Check shortcut route cost before and after a new connector

A shortcut route cost can rise after a new connector becomes available, even when each person switches only to lower their own cost. This original four-user exercise makes that statement precise. Before the connector, stable choices cost each user seventy fictional units. After it is added, the unique stable choice costs each user eighty. The old routes remain available throughout. The difference comes from shared edge loads and the rule for individual switching.

These are arbitrary mathematical cost units. They represent neither travel time nor money, and the exercise contains no observed traffic, product trial or prediction of real behavior. Four distinguishable users each choose one complete directed path from a start S to an end T. Their names can be First, Second, Third and Fourth. The order in a written choice record always follows those four names, so changing one position identifies exactly who moves.

Write the original edges and their load rules

Begin with two intermediate points, J and K. The upper route U follows S to J to T. The lower route L follows S to K to T. The edge from S to J has cost ten times its load. The edge from K to T has the same rule. On either variable edge, load means the number of users whose current chosen paths include that edge. Every user on that edge pays its current cost.

The other two edges have fixed costs. J to T costs fifty units, and S to K costs fifty. A route cost is the sum of its edge costs. Thus, if two users choose U and two choose L, each variable edge has load two and cost twenty. Every upper user pays twenty plus fifty, and every lower user pays fifty plus twenty. All four costs equal seventy, giving a system total of two hundred eighty.

The system total adds the four complete user costs. It does not count each edge just once. If three users share an edge whose current cost is thirty, that edge contributes ninety to the user-cost total. Keep the distinction visible: an edge cost is the charge faced by each user on it; the system total adds what everyone faces. This convention determines every comparison below.

Define a stable record by one-user switches

A choice record is stable when no single user can strictly reduce their own route cost by changing to another available route while the other three keep their choices. Evaluate the proposed new record, including the switching user in the new loads. Equal cost does not qualify as a strict improvement. A stable record need not minimize the sum of all user costs; stability checks the permitted personal deviations separately.

Suppose a users choose U in the original two-route model. The other four minus a choose L. An upper user currently pays ten times a plus fifty. If that user changes to L, the lower load becomes five minus a. Their new cost is 50 + 10(5 - a). The switch strictly helps exactly when a > 5 - a, which for integer a means a is at least three.

The lower side follows the same calculation with its own count. If three or four users occupy either route, one of them can improve by moving to the other. At a two-and-two split, an upper user switching alone would meet three lower users including themselves and pay eighty. They would leave a cost of seventy, so that move does not help. The same holds for a lower user. The stable records are therefore exactly the balanced splits.

There are six such named-user records: choose which two of the four users take U, with the other two taking L. The count is four times three divided by two, because choosing the same pair in reverse order creates no new record. Every stable record has the same four costs of seventy. The six refers to the identities assigned to routes, rather than six different numerical load patterns.

Add one directed connector and retain every old path

Now add J to K with fixed cost zero. Its direction is from J to K only. The old U and L paths still exist, and a third path C becomes available: S to J to K to T. There is no reverse connector and no additional path created by inventing K to J. A C user uses both variable edges and neither fifty-unit edge. Four users now each have three choices, giving eighty-one named choice records.

Let a, b and c be the numbers choosing U, L and C. They are nonnegative integers whose sum is four. The S to J load is a plus c. The K to T load is b plus c. An upper user pays 10(a + c) + 50. A lower user pays 50 + 10(b + c). A connector user pays 10(a + c) + 10(b + c), with zero added for J to K.

The old balanced record still costs seventy per user immediately after the connector is added, because nobody has changed path yet. Availability alone does not alter its loads. But one upper user can now switch to C. The upper variable-edge load stays two, while the lower variable-edge load rises from two to three. The mover pays twenty plus thirty, or fifty. This is a strict personal improvement from seventy, so the old record is no longer stable.

Prove which records can remain stable

Consider any record with at least one U user. That person changes from U to C. Their use of S to J remains, so its load stays a plus c. They stop using the fixed fifty-unit J to T edge and begin using K to T. The latter load becomes b plus c plus one. Their old cost minus new cost is 50 - 10(b + c + 1).

Use a plus b plus c equals four to simplify the reduction. The quantity b plus c plus one equals five minus a. Substituting gives 50 - 10(5 - a), which equals ten times a. Because the moving user was on U, a is positive. Their reduction is strictly positive. Every record containing any upper user consequently admits a permitted improving switch, regardless of how the remaining users are distributed.

For a lower user changing from L to C, K to T retains its load while S to J gains one user. The fifty-unit S to K edge is removed from that user’s path. The same calculation gives a reduction of ten times b, positive whenever a lower user exists. Thus a stable record can contain neither U nor L. Every user must choose C. This argument covers all possible counts, rather than testing only a convenient sequence.

At C, C, C, C, each variable edge has load four and costs forty. Each user pays eighty. A single user changing to U leaves the S to J load at four and replaces the other variable edge with the fixed fifty-unit edge, giving ninety. A single change to L likewise gives ninety. Neither change helps. The all-C record is stable, and the preceding argument proves it is the unique stable named-user record.

Audit an entire sequence of personal improvements

Start at U, U, L, L with costs seventy for everyone. First changes to C, giving C, U, L, L. The variable loads are two and three. First now pays fifty, Second pays seventy, and Third and Fourth each pay eighty. The mover improved from seventy to fifty. The system total remains two hundred eighty, since the mover’s twenty-unit saving is matched by ten-unit increases for each lower user.

Next Third changes from L to C. The record is C, U, C, L, and both variable loads become three. First and Third each pay sixty. Second and Fourth each pay eighty. Third has improved from eighty to sixty. Again the system total is two hundred eighty. A personal improvement has redistributed costs without yet changing their sum, so this step alone would not establish the eventual higher total.

Then Second changes from U to C, producing C, C, C, L. The variable loads become three and four. The three connector users each pay seventy, while Fourth pays ninety. Second improved from eighty to seventy. The total is now three hundred. Finally Fourth switches to C, C, C, C. Both loads become four, all users pay eighty, and Fourth improves from ninety to eighty. The system total rises to three hundred twenty.

Every step strictly reduces its mover’s cost, yet the final stable total exceeds the starting total by forty. The intermediate totals are two hundred eighty, two hundred eighty, two hundred eighty, three hundred and three hundred twenty, including the starting record. This trace is one permitted sequence, not an assertion that real users follow that order. Uniqueness of the stable record came from the separate deviation proof.

Keep the objective and the counterfactual separate

The result does not say that the connector destroys the old routes or makes a lower total impossible. The old balanced record remains feasible and still totals two hundred eighty. What changes is its stability: a user can lower their own cost by leaving it. Holding everyone at that record would require a different choice rule or constraint. Neither is part of this exercise, so it cannot silently be added as an answer.

A lower personal cost on one move also does not mean that every other user benefits. Shared-edge costs change for everyone using the affected edges. Conversely, the final increase from seventy to eighty compares stable records across two different available-route sets. It is not the cost change faced by the final mover, who goes from ninety down to eighty. State which comparison is being made before describing a change as an improvement.

The minimum step cost guide uses fixed-cost moves and an exact-total recurrence. Here costs depend on everyone’s route choices, and the question is unilateral stability. A minimum-step calculation cannot supply that condition.

The mixed strategy security example uses probabilities and an opposing response in a fixed zero-sum table. This model uses deterministic route records and changing shared loads. No expected-security floor is being calculated.

Check every finite record without enlarging the claim

An exact control enumerates all sixteen original U-or-L records and all eighty-one U-or-L-or-C records. For each record it computes loads, each user cost and every one-user alternative using the changed loads. The first enumeration finds exactly the six balanced stable records at seventy per user. The second finds only C, C, C, C at eighty. The improving sequence above is checked against those same definitions.

Enumeration verifies the specified finite case. The count-based deviation argument explains why no other record can be stable in that case. Neither establishes a theorem for every possible network, number of users, connector direction or cost formula. Changing fifty to another constant, allowing a reverse connector, or excluding the mover from load would define a different problem. Recalculate before adopting any conclusion under changed rules.

Keep the two native game accounts within their evidence

The SWOOC Hideaway listing has six images of a folding wall-mounted hook-and-ring game. Its score pegs belong to that game; pictured rooms, people and pool table are not established supplied contents. Material, dimensions, setup, weather, award and tree-planting statements remain unverified marketing claims. None supplies an edge cost in this exercise.

The Hook Rally listing has eight images of a distinct folding tabletop frame with suspended rings and central hooks. The box shows Hook Rally and Fat Brain ToyCo labeling. Dedicated storage straps and attached hardware serve the game. Images do not authenticate material, dimensions, skill benefits or performance, and lifestyle tables and people are not supplied goods.

Retain both stable records and the switching rule

Keep the original two-route balanced witness beside the new all-connector witness. Record the variable-edge rule, the fixed fifty-unit edges, the directed zero-cost connector and the requirement to include the switching user in each new load. Those definitions explain the shortcut route cost result: seventy becomes eighty across the stable outcomes, even though every displayed move improves its mover. Actual route planning or product decisions require their own evidence.

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