Who's the best signing? It depends what you're optimising

The football question

The window is open, there's £10m to spend, and the brief from the board is simple: sign the best players you can.

Best at what, though? Scoring goals? Winning points? Filling the gap the manager keeps complaining about? Those sound like the same question. On the shortlist below they give three different answers, and working out which question you're really asking is the first step of every decision in this series.

The concept

Decision science is the use of maths to make better choices. Much of it comes from operations research (or operational research), which began at the Bawdsey Research Station in England in 1937, as a way to get the most out of Britain's early-warning radar. The problems were military; the method works anywhere there are choices, limits and a goal, which describes football rather well.

Every optimisation problem has the same three parts:

• Decision variables: what you control. Here, which players you sign.
• The objective: the single number you're trying to make as big (or as small) as possible. Goals, points, profit, minutes of rest.
• Constraints: the rules the answer has to obey. The budget, the wage ceiling, "we must sign a centre-back".

Every choice that obeys all the constraints is called feasible, and together they make up the feasible region. Optimisation means finding the feasible choice with the best objective.

A football example

Here's a made-up shortlist. Each player has a fee, and an estimate of what he'd add over a season compared with the player he'd replace: extra goals scored, or goals kept out at the other end.

With five players, each either signed or not, there are 2 × 2 × 2 × 2 × 2 = 32 possible sets of signings, from nobody to all five. Only 15 of them cost £10m or less. That's few enough to check every one, which is exactly what we'll do.

Objective 1: the most goals

If the aim is goals scored, the best affordable set is the striker and the midfielder: £10m, 10 extra goals. No other set within budget adds as many.

Objective 2: the most points

Goals are only worth what they win you, and a goal kept out wins points too. The research piece how many points is a goal worth? found that in Scottish football each goal scored is worth about 0.65 points over a season and each goal conceded costs about 0.60. So the objective becomes

$$\begin{aligned} \text{points} = \;&0.65 \times \text{scored} \\ &+ 0.60 \times \text{kept out} \end{aligned}$$

Now the best set is the winger, the full-back and the midfielder: £10m, only 8 goals scored but 5 kept out, 8.20 points against the striker pair's 7.70. Three cheaper players beat one star, because they cover both ends of the pitch.

Objective 2, with a constraint: we must sign a centre-back

The manager sold his centre-back in the summer and insists on a replacement. That's a constraint: any answer without a centre-back is ruled out. The best set that obeys it is the winger and the centre-back: £9m, 7.45 points.

The centre-back rule costs 0.75 points a season against the best unconstrained choice. That doesn't make it a bad rule: the manager may know something the numbers don't, such as that the full-back can't play centrally. But now the club knows what the rule costs and can decide whether it's worth it. Putting a price on a constraint is one of the most useful things optimisation does.

Why it matters

Clubs argue about signings as if everyone agreed what "best" means. Most of the disagreement is about the objective, not the players. A director of football chasing resale value, a manager chasing points this season and a supporter chasing goals can look at the same shortlist and all be right by their own measure. Writing the objective down turns an argument into a calculation.

Checking every option worked here because there were only 32. Choosing a squad of 25 from 60 candidates has about 52 million billion possible answers: checking a million a second would take over 1,600 years. The rest of this series is about how to find the best answer without trying everything: linear programming, integer programming and more. For another kind of optimisation, finding the best setting by rolling downhill, see gradient descent.

Limitations

• The shortlist is made up. Real estimates of goals added or kept out are uncertain, and a gap of half a point a season is well within that uncertainty. The method, not the answer, is the point.
• Players don't simply add up. Two strikers may need the same service, and a full-back's goals may depend on the winger in front of him. Here each player's value is counted on its own.
• One season, one objective. Wages, age, resale value and injury risk all matter, and choosing between several objectives at once is a later part of this series.
• 0.65 and 0.60 are averages over Scottish Premiership seasons; they will differ a little in other leagues.

Try it yourself

Think of your own club's last window. What was it trying to maximise? Write down the objective in one line, then list the constraints: the budget, the wage ceiling, the positions that had to be filled. Did the signings make sense for that objective, or for a different one?

Reproduce the analysis

This checks all 32 sets of signings and prints every number in the article. Nothing to download.

from itertools import combinations

# A made-up shortlist: fee (£m), goals a season he'd add up front, and goals he'd keep out at the back,
# each compared with the player he'd replace.
shortlist = {
    "Striker":     (7, 8, 0),
    "Winger":      (4, 5, 0),
    "Centre-back": (5, 0, 7),
    "Full-back":   (3, 1, 3),
    "Midfielder":  (3, 2, 2),
}
BUDGET = 10


def points(scored, kept_out):
    return 0.65 * scored + 0.60 * kept_out   # from "How many points is a goal worth?" on this site


# The decision: which players to sign. Every possible set, from nobody to all five.
options = []
for n in range(len(shortlist) + 1):
    for signed in combinations(shortlist, n):
        fee = sum(shortlist[p][0] for p in signed)
        scored = sum(shortlist[p][1] for p in signed)
        kept_out = sum(shortlist[p][2] for p in signed)
        options.append((signed, fee, scored, points(scored, kept_out)))
affordable = [o for o in options if o[1] <= BUDGET]
print(f"{len(options)} possible sets, {len(affordable)} within £{BUDGET}m")


def best(label, candidates, objective):
    signed, fee, scored, pts = max(candidates, key=objective)
    print(f"{label:<32} {' + '.join(signed):<34} £{fee}m  {scored:>2} goals  {pts:.2f} points")
    return pts


best("Most goals scored", affordable, lambda o: (o[2], o[3]))
top = best("Most points", affordable, lambda o: o[3])
forced = best("Most points, with a centre-back", [o for o in affordable if "Centre-back" in o[0]], lambda o: o[3])
print(f"The centre-back rule costs {top - forced:.2f} points")

It prints:

32 possible sets, 15 within £10m
Most goals scored                Striker + Midfielder               £10m  10 goals  7.70 points
Most points                      Winger + Full-back + Midfielder    £10m   8 goals  8.20 points
Most points, with a centre-back  Winger + Centre-back               £9m   5 goals  7.45 points
The centre-back rule costs 0.75 points

Further reading

• Operations research, Wikipedia. The field's history, from wartime radar to airlines and supply chains.
• Mathematical optimization, Wikipedia. Objectives, constraints and feasible regions in general.