Corrected exercises: applied optimization (201-103)

The constraint is $x+y=500$, where $y$ is the share going to treatment; the objective function is the number of cases avoided, $E=40\sqrt{x}+30\sqrt{y}$, to be MAXIMIZED. The five steps: name the variables, write the constraint, write the objective, substitute so that only one variable remains, then differentiate and compare with the endpoints. The constraint serves only to eliminate a variable, and that is its sole role.

Substituting $y=500-x$ gives $E(x)=40\sqrt{x}+30\sqrt{500-x}$. Both square roots require $x\ge 0$ and $500-x\ge 0$, so the domain is $[0, 500]$. It is a CLOSED and BOUNDED interval, so the maximum exists and is found among the critical numbers and the two endpoints.

$E'(x)=\dfrac{40}{2\sqrt{x}}-\dfrac{30}{2\sqrt{500-x}}=\dfrac{20}{\sqrt{x}}-\dfrac{15}{\sqrt{500-x}}$, the minus sign coming from the derivative of $500-x$. Set it to zero: $\dfrac{20}{\sqrt{x}}=\dfrac{15}{\sqrt{500-x}}$, so $20\sqrt{500-x}=15\sqrt{x}$. Squaring: $400\,(500-x)=225x$, that is, $200{,}000=625x$ and $x=320$. The share going to treatment is then $180$.

$E(320)=40\sqrt{320}+30\sqrt{180}$. Since $\sqrt{320}=8\sqrt{5}$ and $\sqrt{180}=6\sqrt{5}$, we get $E(320)=320\sqrt{5}+180\sqrt{5}=500\sqrt{5}\approx 1{,}118.0$ cases avoided. The exact form $500\sqrt{5}$ is worth writing: it shows that the maximum depends only on the total budget and the two coefficients, and it avoids any rounding error.

$E(0)=30\sqrt{500}\approx 670.8$ cases avoided, and $E(500)=40\sqrt{500}\approx 894.4$. Both endpoints are clearly below $1{,}118$, so the critical number is indeed the ABSOLUTE maximum. Management reading: putting everything into prevention, which the coefficient 40 might suggest, would lose more than two hundred avoided cases compared with the 320 versus 180 split. This is the consequence of the diminishing returns of each square root, and it is the quantitative argument against single-pronged policies.

The mathematical domain is $(0, +\infty)$, and the real domain is the strictly positive integers, a point that exercise 6 will come back to. The term $4{,}200g$ is the STRUCTURE cost: rent, staff, equipment; it grows with the number of points. The term $\dfrac{205{,}800}{g}$ is the ACCESS cost: the more points there are, the shorter the users' trips, so it decreases.

$C(g)=4{,}200g+205{,}800\,g^{-1}$, so $C'(g)=4{,}200-\dfrac{205{,}800}{g^{2}}$. Set it to zero: $g^{2}=\dfrac{205{,}800}{4{,}200}=49$, so $g=7$, the negative root being outside the domain.

$C''(g)=\dfrac{2\times 205{,}800}{g^{3}}=\dfrac{411{,}600}{g^{3}}$, strictly positive on the domain. The curve is therefore concave up everywhere, and the critical number is an ABSOLUTE minimum, not merely a local one.

$C(7)=4{,}200\times 7+\dfrac{205{,}800}{7}=29{,}400+29{,}400=58{,}800$ dollars per month. The two terms are EQUAL at the minimum, which is a general property of the form $ag+\dfrac{b}{g}$: setting $a-\dfrac{b}{g^{2}}$ to zero gives $ag=\dfrac{b}{g}$. It is a free check on your work, and a result managers know in the form “at the optimum, the structure cost equals the access cost”.

$C(4)=16{,}800+51{,}450=68{,}250$ dollars and $C(12)=50{,}400+17{,}150=67{,}550$ dollars. Moving three points one way or five the other costs less than 10,000 dollars a month, about 16 percent: the curve is FLAT around its minimum, like all curves of this type. The practical consequence is the same as in the previous chapter: the right recommendation is a range, here 5 to 10 service points, and a debate over the seventh point versus the eighth is about almost nothing.

The term $12{,}000$ is the FIXED cost: questionnaire design, sampling plan, analysis; it is paid whatever $n$ is. The term $9n$ is the COLLECTION cost, nine dollars per respondent. The term $\dfrac{486{,}000}{\sqrt{n}}$ puts a price on IMPRECISION: it decreases like $\sqrt{n}$ because the margin of error of a poll itself decreases like $\dfrac{1}{\sqrt{n}}$, and it is this square root that gives the problem its whole shape.

Rewrite $C(n)=12{,}000+9n+486{,}000\,n^{-1/2}$, so $C'(n)=9-\dfrac{486{,}000}{2}\,n^{-3/2}=9-\dfrac{243{,}000}{n^{3/2}}$, in dollars per respondent.

$9=\dfrac{243{,}000}{n^{3/2}}$ gives $n^{3/2}=\dfrac{243{,}000}{9}=27{,}000$, so $n=27{,}000^{2/3}$. Since $27{,}000=30^{3}$, we have $n^{3/2}=30^{3}$ and $n=30^{2}=900$ respondents. Going through powers of 30 avoids any calculator, and it is the kind of verification a marker notices.

$C(900)=12{,}000+9\times 900+\dfrac{486{,}000}{30}=12{,}000+8{,}100+16{,}200=36{,}300$ dollars. The imprecision cost, $16{,}200$ dollars, remains the heaviest item, and it is exactly TWICE the collection cost: this is the signature of the exponent $\dfrac{1}{2}$, just as the equal terms of exercise 2 were the signature of the exponent 1.

$C''(n)=\dfrac{3}{2}\times\dfrac{243{,}000}{n^{5/2}}>0$ on the domain, so the curve is concave up and $n=900$ is indeed an absolute minimum. Comparisons: $C(400)=12{,}000+3{,}600+\dfrac{486{,}000}{20}=12{,}000+3{,}600+24{,}300=39{,}900$ dollars, and $C(1{,}600)=12{,}000+14{,}400+\dfrac{486{,}000}{40}=12{,}000+14{,}400+12{,}150=38{,}550$ dollars. Dividing the sample by more than two or multiplying it by almost two therefore costs between six and ten percent more: the reasonable range runs from about 600 to 1,400 respondents, and that is what a quote should propose, with the cost range that goes with it.

$N(d)=1{,}600\,d^{1/2}-200d$, so $N'(d)=800\,d^{-1/2}-200=\dfrac{800}{\sqrt{d}}-200$ dollars per day.

$\dfrac{800}{\sqrt{d}}=200$ gives $\sqrt{d}=4$, so $d=16$ days. The critical number is unique on the domain $(0, +\infty)$.

$N(16)=1{,}600\times 4-200\times 16=6{,}400-3{,}200=3{,}200$ dollars per employee. The gross gain is $6{,}400$ dollars and the cost $3{,}200$: at the optimum, the cost is exactly HALF of the gross gain, which is yet another signature of the exponent $\dfrac{1}{2}$.

$1{,}600\sqrt{d}\ge 200d$ is equivalent to $8\sqrt{d}\ge d$, that is, dividing by $\sqrt{d}>0$, to $8\ge\sqrt{d}$, so $d\le 64$. The training stays profitable on $(0, 64]$ days, and beyond that it costs more than it earns. This is also what the figure shows, which comes back down to zero at $d=64$.

Doubling the length does not double the effect, because the gain follows a SQUARE ROOT: going from 16 to 32 days multiplies the gross gain by $\sqrt{2}\approx 1.41$ and not by 2, while the cost doubles exactly. The computation gives $N(32)=1{,}600\sqrt{32}-6{,}400\approx 9{,}051-6{,}400=2{,}651$ dollars, which is 549 dollars LESS than with sixteen days. The proposal therefore loses money while tying up twice as much working time. The general rule, valid for any activity with diminishing returns: increase the effort as long as the MARGINAL gain exceeds the marginal cost, here as long as $\dfrac{800}{\sqrt{d}}>200$, and not an hour longer.

$P'(n)=-0.04n+24$, in thousand dollars per beneficiary. Set it to zero: $0.04n=24$, so $n=600$ beneficiaries. Since $P''=-0.04<0$, it is indeed a maximum.

$P(600)=-0.02\times 360{,}000+24\times 600-1{,}500=-7{,}200+14{,}400-1{,}500=5{,}700$ thousand dollars, that is, 5.7 million.

Yes, it changes everything, because $600$ does not belong to $[0, 450]$. On that interval, $P'(n)=-0.04n+24$ stays POSITIVE, since $-0.04\times 450+24=6>0$: the function is strictly increasing there. The admissible maximum is therefore reached at the RIGHT ENDPOINT, $n=450$, and not at a critical number. This is the situation announced by exercise 7 of the previous chapter: when no critical number lies in the interval, the extremum is necessarily at an endpoint.

$P(450)=-0.02\times 202{,}500+24\times 450-1{,}500=-4{,}050+10{,}800-1{,}500=5{,}250$ thousand dollars. The best admissible choice is therefore to serve exactly 450 beneficiaries, the cap, for a benefit of 5.25 million. Any proposal to serve fewer reduces the benefit, and any proposal to serve more is inadmissible.

The COST OF THE CONSTRAINT is the difference between the free optimum and the constrained optimum: $5{,}700-5{,}250=450$ thousand dollars, that is, 450,000 dollars of social benefit lost each year because of the cap. This number is the most useful one in the whole exercise, because it is what you bring to the negotiating table: it puts a price on what raising the cap would be worth, and it lets you compare that with what raising it would cost the funder. An optimum under constraint therefore never just gives the admissible solution; it also gives the price of the constraint.

$C'(s)=260-\dfrac{9{,}000}{s^{2}}$. Set it to zero: $s^{2}=\dfrac{9{,}000}{260}\approx 34.615$, so $s\approx 5.884$ sessions. Obviously, a program cannot be given in 5.884 sessions.

$C(5)=1{,}300+\dfrac{9{,}000}{5}=1{,}300+1{,}800=3{,}100$ dollars. $C(6)=1{,}560+\dfrac{9{,}000}{6}=1{,}560+1{,}500=3{,}060$ dollars.

Choose $s=6$ sessions, for $3{,}060$ dollars, which is 40 dollars less than with five sessions. Note that the cost at the critical number, $C(5.884)\approx 3{,}059.9$ dollars, is barely lower than the integer cost: since the curve is flat around its minimum, rounding costs less than a dollar.

$D'(s)=1{,}000-\dfrac{30{,}200}{s^{2}}$ vanishes for $s^{2}=30.2$, so $s\approx 5.495$ sessions. $D(5)=5{,}000+\dfrac{30{,}200}{5}=5{,}000+6{,}040=11{,}040$ dollars, and $D(6)=6{,}000+\dfrac{30{,}200}{6}\approx 6{,}000+5{,}033.33=11{,}033.33$ dollars.

The second case is the counterexample you need to know. The critical number is $5.495$, so it ROUNDS to 5; yet it is 6 that gives the lower cost, by nearly 7 dollars. The reason is that the tipping point between two integers is not their arithmetic midpoint $5.5$, but their geometric mean $\sqrt{5\times 6}=\sqrt{30}\approx 5.477$, and $5.495$ is already beyond it. The RULE is therefore: never conclude by rounding, EVALUATE the function at the two neighbouring integers and compare the two values. It is two more computations and the only safe method, even when the difference is seven dollars out of eleven thousand.

$\rho(n)=\dfrac{180n}{0.5n^{2}+40n+1{,}250}$. By the quotient rule, the numerator of $\rho'$ is $180\,(0.5n^{2}+40n+1{,}250)-180n\,(n+40)$, that is, $180\,\bigl(0.5n^{2}+40n+1{,}250-n^{2}-40n\bigr)=180\,\bigl(1{,}250-0.5n^{2}\bigr)$. So $\rho'(n)=\dfrac{180\,(1{,}250-0.5n^{2})}{\bigl(0.5n^{2}+40n+1{,}250\bigr)^{2}}$. The $40n$ terms cancel, which is a sign that the computation is right.

The denominator is a square, so the sign of $\rho'$ is that of $1{,}250-0.5n^{2}$: positive for $n<50$, negative beyond. Solve $0.5n^{2}=1{,}250$, that is, $n^{2}=2{,}500$ and $n=50$ people. The return rises then falls: it is indeed a maximum. Note that $n=\sqrt{\dfrac{c}{a}}$, where $c$ is the fixed cost and $a$ the quadratic coefficient: the linear term $40n$ plays no role, because it is proportional to $n$, like the benefit.

$C(50)=0.5\times 2{,}500+2{,}000+1{,}250=1{,}250+2{,}000+1{,}250=4{,}500$ and $R(50)=9{,}000$, so $\rho(50)=2.00$: each dollar spent produces two dollars of benefit. At $n=30$: $C=450+1{,}200+1{,}250=2{,}900$ and $R=5{,}400$, so $\rho\approx 1.862$. At $n=80$: $C=3{,}200+3{,}200+1{,}250=7{,}650$ and $R=14{,}400$, so $\rho\approx 1.882$. The return is therefore fairly flat between 30 and 80 people.

The net effect is $N(n)=180n-0.5n^{2}-40n-1{,}250=-0.5n^{2}+140n-1{,}250$. Set $N'(n)=-n+140$ to zero, so $n=140$ people, and $N''=-1<0$ confirms the maximum. $C(140)=9{,}800+5{,}600+1{,}250=16{,}650$ and $R(140)=25{,}200$, so $N(140)=8{,}550$ thousand dollars. Compare with $N(50)=9{,}000-4{,}500=4{,}500$ thousand: the net effect almost DOUBLES by serving 140 people rather than 50.

It all depends on what is scarce. If the BUDGET is the constraint, because it is fixed and every dollar not spent here will be used elsewhere, you should maximize the return and serve 50 people: each dollar works hardest there. If NEED is the constraint, the budget being extensible or the money having no better use, you should maximize the net effect and serve 140 people, even though the return falls there to $\dfrac{25{,}200}{16{,}650}\approx 1.51$. Mathematics does not decide between the two criteria, it puts numbers on them; the choice is political, and the professional fault is to present one of the two numbers without saying which one was optimized.

FALSE. The critical number is only a candidate: you must still check that it belongs to the ADMISSIBLE DOMAIN, and compare it with the values at the ENDPOINTS. Exercise 5 gives the typical case: the free optimum is 600 beneficiaries, the cap allows 450, and the answer is 450. Correct statement: the optimum is found among the critical numbers lying in the domain, the points where the function is not differentiable, and the endpoints.

FALSE, and it is the most costly reasoning error of the chapter. The two terms of a total cost almost always oppose each other: minimizing the structure cost of exercise 2 would lead to a single service point, which would make the access cost explode. Correct statement: you minimize the SUM, which leads to a point where neither term is minimal, and where, for the form $ag+\dfrac{b}{g}$, the two are even equal.

FALSE. The tipping point between two integers $n$ and $n+1$ is their GEOMETRIC mean $\sqrt{n(n+1)}$, not their midpoint. For 5 and 6, $\sqrt{30}\approx 5.477$, so $5.495$ is already beyond it and 6 wins, as exercise 6 verifies. Correct statement: you evaluate the function at the two neighbouring integers and compare the values obtained.

FALSE. Exercise 7 puts numbers on it: the return is maximal at 50 people and the net effect at 140. The two criteria answer two different questions, and the gap is no anecdote, since the net effect doubles between the two answers. Correct statement: you maximize the return when the budget is the scarce resource, and the net effect when need is.

FALSE. A constraint that binds changes the value of the optimum, and this loss can be PUT IN NUMBERS: it is the difference between the free optimum and the constrained optimum, that is, $450$ thousand dollars per year in exercise 5. Correct statement: the cost of a constraint is the loss of objective it imposes, and it is the number to bring to a negotiation. If the constraint does not bind, that is, if the free optimum already satisfies it, then its cost is zero and it indeed changes nothing.

$P(p)=9.2p-\bigl(0.04p^{2}+3.0p+180\bigr)=-0.04p^{2}+6.2p-180$ thousand dollars. The coefficient $6.2$ is the NET MARGIN per place before size effects, that is, the subsidy minus the proportional variable cost.

$P'(p)=-0.08p+6.2$, zero for $p=\dfrac{6.2}{0.08}=77.5$ places. Since $P''=-0.08<0$, the curve is concave down and it is indeed a maximum. $P(77.5)=-0.04\times 6{,}006.25+480.5-180=-240.25+480.5-180=60.25$ thousand dollars.

$P(77)=-0.04\times 5{,}929+477.4-180=-237.16+477.4-180=60.24$ and $P(78)=-0.04\times 6{,}084+483.6-180=-243.36+483.6-180=60.24$ thousand. The two values are EQUAL, to the dollar, which is no surprise: $77$ and $78$ are symmetric about $77.5$, and a parabola takes the same value at two points symmetric about its vertex. The service can therefore choose 77 or 78 places indifferently, and decide on another criterion, such as the size of the groups.

The result becomes $P(p)=9.8p-0.04p^{2}-3.0p-180=-0.04p^{2}+6.8p-180$. Setting $-0.08p+6.8$ to zero gives $p=\dfrac{6.8}{0.08}=85$ places. The result is $P(85)=-0.04\times 7{,}225+578-180=-289+578-180=109$ thousand dollars, nearly double the previous one.

The subsidy goes from $9.2$ to $9.8$, an increase of $\dfrac{0.6}{9.2}\approx 6.5$ percent. The optimum goes from $77.5$ to 85 places, an increase of $\dfrac{7.5}{77.5}\approx 9.7$ percent. The shift of the optimum is therefore MORE THAN PROPORTIONAL to the increase in the subsidy. The reason can be read in the formula $p^{*}=\dfrac{s-3.0}{0.08}$: it is not the subsidy $s$ that rules, it is the MARGIN $s-3.0$, which went from $6.2$ to $6.8$, also $9.7$ percent. Every subsidy policy has this leverage effect, and it is stronger the closer the variable cost is to the subsidy: with a variable cost of $3.0$, six percent more subsidy buys ten percent more places.

The total budget is the number of people times the hours received, times the hourly rate: $45\,n\,h=270{,}000$. Dividing by 45: $n\,h=6{,}000$ hours of support in total, so $n=\dfrac{6{,}000}{h}$.

$E(h)=n\,f(h)=\dfrac{6{,}000}{h}\times\dfrac{h^{2}}{h^{2}+100}=\dfrac{6{,}000h}{h^{2}+100}$ successes. Cancelling one $h$ is what makes the problem tractable, and it deserves to be done before any differentiation: differentiating the unsimplified form would require the quotient rule twice.

By the quotient rule: $E'(h)=\dfrac{6{,}000\,(h^{2}+100)-6{,}000h\times 2h}{(h^{2}+100)^{2}}=\dfrac{6{,}000\,(100-h^{2})}{(h^{2}+100)^{2}}$. The denominator is a square, so the sign is that of $100-h^{2}$: positive for $h<10$, negative beyond. The optimal length is $h=10$ hours per person, and it is a maximum.

At the optimum, $n=\dfrac{6{,}000}{10}=600$ people are reached, $f(10)=\dfrac{100}{200}=0.50$, that is, a 50 percent individual success rate, and $E(10)=\dfrac{60{,}000}{200}=300$ successes in total. Neither maximum coverage, which would reach thousands of people with no effect, nor maximum depth, which would give a few dozen almost certain successes, beats this split.

$E(4)=\dfrac{24{,}000}{116}\approx 206.9$ and $E(25)=\dfrac{150{,}000}{725}\approx 206.9$ successes: the two values are EQUAL. This is no accident, since $E(h)=E\!\left(\dfrac{100}{h}\right)$ for every $h$, and $4\times 25=100$. The model is therefore symmetric between coverage and depth around $h=10$: diluting the support to four hours to reach 1,500 people produces exactly as much as concentrating it at twenty-five hours on 240 people, and both lose about a third of the successes compared with the optimum. This symmetry is a result of the MODEL, not of the world: it comes from the form of $f$, whose parameter 100 alone sets the balance point. A serious evaluation would therefore begin by estimating this parameter from data, since it, and it alone, decides the trade-off.