Counter transfer savings with fixed net balances

Counter transfer savings in this invented worksheet means reducing the total amount moved while preserving three net balances. The original plan moves sixteen counters around three labeled nodes. A replacement moves only three. Both leave A down three, B up two and C up one. The question is which movements can change without changing those three final differences.

These are stipulated mathematical amounts, not prices, cryptocurrency, debts or actual accounts. A movement can have any nonnegative real amount. Every directed pair between different nodes is permitted unless a separate version explicitly restricts the arrows. There are no fees, dates, stock limits, funding requirements or permissions to obtain. The model supplies no authority to alter a real person's payment or obligation.

Keep the original arrows and amounts

Write A to B with amount seven, B to C with amount five, and C to A with amount four. All other directed movements initially have amount zero. The gross movement is seven plus five plus four, equal to sixteen. Count each arrow's amount once. An amount arriving at one node and later leaving it can contribute to more than one arrow in this objective.

Define a node's net inflow as its total incoming amount minus its total outgoing amount. A receives four and sends seven, so its net inflow is minus three. B receives seven and sends five, giving plus two. C receives five and sends four, giving plus one. Their net inflows sum to zero: every movement leaves one node and enters another, so the aggregate change across the three nodes cancels.

Separate a net result from gross movement

The three net figures describe what each node gains or loses after the complete plan. They do not list every intervening movement. Here B receives seven while retaining a net gain of only two. C receives five while retaining a net gain of one. Those facts coexist because the outgoing amounts belong in each node's calculation. Reporting only the total sixteen would lose the individual balances; reporting only the balances would lose the total moved.

The replacement problem fixes those net figures at minus three, plus two and plus one. It permits the individual arrows to change. A candidate must preserve all three equations, rather than only their sum zero. For example, giving all three counters to B would preserve the aggregate sum across the system but give B plus three and C zero. That would answer a different question from the one supplied.

Remove an equal cycle and inspect what remains

Subtract four from each of the three original arrows. A to B becomes three, B to C becomes one, and C to A becomes zero. No amount becomes negative. At each node, one incoming and one outgoing arrow have both decreased by four, so their difference is unchanged. The new gross movement is three plus one, equal to four. It saves twelve units of movement relative to the original sixteen.

The remaining arrows form a chain from A through B to C, with no directed cycle. Check its balances directly: A sends three, B receives three and sends one, and C receives one. Their net inflows remain minus three, plus two and plus one. This is a feasible improvement. Being cycle-free, however, does not prove that its gross total four is the smallest permitted total.

Use the allowed direct arrow

Our unrestricted version permits A to C as well as A to B. Send two directly from A to B and one directly from A to C. Set the other four directed amounts to zero. A sends three, B receives two, and C receives one. Thus the replacement preserves every required net balance and has gross movement three. Compared with the original sixteen, the reduction is thirteen.

This is an aggregate movement calculation. It does not claim that three physical tokens have been tracked through a real exchange, that anyone has waived a claim, or that an intermediate step has available funding. The worksheet's arrows and amounts are the full input. Its direct A-to-C arrow is allowed by definition. If that arrow were forbidden, the three-unit construction would be ineligible even though its balance arithmetic remained correct.

Prove the lower bound using every allowed arrow

Name the six directed amounts AB, AC, BA, BC, CA and CB. Each name gives the source first and destination second. Every amount is nonnegative. A's required net loss of three means AB + AC = 3 + BA + CA. Its total outgoing amount exceeds its incoming amount by exactly three. Keep that equation beside the declared sign convention; reversing the definition of net inflow would reverse its presentation.

The gross total is AB + AC + BA + BC + CA + CB. Substitute A's equation for AB + AC. The total becomes 3 + 2BA + 2CA + BC + CB. Each term after the three is nonnegative. Therefore every feasible plan has gross movement at least three. This argument covers noninteger real amounts as well as whole counters and uses all six permitted directions, not just the original three arrows.

Check when the lower bound is attained

For the total to equal three, all four nonnegative extra amounts must be zero: BA, CA, BC and CB. B's balance equation then forces AB to equal two, since B must gain two and has no remaining outgoing amount. C's equation similarly forces AC to equal one. These are exactly the two movements in the direct construction. Thus the gross minimum is attained, and the minimizing six-amount plan is unique.

Keep necessity and sufficiency together. The identity shows that any gross-three solution must have those four zero amounts. The remaining balance equations determine the other two. Conversely, substituting AB = 2 and AC = 1 with the four zeros satisfies every balance and gives gross three. We have both excluded a lower total and exhibited a feasible plan at the bound. A convenient drawing alone would supply neither part.

Count nonzero arrows as a separate objective

The direct plan also uses two nonzero arrows. One nonzero movement changes the balances of only its source and destination. The third node would have net change zero. Since all three required net balances are nonzero here, one arrow cannot suffice. Our two-arrow construction therefore has the minimum number of nonzero movements as well as the minimum gross amount for this example.

Those are separate certificates. The cycle-reduced chain A to B with three and B to C with one also has two nonzero arrows, yet its gross total is four rather than three. Minimizing the number of arrows therefore does not, by itself, choose the minimum-gross plan. A record saying only “two movements” would leave the aggregate amount undecided. State which objective the proposed answer is meant to minimize.

Retain a restricted-arrow version openly

Now make a separate version allowing only the three original directions AB, BC and CA. In particular, AC is forbidden. Let CA equal t. C's required gain of one forces BC = t + 1. B's gain of two then forces AB = t + 3. Every nonnegative t gives a feasible plan, and the required nonnegative amounts imply t is at least zero.

The gross total in this restricted family is t + (t + 1) + (t + 3), or 3t + 4. It is at least four, attained when t is zero. That is the cycle-reduced chain found earlier. The chain is optimal for this restricted graph and suboptimal for the unrestricted graph. Preserve the allowed-arrow list with each answer, rather than calling either calculation wrong after silently switching which movements exist.

Inspect extra movement without changing the balances

Starting with the direct optimum, add an amount r to both AB and BA. For any nonnegative r, the extra outgoing and incoming amounts cancel at A and B. C is unchanged. The plan remains feasible, but its gross total becomes 3 + 2r. Setting r to one gives total five. This simple return pair illustrates why balance preservation alone does not establish a movement saving.

Another feasible family uses BC = s, AB = 2 + s and AC = 1 − s, with the remaining amounts zero. Nonnegativity allows s from zero through one. B still gains two, C still gains one, and A still loses three. The gross total is 3 + s. Its endpoints are the direct minimum at s = 0 and the cycle-free chain at s = 1. The same family makes the extra intermediate movement visible continuously.

A complete integer control examined every six-amount tuple with each amount from zero through seven. Of 262,144 tuples, 1,710 preserved the target net balances. Exactly one had gross total three: AB two, AC one, and four zeros. The restricted original-arrow cases had minimum four. These checks corroborate the printed calculations within their finite range; the nonnegative identity establishes the unrestricted lower bound beyond that range.

Scale the amounts as a new input

For a separate control, double every original movement. AB becomes fourteen, BC ten and CA eight, giving gross movement thirty-two. The target net inflows become minus six, plus four and plus two. Direct AB four and AC two preserve those new balances with gross six. The reduction is twenty-six, not the earlier thirteen. Keep this doubled record separate; the target balances themselves changed along with the movements.

More generally, multiply the three original amounts by a nonnegative real scale k. The target becomes minus 3k, plus 2k and plus k. Repeating A's conservation equation gives gross total 3k + 2BA + 2CA + BC + CB. Its lower bound is 3k, achieved uniquely by AB = 2k and AC = k with the other amounts zero. In the restricted original-arrow version, the corresponding family is CA = t, BC = t + k and AB = t + 3k, giving minimum 4k.

At k = 0, every target net balance is zero and the unique minimum-gross plan has all amounts zero. It uses no nonzero arrows. The earlier two-arrow lower bound therefore applies to positive k, when all three target balances are nonzero; it cannot be copied to the zero case. Negative k is outside this scaled family because it would make the stated original directed amounts negative. Neither a changed scale nor a changed direction should be hidden beneath the original input labels.

Keep neighboring questions and actual items separate

The minimum opening balance guide keeps payments and receipts in a fixed event order and checks whether every running balance stays nonnegative. This model changes aggregate movements and supplies no event order or opening stock. Its minimum gross total does not establish liquidity.

The early path finalization repair guide preserves a directed graph with signed route scores and corrects a premature processing decision. Here the variables are nonnegative movement amounts constrained by net balances. No shortest-route procedure or negative edge supplies the saving.

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The Tough Guy 3PCA9 push broom account shows a wide rectangular head, angled shaft and two attached diagonal braces. Its included-handle wording remains separate from the photograph. No dimensions, material, acid resistance, sweep performance or SA20281 compatibility is verified. Bracing belongs to the pictured broom assembly, without an independently supplied tool established.

Keep the same net balances beside the changed movement plan.

Keep an actual purchase outside the model

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