isoquants, isocosts and cost minimization overheads
DESCRIPTION
Isoquants, Isocosts and Cost Minimization Overheads. We define the production function as. y represents output. f represents the relationship between y and x. x j is the quantity used of the jth input. (x 1 , x 2 , x 3 , . . . x n ) is the input bundle. - PowerPoint PPT PresentationTRANSCRIPT
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Isoquants, Isocosts and Cost Minimization
Overheads
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We define the production function as
f represents the relationship between y and x
xj is the quantity used of the jth input
(x1, x2, x3, . . . xn) is the input bundle
n is the number of inputs used by the firm
f(x) maxy
[y: (x, y) is an element of the production set]
y represents output
f(x1 , x2 , x3 , ) maxy ε P(x)
[y]
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y = f (x1, x2, x3, . . . xn )
0
50
100
150
200
250
300
350
0 2 4 6 8 10 12
Input -x
Ou
tpu
t -y
y
Holding other inputs fixed,the production function looks like this
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Marginal physical product
Marginal physical product is defined as the incrementin production that occurs when an additional unitof a particular input is employed
![Page 5: Isoquants, Isocosts and Cost Minimization Overheads](https://reader035.vdocuments.mx/reader035/viewer/2022081417/56812d54550346895d925aa9/html5/thumbnails/5.jpg)
Mathematically we define MPP as
MPPxj MPPj
Δf(x1 , x2 , x3 , , xn )
Δxj
![Page 6: Isoquants, Isocosts and Cost Minimization Overheads](https://reader035.vdocuments.mx/reader035/viewer/2022081417/56812d54550346895d925aa9/html5/thumbnails/6.jpg)
Graphically marginal product looks like this
-40
-30
-20
-10
0
10
20
30
40
50
60
1 2 3 4 5 6 7 8 9 10 11
Input -x
Ou
tpu
t -y
A MPP
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The Cost Minimization Problem
C(y, w1,w2 , ) minx1 ,x2 , ,xn
Σn
i 1wixi such that y f(x1,x2 , xn)
Pick y; observe w1, w2, etc;choose the least cost x’s
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Isoquants
An isoquant curve in two dimensions representsall combinations of two inputs that produce the same quantity of output
The word “iso”means same
while “quant” stand for quantity
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Isoquants are contour lines of the production function
If we plot in x1 - x2 space all combinations of x1 and x2 that lead to the same (level) height for the production function, we get contour lines similar to those you see on a contour map
Isoquants are analogous to indifference curves
Indifference curves represent combinations of goodsthat yield the same utility
Isoquants represent combinations of inputs that yieldthe same level of production
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Production function for the hay example
0 10 20 30 40 50 60 70
x1
015
3045
6075
x2
0100000200000300000
y
0 10 20 30 40 50 60 70
x1
015
3045
6075
x2
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Another view
0
10
20
30
40
50
60
70
x1
0
15
30
45
60
75
x2
0200000
y
0
10
20
30
40
50
60
70
x1
0
15
30
45
60
75
x2
0200000
y
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Yet another view (low x’s)
05
1015
2025
x1
0
5
10
15
20
25
x2
01500030000
75000
y
05
1015
2025
x1
0
5
10
15
20
25
x2
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With a horizontal plane at y = 250,000
0 10 20 30 40 50 60 70
x1
015
3045
6075 x2
0100000200000300000
y
0 10 20 30 40 50 60 70
x1
015
3045
6075 x2
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0 10 20 30 40 50 60 70
x1
015
3045
6075 x2
0100000200000300000
y
0 10 20 30 40 50 60 70
x1
015
3045
6075 x2
With a horizontal plane at 100,000
![Page 15: Isoquants, Isocosts and Cost Minimization Overheads](https://reader035.vdocuments.mx/reader035/viewer/2022081417/56812d54550346895d925aa9/html5/thumbnails/15.jpg)
Contour plot
0 10 20 30 40 50 60 70x1
0
20
40
60
802x
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0 10 20 30 40 50 60 70x1
0
20
40
60
80
2xAnother contour plot
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There are many ways to produce2,000 bales of hay per hour
Workers Tractor-Wagons Total Cost AC 10 1 80 0.04 6.45 1.66 71.94 .03597
5.48 2 72.8658 0.03643.667 3 82.0015 0.0412.636 4 95.8167 0.04791.9786 5 111.872 .0559
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Plotting these points in x1 - x2 space we obtain
0 2 4 6 8 10x1
0
2
4
6
8
10
2x
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Or
Isoquant y = 2,000
0
2
4
6
8
10
12
0 2 4 6 8 10 12 14
X2
X1
Isoquant y = 2000
x2 = 4, x1 = 2.636
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0 10 20 30 40 50 60 70
x1
015
3045
6075 x2
0100000200000300000
y
0 10 20 30 40 50 60 70
x1
015
3045
6075 x2
Cutting Plane for y = 10,000
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0 10 20 30 40x1
0
10
20
30
40
2xIsoquant for y = 10,000
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Only the negatively sloped portionsof the isoquant are efficient
0 10 20 30 40x1
0
10
20
30
40
2x
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Isoquant for y = 10,000
x1 x2 output y-- 1 10,000-- 2 10,000-- 3 10,00012.469 4 10,0009.725 5 10,0008.063 6 10,0006.883 7 10,0005.990 8 10,0005.290 9 10,000
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y = 10,000
0 2 4 6 8 10x1
0
2
4
6
8
102x
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0 2 4 6 8 10x1
0
2
4
6
8
10
2x
y = 2, 000 y = 10,000
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Graphical representation
Isoquants y = 2,000, y = 10,000
0
2
4
6
8
10
12
14
0 2 4 6 8 10 12 14
x 2
Isoquant y = 2000
Isoquant y = 10000
x1
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y = 2,000, y = 5,000, y = 10,000
0 2 4 6 8 10x1
0
2
4
6
8
10
2x
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More levels
0 5 10 15 20 25 30x1
0
5
10
15
20
25
30
2x
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And even more
0 10 20 30 40 50 60x1
0
10
20
30
40
50
60
2x
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Comparison to full map
0 10 20 30 40 50 60 70x1
0
20
40
60
80
2x
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Slope of isoquants
An increase in one input (factor) requires a decrease in theother input to keep total production unchanged
Therefore, isoquants slope down (have a negative slope)
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Properties of Isoquants
Isoquants are convex to the origin
This means that as we use more and more of an input,its marginal value in terms of the substituting forthe other input becomes less and less
Higher isoquants represent greater levels of production
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Slope of isoquants
The slope of an isoquant is called the marginal rate of (technical) substitution [ MR(T)S ] between input 1 and input 2
The MRS tells us the decrease in the quantity of input 1 (x1)that is needed to accompany a one unit increasein the quantity of input 2 (x2),
in order to keep the production the same
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0
2
4
6
8
10
12
14
0 2 4 6 8 10 12 14
x2
Isoquant y = 2000
Isoquant y = 10000
x1
The Marginal Rate of Substitution (MRTS)
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Algebraic formula for the MRS
MRSx1,x2
Δx1
Δx2
y constant
The marginal rate of (technical) substitution ofinput 1 for input 2 is
We use the symbol - | y = constant - to remind us that the measurement is along a constant production isoquant
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Example calculations y = 2,000
MRSx1,x2
Δx1
Δx2
y constant
10 5.481 2
4.52 1
4.52
Change x2 from 1 to 2Workers Tractor-Wagons
x1 x2
10 15.48 23.667 32.636 41.9786 5
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Example calculations y = 2,000
Change x2 from 2 to 3
MRSx1,x2
Δx1
Δx2
y constant
5.48 3.6672 3
1.813 1
1.813
Workers Tractor-Wagons x1 x2
10 15.48 23.667 32.636 41.9786 5
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More example calculations y = 10,000
MRSx1,x2
Δx1
Δx2
y constant
9.725 8.0635 6
1.662 1
1.662
Change x2 from 5 to 6
x1 x2
12.469 49.725 58.063 66.883 75.990 8
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A declining marginal rate of substitution
The marginal rate of substitution becomes larger in absolute value as we have more of an input.
When the firm is using 10 units of x1, it can give up 4.52 units with an increase of only 1 unit of input 2, and keep production the same
But when the firm is using only 5.48 units of x1, it can onlygive up 1.813 units with a one unit increase in input 2 and keep production the same
The amount of an input we can to give up and keep production the same is greater, when we already have a lot of it.
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Slope of isoquants and marginal physical product
Marginal physical product is defined as the incrementin production that occurs when an additional unitof a particular input is employed
MPPxj MPPj
Δf(x1 , x2 , x3 , , xn)
Δxj
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Marginal physical product and isoquants
All points on an isoquant are associated with the same amount of production
Hence the loss in production associated with x1
must equal the gain in production from x2 , as we increase the level of x2 and decrease the level of x1
MPPx1Δx1 MPPx2
Δx2 0
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MPPx1 Δx1 MPPx2 Δx2
Rearrange this expression by subtracting MPPx2 x2
from both sides,
MPPx1 Δx1 MPPx2 Δx2 0
Then divide both sides by MPPx1
Δx1 MPPx2 Δx2
MPPx1Then divide both sides by x2
Δx1
Δx2
MPPx2MPPx1
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The left hand side of this expression is the marginal rate of substitution of x1 for x2, so we can write
MRSx1x2
Δx1
Δx2
MPPx2
MPPx1
So the slope of an isoquant is equal tothe negative of the ratio of the marginal physicalproducts of the two inputs at a given point
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The isoquant becomes flatter as we moveto the right, as we use more x2 (and itsMPP declines) and we use less x1 ( and itsMPP increases)
So not only is the slope negative, but the isoquantis convex to the origin
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0
2
4
6
8
10
12
14
0 2 4 6 8 10 12 14
x2
Isoquant y = 2000
x1
The Marginal Rate of Substitution (MRTS)
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Approx x1 x2 MRS MPP1 MPP2
12.4687 4.0000 --- 664.6851 2585.740011.8528 4.1713 -3.5946 739.5588 2465.2134 9.7255 5.0000 -2.5672 1010.5290 2050.0940 9.3428 5.1972 -1.9411 1063.1321 1975.4051 8.0629 6.0000 -1.5941 1254.9695 1724.5840 6.9792 6.9063 -1.1959 1447.3307 1508.9951 6.8827 7.0000 -1.0291 1466.3867 1489.5380
MRSx1,x2
Δx1
Δx2
y constant
9.3428 8.06295.1972 6
1.2799 0.8028
1.594
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Approx x1 x2 MRS MPP1 MPP2
12.4687 4.0000 --- 664.6851 2585.740011.8528 4.1713 -3.5946 739.5588 2465.2134 9.7255 5.0000 -2.5672 1010.5290 2050.0940 9.3428 5.1972 -1.9411 1063.1321 1975.4051 8.0629 6.0000 -1.5941 1254.9695 1724.5840 6.9792 6.9063 -1.1959 1447.3307 1508.9951 6.8827 7.0000 -1.0291 1466.3867 1489.5380
MRSx1x2
Δx1
Δx2
MPPx2
MPPx1
x2 rises and MRS falls
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Isocost lines
Quantities of inputs - x1, x2, x3, . . .
Prices of inputs - w1, w2, w3, . . .
An isocost line identifies which combinations of inputsthe firm can afford to buy with a given expenditureor cost (C), at given input prices.
w1x1 w2x2 w3x3 wnxn C
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Graphical representation
Cost = 120 w1 = 6 w2 = 20
02468
10121416182022
0 1 2 3 4 5 6 7
x2
x1
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Slope of the isocost line
w1x1 w2x2 C
w1x1 C w2x2
x1 Cw1
w2
w1
x2
So the slope is -w2 / w1
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Example
C = $120, w1 = 6.00, w2 = 20.00
6x1 20x2 120
6x1 120 20x2
x1 1206
206x2
20 3 13x2
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Intercept of the isocost line
x1 Cw1
w2
w1
x2
So the intercept is C / w1
With higher cost, the isocost line moves out
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Isoquants and isocost lines
We can combine isoquants and isocost lines to help us determine the least cost input combination
The idea is to be on the lowest isocost linethat allows production on a given isoquant
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Combine an isoquant with several isocost lines
Isocost lines for $20, $60, $120, $180, $240, $360
0 2 4 6 8 10 12 14x1
0
5
10
15
202x
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Consider C = 120 and C = 180
At intersection there are opportunities for trade
0
4
8
12
16
20
24
3 4 5 6 7 8 9 10 11 12 13
X2
X1
Isoquant y = 10000
Isocost 120
Isocost 180
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Isocost 160
Isoquant y = 10000
0
4
8
12
16
20
24
4 5 6 7 8 9 10 11 12 13
X2
X1 Isocost 120
Isocost 180
Add C = 160
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Isocost 160
Isocost 154.6
Isoquant y = 10000
0
4
8
12
16
20
24
4 5 6 7 8 9 10 11 12 13
X2
X1 Isocost 120
Isocost 180
Add C = 154.6
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Isocost 160
Isocost 154.6
Isoquant y = 10000
0
4
8
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24
4 5 6 7 8 9 10 11 12 13
X2
X1 Isocost 120
Isocost 180
In review
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The least cost combination of inputs
The optimal input combination occurs wherethe isoquant and the isocost line are tangent
Tangency implies that the slopes are equal
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Slope of the isocost line
-w2 / w1
Slope of the isoquant
MRSx1x2
Δx1
Δx2
MPPx2
MPPx1
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w2
w1
MPPx2
MPPx1
Optimality conditions
w2
w1
MRSx1x2
Δx1
Δx2
Substituting we obtain
w2
w1
MRSx1x2
Δx1
Δx2
MPPx2
MPPx1
The price ratio equals the ratio of marginal products
Slope of the isocost line = Slope of the isoquant
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We can write this in a more interesting form
Multiply both sides by MPPx1
w2
w1
MPPx2MPPx1
and then divide by w2
MPPx1
w1
MPPx2x2
MPPx1 w2
w1
MPPx2
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Graphical representation
0 2 4 6 8 10 12 14x1
0
5
10
15
20
2x
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Statement of optimality conditions
a. The optimum point is on the isocost line
b. The optimum point is on the isoquant
c. The isoquant and the isocost line are tangent at the optimum combination of x1 and x2
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d. The slope of the isocost line and the slopeof the isoquant are equal at the optimum
w2
w1
MRSx1x2
MPPx2MPPx1
e. The ratio of prices is equal to the ratio ofmarginal products
w2
w1
MPPx2MPPx1
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f. The marginal product of each input divided byits price is equal to the marginal product ofevery other input divided by its price
MPPx1
w1
MPPx2
w2
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Approx x1 x2 MRS MPP1 MPP2 MPP1/w1 MPP2/w2 MRS -w2 / w1
1.0000 -3.33332.0000 -3.33333.0000 -3.3333
12.4687 4.0000 664.6851 2585.7400 110.7809 129.2870 -3.8902 -3.333311.8528 4.1713 -3.5946 739.5588 2465.2134 123.2598 123.2607 -3.3334 -3.33339.7255 5.0000 -2.5672 1010.5290 2050.0940 168.4215 102.5047 -2.0287 -3.33339.3428 5.1972 -1.9411 1063.1321 1975.4051 177.1887 98.7703 -1.8581 -3.33338.0629 6.0000 -1.5941 1254.9695 1724.5840 209.1616 86.2292 -1.3742 -3.33336.9792 6.9063 -1.1959 1447.3307 1508.9951 241.2218 75.4498 -1.0426 -3.33336.8827 7.0000 -1.0291 1466.3867 1489.5380 244.3978 74.4769 -1.0158 -3.33335.9898 8.0000 -0.8929 1663.9176 1305.9560 277.3196 65.2978 -0.7849 -3.33335.2904 9.0000 -0.6994 1855.3017 1155.0760 309.2169 57.7538 -0.6226 -3.33334.7309 10.0000 -0.5595 2044.1823 1026.1700 340.6971 51.3085 -0.5020 -3.33334.2773 11.0000 -0.4535 2232.4134 912.4680 372.0689 45.6234 -0.4087 -3.33333.9071 12.0000 -0.3702 2420.9761 809.4240 403.4960 40.4712 -0.3343 -3.33333.6042 13.0000 -0.3029 2610.3937 713.8360 435.0656 35.6918 -0.2735 -3.3333
Example Table w1 = 6, w2 = 20
To get an x1, I can give up 3.33 x2 in terms of cost
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Intuition for the conditions
The isocost line tells us the rate at which the firmis able to trade one input for the other,given their relative prices and total expenditure
For example in this case the firm must give up 3 1/3 units of input 1 in order to buy a unit of input 2
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0
4
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4 5 6 7 8 9 10 11 12 13
X2
X1
Isocost 180
w1 = 6w2 = 20C = 180
3
6.66
10
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Isoquant y = 10000
0
4
8
12
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24
4 5 6 7 8 9 10 11 12 13x2
x1
The isoquant tells us the rate at which the firmcan trade one input for the other and remainat the same production level
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If there is any difference between the rate atwhich the firm can trade one input for anotherwith no change in production and the rateat which it is able to trade given relative prices,the firm can always make itself better off bymoving up or down the isocost line
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Isocost 160
Isoquant y = 10000
0
4
8
12
16
20
24
4 5 6 7 8 9 10 11 12 13x2
x1
Isocost 180
The isoquant tells us the rate at which the firmcan trade one input for the other and remainat the same production level
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When the slope of the isoquant is steeperthan the isocost line, the firm will move down the line
When the slope of the isoquant is less steepthan the isocost line, the firm will move up the line
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0
4
8
12
16
20
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4 5 6 7 8 9 10 11 12 13x2
x1
Isoquant y = 10000
When the slope of the isoquant is steeperthan the isocost line, the firm will move down the line
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The End