ECO206 Study Guide 02 solution PDF

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ECO206Y – Study Guide 02 – OptimizationSolutionWarning about usage of solutionsThe questions on the study guide are carefully selected not only for the deepened understanding the solution yields but also for the skills and insights you develop while working out the solution. One skill especially imp...

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ECO206Y – Study Guide 02 – Optimization Solution Warning about usage of solutions The questions on the study guide are carefully selected not only for the deepened understanding the solution yields but also for the skills and insights you develop while working out the solution. One skill especially important to learn in ECO206 is how to "unstuck" yourself after you have gotten stuck in a problem’s solution. To practice this skill, you need to get stuck in the first place. Looking at the solution does not tell you how to figure out the solution, and in most cases prevents you from finding out in the future. Therefore, I want to caution you against looking at the solutions too early. Ideally, you want to look at the solutions only after you have a solution written out that you are convinced is right and you want to double check your answer. Short of having an answer of your own, please make sure you have tried at least four of the following approaches before looking at the solution: • Attempted to solve the problem. • Tried to solve the problem with a different approach (e.g. graphically instead of with calculus). • Wrote down the definitions of items involved in the problem and explained the problem in English. • Tried to find and solve a special case. Made up numbers for all parameters and all but one of the variables, and solve this problem. • Made up numbers for all parameters and solved this problem. • Changed the problem into a simpler one, e.g., with two consumers instead of three, with a discrete or uniform distribution instead of a complicated continuous one, and tried to solve the simpler problem. • Talked to a class mate about the problem. • Talked to a peer mentor at the Economic Study Center about the problem. • Talked to the TA about the problem after tutorials or in office hours. • Talked to the professor after class/in office hours.

Graphical question 1. Assume the following Cobb-Douglas describes your utility function for candy c and soda s: U(c, s) = c .5 s.5 (a) What does this utility function imply about your consumption preferences for candy and soda? Are they substitutes? Complements? How do you know? Answer: We can see from the utility function, that the indifference curve does not have shape of neither a straight line (perfect substitutes) and nor L-shape (perfect complements). Note: After the formal definition is introduced in week 4/chapter 5 in the textbook. Come back ∗ ∗ to this example. Check ∂∂ pc and ∂∂ ps to explain that candy and soda are neither complements nor c s substitutes.

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(b) On thin paper, draw the indifference lines for U=1, U=2, U=3, U=4, and U=5. Answer:

Figure 1: Candy vs. Soda - Utility (c) Now assume that candy costs 20¢and soda costs 80¢. On an new piece of thin paper draw a diagram with budget lines for I = $1, I = $2.40, I = $3.20, and I = $4.00. Answer:

Figure 2: Candy vs. Soda - Budget Constraint (d) Slide the two pieces of paper on top of each other, the utility diagram on the bottom, the budget curves on the top. What budget do you need at least if you want to achieve a utility of 4? Which consumption bundle achieves U=4 at the lowest cost? Answer: We need at least I=$3.20 to achieve U=4. Approximately, (c, s) = (8, 2) is the cheapest consumption bundle that achieves U=4.

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Figure 3: Candy vs. Soda - Utility Level of 4 (e) Slide the two pieces of paper on top of each other, now with the budget curves on the bottom, and the utility diagram on the top. What utility can you achieve at most with a budget of $3.20? Which consumption bundle yields the highest utility along the I=$3.20 budget line? Answer: The budget of I=$3.20 will achieve U=4 at most. Approximately, (c, s) = (8, 2) is the consumption bundle along the I=$3.20 budget line that achieves this.

Figure 4: Candy vs. Soda - I=$3.20 (f) Compare your answers in (d) and (e). Coincidence? Answer: The answers from (d) and (e) coincide. As we would expect given the duality between the utility maximization problem and cost minimization problems are identical.

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Modelling 2. Assume that high and low oranges are substitutes, i.e., consuming one high quality orange yields the same utility as consuming γ low quality oranges, γ ≥ 1. Formally, we can represent this with U(h, l) = h+ γ1 l where h denotes the number of high-quality oranges consumed and l denotes the number of low-quality oranges consumed. We now want to consider consumers in Florida and Vermont and compare their purchasing behavior. Vermont does not produce any oranges. All oranges sold and purchased in Vermont are brought in from Florida. The per-unit transportation cost t is the same for low and high quality oranges. (a) Assume all consumers in Florida spend the same expenditure I on oranges, and the price for low and high quality oranges is ph and pl , respectively. Answer: In general, we approach such a utility maximization problem by forming the Lagrangian and solving the corresponding unconstrained problem. Write down the budget constraint for Florida consumers and solve their optimization problem. We will see that this approach does not work in this case. So we’ll get stuck and need to find a different way to solve the problem. Budget constraint: ph h + pl l = I F Florida consumers’ problem: max U(h, l) = h + 1γ l subject to ph h + pl l = I F Use the Lagrangian: L = h + γ1l + λ(I F − ph h − pl l) FOC: ∂L ∂ h : 1 − ph λ = 0 ∂L 1 ∂ l : γ − pl λ = 0 ∂L ∂λ

: I F − ph h − pl = 0

Note that only the last equation contains l and h. We can thus not solve for h∗ over l ∗ . Thus we find that the Lagrangian Multiplier Approach does not work here. To get a better understanding of the nature of this optimization problem, let us consider the indifference curves and budget constraints. To get a sense of the indifference curves, let’s find a selection of (h, l) that yield the same utility, say 10. The following combinations all yield the same utility 10: h 10 9 8 6 4 2 1 0 l 0 γ 2γ 4γ 6γ 8γ 9γ 10γ We see that the indifference curve for U0 = 10 is a straight line. We can verify that the same is the case for other utility values. Solving U0 = h + γ1 · l for l as a function of h, we find l = U0 γ − γ · h.

Similar, we can find numbers that make up the budget line and solve pl l + ph h = I for l as a function of h: l = pI − pph · h. l l Plotting these lines, there are two options: The budget line can be flatter than the indifference p curve, i.e., ph < γ: l

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@ Figure 5: Indifference curve and budget line; High Quality vs. Low Quality Oranges or the budget line can be steeper than the indifference curve, i.e.,

ph pl

> γ:

Figure 6: Indifference curve and budget line; High Quality vs. Low Quality Oranges If the indifference curves and the budget constraints are straight lines coincide, then all bundles p will be optimal. (This is the case when pl h = γ) If the two lines do not coincide, then we see from the graphs that there are two extremes: Consuming only high or only low quality oranges. One of these two options will be the solution, so the solution will be a corner solution. Let us figure out which of the extremes is the optimal consumption bundle. We can do this graphically or from first principles. Graphically, we see that in the first of the two cases depicted above, p the consumer optimally only consumes high quality oranges. So if p h < γ, then h∗ = I/ph , l∗ = 0. l In the second of the two cases depicted above, the consumer optimally only consumes low quality p oranges. So if p h > γ, then h∗ = 0, l∗ = I/pl . l Alternatively, from first principles: Once we realize that the consumer will spend all money on either low or high quality oranges, we can compare the utility derived from these two options. Spending all income on high quality means h∗ = I/ph , l∗ = 0 and yields utility of pI . Spending all income on h I low quality oranges means h∗ = 0, l∗ = I/pl and yields utility γ1 · pl . So the consumer will spend all

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money on high quality oranges if and only if 1 I I > · ph γ pl i.e., if γ>

ph . pl

To summarize, the optimal consumption in Florida is given by  p 0 ifγ > phl l F∗ = I otherwise pl hF∗ =



I ph

0

p

if γ > phl otherwise

(b) Assume that all consumers in Vermont spend the same amount of money I on oranges. Write down the budget constraint for Vermont and solve their optimization problem. Answer: Now, there is transportation cost and this is added to the price of oranges. The prices become: · High quality oranges: ph + t · Low quality oranges: pl + t Budget constraint: (ph + t)h + (pl + t)l = I Vermont consumers’ problem: max U(h, l) = h + 1γ l subject to (ph + t)h + (pl + t)l = I Both indifference curve and budget constraint are still straight lines. We proceed as we did in ph +t = γ) If part (a). If they coincide then all bundles will be optimal. (This is the case when pl +t the two lines do not coincide, there are two extremes: Consuming only high or only low quality oranges. (see figures). One of these two options will be the solution. As we have seen in part (a), we have corner solutions: either h = 0 or l = 0. When h = 0, (ph + t)0 + (pl + t)l = I 0 + (pl + t)l = I l=

I pl + t

I The utility derived from this consumption bundle, is γ1 · pl +t . When l = 0, (ph + t)h + (pl + t)0 = I

(ph + t)h + 0 = I h=

I ph + t

The utility derived from this consumption bundle, is γ1 · p I+t . h So the consumer will spend all money on high quality oranges if and only if 1 I I > · ph + t γ pl + t

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i.e., if γ>

ph + t

. pl + t To summarize,the optimal consumption in Vermont is given by  p +t 0 if γ > plh+t l ∗V = I otherwise pl +t hV∗ =



p +t

I ph +t

if γ > phl +t otherwise

0

Note that in the above utility function γ is a preference parameter that represents the extent of which consumers prefer high quality oranges over low quality oranges. The higher the γ the less that consumer benefits from low-quality oranges, and enjoys high-quality oranges more, relatively speaking. So consumers with high γ are more likely to consume high quality oranges. Now assume that both in Vermont and in Florida, the parameter γ is distributed over the population according to the probability distribution: f (γ) = e 1−γ between 1 and +∞. The corresponding cumulative distributive function is F(γ) = 1 − e 1−γ for γ ≥ 1. That is to say, for any threshold γ0 ≥ 1, a proportion 1 − e 1−γ0 of the population have a taste parameter γ smaller than γ0 , and a proportion e 1−γ0 a have a taste parameter γ higher than γ0 . For instance F(2) ≈ 0.62 so in each state, about 62% of the population have a γ smaller than 2. (No one has a parameter ? strictly lower than 1.) (c) What proportion of people in Florida will predominantly buy low quality oranges? Answer: In part (a), we saw that consumers in Florida will buy only Low quality oranges if their preference p parameter γ is such that γ ≤ plh . Otherwise, the consumer will buy only high quality oranges. p

Denote the cut-off preference parameter as γˇ = ph . l By definition of the cumulative distribution function (CDF) of γ, we know that a proportion F(ˇ γ) have a taste parameter γ smaller than γˇ and will buy low quality oranges. So the proportion of people buying low quality oranges in Florida is simply:   ph People FL = F(ˇγ) = F pl = 1−e

1−

ph pl

All other people will buy high quality oranges. It will be useful to have an expression for people who buy only high quality oranges. ph

PeopleFH = 1 − People LF = e 1− pl

If we were not sure (or didn’t know the CDF, only the probability distribution), we could compute these numbers using the definition of a probability distribution function (PDF): for every value of γ there are f (γ) many people. We want to add up all consumers that only buy low quality oranges, so we sum up all consumers with γ less than γˇ: Z ˇγ L PeopleF = f (γ)dγ 1

=

Z

ph pl

e 1−γ dγ

1

   pph = − e 1−γ 1 l   p 1− h = −e pl + e 0 = 1 − (e

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1−

ph pl

)

And for people buying high-quality, we sum up consumers for which γ ≥ γˇ: People HF

= =

Z

∞

f (γ)dγ

ˇγ

Z

∞

e 1−γ dγ

ph pl

 ∞ = −(e 1−γ ) ph pl   p 1− p h = −0 + (e l ) p

=e

1− p h l

Note: The sum of PeopleFL and People HF should equal 1, and it does: pl

pl

People FL + People HF = 1 − (e 1− ph ) + e 1− ph =1

This makes sense because consumers in Florida either choose high or low quality oranges and no consumer chooses neither. The sum of the share of each type of consumers must equal 1. (d) What proportion of people in Vermont will predominantly buy low quality oranges? Answer: Similarly, we saw in part (b), that consumers in Vermont will buy only Low quality oranges if γ < ph +t p +t ˆ = plh +t . Using either the CDF or the PDF to get: p +t . Denote cut-off preference parameter as γ l

PeopleVL = 1 − (e

p +t

1− ph+t l

)

All other people will buy high quality oranges: People HV = e

p +t

1− ph+t l

(e) Compute and compare the average quality of oranges in Florida and in Vermont. Answer: Since everyone in Florida and Vermont has the same income I, all consumers who buy only low quality oranges will buy pI in Florida and p I+t in Vermont. l l We have, low quality oranges: I · People LF pl p I 1− l = · [1 − (e pl )] pl

Q LF =

I · People VL pl + t p +t I 1− h · [1 − (e pl +t )] = pl + t

Q LV =

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Similarly, high quality oranges: I

Q HF =

· People FH ph p I 1− h · [e pl ] = ph

I · People VH ph + t p +t I 1− h · [e pl +t ] = ph + t

Q HV =

We next want to compute the average quality of oranges in Florida. To do that we might (1) assign a quality of 0 to low quality oranges and a quality of 1 to high quality oranges, or any other two numbers that you consider appropriate. Alternatively, we might (2) ask what percentage of oranges consumed (bought) is a high quality orange. For either (1) or (2) we compute Q¯ F =

Q HF QFL + Q HF ph

1− · [(e pl )]    p  p 1− h 1− h · 1 − (e pl ) + pI · e pl I ph

=

I pl

h

p

1− p h

1 ph

=

· [(e l )]    p p  1 1− h 1− h · 1 − (e pl ) + ph · e pl

1 pl

p

[(e

=

1− p h

l )]    p  p 1− h 1− ph · 1 − (e l ) + e pl

ph pl

1

Q¯ F =



ph pl

· e

ph pl

 −1 +1

−1

Similarly in Vermont: ¯V = Q

QVH QVL + Q HV

= I pl +t

=

=

¯V = Q

1 pl +t

I ph +t

1−

1 ph +t

1−

ph +t pl +t

ph +t

· [(e pl +t )] i i h h p +t ph + t 1 1− 1− h · 1 − (e pl +t ) + p +t · e pl +t h

[(e ph +t pl +t

ph +t

· [(e pl +t )] h h p +t i ph + t i 1− 1− h I · 1 − (e pl +t ) + ph +t · e pl +t

h

· 1 − (e h

· e

1−

1−

1 ph +t p l +t

9

−1

ph + t p l +t

ph +t p l +t

)] i i h p +t 1− h ) + e pl +t

i −1 +1

¯ F and Q¯V and see how the increase in price (due to additional transNow, we want to compare Q portation cost in Vermont) affects average quality of oranges. We can consider the price ratio in ¯ p +t dQ Florida as a special case of p l+t where t = 0. So we want to look at d tV . h

p +t

p +t As a first step let us see how the price ratio plh+t changes when t changes. Denote x = plh+t .

p +t dx = h pl + t dt (1)( pl + t) − (ph + t )(1) = (pl + t)2 p + t − ph + t = l (pl + t)2 pl − ph = (pl + t)2 ph > pl , so the numerator is negative. The denominator is (ph + t)2 is always positive. Hence, we p +t the price ratio plh+t decreases as t increases. Since we have ddtx < 0 already, let us write Q¯V = and compute

¯V dQ dx

. (We can then find

dQ¯V dt

=

1 x · e x −1 − 1

dQ¯V dx

·

dx d t .)

¯

dQ Note informally, we expected d xV < 0. As x increases, so does e x −1 and hence x · e x −1 . Thus the denominator increases and the overall fraction has to decrease. Let’s see whether that intuition bears out formally: The derivative of the denominator is

Thus,

 d  x · e x −1 − 1 = e x −1 + x · e x −1 = (x + 1) · e x −1 . dx

¯V dQ dx

=

−(x + 1) · e x −1 (x · e x −1 − 1)2

< 0.

Putting everything together, we find ¯V ¯V d x dQ dQ = > 0. · dt dx dt This result implies that the percentage of oranges consumed that are high quality oranges, i.e., the average quality of oranges consumed, increases with t. If Vermont has a higher transportation cost than Florida, it must have a higher average quality of oranges consumed. Note: We can also write down the complete expression for how the percentage of oranges consumed that are high quality oranges changes as t increases, but it is not clear that this yields any additional insights:

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¯V dQ dQ¯V d x · = dx dt dt −(x + 1) · e x −1 pl − ph = · (x · e x −1 − 1)2 (pl + t)2 (x + 1) · e x −1 ph − pl · = (x · e x −1 − 1)2 (pl + t)2 ph +t

p +t

−1 ( p h+t + 1) · e pl +t ph − pl = l 2 · ph +t 2 (p ph + t −1 l + t) p l +t −1 pl +t · e

Analytical Problems 3. (a) Suppose that f (x, y) = x y. Find the maximum value for f if x and y are constrained to sum to 1. Solve this problem in two ways: by substitution and by using the Lagrangian multiplier method. Answer: We’re given that x and y are constrained to sum to 1. This means, x + y = 1. First, let’s solve this by substitution. We can rewrite the constraint as y = 1 − x. If we plug this into f (x , y), we have f (x , y) = x (1 − x ) = x − x 2 . Take FOC: ∂f ∂x

= 1 − 2x = 0 2x = 1 x∗ =

1 2

y∗ = 1 − x∗ =

1 2

Now, let’s use the Lagrangian multiplier approach. L = x y + λ(1 − x − y) Take FOC: ∂L ∂x = y −λ = 0 ∂L ∂y = x −λ=0 Equate two conditions to get: y = λ = x, thus x = y 1 Plug this into the constraint: x + (x) = 1 =⇒ x ∗ = 21 and y ∗ = 2 . In both cases, we obtain the same answer. Note: feel free to use whichever method you find easier.

(b) The dual problem to the one described in the previous problem is minimize x + y subject to x y = 0.25 Solve this problem using the Lagrangian technique. Then compare the value you get for the Lagrangian multiplier with the value you got in Problem 2.3. Explain the relationship between the two solutions. Answer: First, write down the Lagrangian: L = x + y + λ(0.25 − x y)

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Take FOC: ∂L ∂ x = 1 − λy ∂L ∂ y = 1 − λx

Equate these two conditions:

1 y

=λ=

1 x

to get x = y 1

Plug into the constraint: x ( x ) = 0.25 =⇒ x = 0.252 =⇒ x ∗ =

1 2

and y ∗ = 21 .

Problem a. is a constrained utility maximization problem and problem b. is a constrained cost minimization problem. The solutions from two exercises are identical because any constrained maxim...