Introduction
A rate can be positive during one part of a time interval and negative during another. A signed rectangle sum keeps track of both directions: contributions from positive rates add to the total, while contributions from negative rates subtract from it.
This means the final signed sum may be small even when the individual contributions are large. The positive and negative parts can cancel. To understand what a rectangle sum says, it is therefore important to distinguish the net signed estimate from the total magnitude of the estimated contributions.
Signed Contributions from Rectangles
Suppose an interval is divided into subintervals. On subinterval \(i\), choose a sample time and let \(r_i\) be the rate at that time. Let \(\Delta t_i\) be the width of that subinterval, measured as a positive duration. The rectangle's signed contribution is its sampled rate multiplied by its width.
Since each \(\Delta t_i\) is positive, \(c_i\) has the same sign as \(r_i\). A positive sampled rate gives a positive contribution; a negative sampled rate gives a negative contribution; a zero sampled rate gives zero. The signed sum adds the contributions with their signs. The total magnitude adds their absolute values instead.
When \(r_i\) is a rate per unit time, multiplying by \(\Delta t_i\) gives an amount of the quantity being tracked. For example, if the rate is measured in litres per minute and the width in minutes, the contribution is measured in litres. The signed sum describes the net estimated change, not necessarily the amount that moved in both directions combined.
Worked Example: Positive and Negative Rates on Unequal Intervals
Worked Example: Positive and Negative Rates on Unequal Intervals
A rate is sampled on three consecutive intervals. The interval widths are \(2\), \(1\), and \(3\) minutes, and the sampled rates are \(4\), \(-6\), and \(2\) litres per minute. Calculate the signed rectangle sum and the total magnitude of the rectangle contributions.
First calculate each contribution as rate multiplied by interval width:
Add the contributions with their signs to obtain the signed sum:
For total magnitude, take the absolute value of each contribution before adding:
Check: The positive contributions total \(8+6=14\) L, and the negative contributions total \(-6\) L. Their signed combination is \(14-6=8\) L, as calculated. Their magnitudes total \(14+6=20\) L. Thus \(S=8\) L and \(G=20\) L.
The signed answer is not the sum of all movement in both directions. It is the positive contribution after accounting for the negative one. The magnitude answer records both directions without allowing one to subtract from the other.
Cancellation in the Sum
The relationship between the signed sum and the total magnitude can be made precise. Separate the contributions into positive, negative, and zero terms. Let \(P\) be the sum of the positive contributions, and let \(N\) be the sum of the absolute values of the negative contributions. Both \(P\) and \(N\) are nonnegative. Zero contributions affect neither sum.
For example, if positive contributions total \(12\) units and negative contributions have magnitudes totaling \(11\) units, the signed sum is \(12-11=1\) unit, while the total magnitude is \(12+11=23\) units. The small net result comes from cancellation, not from small contributions.
Why the Cancellation Formula Holds
We can derive the result by considering which of \(P\) and \(N\) is larger. If \(P\geq N\), then \(S=P-N\geq 0\), so \(|S|=P-N\). Therefore:
If \(N\geq P\), then \(S=P-N\leq 0\), so \(|S|=N-P\). Therefore:
In either case, \(G-|S|=2\min(P,N)\), which is nonnegative. Rearranging gives \(|S|\leq G\). Also, the difference is zero precisely when the smaller of \(P\) and \(N\) is zero. That is exactly the case in which contributions do not include both a positive and a negative total.
Worked Example: Large Contributions Can Produce a Zero Sum
Worked Example: Large Contributions Can Produce a Zero Sum
Over three consecutive intervals, a sampled rate is \(5\) units per hour for \(2\) hours, \(-3\) units per hour for \(2\) hours, and \(-2\) units per hour for \(2\) hours. Calculate the signed sum and total magnitude.
Calculate the contributions one at a time:
Add the signed contributions:
Now add their magnitudes:
Check: The positive contributions total \(P=10\), and the negative contributions have total magnitude \(N=6+4=10\). Thus \(S=P-N=10-10=0\), and \(G=P+N=10+10=20\). The cancellation formula also gives \(G-|S|=20-0=20=2\min(10,10)\).
A zero signed sum does not mean there were no contributions. Here, \(20\) units of estimated contribution are present in total magnitude, with \(10\) units in each direction. They balance exactly in the net sum.
Worked Example: Same Net Sum, Different Total Magnitudes
Worked Example: Same Net Sum, Different Total Magnitudes
Compare two two-interval rectangle sums. Each interval has width \(1\) minute. The sampled rates for the first sum are \(7\) and \(-5\) units per minute. The sampled rates for the second sum are \(2\) and \(0\) units per minute. Find \(S\) and \(G\) for each.
For the first sum, the contributions are:
Therefore:
For the second sum, the contributions are:
Therefore:
Check: Both signed sums equal \(2\). For the first, \(P=7\) and \(N=5\), so \(P-N=2\) and \(P+N=12\). For the second, \(P=2\) and \(N=0\), so \(P-N=2\) and \(P+N=2\). The same net estimate can therefore accompany very different amounts of cancellation.
A Common Mistake: Taking Absolute Values Too Soon
Suppose the sampled contributions are \(10\) units and \(-6\) units. A common error is to take absolute values before calculating the requested signed sum:
This calculation is valid for the total magnitude, but it is not the signed sum. For the signed sum, preserve the signs when adding:
Correction: First identify which quantity is requested. Calculate \(S=\sum c_i\) for net signed accumulation. Calculate \(G=\sum |c_i|\) for total magnitude. Taking absolute values changes the question; it does not merely change the presentation of the same answer.
A related mistake is to add the rates without multiplying each by its own interval width. Rates alone are not contributions. If intervals have different widths, each sampled rate must be multiplied by the width of the interval it represents.
Interpreting a Signed Rectangle Sum Carefully
A signed rectangle sum is an estimate built from sampled rates and interval widths. Its sign indicates the net direction represented by those rectangles: positive if the positive contributions exceed the negative contributions, and negative if the negative contributions have greater magnitude. A zero sum means the two sides balance in the rectangle calculation.
The total magnitude \(G\) answers a different question: how large the rectangle contributions are when both directions count positively. It is always at least \(|S|\), and the gap between them records cancellation. Neither number should be described as the other. In particular, a small signed sum alone does not show that the individual contributions were small.
These calculations concern the chosen rectangles. They do not by themselves assert that the rectangles give the exact accumulated change of an underlying varying rate. To interpret the result, keep track of the sample rates, widths, units, and whether the requested quantity is net signed accumulation or total magnitude.
Check Your Understanding
For each question, keep signed contributions distinct from their magnitudes.
- A rate is sampled as \(3\) units per minute for \(2\) minutes, then \(-4\) units per minute for \(1\) minute. What are the signed sum and total magnitude?
- Can a signed rectangle sum equal zero when some contributions are nonzero? Explain using positive and negative totals.
- For contributions \(8\), \(-3\), and \(-2\), calculate \(P\), \(N\), \(S\), and \(G\).
- What does the inequality \(|S|\leq G\) say about a signed sum and the total magnitude?
- When does equality \(|S|=G\) hold?
