# Tag Info

33

Is it simply for saving warehouse costs? Probably yes, holding onto inventory is very expensive. You have to pay for warehousing of the good, it takes the spot of some other inventory that might be in high demand. Food is also perishable so it cannot be stored indefinitely. Stores have to always guess what demand for their products will be, sometimes they ...

29

prices should have already been set to maximize the trade off between profit-per-sale and volume sold But profit-per-sale depends on costs, which depends on the theft numbers, so if theft increases, the equation changes.

19

What you're describing is retail shrink. It is taken into consideration when setting prices. A business will typically have consultants come in, measure their shrink to be X percent, and prices will be adjusted accordingly. Back when I worked in retail, there was a big printout in the break room informing everyone on shrink. The 5 kinds of shrink outlined on ...

14

It's not about saving warehousing cost. They can't sell 12 months old Christmas chocolate next year, so they have to sell it in the next few months. Their options are to either sell it to customers, sell it to a business, or pay to throw it away. The price indicates that they think selling it very cheaply to customers is the most profitable option, all ...

13

"I am not given wealth $w$ although I suppose I could assume any firm who is purchasing has some budget." No. This is exactly where the fundamental microeconomic theory of the firm differs from the microeconomic Consumer Theory: the firm is not constrained by a budget. The reason is that this fundamental theory deals most and foremost with the "long-term" ...

10

It is the first one, $TC(0) = FC$. This is the definition. Also consider that it is not clear what is "transformed by $q$ in some way". In case of $$\frac{5q}{q+1} + \frac{5}{q+1}$$ are the two fractions transformed by $q$, or should I just sum them up to 5? With your function, one can rearrange it to $$TC(q) = \frac{5}{q+1} + 5 + 5q + q^2 = -\frac{5q}{... 9 What he's saying is that once the burger is already made, the cost of making the burger is a sunk cost, and thus the marginal cost of the burger is just the cost of the tiny labor involved in picking it up and selling it to the customer. That's sort of an odd position to take. Obviously once the burger is cooked you can't recuperate the costs, but in the ... 9 Price is set by the competition In general, the prices are set by the supply and demand for the whole market. If a merchant sells the same goods for a higher price than competitors without a corresponding advantage (location, better service, convenience) then people won't buy these goods and the merchant will earn less profit as the decrease in volume will ... 8 This question really forces one to think about the role that quantity plays in the competitive equilibrium. The two main points that, I think, explain the way this works are: The market quantity is endogenous In competitive equilibrium, the market clears I think the thing that is perhaps causing confusion here is that, recalling that it is a true statement ... 7 As said in my comment, The fixed cost is usually defined as the cost when quantity is equal to zero, and the variable cost as the total cost minus the fixed cost. Hence, if TC(q) is the total cost for the given level of quantity q, then FC=TC(0) is the fixed cost, which is a constant independent of q; and VC(q)=TC(q)-FC is the variable cost. ... 6 Your are right. You have to minimize the average cost.$$c(Q)=\frac{C(Q)}{Q}=6000 +40Q+Q^2$$Calculate the first derivative and set it equal to zero:  c'(Q)=40+2Q=0  Solve this equation for Q. Denote the optimal value as Q^*. Q^* can be a local maximum or a local minimum If c''(Q^*)>0, then you have found the local minimum. The local ... 6 Essentially, the question of electrification is a trade-off: electrified railways are more energy efficient, the engines often supply more power and they are quieter and without local emissions while at the same time requiring a large up-front cost and additional maintenance. The benefits of electric railways are best played out when running many trains on ... 5 The derivative is used in some contexts, but not all, when the cost function is differentiable. In those contexts, it tends to be assumed that supply is continuous, not discrete. This is a matter of convention and of analytic convenience. It has the advantage of being consistent, whether you're approaching the supply point from above or from below. But in ... 5 Since it was mentioned in an another answer let's clear this first: whether the transportation (and its time and monetary costs) should be associated with the intended consumption of the good you are going to purchase, or it can be considered as consumption on its own, depends on your subjective view of it: do you derive any form of pleasure by the trip ... 5 You might want to read up on repeated games. You are right, in a one-period model, once produced, the seller has little marginal cost, so could potentially sell at any price. However, his price at t will affect behavior at t+1. He needs to credibly commit (or signal) that he will not do this again at t+1, otherwise he will be stuck in the same ... 5 You're making this way more complicated than it needs to be. Edit: Okay it's a little more complicated that I thought but hey! What a cool result! AC = \frac{C(q)}{q} \\ MC = C'(q) When you minimize AC with respect to q,$$\frac{\partial AC}{\partial q} = \frac{C'(q) \cdot q - C(q)}{q^2} = 0\implies C'(q) \cdot q - C(q) = 0\implies C'(q) \...

5

The people who do the planning do explicitly account for the value of time and convenience to transit users. At least, in all the systems I've worked on. (source: I have had jobs where I worked on exactly this, for London, for the Netherlands, and for international train travel) Interchanges are modelled with penalties which cover: the time taken to move ...

5

Your intuition is correct. First, you're right that "marginal cost only depends on variable cost", since \begin{equation} MC(q)=\frac{\mathrm dTC(q)}{\mathrm dq}=\frac{\mathrm d(FC+VC(q))}{\mathrm dq}=\frac{\mathrm dVC(q)}{\mathrm dq}. \end{equation} Next, if marginal cost is some constant $k$, then variable cost must be $VC(q)=kq$, because we can ...

4

Suppose initially there are no fixed costs. What does it mean to take an average? Consider a cost function $C(y)$. What does it mean to take the “average” of this function? Mathematically, it is just $$A(y) = \frac{c(y)}{y}$$ Let’s suppose we are considering $C(y) = y^3$. Suppose we now consider $y = 5$. Then $$A(5) = \frac{5^3}{5} = 25$$ This is just ...

4

Yes, if there are non-zero fixed costs, and constant marginal cost, then average cost decreases strictly monotonically with quantity, asymptotic to the marginal cost.

4

Short answer: Yes, it is possible. Decreasing average cost implies that marginal cost is less than average cost ($MC<AC$, which can be proved by simply taking the first derivative of $C(q)/q$). With constant marginal cost, there exists a simple linear cost function $C(q)=F+a\times q$ that satisfies the constant $MC$ condition, where the constant $F$ is ...

4

You are missing the average cost curve in the same diagram. Basic algebra gives us the following. Let's find the minimum of the $AC = C/Q$. We have $$\frac {\partial AC}{\partial Q} = \frac {MC\cdot Q - C}{Q^2}$$ For this to be equal to zero, we must have $MC \cdot Q = C \implies MC = AC$. So when $AC$ is at its minimum, it equals $MC$. But we also ...

4

I'll offer a less algebraic alternative to Alecos's answer. In short, yes and no. The "no" part Normally the MC and AC curves would look like the following, with MC intersecting AC from below AC's minimum point. Suppose price $P_0$ were below this point. Then the firm would sell at a quantity below $Q_1$. But what does this imply for the firm's ...

4

Hint For profit maximization, either $x_1$ or $x_2$ (but not both) must be zero. If not, say $x_1^*>x_2^*>0$ at the optimum, then one could increase profit by lowering cost by reducing $x_2^*$ without affecting output and thus revenue. Let $z=\max\{x_1,8x_2\}$. The profit function can be written as \begin{equation} p[3(x_3)^{1/3}(z)^{1/3}]-w_3x_3-...

3

To help you discern the two, let's try to explain with words and understand what information are we getting from the derivative and from the difference, respectively: The derivative gives you information about the change in cost relative to change in quantity produced, in a specific, local, point(quantity)1. In other words you are measuring the change in ...

3

The total cost was 291 m.u. for 1 unit given $MC(q) \equiv \frac{dC(q)}{q} = 3 q^2 - 40 q + 220$. First, the area under the Marginal Cost gives the Total Variable Cost, $TVC(q)$. To find this area, we will integrate the Marginal Cost: \begin{align} TVC(q) &= \int_0^q\frac{dTVC(q)}{dq} dq = \int_0^q (3q^2 - 40 q + 220) dq\\ &= \left. \left(q^3 -... 3 Marginal cost is always the same as the shadow price in the cost minimization problem \begin{eqnarray*} \min_x && w \cdot x \\ s.t. && f(x) = y. \end{eqnarray*} In optimum the shadow price (Lagrange multiplier) belonging to the condition f(x) = y is the marginal cost. However there are other optimization problems where the shadow price is ... 3 The diagram is fine as an illustration of a possible relation between the long run average cost curve and one particular short run average cost curve. Long run in the context of a firm's cost functions means that all inputs are free to vary, so it makes sense to refer to the long run cost curve. Short run means that one or more inputs (eg capital equipment)... 3 \newcommand{\fone}{\color{red}{f_1(q)}} \newcommand{\ftwo}{\color{blue}{f_2(q)}} For the sake of simplicity, call f(q) = \frac{w}{k}q + \frac{rk}{q} = rk\left(\underbrace{\frac{1}{q}}_{\fone} + \underbrace{\frac{w}{rk^2}q}_{\ftwo} \right) = rk (\fone + \ftwo)  where I have factored $rk$ out of the expression. Now you want to understand each term ...

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