” or ”no“, and justify your answer. (a) Suppose f (x) is an increasing function of one variable. Is f (x) quasiconcave? Answer Yes, an increasing function of one variable is quasiconcave. Any convex combination of two points on this function will be at least as large as the smallest of the two points. Using the differential-based approach, f is quasiconcave, if for any x0 and x1 , f (x1 ) ≥ f (x0 ) ⇒ ∂f (x0 )/∂x(x1 − x0 ) ≥ 0. This must be true for any increasing function. (b) Suppose f (x) is a decreasing function of one variable. Is f (x) quasiconcave? Answer Yes, a decreasing function of one variable is quasiconcave. Similarly to (a), f is quasiconcave if for any x0 , x1 and t ∈ [0, 1], it is true that f (tx0 + (1 − t)x1 ) ≥ min[f (x0 ), f (x1 )]. (c) Suppose f (x) is a function of one variable and there is a real number b such that f (x) is decreasing on the interval (− inf, b] and increasing on [b, + inf). Is f (x) quasiconcave? Answer No, if f is decreasing on (− inf, b] and increasing on [b, + inf) then f (x) is not quasiconcave. Proof: Let a < b < c, and let tb = c−b ∈ [0, 1], tb a + (1 − tb )c = b. Given the nature c−a of f , f (b) < min[f (a), f (c)]. Then f (tb a + (1 − tb )c) < min[f (a), f (c)], so f is not quasiconcave. (d) Suppose f (x) is a function of one variable and there is a real number b such that f (x) is increasing on the interval (− inf, b] and decreasing on [b, + inf). Is f (x) quasiconcave? Answer Yes. Proof: Let a < b < c, for x ∈ [a, b], f (x) ≥ f (a) and for x ∈ [b, c], f (x) ≥ f (c). Hence, for any x ∈ [a, c], f (x) ≥ min[f (a), f (c)].

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1 Mathematical Appendix (e) You should now be able to come up with a characterization of quasiconcave functions of one variable involving the words “increasing” and “decreasing”. Answer Any function of one variable f (x) is quasiconcave if and only if is either continuously increasing, continuously decreasing or first increasing and later decreasing.

1.2 Chapter A2 A2.1 Differentiate the following functions. State whether the function is increasing, decreasing, or constant at the point x = 2. Classify each as locally concave, convex, or linear at the point x = 2. (a) f (x) = 11x3 − 6x + 8 f1 = 33x2 − 6 increasing locally convex (b) f (x) = (3x2 − x)(6x + 1) f1 = 54x2 − 6x − 1

increasing locally convex

(c) f (x) = x2 −

1 x3

3 x4 increasing locally concave

f1 = 2x +

(d) f (x) = (x2 + 2x)3

f1 = (6x + 6)(x2 + 2x)2 increasing locally convex

x3 − 3x2 + 1 (x3 + 1)3 increasing locally concave  3  1 8 4 1 − − (f) f (x) = [(1/x2 + 2) − (1/x − 2)]4 f1 = +4 x2 x x2 x3 increasing locally convex Z 1 2 2 (g) f (x) = et dt f1 = −ex (e) f (x) = [3x/(x3 + 1)]2

f1 = 18x

x

decreasing locally convex

A2.2 Find all first-order partial derivatives. (a) f (x1 , x2 ) = 2x1 − x12 − x22 f1 = 2 − 2x1 = 2(1 − x1 )

f2 = −2x2

(b) f (x1 , x2 ) = x12 + 2x22 − 4x2 f1 = 2x1 f2 = 4x2 − 4

6

1 Mathematical Appendix (c) f (x1 , x2 ) = x13 − x22 − 2x2 f1 = 3x1

f2 = −2(x2 + 1)

(d) f (x1 , x2 ) = 4x1 + 2x2 − x12 + x1 x2 − x22 f1 = 4 − 2x1 + x2 f2 = 2 − 2x2 + x1 (e) f (x1 , x2 ) = x13 − 6x1 x2 + x32 f2 = 3x22 − 6x1 f1 = 3x12 − 6x2 (f) f (x1 , x2 ) = 3x12 − x1 x2 + x2 f1 = 6x1 − x2 f2 = 1 − x1   (g) g(x1 , x2 , x3 ) = ln x12 − x2 x3 − x23 2x1 −x3 g1 = 2 g2 = 2 2 x1 − x2 x3 − x3 x 1 − x2 x3 − x23 −x2 − 2x3 g3 = 2 x1 − x2 x3 − x23 A2.4 Show that y = x12x2 + x22 x3 + x32 x1 satisfies the equation ∂y ∂y ∂y = (x1 + x2 + x3 )2 . + + ∂x1 ∂x2 ∂x3 The first-order partial derivatives are ∂y/∂x1 = 2x1 x2 + x32, ∂y/∂x2 = x12 + 2x2 x3 , and ∂y/∂x3 = x22 + 2x3 x1 . Summing them up gives ∂y ∂y ∂y = x21 + x22 + x32 + 2x1 x2 + 2x1 x3 + 2x2 x3 = (x1 + x2 + x3 )2 . + + ∂x1 ∂x2 ∂x3 A2.5 Find the Hessian matrix and construct the quadratic form, zT H(x)z, when (a) y = 2x1 − x12 − x22   −2 0 H= 0 −2

zT H(x)z = −2z12 + 2 ∗ 0z1 z2 − 2z 22 (b) y = x12 + 2x22 − 4x2   2 0 H= 0 4 T z H(x)z = 2z 12 + 2 ∗ 0z1 z2 + 4z22

7

1 Mathematical Appendix (c) y = x13 − x22 + 2x2   6x1 0 H= 0 −2 T z H(x)z = 6x1 z12 − 2z 22 (d) y = 4x1 + 2x2 − x12 + x1 x2 − x22   −2 1 H= 1 −2 T z H(x)z = −2z12 + 2z1 z2 − 2z22 (e) y = x31 − 6x1 x2 − x32   6x1 −6 H= −6 6x2

zT H(x)z = 6x1 z12 − 12z1 z2 + 6x2 z22 p A2.8 Suppose f (x1 , x2 ) = x21 + x22. (a) Show thatpf (x1 , x2 ) is homogeneous of degree p 1. p f (tx1 , tx2 ) = (tx1 )2 + (tx2 )2 = t2 (x12 + x22) = t x21 + x22 . (b) According to Euler’s theorem, we should have f (x1 , x2 ) = (∂f /∂x1 ) x1 +(∂f /∂x2 ) x2 . Verify this. x2 + x22 x1 x2 x2 = p 1 = 1 · f (x1 , x2 ) = p 2 x 1+ p 2 x21 + x22 x1 + x22 x1 + x22

q

A2.9 Suppose f (x1 , x2 ) = (x1 x2 )2 and g(x1 , x2 ) = (x12x2 )3 . (a) f (x1 , x2 ) is homogeneous. What is its degree? f (tx1 , tx2 ) = t4 (x1 x2 )2 k = 4 (b) g(x1 , x2 ) is homogeneous. What is its degree? g(tx1 , tx2 ) = t9 (x21 x2 )3 k = 9

(c) h(x1 , x2 ) = f (x1 , x2 )g (x1 , x2 ) is homogeneous. What is its degree? h(x1 , x2 ) = (x13 x22)5 h(tx1 , tx2 ) = t25 (x13x22 )5 k = 25

8

x12 + x22

1 Mathematical Appendix (d) k(x1 , x2 ) = g (f (x1 , x2 ), f (x1 , x2 )) is homogeneous. What is its degree? k(tx1 , tx2 ) = t36 (x1 x2 )18 k = 36 (e) Prove that whenever f (x1 , x2 ) is homogeneous of degree m and g(x1 , x2 ) is homogeneous of degree n, then k(x1 , x2 ) = g (f (x1 , x2 ), f (x1 , x2 )) is homogeneous of degree mn. k(tx1 , tx2 ) = [tm (f (x1 , x2 ), f (x1 , x2 ))]n k = mn

A2.18 Let f (x) be a real-valued function defined on Rn+ , and consider the matrix   0 f1 · · · fn f1 f11 · · · f1n    H∗ =  . . ... . . . ...   ..  fn fn1 · · · fnn This is a different sort of bordered Hessian than we considered in the text. Here, the matrix of second-order partials is bordered by the first-order partials and a zero to complete the square matrix. The principal minors of this matrix are the determinants    0 f1 f2       0 f1   f1 f11 f12  , . . . , Dn = |H∗ |.  , D3 = D2 =    f1 f11   f2 f21 f22 

Arrow & Enthoven (1961) use the sign pattern of these principal minors to establish the following useful results:

(i) If f (x) is quasiconcave, these principal minors alternate in sign as follows: D2 ≤ 0, D3 ≥ 0, . . . . (ii) If for all x ≥ 0, these principal minors (which depend on x) alternate in sign beginning with strictly negative: D2 < 0, D3 > 0, . . . , then f (x) is quasiconcave on the nonnegative orthant. Further, it can be shown that if, for all x ≫ 0, we have this same alternating sign pattern on those principal minors, then f (x) is strictly quasiconcave on the (strictly) positive orthant. (a) The function f (x1 , x2 ) = x1 x2 + x1 is quasiconcave on R+2 . Verify that its principal minors alternate in sign as in (ii). Answer The bordered Hessian is   0 x2 + 1 x1 0 1 . H∗ = x2 + 1 x1 1 0

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1 Mathematical Appendix The two principal minors are D2 = −(x2 +1)2 < 0 and D3 = 2x1 x2 +2x1 ≥ 0. Which shows that the function will be quasiconcave and will be strictly quasiconcave for all x1 , x2 > 0. (b) Let f (x1 , x2 ) = a ln(x1 + x2 ) + b, where a > 0. Is this function strictly quasiconcave for x ≫ 0? It is quasiconcave? How about for x ≥ 0? Justify. Answer The bordered Hessian is   a a 0 x1 +x2 x1 +x2   a −a −a H∗ =  x1 +x2 (x1 +x2 )2 (x1 +x2 )2  . a x1 +x2

−a (x1 +x2 )2

−a (x1 +x2 )2

The two principal minors are D2 = −( x1 +a x2 )2 < 0 for x1 , x2 > 0 and D3 = 0. Which shows that the function can not be strictly quasiconcave. However, it can be quasiconcave following (i). For x1 = x2 = 0 the function is not defined. Therefore, curvature can not be checked in this point. A2.19 Let f (x1 , x2 ) = (x1 x2 )2 . Is f (x) Answer The bordered Hessian is  0 H∗ = 2x1 x22 2x21 x2

2 concave on R+2 ? Is it quasiconcave on R+ ?

 2x1 x22 2x21 x2 2x22 4x1 x2  . 4x1 x2 2x12

The two principal minors are D2 = −(2x1 x2 )2 < 0 and D3 = 16x14x42 ≥ 0. Which shows that the function will be strictly quasiconcave. Strict quasiconcavity implies quasioncavity. A2.25 Solve the following problems. State the optimised value of the function at the solution. (a) minx1 ,x2 = x12 + x22 s.t. x1 x2 = 1 x1 = 1 and x2 = 1 or x1 = −1 and x2 = −1, optimised value= 2 (b) minx1 ,x2 = x1 x2 s.t. x12 + x22 = 1 p p p p x1 = 1/2 and x2 = − 1/2 or x1 = − 1/2 and x2 = 1/2, optimised value= −1/2 2 2 2 2 2 1 (c) max p x1 ,x2 = x1 x2 s.t.px1 /a + x2 /b = p 2 2 2 x1 = a /3 and x2 = 2b /3 or x2 = − 2b2 /3, optimised value= 32ab 3 /2 4 4 (d) max 2 s.t. x1 + x2 = 1 √ p x1 ,x2 = x1 + xp 4 x1 = 4 1/2 and x2 = 4 1/2, optimised value= 23 = 23/4 (e) maxx1 ,x2 ,x3 = x1 x22x33 s.t. x1 + x2 + x3 = 1 x1 = 1/6 and x2 = 1/3 = 2/6 and x3 = 1/2 = 3/6, optimised value= 1/432 = 108/66

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1 Mathematical Appendix A2.26 Graph f (x) = 6 − x2 − 4x. Find the point where the function achieves its unconstrained (global) maximum and calculate the value of the function at that point. Compare this to the value it achieves when maximized subject to the nonnegativity constraint x ≥ 0. Answer This function has a global optimum at x = −2. It is a maximum as the secondorder partial derivative is less than zero. Obviously, the global maximum is not a solution in the presence of a nonnegativity constraint. The constrained maximization problem is L(x, z, λ) = 6 − x2 − 4x + λ(x − z) The first order conditions and derived equations are: ∂L ∂L = −2x − 4 + λ = 0 = −λ ≤ 0 ∂x ∂z λ= x−z z = x

∂L = x−z = 0 ∂λ λx = 0

If λ = 0, then x = −2 would solve the problem. However, it does not satisfy the nonnegativity constraint. If λ 6= 0, then x = 0. As the function is continuously decreasing for all values x ≥ 0, it is the only maximizer in this range. y ✻

✲

x

Figure 3: Graph to Exercise A2.26

11

2 Consumer Theory

2 Consumer Theory 2.1 Preferences and Utility 1.6 Cite a credible example were the preferences of an ‘ordinary consumer’ would be unlikely to satisfy the axiom of convexity. Answer: Indifference curves representing satiated preferences don’t satisfy the axiom of convexity. That is, reducing consumption would result in a higher utility level. Negative utility from consumption of ‘bads’ (too much alcohol, drugs etc.) would rather result in concave preferences. 1.8 Sketch a map of indifference sets that are parallel, negatively sloped straight lines, with preference increasing northeasterly. We know that preferences such as these satisfy Axioms 1, 2, 3, and 4. Prove the they also satisfy Axiom 5′ . Prove that they do not satisfy Axiom 5. Answer: Definition of convexity (Axiom 5′ ): If x1 % x0 , then tx1 + (1 − t)x0 % x0 for all t ∈ [0, 1]. Strict convexity (Axiom 5) requires that, if x1 6= x0 and x1 % x0 , then tx1 + (1 − t)x0 ≻ x0 for all t ∈ [0, 1]. The map of indifference sets in the figure below represent perfect substitues. We know that those preferences are convex but not stricly convex. Intuitively, all combinations of two randomly chosen bundles from one indifference curve will necessarily lie on the same indifference curve. Additionally, the marginal rate of substitution does not change by moving from x0 to x1 . To prove the statement more formally, define xt as convex combination of bundles x0 to x1 : xt = tx0 + (1 − t)x1 . Re-writing in terms of single commodities gives us: xt = (tx10, tx02 ) + ((1 − t)x11, (1 − t)x12 ). A little rearrangement and equalising the two definitions results in the equality tx0 + (1 − t)x1 = (tx10 + (1 − t)x11), tx20 + (1 − t)x12 ). That is, the consumer is indifferent with respect to the convex combination and the original bundles, a clear violation of strict convexity.

12

2 Consumer Theory

x2

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❍

❍

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❍ ❍

❍ ❍

❍x r

❍

t

❍

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