Weighted assignment cost savings: reverse the pairing

Weighted assignment cost savings starts with fixed inputs

Three fixed weights, 1, 2 and 4, must each receive one of three fixed rates, 2, 5 and 9. Every weight receives exactly one rate, and every rate is used once. Define the total as the sum of each weight multiplied by its assigned rate. Pairing the lists in the same increasing order gives 48. Pairing them in opposite order gives 27. Under these exact rules, 27 is the minimum over every permitted pairing.

These are invented, nonmonetary cost units for a mathematical worksheet. The weights and rates describe no actual purchase, person, job or service. They remain fixed while their associations change. Every possible weight-rate pairing is permitted in the original model, with no setup charges or extra constraints. Those assumptions matter because the proof will improve an assignment by exchanging two rates. An exchange forbidden by another model cannot be used to justify its optimum.

Write the objective before deciding which pairings look good

Keep the weights in the order 1, 2, 4 and write the assigned rates underneath them. For the rate row 2, 5, 9, the contributions are 1×2=2, 2×5=10 and 4×9=36. Their sum is 48. For the rate row 9, 5, 2, the contributions are 9, 10 and 8, totaling 27. Both rows use the same three weights and the same three rates exactly once.

The comparison changes neither a weight nor a rate. It changes which two numbers are multiplied together. In particular, sorting both lists in the same direction is not automatically economical under a sum-of-products objective. The largest contribution in the first row is 4×9=36. Moving rate 9 to weight 1 while moving rate 2 to weight 4 reduces that concentration. We will prove the precise inequality behind the change rather than relying on that visual impression.

Check all six assignments in the small example

With the weights held at 1, 2, 4, the rate rows 2,5,9 and 2,9,5 produce totals 48 and 40. The rows 5,2,9 and 5,9,2 produce 45 and 31. The remaining rows 9,2,5 and 9,5,2 produce 33 and 27. These are all six permutations of three distinct rates, so the example has no unexamined seventh assignment. The smallest listed total is 27 and the largest is 48.

The enumeration establishes the minimum for these particular inputs. It also provides checks on the arithmetic in the later proof. For instance, 5,9,2 gives 1×5+2×9+4×2=5+18+8=31. Swapping its first two rates gives 9,5,2 and total 27. The improvement is four cost units. A proof for arbitrary list lengths needs a reason that this kind of exchange is always safe under the declared assumptions.

Derive the two-position exchange identity

Take two weights a and b with a≤b, and two rates r and s with r≤s. Pairing them in the same order contributes ar+bs. Pairing them in opposite order contributes as+br. Subtract the opposite-order contribution from the same-order contribution: ar+bs−as−br. Grouping terms gives a(r−s)+b(s−r), which is (b−a)(s−r). Both factors are nonnegative, so their product is nonnegative.

Therefore ar+bs is at least as+br. Assigning the larger rate to the smaller of these two weights never costs more in this fixed pair-product model. If both inequalities are strict, the product is positive and the exchange strictly lowers the total. If either the weights or the rates are equal, the product is zero and the exchange leaves the contribution unchanged. This identity supplies the improvement amount as well as its direction.

Apply it to the first two positions of 5,9,2. Here a=1, b=2, r=5 and s=9. The reduction is (2−1)(9−5)=4, agreeing with the direct comparison between totals 31 and 27. The third contribution, 4×2=8, is unchanged. Keeping that untouched contribution visible shows why the two-position calculation also describes the change in the complete three-position objective.

Turn the local identity into a global minimum proof

For a general finite list, put the fixed weights in nondecreasing order. They can be written a₁≤a₂≤…≤aₙ. Any allowed assignment corresponds to some ordering of the fixed rates below them. If two neighboring rates rise from left to right, exchange those rates. The two weights are already nondecreasing, so the identity proves that this exchange cannot increase the total. All other contributions remain exactly as they were.

Continue exchanging neighboring rising rates until the rate row is nonincreasing. This sorting process is finite: each exchange removes the out-of-order relationship of that neighboring pair for descending order, while preserving its order relative to every other entry. There are only finitely many pairs of entries. Equal rates need not be exchanged. When no rising neighbors remain, the entire row is nonincreasing, so it is the fixed rate list in descending order.

Starting from any allowed assignment, we have reached the opposite-order pairing without increasing its cost. The opposite-order pairing is itself allowed because every pairing was permitted. Its total is therefore no greater than the total of any starting assignment. That is a global minimum proof. The conclusion does not rest on trying a few promising swaps or assuming that every local optimum in every optimization problem must also be global.

The weights were sorted only to express the argument. If the original worksheet gives them in another order, retain their identities and attach the resulting assigned rates to those same weights when reporting the answer. A change of row order for presentation must not silently change which weight receives which rate. The mathematical object is the association between the fixed entries, rather than the left-to-right appearance of one convenient table.

Keep attainment separate from uniqueness

For the original distinct weights 1,2,4 and distinct rates 2,5,9, any rate row that is not descending has a neighboring rise. The corresponding weight inequality is strict and the rate inequality is strict, so exchanging that rise strictly lowers the total. Such a row cannot be minimum. The descending row 9,5,2 is therefore the unique minimizing rate assignment for these distinct inputs, with total 27.

Do not transfer that uniqueness statement to lists with ties. In a separate two-position example, use weights 1 and 1 with rates 2 and 5. Both assignments have total 1×2+1×5=7. Reversing the rates produces a different labeled assignment but the same total, because the weight difference in the exchange identity is zero. The minimum still exists; it simply is not attained by only one association.

Equal rates create the same zero difference numerically. If the rates carry separate names despite equal values, exchanging their identities can create another labeled record without changing cost. Whether those records count as different answers depends on the question being asked. Keep the original input identities when reporting assignments, and distinguish a minimum value from a claim that only one assignment reaches it. The proof of the value does not require an unnecessary uniqueness claim.

A forbidden pairing changes what the proof may do

Consider another separate model with weights 1 and 2 and rates 2 and 5. When all pairings are allowed, the same-order assignment costs 1×2+2×5=12, while the opposite assignment costs 1×5+2×2=9. Now forbid assigning rate 2 to weight 2. The formerly cheaper assignment is unavailable. Of the two possible assignments, only the same-order one remains legal, and its total is 12.

The exchange identity is still algebraically true. What fails is permission to perform the exchange inside the restricted problem. Sorting a list does not establish that every resulting association satisfies a new eligibility rule. This boundary example prevents an unrestricted minimum from being presented as the answer to a different assignment problem. A restricted set of allowed matches needs its own feasible-solution analysis and minimum proof, rather than an unexamined appeal to opposite sorting.

Use the comparison without turning it into a retail promise

The original worksheet's total falls from 48 to 27 when its same-order assignment is replaced by the minimum assignment. That is a difference of 21 fictional cost units between two stated rows. It does not establish a discount, a current quote or savings on a product. A real decision would need its actual objective, allowed associations and applicable costs before this model could even be considered relevant.

The neighboring optimization questions have different decision objects. Choosing one real meeting location to minimize absolute distances is different from permuting a fixed rate list. Covering target points with several fixed-length intervals is different again. A shared word such as “minimum” does not transfer either method to this assignment. Here the decisive calculation is the product difference for two fixed weights and two fixed rates, together with permission to exchange the rates.

Keep the supporting references separate

The alien-shaped eraser and claw-container listing shows character-shaped pieces and a complete red-and-clear rocket container with a visible claw assembly. The supplied novelty-toy role is retained alongside the erasers. Printed markings do not authenticate licensing or a maker relationship. No working-claw, material, erasing, quantity or stock result is established.

The pink printed eraser listing shows complete beveled blocks with the same printed legend described by its native account. The catalog maker and model are not visually authenticated. The pictures supply no independent delivered-count, dimensions, material or erasing-performance verification. Neither optional product account supplies the weighted-assignment proof.

The holiday meeting-point guide minimizes an equal-weight sum of absolute distances by choosing a real coordinate. This worksheet holds its weights and rates fixed and changes only their one-to-one associations. Its minimum follows from a product-exchange identity, not from a median location.

The fixed interval cover cost guide minimizes the number of equal-cost covers under its containment rule. Here no target locations are covered and no interval is moved. The allowed change exchanges two assigned rates while preserving every weight and rate.

Keep any payment question attached to its own order

If you later consider buying a worksheet supply, treat that as a separate decision with its own evidence. A correct assignment calculation does not confirm a pack count, authenticate a maker or establish that a pictured mechanism works. Inspect the current offer on republeq.com and resolve the item, variant, quantity, delivery and return questions relevant to the purchase you actually intend. If cryptocurrency is among the payment choices presented for that order, read those current instructions before deciding whether to use it. The mention of USDT in this article does not establish token acceptance, a supported network or a suitable wallet. Where an appropriate cryptocurrency option is available, verify the requested asset, network, receiving destination, amount and order reference against the current transaction information. Keep any clarification with the exact offer it concerns. A change in the basket or payment instruction calls for checking the changed information, rather than assuming that an earlier answer still applies. Save any eventual confirmation with the order and transaction records so a later enquiry can identify what actually occurred. If an attempted payment or order remains uncertain, reconcile that specific status before considering another payment. This conditional note reports no transaction, guarantees no acceptance, promises no discount and establishes no refund right. It also gives no assurance about fees, speed, price stability or recovery from an error. The mathematical example can be completed without buying anything or setting up an account. Its fixed lists, six assignments and exchange identity remain fictional worksheet inputs; they are not quotes, token amounts or instructions to send funds.

Retain the assumptions with the minimum

For weights 1,2,4 and rates 2,5,9, record the minimizing association 1→9, 2→5 and 4→2 with total 27. Keep the two-position identity and the unrestricted-pairing rule beside it. They explain why every allowed alternative costs at least as much. Preserve the tied-input and forbidden-pair examples separately: one limits uniqueness, while the other changes feasibility. Neither should silently replace the original model or its proved answer.