Deriving Market Demand From Individual Equations

From Individual Equations to Market Demand

An individual demand equation describes how much one buyer plans to purchase at each price, with other relevant conditions held fixed. Market demand answers the same question for all buyers in the market together. The key operation is to add quantities at a common price: at any particular price, each buyer chooses a quantity, and those quantities form the total demanded.

When individual equations are linear, adding them can produce a linear market equation over a range of prices. But the same equation may not apply at every price. If one buyer's quantity reaches zero before another's, that buyer contributes nothing at higher prices. The market equation must then be written in pieces, according to which buyers are still demanding the good.

Definition: Suppose buyer $i$'s quantity demanded at price $p$ is $q_i(p)=a_i-b_i p$, where $a_i>0$ and $b_i>0$. Under the assumption that quantities cannot be negative, the buyer's demand is $q_i(p)=\max(0,a_i-b_i p)$. For $n$ buyers, market demand at price $p$ is the horizontal sum $Q_D(p)=\sum_{i=1}^{n}q_i(p)$. The price at which buyer $i$'s linear expression reaches zero is $a_i/b_i$.

Here, “horizontal sum” means that quantities are added across buyers at the same price. It does not mean adding prices or combining quantities observed at different prices. For example, if two buyers demand 6 and 3 units, respectively, at a price of $4, market demand at $4 is 9 units.

The equations below assume a single good with the same price for every buyer, nonnegative prices, and individual linear demand expressions that apply over the stated ranges. Other determinants of demand are held fixed. If an expression $a_i-b_i p$ becomes negative, it is not a feasible negative purchase: that buyer's quantity is zero instead. This is why different zero-demand prices can make aggregate demand piecewise linear.

A Reliable Method for Combining Equations

Find each buyer's zero-demand price.
Set $a_i-b_i p=0$ and solve $p=a_i/b_i$. These prices mark where a buyer stops contributing a positive quantity.

Order the zero-demand prices.
Use them to divide the possible price range into intervals. Within each interval, the set of buyers with positive demand is fixed.

Add only active buyers' equations.
For each interval, sum the quantities of buyers whose demand is positive there. Do not include a linear expression that would be negative at that price.

Check the result.
At selected prices, calculate each buyer's nonnegative quantity directly and compare their sum with the relevant market equation.

Worked Example: Two Buyers With Different Zero-Demand Prices

Two buyers' weekly demands for a service are $q_A(p)=20-2p$ and $q_B(p)=12-p$, where $p$ is the price in dollars per service. Derive market demand for $p\geq0$, then verify it at prices of $4 and $11.

Method: identify the price ranges first. For buyer A, solve $20-2p=0$, giving $p=10$. For buyer B, solve $12-p=0$, giving $p=12$. Both buyers demand positive quantities below $10; only B does so from above $10 through $12; neither demands a positive quantity above $12.

For $0\leq p\leq10$, add both equations:

$$ Q_D(p)=(20-2p)+(12-p)=32-3p. $$

For $10<p\leq12$, A's quantity is zero, so only B contributes:

$$ Q_D(p)=12-p. $$

For $p>12$, both buyers' quantities are zero. Thus the complete market equation is

Q_D(p)= \begin{cases} 32-3p, & 0\leq p\leq10,\\ 12-p, & 10<p\leq12,\\ 0, & p>12. \end{cases}

Check at $p=4$. Buyer A demands $20-2(4)=12$ services, and buyer B demands $12-4=8$. The total is $12+8=20$. The first market equation gives $32-3(4)=32-12=20$, as required.

Check at $p=11$. Buyer A's linear expression gives $20-2(11)=-2$, which is not a feasible quantity, so A demands zero. Buyer B demands $12-11=1$, making the total $0+1=1$. The applicable market equation gives $12-11=1$, matching the direct sum. The check also shows why extending the first equation to this price would be wrong.

Why the Market Equation Is Piecewise

The aggregate equation follows directly from adding each buyer's actual quantity, not from adding every buyer's linear expression at every price. A buyer contributes $a_i-b_i p$ only while that expression is positive. Once it reaches zero, the buyer contributes zero. Formally, the market equation is the sum of these nonnegative quantities:

Q_D(p)=\sum_{i=1}^{n}\max(0,a_i-b_i p).

Between two consecutive zero-demand prices, the set of positive terms in this sum does not change. Each active term is linear in $p$, while each inactive term is zero. Adding the active terms therefore gives a linear expression on that interval. When price passes a zero-demand price, at least one term becomes inactive; the expression for the market total may change. This establishes why a market-demand equation with different buyer thresholds is generally piecewise linear.

At a buyer's zero-demand price, including or excluding that buyer's term gives the same total, because the term equals zero at that exact price. The intervals can therefore be assigned boundary points consistently, as long as the equation used at each boundary gives the correct total.

Worked Example: A Group of Similar Buyers and One Smaller Group

In a hypothetical market, each of three households demands $18-3p$ units per week, and each of two other households demands $10-2p$ units per week. The price $p$ is in dollars. Derive market demand and verify the total at $p=4$ and $p=5.5$.

Method: multiply each representative equation by the number of households in that group, then account for zero-demand prices. For the first group, $18-3p=0$ at $p=6$. For the second, $10-2p=0$ at $p=5$. Thus both groups are active through $5, only the first group is active above $5 through $6, and demand is zero above $6.

For $0\leq p\leq5$, the three households in the first group together demand $3(18-3p)=54-9p$. The two in the second group demand $2(10-2p)=20-4p$. Adding gives

$$ Q_D(p)=(54-9p)+(20-4p)=74-13p. $$

For $5<p\leq6$, the second group's linear expression would be negative, so that group contributes zero. The first group contributes $54-9p$. Above $6, both groups contribute zero. Therefore,

Q_D(p)= \begin{cases} 74-13p, & 0\leq p\leq5,\\ 54-9p, & 5<p\leq6,\\ 0, & p>6. \end{cases}

Check at $p=4$. Each first-group household demands $18-3(4)=6$, so that group demands $3(6)=18$. Each second-group household demands $10-2(4)=2$, so that group demands $2(2)=4$. The total is $18+4=22$. The first market equation gives $74-13(4)=74-52=22$.

Check at $p=5.5$. Each first-group household demands $18-3(5.5)=18-16.5=1.5$, giving $3(1.5)=4.5$. The second group's expression is $10-2(5.5)=10-11=-1$, so its feasible demand is zero. The total is $4.5$. The middle equation gives $54-9(5.5)=54-49.5=4.5$. Both checks agree.

Verifying Multiple Breakpoints

With more than two types of buyers, there can be several price thresholds. A careful solution lists all thresholds in order and recomputes the sum each time the active set changes. A quick check at each threshold is useful: the equation on either side should produce the same total at the threshold, because the buyer entering or leaving the sum has quantity zero there.

Worked Example: Three Individual Equations and Three Price Ranges

Three buyers have weekly demands $q_A(p)=24-4p$, $q_B(p)=21-3p$, and $q_C(p)=10-2p$, with nonnegative quantities and $p\geq0$. Derive market demand and verify it at $p=5$ and $p=6.5$.

Method: order all three zero-demand prices. Buyer A reaches zero at $24/4=6$; B reaches zero at $21/3=7$; and C reaches zero at $10/2=5$. Thus all three buyers are active from zero through $5. Above $5 through $6, A and B remain active. Above $6 through $7, only B remains active. Above $7, no buyer demands a positive quantity.

For $0\leq p\leq5$, add all three equations:

$$ Q_D(p)=(24-4p)+(21-3p)+(10-2p)=55-9p. $$

For $5<p\leq6$, buyer C contributes zero, so the sum is $(24-4p)+(21-3p)=45-7p$. For $6<p\leq7$, only B contributes, so the sum is $21-3p$. The complete result is

Q_D(p)= \begin{cases} 55-9p, & 0\leq p\leq5,\\ 45-7p, & 5<p\leq6,\\ 21-3p, & 6<p\leq7,\\ 0, & p>7. \end{cases}

Check at $p=5$. A demands $24-4(5)=4$, B demands $21-3(5)=6$, and C demands $10-2(5)=0$. The total is $4+6+0=10$. The first equation gives $55-9(5)=55-45=10$. The equation just above $5 would also give $45-7(5)=10$, confirming the boundary total.

Check at $p=6.5$. A's expression is $24-4(6.5)=24-26=-2$, so A demands zero. B demands $21-3(6.5)=21-19.5=1.5$. C's expression is $10-2(6.5)=10-13=-3$, so C also demands zero. The total is $1.5$, and the applicable equation gives $21-3(6.5)=1.5$. The direct calculation and market equation agree.

Common Mistake: Adding Inactive Buyers

A frequent error is to add all the original linear expressions even after one becomes negative. That calculation can make market demand too large or even negative. For example, in the first worked example at $p=11$, mechanically adding the two expressions gives

$$ (20-2(11))+(12-11)=-2+1=-1. $$

The result of $-1$ is not a feasible market quantity. The correction is to replace buyer A's negative expression with zero: market demand is $\max(0,-2)+\max(0,1)=0+1=1$. Equivalently, on the range $10<p\leq12$, use only buyer B's equation, $Q_D(p)=12-p$.

Another related error is to multiply a representative equation by the number of buyers in a group without checking whether the group is still active at the price in question. Group size scales the group's total only while its individual members have positive demand. First determine the applicable price range, then add or multiply.

Key takeaway: Market demand at a given price is the sum of all buyers' nonnegative quantities at that price. Find each buyer's zero-demand price, divide the price range at those thresholds, and add only the equations of buyers who are active in each range. Verify the result by direct substitution into individual demands.

Check Your Understanding

For each problem, identify the active buyers in the relevant price ranges and check the final total against the individual quantities.

Buyer A demands $16-2p$, and buyer B demands $9-p$, with nonnegative quantities and $p\geq0$. Derive the piecewise market equation and calculate demand at $p=5$. (Answer: $Q_D=25-3p$ for $0\leq p\leq8$, $Q_D=9-p$ for $8<p\leq9$, and $Q_D=0$ for $p>9$; at $p=5$, $Q_D=10$.)
Two buyers each demand $14-2p$, and three other buyers each demand $12-3p$. Quantities cannot be negative. Derive market demand and calculate it at $p=5$. (Answer: $Q_D=64-13p$ for $0\leq p\leq4$, $Q_D=28-4p$ for $4<p\leq7$, and $Q_D=0$ for $p>7$; at $p=5$, $Q_D=8$.)
Three buyers demand $30-5p$, $18-2p$, and $20-4p$, respectively. Derive the market equation and calculate total demand at $p=7$. (Answer: $Q_D=68-11p$ for $0\leq p\leq5$, $Q_D=48-7p$ for $5<p\leq6$, $Q_D=18-2p$ for $6<p\leq9$, and $Q_D=0$ for $p>9$; at $p=7$, $Q_D=4$.)
Five households each demand $8-p$, and two households each demand $20-4p$, with nonnegative quantities. Derive market demand and find it at $p=3$ and $p=6$. (Answer: $Q_D=80-13p$ for $0\leq p\leq5$, $Q_D=40-5p$ for $5<p\leq8$, and $Q_D=0$ for $p>8$; demand is $41$ at $p=3$ and $10$ at $p=6$.)

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