Why the Short-Run Supply Slope Matters
A change in aggregate demand does not have a fixed effect on real output. In the short run, the result depends in part on how much firms increase production when the price level rises. If short-run aggregate supply (SRAS) is relatively flat, a demand increase tends to produce a larger output response and a smaller price-level response. If SRAS is relatively steep, the same demand increase tends to produce a smaller output response and a larger price-level response.
This comparison is useful when considering a demand shift intended to reduce a recessionary output gap. The shift can raise output, but its effect on output depends on the slope of SRAS. The analysis below isolates that relationship: it holds the initial equilibrium, aggregate-demand slope, and demand shift fixed, and varies only the short-run supply slope.
A Linear Model for Comparing Slopes
Use a local linear model centered on an initial short-run equilibrium, with initial output \(Y_0\) and initial price level \(P_0\). Let \(b>0\) be the absolute slope of inverse AD and \(s>0\) the slope of SRAS. Let \(D>0\) denote an upward shift in inverse AD, measured in price-level index points. The relationships after the shift are:
Here, \(Y\) is real output, and \(P\) is the price level. The initial equilibrium is at \(Y=Y_0\), \(P=P_0\). A positive \(D\) shifts AD upward: at any given output, the price level consistent with AD is higher than before. The model treats both curves as straight lines over the range being analyzed. It holds SRAS fixed and assumes there are no simultaneous supply shifts or changes in other determinants of the curves.
To find the new equilibrium, set the two expressions for \(P\) equal. The resulting changes from the initial equilibrium are:
Because \(b\) and \(s\) are positive, both changes are positive for an upward demand shift. The denominator shows that a steeper SRAS reduces the output response. The price response, by contrast, increases with \(s\), holding \(b\) and \(D\) fixed.
Why the Shift Is Divided Between Output and Prices
The equations also show why the demand shift does not generally translate entirely into either output or the price level. Starting from the initial equilibrium, write the changes as \(\Delta Y=Y-Y_0\) and \(\Delta P=P-P_0\). Along the shifted AD curve, the price level changes by the shift \(D\) minus the decline in price associated with moving to higher output. Along SRAS, the price level rises with higher output.
Equating the right sides gives \(D-b\Delta Y=s\Delta Y\), so \(D=(b+s)\Delta Y\). Therefore, \(\Delta Y=D/(b+s)\), and substituting into the SRAS relationship gives \(\Delta P=sD/(b+s)\). This is the comparative-static result: the demand shift is accommodated partly by higher output and partly by a higher price level, with their relative sizes governed by the slopes.
The output share of the vertical shift is \(b\Delta Y/D=b/(b+s)\), while the price-level share is \(\Delta P/D=s/(b+s)\). These shares sum to one. As \(s\) becomes smaller, the output share grows; as \(s\) becomes larger, the price-level share grows. This statement concerns the specified linear model, not every possible shape of AD and SRAS.
Worked Example: Same Demand Increase With Flat and Steep Supply
An economy begins at \(Y_0=100\) output-index units and \(P_0=100\) price-index points. In both comparisons, inverse AD has slope parameter \(b=2\), and AD shifts upward by \(D=12\) price-index points. Compare a relatively flat SRAS with slope \(s=0.5\) to a relatively steep SRAS with slope \(s=4\). Find the output and price-level changes, then verify each new equilibrium using the curve equations.
Method. Apply the comparative-static formulas with the same \(b\) and \(D\) in each case; change only \(s\).
Flat SRAS. The denominator is \(b+s=2+0.5=2.5\). Thus,
The new equilibrium is \(Y=100+4.8=104.8\) and \(P=100+2.4=102.4\). Check SRAS: \(100+0.5(104.8-100)=100+2.4=102.4\). Check shifted AD: \(100-2(104.8-100)+12=100-9.6+12=102.4\). Both curves give the same price level.
Steep SRAS. Now \(b+s=2+4=6\), so
The new equilibrium is \(Y=100+2=102\) and \(P=100+8=108\). Check SRAS: \(100+4(102-100)=108\). Check shifted AD: \(100-2(102-100)+12=100-4+12=108\).
Conclusion and trap avoided. With flat SRAS, output rises by 4.8 units and the price level by 2.4 points. With steep SRAS, output rises by 2 units and the price level by 8 points. A common mistake is to treat the demand shift of 12 as a 12-point increase in the equilibrium price level in both cases. The shift raises the AD schedule by 12 at a given output; it does not by itself determine the equilibrium price change.
Using the Comparison to Think About Output Stabilization
Suppose output begins below potential output, \(Y^*\), so the economy has a negative output gap \(Y_0-Y^*<0\). An upward demand shift raises short-run equilibrium output in this model. If the goal is to reduce the gap, the same shift produces a larger immediate output increase under flatter SRAS. With steeper SRAS, more of the shift appears as a price-level increase and less as an output increase.
This is a comparison of the model's short-run equilibrium effects, not a conclusion that any particular demand shift is necessarily desirable or sufficient. The calculation presumes the same initial output and price level and the same AD slope across the cases. It does not incorporate a changing SRAS, a different AD slope, or other simultaneous changes. Potential output is used to describe the gap; it does not enter the two change formulas directly.
Worked Example: Comparing the Reduction in a Recessionary Gap
An economy has potential output \(Y^*=500\), initial equilibrium output \(Y_0=480\), and initial price level \(P_0=120\). Inverse AD has \(b=1.5\), and an upward demand shift has size \(D=8\). Compare SRAS slopes \(s=0.5\) and \(s=3.5\). For each case, calculate the new equilibrium and the remaining output gap.
Method. First calculate the equilibrium changes, then compare the new output with potential output. The gap is output minus potential output, so a negative result indicates output remains below potential.
Flat SRAS, \(s=0.5\). The denominator is \(1.5+0.5=2\). Therefore,
The new equilibrium is \(Y=480+4=484\) and \(P=120+2=122\). The output gap is \(484-500=-16\). The initial gap was \(480-500=-20\), so the shift narrows it by 4 output units without closing it.
Steep SRAS, \(s=3.5\). The denominator is \(1.5+3.5=5\). Thus,
The new equilibrium is \(Y=480+1.6=481.6\) and \(P=120+5.6=125.6\). The output gap is \(481.6-500=-18.4\), a narrowing of 1.6 units from the initial gap of \(-20\).
Check. In the flat case, SRAS gives \(120+0.5(4)=122\), and shifted AD gives \(120-1.5(4)+8=122\). In the steep case, SRAS gives \(120+3.5(1.6)=125.6\), and shifted AD gives \(120-1.5(1.6)+8=125.6\). The flat-SRAS case produces the larger output increase, while the steep-SRAS case produces the larger price-level increase.
Trap avoided. Do not report the change in the output gap as though it were the new output gap. The flat case reduces the gap by 4 units, but the remaining gap is still \(-16\), not \(-4\).
Solving for a Shift That Reaches Potential Output
The same relationships can be rearranged to ask how large a demand shift would be required to achieve a specified output change. If output is initially \(g\) units below potential and the target is to close that gap in the model, then \(\Delta Y=g\). Since \(D=(b+s)\Delta Y\), the required shift is \(D=(b+s)g\). A steeper SRAS requires a larger upward shift for the same output increase, and the corresponding price-level change is \(sg\).
Worked Example: Required Shifts Under Two Supply Slopes
An economy begins at \(Y_0=580\) with potential output \(Y^*=600\) and price level \(P_0=150\). Inverse AD has slope parameter \(b=0.5\). Calculate the upward AD shift required to reach potential output if SRAS has slope \(s=0.5\), and then if it has slope \(s=2.5\). Also calculate the price-level change in each case.
Method. The required output increase is \(600-580=20\). Use \(D=(b+s)\Delta Y\), then use \(\Delta P=s\Delta Y\).
Flat SRAS, \(s=0.5\). The required shift is \(D=(0.5+0.5)(20)=1(20)=20\) price-index points. The price-level change is \(\Delta P=0.5(20)=10\) points. The resulting equilibrium is \(Y=600\), \(P=150+10=160\).
Steep SRAS, \(s=2.5\). The required shift is \(D=(0.5+2.5)(20)=3(20)=60\) price-index points. The price-level change is \(\Delta P=2.5(20)=50\) points. The resulting equilibrium is \(Y=600\), \(P=150+50=200\).
Check. In the flat case, the change equations give \(\Delta Y=20/(0.5+0.5)=20\) and \(\Delta P=10\). In the steep case, they give \(\Delta Y=60/(0.5+2.5)=20\) and \(\Delta P=50\). Both reach potential output, but the steeper SRAS case requires a larger AD shift and yields a larger price-level change.
Trap avoided. The required shift is not just the desired output increase. It depends on both the AD slope and the SRAS slope through \(b+s\); using \(D=\Delta Y\) would ignore the model's price response.
Common Errors and Interpretation
- Using the SRAS slope in the wrong place. The output change is \(D/(b+s)\), not \(sD/(b+s)\). The latter expression gives the price-level change.
- Calling a larger SRAS slope “flatter.” With price level on the vertical axis and output on the horizontal axis, a larger positive slope is steeper.
- Changing more than one assumption in the comparison. To isolate the effect of SRAS slope, keep the initial equilibrium, the AD slope, and the size of the AD shift fixed.
- Confusing a shifted curve with a new equilibrium. A shift changes the curve at each given output; the intersection with the other curve determines the new output and price level.
- Confusing price-level change with inflation. The calculation gives a change in the price-level index between two equilibria. It does not, by itself, calculate an inflation rate over a specified period.
The slope comparison is most useful as a controlled “what if” exercise. Under the stated linear assumptions, an identical upward AD shift raises output more when SRAS is flatter and raises the price level more when SRAS is steeper. Always identify what is held constant, use the same units for both comparisons, and verify that the calculated point lies on both curves.
Check Your Understanding
For each problem, use the stated linear model and report changes in the units given.
- Initial output is 200 units and the initial price level is 90 index points. Inverse AD has \(b=1\), and it shifts upward by \(D=10\). Calculate the output and price-level changes for SRAS slopes \(s=0.25\) and \(s=2\). Which case has the larger output response? (Answer: For \(s=0.25\), \(\Delta Y=8\), \(\Delta P=2\); for \(s=2\), \(\Delta Y=10/3\), \(\Delta P=20/3\). The flatter case has the larger output response.)
- At an initial equilibrium of \(Y_0=300\), \(P_0=110\), inverse AD has \(b=0.5\) and shifts upward by \(D=9\). For \(s=0.5\), find the new equilibrium and verify the price level from both curves. (Answer: \(\Delta Y=9\), \(\Delta P=4.5\); new equilibrium \(Y=309\), \(P=114.5\). Both curves give \(114.5\).)
- An economy has \(Y_0=240\), \(P_0=102\), \(Y^*=250\), \(b=1\), and \(s=3\). AD shifts upward by \(D=16\). Find the new equilibrium and remaining output gap. (Answer: \(\Delta Y=4\), \(\Delta P=12\); new equilibrium \(Y=244\), \(P=114\); remaining gap \(244-250=-6\).)
- Output is initially 20 units below potential, and inverse AD has \(b=0.5\). Find the required AD shift and resulting price-level change to reach potential under \(s=0.5\) and \(s=2.5\). (Answer: For \(s=0.5\), \(D=20\) and \(\Delta P=10\); for \(s=2.5\), \(D=60\) and \(\Delta P=50\).)
- With \(b=1\), the same \(D=12\) shift produces an output increase of 8 units under a flat SRAS. Under a second, steeper SRAS, the price-level increase is 9 points. Find each SRAS slope and the output increase in the steeper case. (Answer: The flat slope is \(s=0.5\); the steeper slope is \(s=3\); the steeper-case output increase is \(3\) units.)