Production Economics 2

Production Economics 2 - Multiple Input Production...

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Unformatted text preview: Multiple Input Production Economics for Farm Management AAE 320 Paul D. Mitchell Multiple Input Production Most agricultural production processes have more than one input, e.g., capital, labor, land, or N, P, K fertilizer, plus herbicides, insecticides, tillage, water, etc. How do you decide how much of each input to use when you have more than one input? Derive the Equal Margin Principle and show its use to answer this question Equal Margin Principle Will derive using calculus so you see where it comes from Applies whether you use a function or not, so can apply Equal Margin Principle to the tabular form of the multiple input production schedule Calculus requires Partial Derivatives Partial Derivatives Derivative of function that has more than one variable What's the derivative of f(x,y)? Depends on which variable you are talking about. Remember a derivative is the slope Derivative of f(x,y) with respect to x is the slope of the function in the x direction Derivative of f(x,y) with respect to y is the slope of the function in the y direction Partial Derivatives Think of a hill; its elevation (z) is a function of the location in latitude (x) and longitude (y): z = f(x,y) At any spot on the hill (defined by a latitudelongitude pair (x,y)) the hill will have a slope in the x direction and in the y direction Slope in x direction: dz/dx = fx(x,y) Slope in y direction: dz/dy = fy(x,y) Q Q=f(X,Y) Y X Source: "Neoclassical Theories of Production" on The History of Economic Thought Website: http://cepa.newschool.edu/het/essays/product/prodcont.htm Partial Derivatives Notation: if have q = f(x,y) First Partial Derivatives Second (Own) Partial Derivative dq/dx = fx(x,y) and dq/dy = fy(x,y) Second Cross Partial Derivative d2q/dx2 = fxx(x,y) and d2q/dy2 = fyy(x,y) d2q/dxdy = fxy(x,y) Partial Derivatives Partial derivatives are the same as regular derivatives, just treat the other variables as constants q = f(x,y) = 2 + 3x + 6y 2x2 3y2 5xy When you take the derivative with respect to x, treat y as a constant and vice versa fx(x,y) = 3 4x 5y fy(x,y) = 6 6y 5x Think Break #5 Give the 1st and 2nd derivatives [fx(x,y), fy(x,y),, fxx(x,y), fyy(x,y), fxy(x,y)] of each: f(x,y) = 7 + 5x + 2y 5x2 4y2 11xy f(x,y) = 5 2x + y x2 3y2 + 2xy Given production function q = f(x,y), find (x,y) to maximize = pf(x,y) rxx ryy K FOC's: d /dx = 0 and d /dy = 0 and solve for pair (x,y) d /dx = pfx(x,y) rx = 0 d /dy = pfy(x,y) ry = 0 Just p x MPx = rx and p x MPy = ry Just MPx = rx/p and MPy = ry/p These still hold, but we also have more Equal Margin Principle Equal Margin Principle Profit Maximization again implies p x MPx = rx and p x MPy = ry p x MPy = ry depends on x Note that p x MPx = rx depends on y and Note that two equations and both must be satisfied, so rearrange (make ratio) pMPx rx = pMPy ry pMPx pMPy = rx ry Equal Margin Principle Equal Margin Principle is expressed mathematically in two ways 1) MPx/rx = MPy/ry 2) MPx/MPy = rx/ry Ratio of MPi/ri must be equal for all inputs Ratio of MP's must equal input price ratio Intuition: Corn Example MPi is bu of corn per lb of N fertilizer (bu/lb) ri is $ per lb of N fertilizer ($/lb) MPi/ri is bu corn per $ spent on N fertilizer (bu/lb)/ ($/lb) = bu/$ MPi/ri is how many bushels of corn you get for the last dollar spent on N fertilizer MPx/rx = MPy/ry means use inputs so that the last dollar spent on each input gives the same output Intuition: Corn Example MPx is bu of corn from last lb of N fert. (bu/lb N) MPy is bu of corn from last lb of P fert. (bu/lb P) MPx/MPy = (lbs P/lbs N) is how much P need if cut N by 1 lb and want to keep output constant Ratio of marginal products is the substitution rate between N and P in the production process Intuition: Corn Example rx is $ per lb of N fertilizer ($/lb N) ry is $ per lb of P fertilizer ($/lb P) rx/ry is ($/lb N)/($/lb P) = lbs P/lbs N, or the substitution rate between N and P in the market MPx/MPy = rx/ry means use inputs so that the substitution rate between inputs in the production process is the same as the substitution rate between inputs in the market Marginal Rate of Technical Substitution The ratio of marginal products (MPx/MPy) is the substitution rate between inputs in the production process MPx/MPy is called the Marginal Rate of Technical Substitution (MRTS): the input substitution rate at the margin for the production technology If cut X by one unit, how much must you increase Y to keep output the same Optimality condition MPx/MPy = rx/ry means set substitution rates equal Equal Margin Principle Intuition MPx/rx = MPy/ry means use inputs so the last dollar spent on each input gives the same output at the margin MPx/MPy = rx/ry means use inputs so the substitution rate at the margin between inputs is the same in the production process as in the market place Compare to p x MP = r or VMP = r Compare to MP = r/p Equal Margin Principle Graphical Analysis via Isoquants Isoquant ("equalquantity") plot or function representing all combinations of two inputs producing the same output quantity Intuition: Isoquants are the two dimensional "contour lines" of the three dimensional production "hill" First look at in Theory, then a Table Isoquants in Theory Input Y ~ Output Q = Q Input X Swine Feeding Operation: Output (hogs sold/year) Capital ($1,000) 1 250 500 750 1000 1250 1500 1750 2000 0.5 1.5 3 5 Labor (persons/year) 4 3 5 8 6 14.5 27.5 30 7 22 29 31 2 1 3 3 2 8 8 9 10 26 28 29.5 31 32 32.5 33 33.5 25 27.5 30 29.5 31 30.5 14.5 21.5 22 27.5 27.5 30 8 14.5 12 22 31 31.5 31.5 31.5 31.5 31.5 31.5 31 32 32 32 8 14.5 27.5 8 14.5 27.5 8 6.5 22 27.5 22 25 29 30.5 30 30 27.5 31 31 30 32 32.5 32.5 32 32.5 32 32.5 33 33 Swine feeding operation: Output (hogs sold/year) Capital ($1,000) 1 250 500 750 1000 1250 1500 1750 2000 0.5 1.5 3 5 Labor (persons/year) 4 3 5 8 6 14.5 27.5 30 7 22 29 31 2 1 3 3 2 8 8 9 10 26 28 29.5 31 32 32.5 33 33.5 25 27.5 30 29.5 31 30.5 14.5 21.5 22 27.5 27.5 30 8 14.5 12 22 31 31.5 31.5 31.5 31.5 31.5 31.5 31 32 32 32 8 14.5 27.5 8 14.5 27.5 8 6.5 22 27.5 22 25 29 30.5 30 30 27.5 31 31 30 32 32.5 32.5 32 32.5 32 32.5 33 33 35 30 25 Hogs/year (1,000) 20 15 10 5 0 250 500 750 1000 1250 1500 1750 2000 S8 S7 S6 S5 S4 Labor (persons) S3 S2 S1 Capital ($1,000) Isoquants and MPx/MPy Isoquant Slope = Y/ X = Substitution rate between X and Y at the margin. Y X If reduce X by the amount X, then must increase Y by the amount Y to keep output fixed ~ Output Q = Q Input Y Input X Isoquants and MPx/MPy Isoquant slope = dY/dX dY/dX = (dQ/dX)/(dQ/dY) dY/dX = MPx/MPy = Ratio of MP's = Isoquant Slope = MRTS Need minus sign since marginal products are positive and slope is negative ~ Output Q = Q Input X Input Y Y X Isoquants and MPx/MPy = rx/ry Isoquant slope ( MPx/MPy) is (minus) the substitution rate between X and Y at the margin Thus define minus the isoquant slope as the "Marginal Rate of Technical Substitution" (MRTS) Price ratio rx/ry is the substitution rate between X and Y (at the margin) in the market place Economically optimal use of inputs sets these substitution rates equal, or MPx/MPy = rx/ry Isoquants and MPx/MPy = rx/ry Find the point of tangency between the isoquant and the input price ratio Kind of like the MP = r/p: tangency between the production function and inputoutput price ratio line ~ Output Q = Q Slope = MPx/MPy = rx/ry Input X Input Y Soybean Meal (lbs) 10 15 20 25 30 35 40 45 50 55 60 65 70 Corn (lbs) 376.8 356.3 339.0 326.0 315.0 307.5 300.6 294.6 289.2 284.4 280.2 276.4 272.9 Soybean meal and corn needed for 125 lb feeder pigs to gain 125 lbs 375 350 325 Corn (lbs) 300 275 250 225 200 0 20 40 Soybean Meal (lbs) 60 80 Which feed ration do you use? Soyb Meal Corn 10 15 20 25 30 35 40 45 50 55 60 65 70 376.8 356.3 339.0 326.0 315.0 307.5 300.6 294.6 289.2 284.4 280.2 276.4 272.9 MRTS 4.10 3.46 2.60 2.20 1.50 1.38 1.20 1.08 0.96 0.84 0.76 0.70 MRTS = As increase Soybean Meal, how much can you decrease Corn and keep output constant? Can't use MRTS = Can't use MRTS = MPx/MPy since Q = 0 on an since isoquant Use MRTS = Y/ X X = (SoyM2 SoyM1) = (294.6 300.6)/(45 40) Y = (Corn2 Corn1) = = (284.4 289.2)/(55 50) Interpretation: If increase soybean meal 1 lb, decrease corn by 1.20 or 0.96 lbs and keep same gain on hogs Soy Meal (lbs) 10 15 20 25 30 35 40 45 50 55 60 65 70 Corn MRTS (lbs) 376.8 356.3 339.0 326.0 315.0 307.5 300.6 294.6 289.2 284.4 280.2 276.4 272.9 4.10 3.46 2.60 2.20 1.50 1.38 1.20 1.08 0.96 0.84 0.76 0.70 P ratio 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 2.20 Economically Optimal input use where MRTS = input price ratio, MPx/MPy = rx/ry Soybean Meal Price $176/ton = $0.088/lb Corn Price $2.24/ bu = $0.04/lb Ratio: 0.088/0.04 = 2.20 lbs corn/lbs soy Keep straight which is X and which is Y!!! Intuition What does Y/ X = rx/ry mean? Ratio Y/ X is how much less Y you need if you increase X, or the MRTS Cross multiply to get Yry = Xrx Yry = cost savings from decreasing Y Xrx = cost increase from increasing X Slide down the isoquant (decreasing Y and increasing X) as long as cost savings exceeds cost increase Main Point on MRTS When you have information on inputs needed to generate the same output, you can identify the optimal input combination Theory: (MRTS) MPx/MPy = rx/ry Practice: (isoquant slope) Y/ X = rx/ry Note how the input in the numerator and denominator changes between theory and practice Think Break #6 Grai Hay MRT pric n (lbs) S e (lbs) ratio 825 900 975 1050 1125 1200 135 0 113 0 935 770 625 525 2.20 1.93 2.93 The table is rations of grain and hay that put 300 lbs of gain on 900 lb steers. 1) Fill in the missing MRTS 2) If grain is $0.04/lb and hay is $0.03/lb, what is the economically optimal feed ration? Summary Can identify economically optimal input combination using tabular data and prices Requires input data on the isoquant, i.e., different feed rations that generate the same amount of gain Again: use calculus to fill in gaps in the tabular form of isoquants Types of Substitution Perfect substitutes: soybean meal & canola meal, corn & sorghum, wheat and barley Imperfect substitutes: corn & soybean meal Nonsubstitutes/perfect complements: tractors and drivers, wire and fence posts Perfect Substitutes MRTS (slope of isoquant) is constant Examples: canola meal and soybean meal, or corn and sorghum in a feed ration Constant conversion between two inputs 2 pounds of canola meal = 1 pound of soybean meal: Cnla/ Soyb = 2 1.2 bushels of sorghum = 1 bu corn Srgm/ Corn = 1.2 MRTS = rx/ry still applies MRTS = k: Y/ X = Cnla/ Soyb = 2 If rx/ry < k, use only input x Economics of Perfect Substitutes If rx/ry > k, use only input y soybean meal < twice canola meal price soyb meal $200/cwt, cnla meal $150/cwt soybean meal > twice canola meal price soyb meal $200/cwt, cnla meal $90/cwt If rx/ry = k, use any xy combination Use only Canola Meal rx/ry > 2 2 Perfect Substitutes: Graphics Canola Meal 1 Use either Canola or Soybean Meal y = 2 rx/r Use only Soybean rx/ry < 2 Meal Soybean Meal Economics of Imperfect Substitutes This the case we already did! Input Y Input X Perfect Complements, or NonSubstitutes No substitution is possible The two inputs must be used together Without the other input, neither input is productive, they must be used together Tractors and drivers, fence posts and wire, chemical reactions, digger and shovel Inputs used in fixed proportions 1 driver for 1 tractor Economics of Perfect Complements MRTS is undefined, the price ratio rx/ry does not identify the optimal combination Leontief Production Function Marginal products of inputs = 0 Either an output quota or a cost budget, with the fixed input proportions, define how much of the inputs to use Q = min(aX,bY): Perfect Complements: Graphics Price ratio does not matter rx/ry > 2 Drivers rx/ry < 2 Tractors Multiple Input Production with Calculus Use calculus with production function to find the optimal input combination General problem we've seen: Find (x,y) to maximize = pf(x,y) rxx ryy K Will get FOC's, one for each choice variable, and SOC's are more complicated Multiple Input Production with Calculus Max 10(7 + 9x + 8y 2x2 y2 2xy) 2x 3y 8 FOC1: 10(9 4x 2y) 2 = 0 FOC2: 10(8 2y 2x) 3 = 0 Find the (x,y) pair that satisfies these two equations Same as finding where the two lines from the FOC's intersect FOC1: 10(9 4x 2y) 2 = 0 FOC2: 10(8 2y 2x) 3 = 0 1) Solve FOC1 for x 88 40x 20y = 0 40x = 88 20y x = (88 20y)/40 = 2.2 0. 5y 2) Substitute this x into FOC2 and solve for y 77 20y 20x = 0 77 20y 20(2.2 0.5y) = 0 77 20y 44 + 10y = 0 33 10y = 0, or 10y = 33, or y = 3.3 3) Calculate x: x = 2.2 0. 5y = 2.2 0.5(3.3) = 0.55 Second Order Conditions SOC's are more complicated with multiple inputs, must look at curvature in each direction, plus the "cross" direction (to ensure do not have a saddle point). 1) Own second derivatives must be negative for a maximum: fxx < 0, fyy < 0 (both > 0 for a minimum) 2) Another condition: fxxfyy (fxy)2 > 0 FOC1: 10(9 4x 2y) 2 = 0 FOC2: 10(8 2y 2x) 3 = 0 Second Derivatives 1) fxx: 10( 4) = 40 2) fyy: 10( 2) = 20 3) fxy: 10( 2) = 20 SOC's 1) fxx = 40 < 0 and fyy = 20 < 0 ( 40)( 20) ( 20)2 = 800 400 > 0 Solution x = 0.55 and y = 3.3 is a maximum 2) fxxfyy (fxy)2 > 0 Milk Production M = 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH M = milk production (lbs/week) G = grain (lbs/week) H = hay (lbs/week) Milk $14/cwt, Grain $3/cwt, Hay $1.50/cwt What are the profit maximizing inputs? (Source: Heady and Bhide 1983) Milk Profit Function = 0.14( 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH) 0.03G 0.015H FOC's 0.14(2.56 0.0101G 0.00352H) 0.03 = 0 0.14(1.05 0.00218H 0.00352G) 0.015 = 0 Alternative: jump to optimality conditions MPG = rG/p and MPH = rH/p MPG = 2.56 0.0101G 0.00352H = rG/p = 0.214 MPH = 1.05 0.00218H 0.00352G = rH/p = 0.107 Solve FOC1 for H: 2.56 0.0101G 0.00352H = 0.214 2.346 0.0101G = 0.00352H H = (2.346 0.0101G)/0.00352 H = 666 2.87G Substitute this H into FOC2 and solve for G: 1.05 0.00218H 0.00352G = 0.107 0.943 0.00218(666 2.87G) 0.00352G = 0 0.943 1.45 + 0.00626G 0.00352G = 0 0.00274G 0.507 = 0 G = 0.507/0.00274 = 185 lbs/week Substitute this G into the equation for H H = 666 2.87G H = 666 2.87(185) H = 666 531 = 135 lbs/week Check SOC's GG = (0.14)0.0101 = 0.001414 < 0 GH = (0.14)0.00352 = 0.000493 SOC's satisfied for a maximum GG HH ( GH)2 = 1.88 x 107 > 0 HH = (0.14)0.00218 = 0.000305 < 0 Milk Output M = 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH M = 25.9 + 2.56(185) + 1.05(135) 0.00505(185)2 0.00109(135)2 0.00352(185)(135) = 309 lbs/week 309 x 52 = 16,068 lbs/year (It's 1983!) Profit = 0.14(309) 0.03(185) 0.015(135) = $41.20/week Think Break #7 Repeat the previous problem, but now use a milk price of Milk $10/cwt M = 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH Prices: Grain $3/cwt, Hay $1.50/cwt What are the profit maximizing inputs? Think Break #7 Discussion Summary: Price changed from $14/cwt to $10/cwt, a 29% decrease Grain changed from 185 to 182.5, a 1.35% decrease Hay changed from 135 to 118.2, a 12.4% decrease Milk output changed from 309 to 306, a 0.9% decrease Think Break #7 Discussion Milk price decreased from $14/cwt to $10/ cwt, a 29% decrease Milk output changed from 309 to 306, a 0.9% decrease The standard agricultural problem: output nonresponsive (inelastic) to price P S changes P Q D Q Summary Optimality Conditions for multiple inputs have several expressions, all imply Equal Margin Principle MRTS, isoquant, and optimal input use are all connected, know the graphics Different types of input substitution Be able to identify optimal input use Tabular form Calculus of optimization ...
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This note was uploaded on 08/08/2008 for the course AAE 320 taught by Professor Mitchell during the Spring '08 term at Wisconsin.

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