HN16 - Calculating with bounds

Bounds and attainable endpoints

A lower bound is a value below which the quantity cannot fall; an upper bound is a value above which it cannot rise. We seek the tightest bounds supplied by the information. A minimum or maximum must actually be attainable. An excluded endpoint can be a tight bound even though no allowed value equals it.

Worked example 1: a sum

Question. Find the bounds of $a+b$, where $12.35\le a<12.45$ and $7.15\le b<7.25$.

Plan. Add the two lower endpoints and then the two upper endpoints.

Both lower endpoints can occur together. Neither upper endpoint is included.

Answer. $19.50\le a+b<19.70$. The minimum is $19.50$; there is no attained maximum, but values can approach $19.70$ as closely as desired.

Check. Taking $a=12.449$, $b=7.249$ gives $19.698$, below the upper bound. Adding the reported values gives $19.6$, inside the interval.

Worked example 2: a difference

Question. Find the bounds of $a-b$ for the same intervals.

Plan. Use the smallest $a$ and largest limiting $b$ for the lower bound; reverse those choices for the upper bound.

The first calculation needs the excluded value $b=7.25$; the second needs the excluded value $a=12.45$. Neither limiting difference is attained.

Answer. $5.10<a-b<5.30$. These are limiting bounds, not an attained minimum and maximum.

Check. $12.35-7.249=5.101$ and $12.449-7.15=5.299$ approach the two bounds from inside the interval.

Watch out

Subtracting the two lower endpoints does not give the lower bound of a difference. Here $12.35-7.15=5.20$, but $a-b$ can be smaller. To minimise a difference, subtract as much as possible.

Worked example 3: multiplying positive measurements

Question. Positive measurements $a=8.0$ and $b=3.0$ are each correct to the nearest $0.1$. Find the bounds of $ab$.

Plan. Form the input intervals, then multiply matching extremes.

Answer. $23.4525\le ab<24.5525$. Both lower inputs are allowed, so the lower product is attained; the upper product is excluded.

Check. The reported-input calculation $8\times3=24$ lies inside the bounds. Using $8.049\times3.049=24.541401$ gives a product close to, but below, the upper bound.

Worked example 4: dividing positive measurements

Question. Find the bounds of $a/b$ for $7.95\le a<8.05$ and $2.95\le b<3.05$.

Plan. For the lower bound divide the smallest numerator by the largest limiting denominator; reverse the choices for the upper bound.

Answer. $\dfrac{159}{61}<\dfrac ab<\dfrac{161}{59}$. The limiting values are approximately $2.606557$ and $2.728814$. Neither is attained: the lower needs excluded $b=3.05$, and the upper needs excluded $a=8.05$.

Check. The central quotient $8/3=2.6666\ldots$ lies inside. Also, $3.05\times(159/61)=7.95$ and $2.95\times(161/59)=8.05$, verifying the endpoint arithmetic by multiplication.

Problem-solving

Describe the direction before choosing endpoints: “larger numerator makes it larger; larger positive denominator makes it smaller.” This explains the choices and helps prevent a reversed quotient bound.

Worked example 5: a guaranteed decimal answer

Question. A positive measurement is reported as $x=10.30$, correct to the nearest $0.01$. A result is $r=x/2$, with $2$ exact. Find the greatest number of decimal places to which $r$ can be guaranteed.

Plan. Divide both measurement bounds by the exact positive number, then compare the possible results at successive rounding accuracies.

Every number in this interval rounds to $5.15$ to two decimal places: the whole interval lies between the rounding thresholds $5.145$ and $5.155$. Three decimal places are not guaranteed; permitted values $5.148$ and $5.152$ round differently.

Answer. $r=5.15$ is guaranteed to two decimal places, but not three.

Check. The threshold for rounding up to $5.16$ is $5.155$, beyond the upper bound. The threshold below which the result would round to $5.14$ is $5.145$, below the lower bound.

Worked example 6: uncertainty in the denominator

Question. Now $x=10.30$ and $y=2.00$ are each correct to the nearest $0.01$, and $r=x/y$. Find the greatest number of decimal places to which $r$ is guaranteed.

Plan. Use opposite quotient extremes before testing the rounding accuracy.

The limiting values are approximately $5.134663$ and $5.165414$. Values near the lower limit round to $5.1$ to one decimal place; values near the upper round to $5.2$. However, every allowed result lies between $4.5$ and $5.5$, so all round to $5$ to the nearest whole number.

Answer. Only zero decimal places, giving $5$, is guaranteed. The nominal value $10.30/2.00=5.15$ is not a guaranteed two-decimal-place answer.

Check. Allowed inputs $x=10.295,y=2.004$ give about $5.1372$, while $x=10.304,y=1.995$ give about $5.1649$. Their one-decimal-place results differ, confirming the failure at that accuracy.

Worked example 7: a possible whole-number count

Question. A tank contains $12\,\mathrm L$, correct to the nearest litre. Each bottle holds exactly $0.8\,\mathrm L$. What is the maximum possible number of full bottles?

Plan. Use the upper volume limit, divide by the exact bottle size, then check which whole count can actually occur.

Sixteen bottles would need $16\times0.8=12.8\,\mathrm L$, outside the allowed volume. Fifteen need $15\times0.8=12\,\mathrm L$, which is allowed.

Answer. The maximum possible count is $15$ full bottles. Do not round $15.625$ to $16$.

Check. An allowed tank volume of $12\,\mathrm L$ fills exactly $15$ bottles, so this integer maximum is attained. It is not guaranteed for every tank: $11.5\,\mathrm L$ fills only $14$ full bottles.

Key idea

Guaranteed accuracy means the entire allowed interval gives the same rounded result. A possible maximum count is a different question: check both that the count fits and that the next one cannot fit.