Bin count: assign indivisible items by weight

A bin count calculation assigns every supplied indivisible item to a capacity-limited group and finds the fewest nonempty groups that work. Use this independent four-item exercise to save every grouping, every minimum solution and a reason fewer bins cannot suffice. Weight feasibility does not establish physical fit or permission to ship.

Content checked: 18 October 2026.

What does a bin count calculation actually establish?

Imagine a small shop preparing an internal allocation note before considering actual packaging. The weights below are fictional teaching inputs in one abstract unit. The capacity is a stipulated number, not a carrier allowance, certified load rating or claim about a particular container.

Every item is mandatory and must appear in exactly one group. You cannot split a heavy item between containers, drop an inconvenient identifier or count one item twice. Containers are identical and available without a numerical limit. The objective is minimum used-container count, not maximum value, minimum price or a delivery promise.

Dimensions, protective material, empty-container weight, closure, fragility and compatibility are not supplied. Those omitted conditions matter in real work and require their own evidence. Keep those missing conditions visible.

The artificial intelligence guide covers broader workflows. This manual note needs no assistant, solver installation, upload or account.

Keep four item identities and their weights together

Copy the contract: “Practice V1. P weighs 7, Q weighs 5, R weighs 3 and S weighs 3 abstract units. Every container has capacity 10. All four items are required, indivisible and assigned exactly once. Containers are identical; no additional constraint is supplied.” Write spaces between labels and values when preparing your own source table.

R and S have equal weights but remain different items. Replacing them with one anonymous three-unit item would lose a mandatory record and hide a tied solution. Preserve each identifier with its weight. Recalculate a corrected source as a new version.

The total is 7+5+3+3=18. A single capacity 10 container cannot hold that total, so at least two are necessary. That establishes a lower bound, not an assignment. You still need to show how intact items can be divided into groups whose individual totals each respect 10.

The minimum order worksheet rounds a homogeneous order quantity to whole-pack constraints. Our item quantities are fixed and heterogeneous. Do not resize, split or add items to make a grouping work.

Inspect fifteen distinct partitions, not container labels

A partition divides all item identifiers into nonempty groups with every identifier appearing exactly once. P+R / Q+S means one container has P and R, while another has Q and S. Reversing the container order produces the same partition. Item order within a container is irrelevant.

Four named items have fifteen such partitions: one using one group, seven using two, six using three and one using four. The control count is 1+7+6+1=15. Empty containers are not counted. An empty selection is not an alternative because all four supplied items must be assigned.

First inspect identifier conservation in each row. Then add the weights inside each group. Capacity minus that group's total is its remaining capacity, or slack. A negative slack rejects the entire partition. Spare capacity in another container cannot compensate, because the offending item combination remains together in the overloaded group.

The slash-separated totals and slacks follow the same group order as the membership column. For P+R / Q+S, totals are 10 / 8 and slacks 0 / 2. Zero satisfies this simplified inequality; it does not establish measurement tolerance or safe loading. Read a pass as weight-feasible under the supplied contract only.

Why are there two equally small assignments?

P with R totals 7+3=10, while Q with S totals 5+3=8. Both groups pass, so two containers are feasible. Since the source total 18 exceeds one container's capacity 10, one container is impossible. Together, the lower-bound argument and this constructive witness prove that the minimum is exactly two.

There is a second equally small partition: P+S / Q+R. Its totals are also 10 and 8, but the item identifiers differ. Both must remain in a statement claiming all minimum assignments. Container labels cannot break that tie, and equal item weights are not permission to erase identity.

Inspect all seven two-group rows. These are the only two that pass. For example P+Q / R+S overloads the first group with 12; P alone with Q+R+S overloads the other with 11. A total capacity of 20 across two containers does not automatically make every two-group partition valid.

Other three- or four-container partitions may pass, but they cannot improve a known minimum of two. Feasibility and minimum count are separate claims. One witness proves feasibility; a matching lower bound proves optimality. Inspect every tied-size partition before claiming all minima.

The product mix exercise chooses optional quantities for additive contribution. Here every item is mandatory and each group has its own limit. The minimum demo set selects overlapping feature coverage; duplicating an item here is an error, not harmless repeated coverage.

Why can rounding up total weight still underestimate bins?

Consider a separate counterexample, not a revision to the English source: four items weigh 6,6,6 and 2, with capacity 10. Their total 20 gives a rounded-up total bound of two. Nevertheless each of the three six-unit items must occupy a different container, because any pair totals 12. Thus at least three containers are necessary.

The two-unit item can join any one of those six-unit items, producing loads 8,6 and 6 in three containers. That proves the true minimum is three, despite the total bound of two. The three equal heavy weights still represent three distinct identifiers. Do not replace this counterexample with a claim that the total-weight bound is useless: it remains a valid lower bound.

A known item weighing 11 with capacity 10 is a different failure. That indivisible item cannot fit any container even by itself, so more identical containers cannot fix it. Report infeasible under the contract instead of splitting or deleting the record. Unknown weight or capacity is not a known oversize item; preserve the missing input and ask its source owner.

For an independently supplied empty item list, zero nonempty containers are needed. Nonpositive capacity, negative weights, duplicate identifiers or conflicting source versions are outside our positive-input exercise and require clarification. Do not turn an absent weight into zero or carry the old optimum into a revised unknown-input card.

The box dimension card shows a separate geometric test. A weight-feasible grouping may fail physical dimensions or protection requirements. The fixed-kit calculation concerns component quantities, not assigning these complete items. None of these arithmetic notes authorises dispatch or a safety decision.

Save the source, load evidence and complete tie list

Open a plain-text editor already available on your device and create a document headed “Bin count V1: fictional weight-only assignment”. Copy all four source records and the contract. Add the fifteen-row ledger, each group total and slack, then the conclusion “Minimum two; all minimum partitions P+R / Q+S and P+S / Q+R.”

Keep the lower-bound sentence beside that conclusion: total 18 cannot fit one capacity 10 bin, while the named two-bin witnesses do fit by weight. Put the separate 6,6,6,2 counterexample under its own heading so its total and minimum are not confused with the main card. Note that dimensions and real dispatch conditions remain unchecked.

Use the editor's save command, select a folder you recognise and name the file bin-count-V1.txt. Close the document. In your file manager, open that folder and reopen the saved file. Confirm that P,Q,R,S each appear once in every partition, negative slacks remain visible and both minimum assignments are still present.

Allow two minutes for rules, three for one- and two-group rows, four for remaining partitions, three for the decision and three for saving. This is a proposed allowance, not a measured reader result. The article's arithmetic was cross-checked by independent enumerations and supporting files were saved and read back through filesystem operations; no graphical interface, real packing or solver software was tested.

• Each item appears exactly once in every partition.
• Each group retains its own load and slack.
• A lower bound is not labelled a feasible assignment.
• The minimum and every tied minimum remain visible.