Solving Linear AD-AS Equilibrium Problems

Why Solve the AD–SRAS Equations?

An aggregate-demand and aggregate-supply diagram represents output and the price level together. When each curve is specified by a linear equation, their intersection can be found algebraically rather than estimated from a graph. The calculation gives the model’s short-run equilibrium output and short-run equilibrium price level. If potential output is also provided, the equilibrium output can then be classified as above, below, or equal to potential.

The earlier material on aggregate demand and aggregate supply establishes the basic directions of the curves: aggregate demand slopes downward in a price-level/output diagram, while short-run aggregate supply slopes upward. This tutorial focuses on solving for their intersection. All numerical models below are hypothetical; their coefficients are chosen for illustration and are not estimates of an actual economy.

Definition: In a fixed short-run AD–SRAS model, an equilibrium is a pair \((Y_e,P_e)\) that satisfies both the aggregate-demand equation and the short-run aggregate-supply equation. Here, \(Y_e\) is equilibrium real output and \(P_e\) is the equilibrium price level. Potential output, written \(Y_p\), is the level of output the model identifies as potential; it must be expressed in the same units as \(Y_e\) to make a valid comparison.

Set Up the Equations and Units

A convenient common form writes the price level as a linear function of output:

\text{AD: } P=A-bY,\qquad \text{SRAS: } P=C+dY, \qquad b>0,\ d>0.

The AD slope is negative because \(b\) is positive and appears with a minus sign. The SRAS slope is positive because \(d>0\). The constants \(A\) and \(C\) are intercepts, and the equations must use the same units for output and the same price-level scale. For example, if \(Y\) is measured in billions of real dollars and \(P\) is a price index, the intercepts and slopes must be interpreted consistently with those units.

The model treats both curves as fixed while the equilibrium is calculated. Potential output is a comparison benchmark, not an additional equation needed to find the intersection. If a problem supplies AD or SRAS in a different form, first identify what each equation says and then use the condition that equilibrium output and price must satisfy both curves.

The Intersection Formula

At equilibrium, the price demanded at output \(Y_e\) must equal the price associated with supplying that same output. Thus \(A-bY_e=C+dY_e\). Collecting the output terms gives the general solution:

Y_e=\frac{A-C}{b+d}, \qquad P_e=C+dY_e =\frac{dA+bC}{b+d}.

The denominator is \(b+d\), rather than \(d-b\), because the AD slope is \(-b\) while the SRAS slope is \(+d\). With \(b,d>0\), the denominator is positive, so this linear system has one intersection. The intersection need not be economically meaningful if it implies an impossible value or lies outside a problem’s stated range; in that case, the equations and assumptions require further scrutiny.

Why the formula holds: At the intersection, both curves assign the same \(P\) to the same \(Y\). Equating \(A-bY_e=C+dY_e\) and moving the output terms to one side gives \(A-C=(b+d)Y_e\), hence \(Y_e=(A-C)/(b+d)\). Substituting this value into SRAS gives \(P_e=C+d(A-C)/(b+d)\). Putting the terms over the common denominator yields \(P_e=[C(b+d)+d(A-C)]/(b+d)=(bC+dA)/(b+d)\). The formulas therefore follow directly from the requirement that the two equations hold simultaneously.

A practical alternative is to solve the two equations by substitution or elimination without memorizing the formulas. Either method should produce the same pair. After solving, substitute the proposed output into both curves: their implied price levels must match. This check often catches an algebra error before the result is compared with potential.

Worked Solutions

Worked Example: Finding the Intersection and Comparing with Potential

A hypothetical economy has real output \(Y\) measured in billions of dollars and price level \(P\) measured in index points. Its curves are \(P=160-0.5Y\) for AD and \(P=40+0.25Y\) for SRAS. Potential output is \(Y_p=150\) billion dollars. Find the short-run equilibrium and determine whether its output is above or below potential.

Method. Equate the two expressions for the price level. The equations are defined in matching output and price-level units, and the specified slopes give downward-sloping AD and upward-sloping SRAS.

At the intersection:

$$ 160-0.5Y=40+0.25Y $$

Subtract 40 from both sides and add \(0.5Y\) to both sides:

120=0.75Y,\qquad Y_e=\frac{120}{0.75}=160.

Substitute \(Y_e=160\) into AD to find the price level:

$$ P_e=160-0.5(160)=160-80=80. $$

Check SRAS independently: \(40+0.25(160)=40+40=80\). Both curves therefore give the same price level at output 160. Finally, compare output values in billions: \(Y_e=160>Y_p=150\). Equilibrium output is above potential output.

Common mistake avoided. The intercept difference is \(160-40=120\), not \(160+40=200\). Using 200 would give the wrong output. The check in both original equations confirms the correct intersection.

Worked Example: An Equilibrium Below Potential

In another hypothetical model, output is measured in billions of real dollars and the price level in index points. AD is \(P=210-0.6Y\), SRAS is \(P=30+0.3Y\), and potential output is \(220\) billion dollars. Calculate the equilibrium and classify its output relative to potential.

Method. Set the price levels equal. The positive magnitudes of the slopes are \(0.6\) and \(0.3\); the corresponding AD slope is negative and the SRAS slope is positive.

$$ 210-0.6Y=30+0.3Y $$

Subtract 30 from both sides and add \(0.6Y\) to both sides:

180=0.9Y,\qquad Y_e=\frac{180}{0.9}=200.

The equilibrium price level from AD is:

$$ P_e=210-0.6(200)=210-120=90. $$

Check SRAS: \(30+0.3(200)=30+60=90\), so the same price level satisfies both equations. Since \(200<220\), equilibrium output is below potential output.

Common mistake avoided. Potential output is not substituted into either curve to obtain the short-run equilibrium. It is used only after solving the intersection. Substituting \(Y_p=220\) would answer what each curve implies at potential output, not where the two curves intersect.

Worked Example: Equations Written as Output Functions

A planning exercise gives AD as \(Y=480-3P\) and SRAS as \(Y=80+P\), with output in billions of real dollars and the price level in index points. Potential output is \(175\) billion dollars. Find the intersection and compare output with potential.

Method. Both equations give output as a function of price. At equilibrium, the output demanded at a given price equals the output supplied at that same price. Equate the right-hand sides:

$$ 480-3P=80+P $$

Subtract 80 from both sides and add \(3P\) to both sides:

400=4P,\qquad P_e=\frac{400}{4}=100.

Substitute the price level into either equation. Using SRAS gives \(Y_e=80+100=180\). Using AD gives \(Y_e=480-3(100)=480-300=180\). Both equations agree. Since \(180>175\), equilibrium output is above potential output.

Common mistake avoided. In this form, equating the equations gives the price level first, not output. Report both coordinates, in their stated units, and verify by substitution into both curves.

A Reliable Procedure and Common Traps

For a linear problem, a short sequence of checks keeps the algebra and interpretation separate:

Read the model and units.
Identify which equation is AD, which is SRAS, the units for \(Y\) and \(P\), and whether potential output is supplied in the same output units.

Impose the intersection condition.
Set the two price expressions equal, or set the two output expressions equal, depending on how the equations are written.

Solve and recover the other coordinate.
Find either \(Y_e\) or \(P_e\), then substitute into one curve to find the remaining value.

Verify and compare.
Substitute the result into both equations. Only then compare \(Y_e\) with \(Y_p\).

Several mistakes recur in exam-style calculations:

Using the wrong sign for a slope. In \(P=A-bY\), the AD slope is \(-b\), not \(+b\). In \(P=C+dY\), the SRAS slope is \(+d\). Equating the stated equations directly is safer than rewriting signs from memory.
Mixing units. If one output figure is in billions and another is in millions, convert before comparing. Likewise, do not treat a price index as a dollar amount.
Stopping after finding one coordinate. An equilibrium answer requires both \(Y_e\) and \(P_e\), unless the question explicitly requests only one.
Checking only one curve. A value found from AD alone is not necessarily an equilibrium. Both curves must produce the same coordinate pair.
Confusing equilibrium with potential. Potential output does not generally equal short-run equilibrium output. Compare the two after solving; do not assume they coincide.

What the Comparison Tells You

The comparison is a classification, not a second equilibrium calculation. If \(Y_e>Y_p\), short-run equilibrium output is above potential; if \(Y_e<Y_p\), it is below potential; and if \(Y_e=Y_p\), it equals potential. This statement depends on comparable definitions and units for the two output values. No claim about the size of a difference is needed to make the classification.

Solving the equations is useful because it replaces a visual estimate with a reproducible result. The graph remains valuable for seeing that the solution is an intersection and for checking whether the curve directions are sensible. Algebra supplies exact coordinates for the specified linear model; it does not make the model’s assumptions or inputs more accurate than they are.

Key takeaway: A short-run AD–SRAS equilibrium is the pair that satisfies both equations. Set their expressions equal, solve for one variable, substitute to find the other, verify both equations, and only then compare equilibrium output with potential output.

Check Your Understanding

For each original problem, find the equilibrium price level and output, check the intersection, and compare output with potential where given.

A hypothetical economy has AD \(P=180-0.4Y\), SRAS \(P=60+0.2Y\), and potential output \(Y_p=210\). Output is in billions of real dollars and \(P\) is in index points. (Answer: \(Y_e=200\), \(P_e=100\); output is below potential.)
AD is \(Y=520-2P\) and SRAS is \(Y=100+P\), with output in billions and the price level in index points. Potential output is \(230\) billion. (Answer: \(P_e=140\), \(Y_e=240\); output is above potential.)
AD is \(P=250-0.5Y\), SRAS is \(P=70+0.25Y\), and potential output is \(240\). Output is measured in billions of real dollars. (Answer: \(Y_e=240\), \(P_e=130\); output equals potential.)
A hypothetical market-wide model gives AD \(P=300-0.8Y\) and SRAS \(P=60+0.4Y\). Output is in billions of real dollars and potential output is \(190\) billion. (Answer: \(Y_e=200\), \(P_e=140\); output is above potential.)
AD is \(Y=700-5P\), SRAS is \(Y=100+P\), and potential output is \(215\) billion real dollars. Output is measured in billions and \(P\) in index points. (Answer: \(P_e=100\), \(Y_e=200\); output is below potential.)

What You'll Learn