CMA Intermediate · Operations Management and Strategic Management

Job Evaluation, Job Allocation - Assignment: formula sheet

Full chapter guide

Key formulas

Point rating: total job points: Total points of a job = Σ (points assigned to each factor for that job); Factors usually include skill, effort, responsibility and working conditions, each with weighted degrees.
Point rating: wage per point: Money value of a point = Total wage bill for the benchmark jobs ÷ Total points of those jobs; Multiply by a job's points to get its wage. Use only if the question gives a total pay and points basis.
Factor comparison: job wage: Job wage = Σ (money value allotted to each factor for that job); Each key job's existing wage is split across factors, and other jobs are compared factor by factor with key jobs.
Ranking method: Rank jobs from highest to lowest by overall difficulty and worth; No factors or points. Simple, but subjective.
Objective function (minimisation): Minimise Z = Σi Σj cij xij, for i, j = 1 to n; cij is the cost of giving job j to worker i. Use maximise if cij is profit.
Worker constraint: Σj xij = 1 for every worker i; Each worker gets exactly one job.
Job constraint: Σi xij = 1 for every job j; Each job is done by exactly one worker.
Binary condition: xij = 0 or 1; xij = 1 means worker i is assigned job j.
Balanced condition: Number of workers = Number of jobs; If not equal, add dummy rows or columns with zero cost to make the matrix square.
Number of variables and constraints: n × n variables; 2n constraints; A 4 × 4 problem has 16 variables and 8 constraints.
Row reduction: New cell = cell − (smallest value in its row); Do this for every row first. Every row now has at least one zero.
Column reduction: New cell = cell − (smallest value in its column); Do this on the row-reduced matrix. Skip any column that already has a zero.
Optimality test: Minimum number of lines covering all zeros = n ⇒ optimal; Lines are horizontal or vertical only. If the count is less than n, the solution is not yet optimal.
Matrix adjustment: Uncovered cells: subtract k. Cells at line intersections: add k. Singly covered cells: unchanged. Here k = smallest uncovered value.; This creates at least one new zero. Then redo the line test.
Total cost: Total cost = Σ original cost of the assigned cells; Always add from the original matrix, not the reduced one.
Maximisation to minimisation: Regret = (largest value in the matrix) − (cell value); Apply to every cell, then run the Hungarian method on the regret matrix. Total profit is read from the original matrix.
Balancing rule: Number of rows = number of columns; If rows < columns, add dummy rows. If columns < rows, add dummy columns. Add as many as the difference.
Dummy entries: Dummy row or column cells = 0; A dummy has zero cost or zero profit. Add it before converting a maximisation problem.
Optimality test: Minimum lines covering all zeros = order of the matrix (n); If lines are fewer than n, revise the matrix. If equal to n, an assignment on zeros exists.
Revision step: Subtract the smallest uncovered value from uncovered cells; add it to cells at line intersections; Cells covered by one line stay unchanged.
Prohibited assignment: Cost of blocked cell = M (a very large number); For minimisation. M is never chosen. Treat M as unchanged by subtraction: M − k is still very large.
Valid assignment test: Number of lines = n ⇒ optimal assignment possible; If the minimum lines covering all zeros is less than n, revise the matrix and repeat.
Revision step: Subtract the smallest uncovered element from uncovered cells; add it to cells at line intersections; Cells covered by one line stay unchanged.
Total cost: Total cost = Σ original cost of each assigned cell; Always read the cost from the original matrix, not the reduced one.
Multiple optima signal: More than one way to choose n independent zeros ⇒ multiple optimal solutions; All such solutions have the same total cost.
TSP set-up: Diagonal cells (i to i) = M; tour must be one closed cycle through all n cities; Assignment answer with sub-tours is not a valid tour.

Quick revision

Job evaluation rates the job; merit rating rates the employee.
Job evaluation helps fix fair relative pay for jobs.
Merit rating assesses an employee's performance and is used for things like promotion and increments.
Assignment problem: n workers, n jobs, one job per worker, one worker per job.
It is a special case of the transportation problem where every supply and demand is 1.
Hungarian steps: subtract row minimum, subtract column minimum, cover zeros with the fewest lines.
If the minimum number of lines equals n, an optimal assignment exists among the zeros.
If lines are fewer than n, subtract the smallest uncovered value from uncovered cells and add it to cells at line intersections.
Maximisation: subtract every entry from the largest entry, then minimise.
Unbalanced matrix: add a dummy row or column with zero costs to make it square.
Restricted assignment: put a very large cost in the prohibited cell.
Report total cost from the original matrix; several zero choices may give alternative optimal solutions with the same total.

Common mistakes

Saying job evaluation assesses the employee's performance. Fix: Write that job evaluation rates the job's worth; merit rating rates the person's performance.
Mixing up ranking and grading. Fix: In ranking, you order the jobs against each other. In grading, you first define grade descriptions and then place each job in a grade.
Solving an unbalanced matrix without adding a dummy Fix: Count rows and columns first. Add a dummy row or column with zero costs to make it square.
Adding the reduced costs to find the total cost Fix: Always return to the original matrix. Add the original cost of each assigned pair.
Doing column reduction first or skipping row reduction. Fix: Always reduce rows first, then columns. Reducing columns that already contain a zero changes nothing, so skip those.
Adding the cost from the reduced matrix. Fix: Mark your chosen cells and go back to the original matrix to add the costs.
Subtracting from the wrong number, such as each row's largest value. Fix: For conversion, use one single largest value from the whole matrix. Use row minimums only in the reduction step.
Reporting the total from the regret matrix. Fix: Always return to the original profit matrix and sum the cells at the chosen positions.
Putting 0 or a small number in a prohibited cell instead of M. Fix: A prohibited cell must be M in minimisation. Write it before you start reducing.
Subtracting from M and treating the result as a small number. Fix: Treat M minus any number as still M. Never let it become a zero or the smallest uncovered element.

Exam tips

Questions often ask for the difference between the two terms. Prepare five clear points and write them as a list.
For method questions, always state whether the method is analytical or non-analytical.
In point rating numericals, show the sum of points per job before converting to money.
Keep a one-line advantage and limitation ready for each method. They add step marks.
In MCQs, check whether the statement refers to the job or the person. That alone decides many options.
For a formulation question, write the variable definition, objective function, both sets of constraints and the binary condition. Each earns marks.
In MCQs, remember that the assignment problem needs a square matrix and that its constraints are equalities with right-hand side 1.
The difference between assignment and transportation is a favourite short-answer question. Prepare a four-point comparison on supply and demand, matrix shape, variable values and method.