(objective-a) the 'work' done searching for a good parking spot, as well as

(objective-b) the total physical work done after parking. This includes the work done while walking to the entrance, as well as the work done after shopping and exiting the building. Let's focus on objective-b here, given that objective-a has been analyzed before, using the following notation.

Parking coordinates (O, D), where

O = walking distance from the parking spot to the entrance, and

D = walking distance from the exit to the parking spot.

M = mass of the person,

S = expected mass of purchased items, and

(μ, g) = constants denoting the coefficient of friction during walking, and the acceleration due to gravity, respectively.

Work done = force * distance = μg * mass * distance (A prior blog shows how to optimize our in-store shopping route.)

Objective is to minimize: M*O + (M+S)*D

A minimal work objective suggests these simple rules:

1) Heavy shopping: park close to the exit.

2) Light shopping: park to minimize total walk.

Thus, the goodness of a parking spot depends on the shopping context. Let us figure out a good parking spot for the shopping scenario below.

**Linear Store, Manhattan Distances**

Consider a long, 'linear' store along the x-axis, with the exit @ x = 0, and entrance @ x = L. Assume 'Manhattan' walking distances (

*ℓ*

^{ 1}) and that prior customers selected spots nearest the building along y-axis, and picked their spots along the x-axis based on individual preferences. Given this, the distance we can expect to walk to/from the store along the y-axis is already (near) minimal. This leaves us with a (non-convex) one-dimensional search for open locations x(i) along the x-axis over three regions. Let us split this task into two steps.

**Step-1: Find the best spot within each region**

A) x(i) = -a ≤ 0

Minimize M(O+D)+S*D = M(L+a +a) + Sa = ML + (2M+S)a*

B) 0 ≤ x(i) = b ≤ L

Minimize M(L-b + b) + Sb = ML +Sb*

C) x(i) = L+c ≥ L

Minimize M(c+L+c) + S(L+c) = ML+SL + (2M+S)c*

The optimal x* within each of these regions is the one closest to the exit. This is independent of S and L, and identifiable by greedy search.

The worst spot in B is no worse than c*, independent of S and L, yielding a binary choice: pick a* or b*.

x(i) ≤ 0: incremental cost = (2M+S)a*, a* ≥ 0

x(i) ≥ 0: incremental cost = Sb*, b* ≥ 0

For small values of S, b* dominates and is optimal for light shopping or window shoppers. As S increases and becomes comparable to M, b* is preferable when:

b*/a* ≤ 1 +2(M/S)

Even if we 'shop till we drop' and carry our own weight, b* can be as much as 3a* and still dominate.

Practically speaking, b* is a solid bet, but when B is full, the shopper is forced to choose between disjoint regions A and C.

x(i) ≤ 0: incremental cost = (2M+S)a*

x(i) ≥ L: incremental cost = SL + (2M+S)c*

Scenarios:

i) Between spots equidistant from the entrance and exit, a* dominates c*, independent of S and L.

ii) For the 'carry our own weight' scenario, a* is preferable when

(a*-c*) ≤ L/3

iii) For unmanageably large S, a* remains preferable if:

(a*-c*) ≤ L

We can now draw some conclusions.

Preferred order of parking is:

b* > a* > c*

which is out of order. The actual search pattern we employ may depend on whether the aisles in the parking lot are along the x- or y-axis.

Clockwise, starting from the exit, scan B, park if feasible. Else scan C, and park if feasible. Else, turn around choose a spot in A.

Clockwise, starting from the exit, scan B, park if feasible. Else turn around, scan A, and park if feasible. Else, choose a spot in C.

**Step-2: Finding the least cost spot among (a*, b*, c*)**The worst spot in B is no worse than c*, independent of S and L, yielding a binary choice: pick a* or b*.

x(i) ≤ 0: incremental cost = (2M+S)a*, a* ≥ 0

x(i) ≥ 0: incremental cost = Sb*, b* ≥ 0

For small values of S, b* dominates and is optimal for light shopping or window shoppers. As S increases and becomes comparable to M, b* is preferable when:

b*/a* ≤ 1 +2(M/S)

Even if we 'shop till we drop' and carry our own weight, b* can be as much as 3a* and still dominate.

Practically speaking, b* is a solid bet, but when B is full, the shopper is forced to choose between disjoint regions A and C.

x(i) ≤ 0: incremental cost = (2M+S)a*

x(i) ≥ L: incremental cost = SL + (2M+S)c*

shoppers would prefer a* to c* unless:

(a*-c*) > L/[2(M/S)+1]

Scenarios:

i) Between spots equidistant from the entrance and exit, a* dominates c*, independent of S and L.

ii) For the 'carry our own weight' scenario, a* is preferable when

(a*-c*) ≤ L/3

iii) For unmanageably large S, a* remains preferable if:

(a*-c*) ≤ L

We can now draw some conclusions.

**Finding a Minimum-Work Parking Spot**Preferred order of parking is:

b* > a* > c*

which is out of order. The actual search pattern we employ may depend on whether the aisles in the parking lot are along the x- or y-axis.

__Light/medium shopping context__Clockwise, starting from the exit, scan B, park if feasible. Else scan C, and park if feasible. Else, turn around choose a spot in A.

__Heavy shopping context__Clockwise, starting from the exit, scan B, park if feasible. Else turn around, scan A, and park if feasible. Else, choose a spot in C.

Great illustration. I really like the fact that you used the approach of linear programming. It made it way to convenient to understand. Hopefully it will be for others aswell

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