BEPP 250 Unit 1 Flashcards
Attachment rate vs EW-product price ratio
Attachment rate is the Percent of the buying population that buys extended warranties.
Ew-product price ratio is the ratio of the extended warranty to the price.
They should be equal in theory, but attachment rate is usually higher because consumers are willing to spend more because they think there’s a higher risk of product breaking than there actually is.
Risk averse
Averse to risk
Doesn’t like risk
Will pay slightly more to minimize the risk.
Suppose you think w% probability the tv needs repair and cost of repair is $L. Let $t be the price of the EW.
Get willingness to pay:
If you don’t buy the extended warranty, then w% of the time you lose $L. 100-w% if the time, you lose nothing
Let U be the function of the price.
wU(-L) + wU(0)is how much you pay if you don’t buy the warranty
If you do buy the warranty, then you just pay t. So U(-t)
Willingesss to pay is when prices are equal
Price t* such that
U(-t*) = wU(-L)+ wU(0)
So you’re willing to pay $w(L)
Willingness to pay is that price
Model for willingness to pay:
Willingness to pay is based on aversion to risk and belief about the likelihood of repair.
Someone’s Risk aversion relation to their estimated probability the tv needs repair
If they have higher aversion to risk, then they think there is a lower chance the tv needs repair.
Even if two people have same willingness to pay for a product, when you introduce a cheaper product, then the willingness to pay for the person with the lower r is lower than the other person.
Decrease in WTP is much steeper for the less risk averse person.
Risk aversion actual relation with buying ew or not
Risk aversion doesn’t explain why people are willing to buy EW at observed prices.
Base decision on expected value of the warranty. Buy EW if price of EW < probability of repair * cost of repair.
People overestimate likelihood of failure by a large amount (actual is 5 percent, estimated 13-15 percent) so retailers sell EW at very high prices
If you tell someone actual failure rate is lower, their WTP goes way down.
If told actual percent tv breaks, then what
Then consumers are way better of, sales and profits go way down. In real world, consumer welfare is negative. When told this, they are slightly positive.
What happens if EW decreases
Consumer welfare goes down, as consumers don’t get any real value from extended warranties. Making it cheaper makes it more likely consumers get scammed
Microeconomics
Body of knowledge that studies indifivudal level behavior and decision making in order to provide tools, results, and ideas to help us
Understand observed phenonomem and guide decisions
Main principles of micro
- Optimization (individual objectives + constraints —>shape decisions and observed behavior.
- Equilibrium (observed outcomes <— interaction across individuals’ decisions, no incentive to change further, system is at rest)
3 assumed properties of preferences (include on cheat sheet)
- Completeness. When facing a choice between 2 bundles, a consumer can rank them so that either: A>B, A<B, or A~B
- Transitivity. Consumers rankings are logically consistent in the sense that if A>B and B>C, then A>C
- Monotonicity. More of a good is better.
All else the same, more of a commodity/ characteristic is better than less.
A good is different from a bad.
We are never satisfied.
U can’t go down as x goes up. U must increase as x goes up. Marginal utility can go down but must remain positive.
Monotonicity
Think of it that we can freely dispose excess beverages in our bundle. Don’t want to drink, then throw it away, doesn’t cost anything
If you have a point C, then you draw vertical and horizontal lines. To the top right, you prefer the point, bottom left you strictly prefer C
Rest you can’t tell, so get indifference curve
Indifference curve
Consider preferences such that A~B.
Graphically represent the set of all bundles that one is indifferent to by the indifference curve
Equally prefer all the bundles on this curve.
Indifference curves and preferences
Because of monotonicity, you strictly prefer points (bundles) above the IC. So, you can draw IC’s next to each other. And you know that you’ll prefer the IC’s further to the top right, as more of a good is better, and more goods to the top right.
Budget constraint
Can’t have unlimited goods, even though we want to, because of a budget constraint.
Budget set is the set of feasible (affordable) bundles given price and income. Set of all bundles that lie on and below the budget line.
Budget line
Suppose we have two goods, x and y. With corresponding prices p1 and p2. And income I
Budget line is p1x+p2y = I
Get it in terms of Y and graph. Bundles below and on it are those that are affordable.
Which bundle do we choose?
Need to be able to afford the bundle, and remember that by monotonicity, more of something is better.
So, we want to be on the highest indifference curve that is still affordable.
So this will be the indifference curve that is tangent to the budget line.
It has to be tangent as if we chose one that intersected, there’d be some point that is still affordable on a higher IC.
Utility function
Utility function is an easier way to show which we prefer.
If (x,y) > (x’,y’), then u(x,y) = U > U’ = u(x’,y’)
Value of the utility function doesn’t matter, it’s an ordinal function.
Just matters relative, ie that you prefer this bundle to another bundle.
Partial derivative
For example take partial deriv WRT x and then WRT y for
2(x-2)^2 -4xy +3y^2
Need partial derivatives for maximization problems with multiple variables.
Partial deriv WRT x —> treat y as a constant
4(x-2) -4y
And for WRT y—> treat x as a constant
-4x +6y
If we want the minimum of this function, equate the partial deriv to 0 and then to each other.
Relationship between IC and utility function
Assign a number, U1 = u(x1,y1) and another U2=u(x2,y2)
We prefer the bundle on the higher IC.
Always will prefer bundle on higher IC, and we represent the IC with the utility function. So the top IC is U2 and bottom is U1 and we strictly prefer bundles on U2
Utility maximization problem in words
Predicting which bundle is chosen (most preferred affordable bundle)is equivalent to finding (x,y) such that:
- u(x,y) = U is the largest AND
- (x,y) satisfies the budget constraint
Utility maximization problem in math
Objective:
Choose bundle (x,y) to maximize utility.
u(x,y)
Subject to budget constraint:
Chosen bundle (x,y)should be affordable
px X + pyY <= I
X and Y are always greater than or equal to 0, as can’t have negative quantity of a good.
Utility maximization graphical solution
The optimal solution is the point where the IC is tangent to the budget line, as if the IC hits the budget line twice, then there’s some point on a higher IC that we would prefer.
Mathematically, this point is captured by equality of slopes of budget line and the IC at the optimal bundle.
Budget line slope is -px/py because we have the EQ px x + py y =I
See next card for slope of IC
Slope of indifference curve
Consider an increase in x given by delta x. To be on the same IC, y must decrease by delta y.
The ratio delta y / delta x tells us how much we need to give up in terms of y if we want to have an incremental increase in x and have utility remain constant.
For a very small increase in x, the ratio delta y /delta x is the first deririvate dy/dx, which is the slope of the IC at (x,y)
This slope, dy/dx measures the trade off between y and x and has a special name, marginal rate of substitution or MRS
So, slope of ic is the deririvite of y / dx
MRS and marginal utility
Change in x given by delta x changed our utility by delta x (U)
Same thing for y
Same IC, so delta U = 0
So, delta x U * delta X + delta y U * delta Y =delta U =0
So, MUx * dx + MUy * dy =dU =0
Where MUx is the marginal utility of x which measures the change in utility for a small change in x
Marginal utility
Additional benefit you get from consuming the marginal unit, which is the very last unit you consume or an additional unit.
MU x =the partial derirative of the utility function with respect to x
= partial deriv u(x,y) / partial deriv x
Assuming monotonicity, utility increases in consumption. MUx must be positive for all x.
However the additional benefit, may decrease as x becomes large (MU still positive though)
This is diminishing marginal utility, MUx is decreasing, meaning the partial deriv of MUx with respect to x which is the second partial deriv of the utility function wrt x is negative.
Bang for buck
We get for the slope of the IC =-MUx/MUy
Rearrange this, and we get equality of bang for buck—>
MUx/px = MUy/py
Bang for buck is this the benefit cost ratio of the marginal unit. The benefit of the last unit / the cost of the last unit.
Want to increase consumption by a tiny bit of the good that has the highest B4B until the B4B’s are equal.
Once equal or you hit a constraint (can’t consume more or can’t consume negative quantities), then you stop.
Utility max problem basic structure and solution
Remember we are trying to maximize utility subject to our budget constraint. Max u (x,y)where x and y are greater than or equal to 0 We also have budget function px X + py Y = I Need to find two unknowns, x* and y*. Equate the bang for bucks for interior solutions and for a corner solution, you consume more of the good if their B4B is strictly greater than the other B4B And the budget constraint.
Interior solution utility max problem
If optimal solution is strictly positive, then B4Bs should all be equal when evaluated at optimal solution. Use equality of B4B to get x* and y*
Make sure you show that x and y are greater than 0
Example of útil Max problem:interior solution: Max sqrt (xy) such that Px X + Py Y = I
Step 1. Examine the bang for bucks.
MU x = partial deriv with respect to x = SQRT (y) / 2SQRT (x)
Divide by px to get bang for buck
MUy =SQRT (x)/2SQRT (y). Divide by py to get B4By
X and Y must be strictly positive, as if x or y were zeros then B4B would go to infinity, which can’t happen.
So, we equate the B4B. Cross multiply and get y = px/py * x
Step 2. Plug this into the budget constraint
PxX + py (px/py x) = I
Now solve for X = I/2px
Then solve for y* by using x* and what we got from step one. Y=px/py * x = px/py * I/2px.
So, X=I/2px and Y=I/2py
Corner solution útil Max
Either zero consumption of one of the goods or hitting some constraint such as a maximum consumption constraint. B4B won’t be equal. Goods you won’t consume will have lowest B4B
Útil Max corner solution ex.
Max x + y such that 2x+y =1
B4Bx = 1/2 and B4By = 1
So B4B y is strictly greater, so we consume all y and no x.
X=0.
Plug this into the budget constraint, y=I
Demand function
Gives us the optimal quantity demanded as a function of prices and income.
Law of demand
If the demand for a good is decreasing as the price goes up
Price and quantity have an inverse relationship.
Normal and inferior good
Normal good is something where demand increases with income.
Inferior good is when demand decreases with income. Consume more of something (McDonald’s)when you’re poorer
Complement and substitute goods
Substitute is when your demand is increases as the price of something else increases.
McDonald’s and Burger King are substitutes. If one price goes up, consume other
Complement is of demand decreases as price of something else goes up. If price of milk goes up, less coffee demanded.
Demand elasticities
Measures the percent change in quantity demanded with respect to a percent change in one factor such as goods own price or ones income.
Units free.
Own price elasticity
If we increase our own price by 1%, how much with the demand of that good increase by?
(dq / q) / (dp/p)
Rearrange to get dq/ dp *p/q
In our example where q1(p1)=I/2p1,
Our own elasticity = -I/2p1^2 * p1 / (I/2p1) = -1
So, as we increase our own price by 1%, demand for that good decreases by 1%
Cross price elasticity
As price of Pepsi goes up by 1%, what happens to quantity demanded of coke?
See cheat sheet
(dq1/q1) / (dp2/p2)
Income elasticity
If we increase income by 1%, what happens to quantity demanded of our good?
Put on cheat sheet
(dq1/ dI) * I/q1
Inverse demand willingness to pay
Inverse demand function gives us info about ones willingness to pay for the marginal unit of a given quantity of the good.
Recall that the demand function tells us that at a given price, say $1, if you demand, say 5 units of Kombucha, the 5th units B4B no longer dominates the other goods B4B (coke)
Bottles 0-4 of Kombucha, higher B4B than coke. So keep consuming, but bottle 5 B4B =B4B coke, so stop consuming more. This is the optimal demand for K=5 bottles
Inverse demand asks what should the price or cost of the last bottle be if you optimally demand 5 bottles and this Kombuchas B4B no longer dominates cokes?
How much are you willing to pay for the 5th bottle if you optimally consume 5 bottles
P1(q1) = the function (1/2q1)in our case
Consumer surplus
How do we measure the net value of consuming a total of q1 quantities of good 1?
Since each point on the inverse demand curve (p1 on y and q1 on x), gives us the WTP or value of that marginal unit, we add up these values from 0 to q1.
Value of consuming total of q1 quantities is the area below the inverse demand curve from 0 to q1.
Consumer surplus =total value -total purchase cost
Top left triangle. Willing to pay much more, got it at a cheaper price.
Be sure to know how to take derivatives
Ok
Uber surge pricing
Dynamic pricing takes advantage of supply and demand.
High demand and low supply, surge pricing discourages low willingness to pay riders from requesting rides while encouraging drivers to accept requests.
Balances out supply and demand
Low willingness to pay people out and higher price results in more drivers
Production functions
A firm transforms inputs (factors of production)such as capital (K)and labor (L)into output (q)
The production function, q=f(K,L) describes the technology for transforming inputs into output.
Short run vs long run
In the short run, at least one input (fixed input)can’t be adjusted. If suppose K is fixed at K bar, then q =f(K bar, L)
In long run all inputs can be adjusted
Depends on horizon
Suppose gas is purchased at start of day, divers hired by week, cars leased by month
Horizon = month, then which inputs fixed? Week?
If month, then car is fixed because lease car by month
Can change gas and driver
If week, then car and driver are fixed, can change gas.
Returns to scale
For some m>1,
Constant returns to scale (CRTS) if:
f(mK,mL)= mf(K,L)
If double both inputs, same as doubling original output.
Increasing returns to scale (IRTS)if:
f(mK, mL) > mf(K,L)
Output from doubling inputs > double of regular inputs
Decreasing returns to scale (DRTS)if:
f(mK,mL) < mf(K,L)
Double inputs produces less than double what you actually produce. For example, too many people, can’t do as much.
Returns to scale Example:
Consider function: f(K,L) = K^a * L^b
What happens to output if we double both inputs?
f(2K,2L)= (2K)^a * (2L)^b = 2^(a+b) * K^a * L^b = 2^(a+b) f(K,L) Compare that to 2f(K,L) If a plus b = 1, then crts If a+b is bigger than 1, then IRTS if less than 1, DRTS
Marginal product of labor
The additional output produced by an additional unit of labor (marginal unit), holding all other factors constant.
Partial deririvatve of q wrt L
=partial deriv f(K,L) / partial deriv L
Average product of labor
Ratio of output to the amount of labor used
= q/L = f(K,L) / L
Marginal and average product relation
Marginal product drives average product
If MPl> APl, the average goes up
If MPl < APl the average goes down
Very last unit pulls the average in that direction
Prove this using calculus —> partial deriv of APL WRT L. We get derivative is positive if MPL is bigger than APL, meaning apl goes up when MPl is bigger
Diminishing marginal returns
Production function is increasing in its inputs (marginal productivity is positive)
But we also would think that is a firm increases its use of an input (holding other inputs fixed),the marginal productivity of this input goes down.
Partial deriv of MPL / partial deriv of L = second partial deriv of f(K,L)/ partial deriv L squared
This should be negative to have diminishing marginal returns
But, we see with an example where q =f(K,L)= K^a * L^b,
MPL = bK^aL^(b-1) because partial deriv with respect to L.
Partial deriv of the MPL = b(b-1)K^a * L^(b-2)
When will this be positive?
Remember b>0
If b=1,then partial deriv of MPL wrt L =0. If b is bigger than 1, then increasing marginal returns and less than 1 then diminishing marginal returns.
Isoquant
Similar to indifference curve.
A collection of input bundles that produce the same amount of output.
q= f(K,L)where K and L are the input bundle.
Car isoquant example where we have red car and blue car
q= f(R,B)
Trade off between the two. Black car is one axes and red car the other. Draw the curve (concave up)
Isoquant slope
Slope measures tradeoff between inputs
Marginal rate of technical substitution
How many black car hours can we give up in exchange for an increase in red car hours and still produce 240 miles?
Slope = MRTS = dB/dR = -MPR / MPB
Where MP R Is marginal product of R
Isocost line
Collection of input bundles that have the same cost
Px (x) + Py(y) = C
y = -(px/py) (x) + C/Py
All the bundles on the line cost C
Isocost trends
Bundles on isocost lines toward bottom left are cheaper. Also will produce less output.
Bundles on each line cost the same.
Cost minimization problem in words
We want to find the cheapest input bundle that will produce at least q units of output, on or above the isoquant.
Optimal bundle is where isoquant is tangent to isocost, slope of isoquant = slope of isocost
Need at least q units of output, want cheapest bundle such that we get q. So, cost minimization.
We want to find the cheapest input bundle (x,y) that produces at least q units of output.
Cost minimization mathematical setup
Isocost slope = -px/py
Slope of isoquant is -MPx/MPy
So equate and rearrange and we get equality of B4B
MPx/px =MPy/py
Min (where x,y>=0) of Px X + Py Y
Subject to production constraint f(x,y) >= q
Interior solution cost minimization
If optimal solution is strictly positive, then B4B should be equal when evaluated at optimal solution.
Use equality of B4B
MPx/ Px = MPy / py
Corner solution cost min
Either zero demand of one of the inputs or hitting some constraint such as a maximum usage constraint. Here, B4Bs aren’t equal
Cost function
Let (x,y) be the solution to our cost min problem for a given quantity, q>0.
Lowest cost of producing at least q is
C(q) =Px X* + Py Y*
C(q) is what we refer to as the cost function
We get the cost function by:
1. solving for optimal input demands given q
2. Plug in optimal input demands to the cost eq. Multiply these input demands by their respective prices
Marginal cost
Cost of producing the marginal unit of output
MC(q) = dC(q) / dq
Average cost
Costs divided by output produced
AC (q) = C(q)/q
Marginal and average cost relation
If MC is bigger then AC is increasing in q
If MC is smaller, then AC decreasing in q
Economics and diseconomies of scale
Economies of scale is if the average cost goes down as output increases.
dAC(q)/dq is negative
Diseconomies of scale is if average cost goes up as output increases.
dAC(q)/dq is positive.
Gets more expensive for later units of q
Main difference cost minimization and utility max
We do the production constraint for cost minimization q= f(x,y)
As opposed to budget constraint for utility max
LR cost function interior example
C(q) = min (x and y greater than or equal to 0) Px X + Py Y
q <= f(x,y)= xy
Step 1 X * and y* > 0 because if either were zero, then q would =0. That can’t happen, as q must be positive. So, we can use equality of B4B. B4Bx = y/px B4By = x/py Equate them, we get y= (px/py) (x)
Step 2.
Remember the constraint is q, not the budget, so we plug this into q=xy
Q = x (px/py)(x)
X* = SQRT(py/px q)
Plus back in where we have Y in terms of x, and we get Y
Step 3. Get the cost function
Plus x* and y* into the original cost equation. C(q)=px X* + Py Y*
LR cost min corner ex
C(q) = min (x and y greater than or equal to 0) Px X + Py Y
q<= f(x,y) =5x + 7y
Also, q>0 and assume 7/py < 5/px
Step 1. B4Bx = 5/px and B4By =7/py
We are given B4Bx is strictly greater, so therefore, y*=0
Step 2. q= 5x+7y, y=0,so q=5x, x=q/5
Step 3. Plug into cost function, C(q) = px (q/5)
Short run cost function
In the short run, we have a fixed input.
So we are just finding one unknown
Short run cost min interior example
C(q) = min (x and y greater than or equal to 0) Px X + Py Y
q= f(x,y) = xy
Y=y bar > 0 where q>0
Step 1. X star must be positive, as need to produce q>0
q=x(y bar)
X*= q/y bar
Step 2.
Plug into cost function,
C(q) = px/ y bar (q) + Py (y bar)
Short run cost min corner ex
C(q) = min (x and y greater than or equal to 0) Px X + Py Y
q<= f(x,y)= 5x+7y
Y=y bar >0 and q>0. Suppose 7/py < 5/px
Step 1. B4B x is bigger than B4B y, but we are forced to use some y bar greater than zero.
We see that y bar alone produces q= 7y bad. So, only use x when q>7 y bar.
So, optimal input demand for x =0 if q<=7y bar and x* = (q-7 y bar)/5 if q>7 y bar
Step 2.
Cost function depends on q.
For q > 7 y bar, we have C(q)= px (q-7y bar)/5 + Py (y bar)
If q is less than or equal to 7y bar, C(q)= py Times y bar
