Application of Mathematics in Real-World Usage · Guide 36 · BTT Mathematics Hub
A supply chain is a network before it is a list of warehouses. Factories inject flow, customers consume flow, intermediate facilities pass flow onward, and every link has a capacity, cost or travel time. Logistics Mathematics asks whether the network can satisfy demand, where bottlenecks appear, and how to choose feasible flows without confusing local savings with whole-system performance.
This guide uses fictional warehouses, demands, costs and routes for Mathematics teaching. It is not commercial logistics, procurement or transport advice. Real networks include uncertain demand, contractual constraints, labour, vehicle regulations, service commitments, disruptions and multi-objective trade-offs beyond these simplified models.
A logistics network is a directed graph with quantities on edges
Represent facilities as nodes and allowed shipments as directed edges. A flow fᵢⱼ is the amount sent from node i to node j during a declared interval.
If Factory F can ship to warehouses A and B, the graph records those possible movements before any quantities are assigned.
Flow conservation balances intermediate nodes
At a pure transshipment node with no production or demand, total inflow equals total outflow.
If a warehouse receives70 units and ships30 to Store1 and40 to Store2, the flow balance is70−30−40=0.
Supply and demand appear as net injections
Suppose one source has supply100 and two stores demand40 and60. Total supply equals total demand, so a balanced feasible plan is possible if network capacities allow it.
If total demand were120 while total supply remained100, no flow assignment could satisfy all demand without adding supply, allowing shortage or redefining the problem.
Capacity constraints bound every route
If edge F→A has capacity50, then0≤fFA≤50. A plan sending55 along that route is mathematically infeasible even if it is cheap.
Optimisation happens only inside the feasible set defined by conservation and capacities.
A bottleneck limits end-to-end throughput
A serial route has link capacities100,80 and120 units/hour. Its maximum steady path throughput cannot exceed80 units/hour because every unit must cross the narrowest link.
Increasing the120-capacity edge to200 does not raise end-to-end capacity while the80-capacity bottleneck remains unchanged.
Parallel routes add capacity when they are independently usable
Two parallel source-to-destination routes with independent capacities40 and60 can carry up to100 units in total if no upstream or downstream cut is tighter.
The network’s maximum flow is controlled by the tightest separating cut, not simply the largest individual edge.
Minimum-cost flow adds a cost per unit to each edge
MIT’s network-optimisation notes define flow cost as the sum Σ cₑfₑ across edges. If20 units use a route costing3 per unit and30 use a route costing5, total cost is20×3+30×5=210.
The cheapest edge cannot simply receive all flow if its capacity or destination balance prevents that assignment.
The transportation problem allocates sources to destinations
Suppose Source A can supply40 and B can supply60. Store X demands50 and Y demands50.
One feasible shipment matrix is A→X40, A→Y0, B→X10, B→Y50. Row sums match source supplies and column sums match store demands.
Allocation cost depends on the whole shipment matrix
Let unit costs be A→X2, A→Y6, B→X4, B→Y3. The feasible allocation above costs40×2+10×4+50×3=270.
Moving units changes several constraints simultaneously, so local cost comparisons need to be checked against row and column balances.
Weighted average transport cost divides total cost by total shipped quantity
For total cost270 across100 units, average transport cost is2.70 per unit.
A simple average of route cost rates would ignore how much volume actually uses each route.
Shortest path minimises accumulated edge weight
If route A→B→D has costs4 and5, total is9. Route A→C→D has costs2 and10, total12. The first route is cheaper even though its first edge is more expensive.
Shortest path is therefore a path-level optimisation, not a greedy choice of the cheapest next edge.
Warehouse throughput is quantity per time
A warehouse processes1,200 cartons in8 hours. Average throughput is150 cartons/hour.
If a two-hour peak processes400 cartons, peak-period throughput is200/hour. Daily average and peak rate answer different capacity questions.
Dock utilisation compares occupied time with available time
Four docks each provide8 hours of capacity, giving32 dock-hours. If loading activity occupies24 dock-hours, utilisation is75%.
High utilisation can create waiting when arrivals and service durations vary, linking logistics capacity to queueing Mathematics.
Space utilisation needs a geometric denominator
A storage zone has10,000m³ usable volume and currently holds6,500m³ of product-equivalent occupied volume. Volumetric utilisation is65% under that definition.
Volume utilisation is not the same as pallet-slot utilisation or floor-area utilisation. The storage metric must match the operational question.
Vehicle capacity creates a bin-packing constraint
A vehicle carries at most100 units. Customer demands are40,35 and30. Their total105 cannot fit in one trip even though each customer individually fits.
At least two trips or vehicles are required before route-distance optimisation begins.
Route consolidation can reduce distance but must preserve feasibility
Two separate out-and-back routes of20km and18km total38km. A combined feasible tour of30km saves8km, or about21.05% relative to38km.
The saving is meaningful only if the combined route still respects vehicle capacity, time windows and other constraints.
Demand-weighted facility location gives a centre-of-gravity approximation
Suppose demand100 occurs at x=0 and demand300 at x=8 on a line. The demand-weighted coordinate is(100×0+300×8)/400=6.
This centre-of-gravity calculation is only a heuristic when transport follows a road network, costs are nonlinear or candidate sites are discrete.
Multicommodity flow shares capacities between product streams
If two product classes share a route of capacity100 and class1 uses60, then class2 can use at most40 on that edge in the simple shared-capacity model.
Optimising each product separately can therefore create a combined plan that exceeds shared capacity.
Lead time along a serial route adds stage times
If transport takes5h, receiving2h and final delivery3h with no overlap, total path lead time is10h.
If stages overlap or waiting is stochastic, simple addition may not represent the actual distribution of completion time.
A complete logistics model states its commodity, interval and objective
State what one unit of flow represents, the time period, supplies, demands, capacities, route costs and whether the objective minimises cost, time, emissions, distance or a weighted combination.
The return path is supply/demand → network → feasible flow → objective value → operational interpretation.
Practice: twenty logistics Mathematics questions
- A warehouse receives70 and ships30+40. Find net imbalance.
- Supply100; demands40 and60. Are totals balanced?
- Supply100; demand120. Find shortage before any policy change.
- An edge capacity is50. Is proposed flow55 feasible?
- Serial capacities100,80,120. Find path bottleneck.
- Parallel route capacities40 and60. Find combined capacity if no other cut is tighter.
- 20 units cost3 each and30 cost5 each. Find total cost.
- For A supply40,B60,X demand50,Y50, check allocation AX40,BX10,BY50.
- With costs AX2,BX4,BY3, find allocation cost from question8.
- Find average cost per unit over100 units.
- Compare path costs4+5 and2+10.
- 1,200 cartons in8h: find average throughput.
- 400 cartons in2h: find peak-period throughput.
- 24 occupied dock-hours out of32: find utilisation.
- 6,500m³ occupied out of10,000m³: find utilisation.
- Vehicle capacity100; demands40,35,30. Can one trip carry all?
- Separate route distance38km; combined route30km. Find distance saving percentage.
- Demand100 atx=0 and300 atx=8. Find weighted coordinate.
- Shared edge capacity100; commodity1 uses60. Find remaining capacity.
- Why can optimising every commodity independently make the combined network plan infeasible?
Worked answers
- 0.
- Yes.
- 20 units.
- No.
- 80 units/time.
- 100 units/time.
- 210.
- Yes: source totals40 and60; destination totals50 and50.
- 270.
- 2.70 per unit.
- 9 versus12; the first path is cheaper.
- 150 cartons/h.
- 200 cartons/h.
- 75%.
- 65%.
- No; total demand105 exceeds capacity100.
- 8/38≈21.05%.
- 6.
- 40.
- Because separate solutions can each consume the same shared capacity, causing their sum to violate the edge constraint.
Sources and connected applications
For shortest paths, maximum flow, minimum-cost flow and multicommodity-flow foundations, see MIT OpenCourseWare: Network Optimization and its lecture notes. All warehouses, routes, demands and costs here are fictional teaching examples.
Continue with Robotics, Kinematics, Coordinate Transforms and Path Planning; Computer Graphics, Vectors, Matrices, Rendering and Interpolation; and Timekeeping, Clocks, Calendars, Drift and Modular Arithmetic. Return to the BTT Mathematics Hub.