Operations Research Using Excel A Case Study Approach

1. 1Name the activities whose level you control and make each one a decision variable. In the Adidas case, x1 is units sold at the mall (location A) and x2 is units sold at the plaza (location B). The case defines both in thousands of units, so fix the unit before you write any coefficient.
2. 2Write the objective as per-unit profit times activity level, summed over activities: Max Z = 80x1 + 40x2. The $80 and $40 are the coefficients (c_j). Profit must behave as a direct function of units sold, so confirm that before you commit to the model.
3. 3Turn each operational limit into a constraint with one coefficient per variable. Accessibility is 9x1 + 4x2 ≤ 14 (average miles travelled against the 14-mile attractiveness range). Population density is 2x1 + 7x2 ≤ 10 (households in thousands against the 10,000 the retailer will serve). Median income is 5x1 + 10x2 ≤ 20 ($ thousands against the $20,000 ceiling). Add x1, x2 ≥ 0.
4. 4Test proportionality before you solve. Set x1 = x2 = 1: Z = $120 and constraint 1 reads 13 ≤ 14, which is feasible. Double both to 2: Z doubles to $240, but constraint 1 reads 26 > 14, which is infeasible. Profit scales with activity, and so does resource use, so the feasibility check runs on every candidate plan.
5. 5With two variables, solve graphically. Plot each constraint line, shade the feasible region, and evaluate Z at each corner point of the region. With more variables or more constraints, move to the simplex method or Excel Solver, because graphing stops being practical.
6. 6Read the result as optimal under constraints, not as the highest profit possible. Profit cannot rise without limit because resources and competition cap every activity. Report the optimum together with the binding constraints that stopped it.
7. 7After solving, run sensitivity analysis. Linear programming returns one feasible optimum, not the range of coefficient or limit values over which that plan stays optimal. Duality gives you the interpretation of those ranges.

Adidas AG chooses sales volumes (in thousands of units) for location A (mall, $80 profit per unit) and location B (plaza, $40 per unit). The model is Max Z = 80x1 + 40x2 subject to 9x1 + 4x2 ≤ 14, 2x1 + 7x2 ≤ 10, 5x1 + 10x2 ≤ 20 and x1, x2 ≥ 0. The corner-point arithmetic below is computed from that model; the passages do not state the optimum. The corners are (0,0); (1.556, 0), where constraint 1 binds on the x1 axis; (0, 1.429), where constraint 2 binds on the x2 axis; and the intersection of constraints 1 and 2. Solving 9x1 + 4x2 = 14 with 2x1 + 7x2 = 10 gives x2 ≈ 1.127 and x1 ≈ 1.055. Constraint 3 reads about 16.5 ≤ 20, so it is slack. Z at the corners is about $124.4 at (1.556, 0), about $57.1 at (0, 1.429) and about $129.5 at the intersection (a profit of roughly $129,450, since units are in thousands). The best plan therefore sells about 1,055 units at A and 1,127 units at B. Accessibility and population density are the binding limits, and income is not.

• If a candidate plan's left-hand side exceeds any right-hand side, then reject it however high its Z. The x1 = x2 = 2 plan returns $240 but breaks the 14-mile limit at 26.
• If the problem has two decision variables, then solve graphically and compare Z at every corner point. If it has several variables or many constraints, then use simplex or Excel Solver.
• If a constraint is slack at the optimum (income in the Adidas case), then relaxing it adds no profit. Direct effort at the binding constraints.
• If profit or resource use does not scale in direct proportion to activity, then linear programming does not apply as written. Check proportionality first.
• If you need the range over which the plan stays optimal when a coefficient or limit changes, then run sensitivity analysis and duality. The base solution does not give that range.

• Mixing units: entering households as 2,000 and 7,000 against a limit of 10 (thousands). The tell is an optimum near zero or constraints that never bind.
• Treating the objective as unbounded profit and ignoring constraints. The tell is a plan nobody can execute, with resource use above the available limits.
• Checking feasibility only once, then scaling the plan up. The tell is a plan that passes at x1 = x2 = 1 and fails at x1 = x2 = 2.
• Reading only the profit figure and skipping which constraints bind. The tell is spending effort or budget loosening a slack limit such as income.
• Stopping at the single optimal answer and presenting it as stable. The tell is no stated range for the coefficients ($80, $40) or limits (14, 10, 20) over which the plan stays optimal, so any small change in them leaves you guessing.

• ▸Activities / decision variables (x_j): list each activity and its unit (e.g. x1 = thousand units at location A).
• ▸Profit per unit (c_j): enter the profit coefficient for each activity (e.g. $80, $40).
• ▸Objective function: write Max Z = c1x1 + c2x2 + ... from the coefficients above.
• ▸Constraint list: name each limiting resource (e.g. accessibility, population density, income).
• ▸Coefficients (a_ij): enter the resource use per unit of each activity, in the same unit as the limit.
• ▸Limits (b_i): enter the maximum available for each resource (e.g. 14, 10, 20).
• ▸Proportionality check: evaluate Z and every left-hand side at x = 1 and x = 2, and record whether feasibility holds.
• ▸Solution and binding constraints: record the optimal x_j and Z, and mark each constraint as binding or slack.
• ▸Sensitivity notes: record the range of each c_j and b_i over which the plan stays optimal.

1. 1Frame the operation as a conversion process: list the inputs (labour, capital, machinery, raw material, data), the activities that combine them, and the output. Efficiency is the output you get from those inputs, so you cannot measure it without all three.
2. 2Choose one measurable output metric before modelling. The passages name project completion time, distance (or travel time) and resource utilization. Write the objective as Z and state whether you are maximizing (profit) or minimizing (cost, time, distance).
3. 3Write every limit as a constraint. The sportswear case uses trade area (maximum 14 miles), population density (10,000, beyond which the store becomes inefficient) and income (median maximum $20,000). Give each constraint a coefficient per unit of activity.
4. 4For a two-variable problem, plot each constraint line and shade the feasible region. Every point inside it is feasible, but feasible does not mean optimal.
5. 5Draw objective-function lines at test values. For Z = 80x1 + 40x2, Z = 40 gives intercepts x2 = 1 and x1 = 0.5, and Z = 60 gives x2 = 1.5 and x1 = 0.75. Lines moving away from the origin mean more activity and more profit, up to the last corner of the feasible region.
6. 6Skip the line-sliding and evaluate Z at every extreme point instead. Read simple corners off the axes. Where two constraints intersect, solve the two equalities simultaneously by algebra, or by Excel when there are many intersections. Select the corner with the best Z.
7. 7For time-based efficiency in projects, use PERT. Estimate optimistic (to), most likely (tm) and pessimistic (tp) times and compute each activity's expected time and variance. Find the critical path from EST, EFT, LST and LFT. Then compute z = (scheduled time – expected project time) / standard deviation of the critical path, and read the probability of on-time completion from the z table.
8. 8For distance or time minimized across pairings (stores to sectors, runners to positions), use the assignment model. Rerun it under each changed condition (a 5 km limit, a store closed for renovation) and compare how the assignments move.

The source's sportswear chain sells x1 and x2 units at two locations to maximize Z = 80x1 + 40x2 under three constraints: accessibility, population density and income. The objective lines show the direction of improvement. At Z = 40 the intercepts are (x1 = 0.5, x2 = 0), (x1 = 0, x2 = 1). At Z = 60 they are 0.75 and 1.5. The feasible region has three extreme points, A, B and C, so there are three candidate optima. Point A is x1 = 0, x2 = 1.43, giving Z = 80(0) + 40(1.43) = 57.2. Point C is x1 = 1.56, x2 = 0, giving Z = 80(1.56) + 40(0) = 124.8. Point B is the intersection of constraints 1 and 3, so you solve those two equalities simultaneously. The passage excerpt does not complete that calculation, and neither do we. The efficient choice is the corner with the highest Z, which cannot be declared until B is evaluated against C's 124.8. Pushing the line past that corner would raise profit on paper, but the plan would then be infeasible.

• If the problem has two decision variables and a few constraints, then use the graphical method and compute Z at every extreme point. If there are many interacting intersections, then solve them in Excel.
• If a point gives a higher Z but lies outside the feasible region, then reject it. The objective line may only be pushed to the last corner of the feasible area.
• If the objective is cost, time or distance, then treat the result as 'optimal under these constraints' rather than a true minimum. No cost reaches zero and no profit reaches an unlimited value.
• If activity times are uncertain, then use PERT: estimate expected time and variance, find the critical path, and report the probability of finishing by the scheduled date instead of a single completion time.
• If a resource or option set changes (a store closes, a distance cap is added), then re-solve the assignment and compare it with the baseline. Do not patch the old assignment by hand.

• Treating a feasible solution as an optimal one. The tell: you stopped at the first point inside the feasible region and never compared Z across corners.
• Evaluating only the easy corners (A and C read off the axes) and skipping the intersection point B. The tell: you name an optimum while one extreme point has no Z value.
• Chasing a higher objective line beyond the last corner. The tell: the proposed plan breaks at least one constraint, such as a trade area over 14 miles or demand above what the store can serve.
• Judging efficiency by outputs alone while ignoring the process. The tell: you compare results without comparing inputs and methods. Ford and its rivals used the same resources, and only the process differed.
• Reporting a PERT project time as certain. The tell: you quote the expected duration (for example, 90 minutes for an activity ranging from 60 to 120) with no variance or completion probability.

• ▸Inputs (resources used): list labour, capital, machinery, raw material and data consumed.
• ▸Output: state the product or service produced and the quantity.
• ▸Efficiency metric: choose one of completion time, distance, or resource utilization.
• ▸Objective (Z) and direction: write the equation and mark maximize or minimize.
• ▸Constraints: list each limit with its coefficient and right-hand-side value.
• ▸Candidate solutions (corner points): list each extreme point with its decision-variable values.
• ▸Z at each candidate: compute the objective at every point, including intersections.
• ▸Selected solution and feasibility check: name the best Z and confirm it meets all constraints.
• ▸Time risk (if a project): record expected duration, critical-path standard deviation, z value and completion probability.

1. 1Write the problem as a gap: state the actual outcome, the desired outcome, and the discrepancy between them. The source defines a problem as exactly this discrepancy. If you cannot name the gap, you cannot size the complexity.
2. 2Count the alternatives. List every choice open to the decision-maker, such as opening new stores, employing more sales people, or reducing prices. The source treats a plethora of choices as the first thing that makes a decision problematic. If only one option fits, the problem is easy.
3. 3Count the criteria. List every observable, measurable criterion (for a discipline choice: job offers, salary package, interest, education background, peer pressure). One criterion makes selection trivial; multiple criteria make it harder but more realistic. Record the number.
4. 4Mark the conflicts between criteria. Check whether the best alternative on one criterion is weak on another. The source's example is a discipline with the best job opportunities that you lack the interest or acumen for. Each conflict is a competing objective, and it means no alternative wins everywhere.
5. 5Map the interdependencies. For each function the decision touches, trace who is affected next. In the source, marketing feedback goes to production, production asks finance for budget, and finance and HR then procure materials and hire people. Count the hand-offs; each is a dependency your solution must respect.
6. 6List the resource limits explicitly: labour, capital, machinery, raw material, data, and time. The source frames operations as inputs converted into outputs. Resources are limited and contested, so the limits define the feasible region.
7. 7Write the model in mathematical form: variables for the choices, an objective for the criterion you most want, and constraints for each resource limit and secondary requirement. Variable count and constraint count become your complexity score.
8. 8Aim for an optimum, not a maximum or minimum. The source states that no best alternative exists, because maximum sales with no added cost is implausible. Set the target as the best attainable trade-off under the constraints, then pick the alternative that delivers it.

The source's vendor-selection problem shows a small, complete complexity count. A firm orders seats as a set of three from two vendors, V1 and V2. Each vendor has a maximum of 21 hours of production time, because the order is placed one day before assembly. V1 takes 0.7 hours per seat and V2 takes 0.3 hours. Unit cost is $30 from V1 and $35 from V2. The objective is cost minimization, and the question is how many units to order from each vendor. Count it as follows: two alternatives (the vendors), one stated criterion (cost) with a hidden competing one (speed, since V1 is slower but cheaper), and a time constraint per vendor. Cost alone would push every order to V1, but its 0.7 hours per seat exhausts the 21-hour limit sooner than V2's 0.3. The constraint is what forces a mix and turns a trivial pick into a formulated LPP. The complexity comes from variables, objectives, and limits interacting, not from the arithmetic. Adding more vendors or more criteria would raise the count further.

• If you have one criterion and one clearly best alternative, decide directly; if you have multiple criteria or alternatives, build a model before choosing.
• If the best alternative on one criterion is weak on another, treat the problem as multi-objective and seek an optimum trade-off, not a single maximum.
• If a decision triggers actions in other functions (production, finance, HR, purchasing), include those dependencies as constraints rather than solving the problem inside one function.
• If resource limits do not bind in your model, the problem is probably under-specified; recheck for labour, capital, machine, material, or time limits.
• If activity or project count keeps growing and manual levelling becomes hugely difficult, move the resource problem to project management software.

• Single-criterion collapse: you pick one metric such as lowest cost and ignore the rest. The tell is a chosen option that looks best on paper but fails on speed, quality, or fit.
• Chasing the maximum: you promise maximum sales at no added cost. The tell is a target that no alternative can meet, which means the trade-off was never modelled.
• Function-silo solving: you optimize inside marketing, production, or finance alone. The tell is that downstream departments receive requests (budget, hiring, procurement) they were never consulted on.
• Unstated constraints: the model leaves out hours, labour, or capital limits. The tell is a solution that allocates everything to the cheapest option and is infeasible in practice.
• Treating the alternatives as exhaustive without listing them: you evaluate only the options already in mind. The tell is that a plethora of choices exists but your analysis compares two.

• ▸Problem gap: state actual outcome versus desired outcome in one line.
• ▸Alternatives: list each choice available and total the count.
• ▸Criteria: list each measurable criterion and total the count.
• ▸Criterion conflicts: note each pair where the best option on one is weak on the other.
• ▸Functions affected: name each department touched and the hand-off between them.
• ▸Resource limits: list labour, capital, machinery, material, data, and time limits with their values.
• ▸Model size: count decision variables and constraints.
• ▸Optimum target: state the acceptable trade-off level rather than a maximum or minimum.
• ▸Complexity verdict: rate low, medium, or high from the counts and decide whether a formal model is needed.

1. 1List every resource the activity converts into output, using the passage's input categories: labour, capital, machinery, raw material, data. Add the production-specific limits (production hours, working hours, capacity, budget). A constraint you never name cannot enter the model.
2. 2Give each resource a single maximum (or minimum) figure, b_i, in one unit with a stated period. The source's examples are '14 miles', '10,000 population', '$20,000 median income', and capacity per 8-hour shift. If a limit has no number and no period, it is not yet a constraint.
3. 3For each decision variable x_j, write the coefficient a_ij: how much of resource i one unit of that activity consumes. In the source, a client travels 9 miles per unit at location A and 4 miles at location B, so the accessibility consumption is 9x1 and 4x2.
4. 4Write each constraint as 'resource used ≤ resource available' (or ≥ for a minimum requirement). Then state the objective, c_j per unit of profit or cost, so the constraints and objective sit in one model.
5. 5Treat every b_i and a_ij as an estimate. The source stresses that parameter values in any linear programming model are only estimates, so record how each figure was obtained and how confident you are in it.
6. 6Solve (simplex, graphical, or Solver for several constraints) and read the slack and surplus. Slack means idle resource on a ≤ constraint; surplus means excess over a ≥ requirement. A resource with no slack is the one limiting the objective.
7. 7Run sensitivity analysis on the limiting resources. It gives the range over which the solution stays optimal and feasible, which tells you whether an extra unit of that resource is worth acquiring.
8. 8Re-run the cycle as constraints change, and stop when the improvement is too small to justify allocating more resources. Mattel did this when it raised plastic-leg capacity from 200 to 2,000 units per 8-hour shift and then had to manage the bottlenecks that followed.

The Adidas retail-store case in the source has two store locations: x1, units sold (thousands) at a shopping mall (location A), and x2, units sold (thousands) at a shopping plaza (location B). Three limits apply. Accessibility: the effective trade area is a maximum of 14 miles. Population density: the store can serve an area of about 10,000 people, and more than that causes inefficiency. Income: the store can serve customers with a median income of at most $20,000. The accessibility coefficients are 9 miles per unit at A and 4 miles per unit at B, so the first constraint reads 9x1 + 4x2 ≤ 14. The other two constraints are built the same way from their own per-unit coefficients. The source concludes that the objective, profit maximisation, has to be achieved within these limits. It also notes that no profit can rise without bound because every productive activity runs under limited resources. The answer is therefore an optimum, not a true maximum. After solving, you read slack on each constraint. A constraint with zero slack is binding, and sensitivity analysis then shows how far its limit can move before the solution changes.

• If a constraint is written ≤ and the solved slack is positive, then that resource is idle and adding more of it will not raise the objective.
• If a constraint is written ≥ and a surplus appears, then you are holding more than required and can reduce it.
• If no solution satisfies all constraints at once, then the model is infeasible. Check the estimated b_i values for the tightest conflict before changing the objective.
• If you face several constraints at different levels, then use the algebraic method or Solver. The graphical method becomes impractical.
• If the next round of improvement is too small to justify allocating further resources, then stop re-optimising and act on the current solution.

• Naming the objective as a true minimum or maximum, such as 'zero cost' or 'highest possible profit'. The tell is a target that no resource limit could permit. Reframe it as the optimum under the stated constraints.
• Treating estimated limits as certain, so the plan breaks when one figure moves. The tell is a single-point answer with no range. Run sensitivity analysis.
• Leaving out a limiting resource, such as machine capacity or a demand-side limit like population served. The tell is a solution that looks feasible on paper but cannot be executed.
• Raising capacity without checking the rest of the line. Mattel's move to mass operation produced bottlenecks. The tell is one section running quickly while another queues.
• Mixing units or periods across constraints, for example hours per shift against hours per week. The tell is coefficients that do not combine into one consistent inequality.

• ▸Resource (labour, capital, machinery, raw material, data, hours): name the input being limited.
• ▸Limit b_i and unit/period: the maximum or minimum available, e.g. per 8-hour shift.
• ▸Direction: ≤ (cap) or ≥ (minimum requirement).
• ▸Consumption per unit a_ij: how much of this resource each activity x_j uses.
• ▸Source and confidence of the figure: how the estimate was obtained and how reliable it is.
• ▸Slack/surplus after solving: idle (≤) or excess (≥) amount; zero means binding.
• ▸Sensitivity range: the limit values over which the solution stays optimal and feasible.
• ▸Bottleneck check: which adjacent step becomes limiting if this resource is expanded.
• ▸Action and stop rule: what you will change, and the improvement size below which you stop.

1. 1State the problem in one clear, concise sentence before any algebra. The source treats problem formulation as the first step of the OR process and says it requires 'clear and concise formulation'. Name the decision maker, the choice they control, and the result they want, for example reducing transportation cost from several warehouses to various retailers.
2. 2Separate causal variables from the outcome. The causal variables are what the manager controls, such as units to be transported. The outcome is what responds, such as transportation cost. Define each causal variable as a symbol with a unit (x = number of units) so every term in the model has a stated meaning.
3. 3Write the objective function as a single equation linking the outcome to the causal variables, and declare it as maximize or minimize. Resource-allocation models normally maximize profit or minimize cost. Simple case: cost C = 10x at $10 per unit.
4. 4List the constraints, one per limited resource. The source names production hours, capacity, working hours and financial limits as the typical ones. Write each as an equation or inequality in the same variables, with the resource limit on the right-hand side.
5. 5Check linearity. The linear-programming setting assumes a unit change in a constraint produces a unit effect on the objective. If doubling x does not double its contribution, the model is not linear and the simplex or graphical methods do not apply as written.
6. 6Test the model against the real situation by running it at simple values. For C = 10x, 1 unit gives $10 and 2 units give $20. If the output does not match what the business would actually see, fix the model first, because conclusions are only the best result when the model represents reality accurately.
7. 7Choose the solution route from the model's size. With two decision variables, use the graphical method. With multiple constraints at different levels, use the simplex method or Excel Solver. Plan to use sensitivity analysis and duality to see how the answer changes as inputs change.
8. 8Read the answer as an optimum, not a true maximum or minimum. The source states that decisions are optimized rather than maximized or minimized, because the ideal of more sales at no extra cost is implausible. Pick the alternative that gives the best attainable result under the constraints.

A logistics company wants to cut the cost of supplying materials from warehouses to retailers. A qualitative approach would simply choose the shortest route. The quantitative approach first formulates the problem: choose how many units to ship so that transportation cost is as low as possible. For one lane, from centre A to centre B, the cost is $10 per unit. Let x be the units shipped; the objective is Cost C = 10x, to be minimized. Test the model: 1 unit costs $10, 2 units cost $20, and so on. This lets the manager vary x, the causal variable, and watch cost, the effect, without costly or risky real-world experiments. With many warehouses and retailers, each lane gets its own variable and cost coefficient, and supply and demand limits become constraints. The manager then compares the costs of the alternatives and decides on evidence, not on route length alone. The conclusion holds only if the $10-per-unit figure reflects what the company actually pays.

• If the problem sentence cannot name the controllable choice and the desired result, stop and rewrite it before defining variables.
• If the goal is profit or revenue, maximize. If it is cost, minimize. State the direction explicitly in the model.
• If a resource is limited (hours, capacity, labour, money), write it as a constraint. If it is not limited, leave it out.
• If a change in a variable does not produce a proportional change in the objective, do not treat the model as linear programming.
• If there are two decision variables, use the graphical method to see the feasible region. If there are more variables or many constraints, move to simplex or Excel Solver.
• If the model's output at a simple test value disagrees with reality, revise the model before solving it.

• Starting with equations before the problem is stated. The tell is a model nobody can explain in a sentence, with variables that have no units or owner.
• Confusing the effect with the cause, for example treating cost as the thing you set rather than units shipped. The tell is an objective function with no controllable variable on the right-hand side.
• Omitting a binding resource limit such as production hours or budget. The tell is a solution that asks for more than the business can supply.
• Assuming linearity that the business does not have. The tell is that doubling an input does not double its cost or profit in practice, yet the model scales it that way.
• Treating the solution as a literal maximum or minimum. The tell is a plan that assumes sales rise with no added cost. The source calls this implausible, so read the result as an optimum.
• Trusting the model without checking it against reality. The tell is a recommendation that fails a simple sanity check, such as a cost per unit that differs from the real $10.

• ▸Problem statement: one clear, concise sentence naming the decision and the desired result.
• ▸Decision variables: each controllable quantity, with a symbol and unit (e.g., x = number of units).
• ▸Outcome measure: the effect that responds to the variables (e.g., transportation cost).
• ▸Objective function: the equation for the outcome in terms of the variables, e.g., C = 10x.
• ▸Direction: maximize or minimize, with a one-line reason (profit up or cost down).
• ▸Constraints: one line per limited resource (production hours, capacity, working hours, finance), written as an equation or inequality.
• ▸Linearity check: does a unit change in each constraint give a unit effect on the objective? Yes or no, with evidence.
• ▸Reality test: model output at 1 and 2 units versus what the business actually sees.
• ▸Solution route: graphical, simplex, or Excel Solver, chosen by the number of variables and constraints.

1. 1Restate the decision in the model's own variables before touching a solver. The source's sales-force case defines x1, x2, x3 as the number of salesmen allocated to sectors 1, 2 and 3. An allocation is the final numeric value of each of these variables, with a unit attached.
2. 2Write the objective and every limit as linear expressions. The source's objective is Maximize Z = 6,000x1 + 5,500x2 + 5,000x3. Its constraints are three distance limits (≤12, ≤12, ≤14) and a salesmen limit (1x1 + 2x2 + 2x3 ≤ 40). Linearity means a one-unit change in a variable moves the objective by a fixed amount, so confirm that holds before solving.
3. 3Choose the solution route by problem shape. Two decision variables can use the graphical method. Several constraints need the simplex method or Excel Solver. Supply-to-demand shipping uses the transportation model, with Least Cost, North-West Corner or Vogel's Approximation (VAM) for the starting plan and then an optimality test. One-to-one job matching uses the assignment model.
4. 4Solve, then read the allocation off the final result. In simplex, the basic variables in the final table carry the quantities and the non-basic variables are zero. In a transportation table, the occupied cells are the routes and quantities to ship. Report every variable with its value, not just Z.
5. 5Check feasibility of the allocation against every original constraint. Substitute the values back in and confirm no limit is exceeded. In the final simplex table, an artificial variable (such as a3 with coefficient M) that remains in the basis signals infeasibility, not an allocation.
6. 6Classify the outcome before acting. The source separates four special cases: degeneracy, unbounded, infeasible and multiple optimal solutions. If several optima exist, choose among them on a practical criterion outside the model. The source notes that alternative optima keep the same objective value but matter in practice.
7. 7Attach sensitivity limits to the allocation. Linear programming gives a feasible optimum but not the range within which it stays valid. Use the allowable RHS range and the objective-coefficient ranges to state how far each limit or coefficient can change before the allocation must be recomputed.
8. 8Treat the result as an optimum, not an absolute maximum or minimum. The source says decisions are optimized, because maximum sales at no extra cost is implausible. Record which limits are binding so the allocation is understood as the best use of scarce resources.

The source's sales-force case assigns 40 salesmen to three sectors. Sales per month are 6,000, 5,500 and 5,000 units for sectors 1, 2 and 3. The model is Maximize Z = 6,000x1 + 5,500x2 + 5,000x3. The constraints are S1: 5.0x1 + 3.5x2 + 1.2x3 ≤ 12; S2: 4.7x1 + 3.0x2 + 5.0x3 ≤ 12; S3: 2.9x1 + 5.2x2 + 2.7x3 ≤ 14; and 1x1 + 2x2 + 2x3 ≤ 40. The passage stops at the formulation, so the optimal x values come from simplex or Solver. What you can do is test any candidate allocation against the constraints. Try one salesman in each sector (1,1,1). S1 gives 9.7 ≤ 12, S3 gives 10.8 ≤ 14, and the salesmen limit gives 5 ≤ 40. S2 gives 4.7 + 3.0 + 5.0 = 12.7, which exceeds 12. The allocation is infeasible, even though its Z of 16,500 looks attractive. An allocation is valid only if it passes all four constraints, and the solver's output must pass the same test before anyone is told where to go.

• If the problem has only two decision variables, solve it graphically. If there are several constraints or variables, use the simplex method or Excel Solver.
• If the task is moving units from supply centres to demand centres at minimum cost, build a transportation table. Start with Least Cost, North-West Corner or VAM, then run the optimality test before releasing quantities.
• If total supply and total demand differ, treat it as an unbalanced transportation problem and add the missing dummy supply or demand before solving. If certain routes cannot be used, mark them unacceptable so they receive no units.
• If an artificial variable stays in the final basis, declare the problem infeasible and revisit the constraints. If the objective can grow without limit, check for an omitted constraint (unbounded).
• If you must commit the allocation to action, first check the allowable ranges. If a limit is likely to move outside its range, re-solve rather than stretch the old answer.

• Reporting only the objective value Z. The tell is a plan with no unit counts per sector, route or job, so nobody knows what to deploy.
• Skipping the back-substitution check. The tell is a candidate like (1,1,1) that looks profitable but breaks the Store 2 distance limit (12.7 > 12).
• Ignoring special cases such as degeneracy, infeasibility or alternative optima. The tell is an artificial variable still in the basis, a tie, or a result the model cannot explain.
• Applying the model's answer after conditions change. The tell is a changed right-hand side or coefficient with no recheck against the sensitivity ranges, and the allocation is reused anyway.
• Chasing the unconstrained maximum instead of the optimum. The tell is a plan that assumes extra sales at no extra cost, which the source calls implausible.

• ▸Decision variables: list each variable (e.g. x1 = salesmen in sector 1) and its unit.
• ▸Objective: write the function and whether it is maximized or minimized (e.g. Z = 6,000x1 + 5,500x2 + 5,000x3).
• ▸Constraints: write each limit with its coefficients and right-hand side.
• ▸Solution method: graphical, simplex, Excel Solver, or transportation/assignment (with the initial-solution method used).
• ▸Proposed allocation: record the final value of every variable, with units.
• ▸Constraint check: substitute the values into each constraint and record left side, right side and pass/fail.
• ▸Special-case flag: mark degenerate, unbounded, infeasible, multiple optima, or none.
• ▸Binding constraints and allowable RHS ranges: list which limits are fully used and how far each can change before re-solving.
• ▸Decision and re-solve trigger: record the approved allocation and the change that would require recomputing it.

1. 1Name the cost you are minimizing and every component of it. In the source's inventory case, total cost is annual holding cost plus annual ordering cost. Write TC = (Q/2)*H + (D/Q)*S before touching any numbers.
2. 2Define the decision variable(s) as the quantities you control (here, batch size Q). Assign each a cost coefficient c_j, so the objective reads Z = c1x1 + c2x2 + ... + cnxn, to be minimized.
3. 3List the constraints as inequalities: resource limits, minimum requirements, and capacity. Add non-negativity (x ≥ 0). Without constraints the problem is unconstrained optimization; with them, it is constrained optimization, and the answer must be feasible as well as optimal.
4. 4Fill in the coefficients from data, not estimates of convenience. In the EOQ case: D = 2,500 units per year, H = 0.15 * 800 = $12 per unit per year, S = $50 per order.
5. 5Solve. For a single-variable cost curve, set the first derivative to zero (slope = 0). For several variables with constraints, use LP (graphical method for two variables, simplex beyond that).
6. 6Confirm you found a minimum, not a maximum: the second derivative must be positive. In the EOQ case, f''(Q) = 250,000/Q³ is always above zero.
7. 7Check feasibility and the LP assumptions. Substitute the solution into every constraint, then test proportionality, additivity, divisibility, and certainty. If a decision variable must be a whole number and the answer is fractional, round it and recompute total cost.
8. 8Compute total cost at the final (rounded) decision, and report it as the optimum under these constraints, not as an absolute minimum.

The source's inventory case: annual demand D = 2,500 units, holding cost H = 0.15 * 800 = $12 per unit, ordering cost S = $50 per order. Total cost is TC(Q) = (Q/2)*12 + (2,500/Q)*50 = 6Q + 125,000/Q. The first derivative is 6 − 125,000/Q². Setting it to zero gives Q = √(125,000/6) = 144.34, rounded to 145 units. The second derivative, 250,000/Q³, is positive, so this is the minimum point. At the optimum, holding cost (about $866) and ordering cost (about $866) are equal, which is the intersection of the two curves in the source's Figure 1.4. Substituting Q = 145 gives 870 holding plus 862.07 ordering, about $1,732 (the source prints $1,732.19; recomputing gives $1,732.07, so verify arithmetic yourself). The lesson is that the minimum is a balance point between two opposing costs, not the smallest value of either.

• If total cost combines two components that move in opposite directions (holding rises with batch size, ordering falls), then look for the point where the slope is zero, which is where the two curves cross.
• If the first derivative is zero, then check the second derivative: positive means a minimum, negative means a maximum. Do not report the result before checking.
• If any resource, capacity, or requirement limits the activity, then treat it as constrained optimization. Test every candidate solution against all constraints, because a cheaper answer that violates one is infeasible.
• If the decision variable must be a whole number but the solution is fractional (e.g., 144.34), then round it and recompute total cost at the rounded value rather than quoting the unrounded cost.
• If cost does not scale proportionally with activity, or coefficients are not known constants, then do not apply LP directly. Proportionality, additivity, divisibility and certainty must hold.

• Chasing zero cost: treating 'minimum' as the lowest imaginable number. The source notes no cost can be minimized to zero because every activity runs under constraints. The tell: a target below what any feasible solution can reach.
• Optimizing one cost component only (e.g., cutting order frequency to save ordering cost). The tell: holding cost balloons and total cost rises, and the two components are far from equal.
• Skipping the feasibility check. The tell: the 'best' solution breaks a constraint. The source shows 9*2 + 4*2 = 26 against a limit of 14, so scaling a feasible solution up can make it infeasible.
• Declaring a stationary point a minimum without the second-derivative test. The tell: no evidence the curve is convex, and no check of f'' > 0.
• Treating estimated coefficients as certain, or assuming costs scale linearly. The tell: the plan fails when real unit costs or demand differ from the inputs used, because LP assumes known constant coefficients.

• ▸Cost being minimized: name the total cost and each component (e.g., holding + ordering).
• ▸Decision variable(s): define each x_j or Q with units.
• ▸Cost coefficients (c_j or H, S, D): list each value and its data source.
• ▸Objective function: write Z = c1x1 + ... + cnxn or TC(Q) in full.
• ▸Constraints: list each as an inequality with its right-hand side (b_i), plus non-negativity.
• ▸Solution method and result: derivative, graphical, or simplex; record the optimal value(s).
• ▸Minimum check: second derivative sign or extreme-point comparison.
• ▸Feasibility check: substitute the solution into every constraint and record pass or fail.
• ▸Final decision and total cost: after any rounding, recompute and record the cost.