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23-Ind-A1 Operations Research · December 2013

Question 1 of 10: LP Formulation — Post Office Full-Time / Part-Time Staff Scheduling

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

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National Exams — December 2013 — 98-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 200 marks across 10 questions and only 100 marks are required, so a candidate would normally answer a subset — all ten are solved below for completeness.

Reference texts: Hillier & Lieberman, Introduction to Operations Research (11th ed., McGraw-Hill) — linear/integer programming, network optimization, dynamic programming, decision analysis and queueing theory; Nahmias, Production and Operations Analysis — inventory models with planned backorders.

Question 1: LP Formulation — Post Office Full-Time / Part-Time Staff Scheduling (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Daily FTE requirement over one week (S–S); full-time (FT) staff work 8 hrs./day at $15/hr = $120/day, 5 consecutive days then 2 consecutive days off; part-time (PT) staff work 4 hrs./day at $10/hr = $40/day, i.e. 0.5 FTE-day each; PT labour capped at 25% of the total weekly labour requirement.

DaySunMonTueWedThuFriSatWeek total
FTE required11171315191416105

Find. A Linear Programming model — decision variables, objective and constraints — that minimizes the weekly labour cost of covering the schedule. Do not solve.

Approach. Model full-time staffing as a cyclic days-off schedule (7 possible 5-consecutive-day shift starts, each covering 5 of the 7 days), add a daily part-time headcount, then impose day-by-day coverage and a union part-time cap.

  1. Decision variables. Index the days $i=1,\dots,7$ (Sun…Sat). Let $x_j\ge0$ integer = number of full-time employees whose 5-consecutive-day work week begins on day $j$ (working days $j,j+1,\dots,j+4$, mod 7, then resting the other 2 consecutive days), for $j=1,\dots,7$. Let $y_i\ge0$ integer = number of part-time employees working on day $i$ (4 hrs. that day only), $i=1,\dots,7$.
  2. Coverage sets. A shift starting on day $j$ covers day $i$ exactly when $j\in\{i,i-1,i-2,i-3,i-4\}\pmod 7$ — i.e. 5 of the 7 possible start-days cover any given day.
  3. Day $i$ coveredFTE$_i$Full-time shifts covering it
    Sun (1)11$x_1,x_4,x_5,x_6,x_7$
    Mon (2)17$x_1,x_2,x_5,x_6,x_7$
    Tue (3)13$x_1,x_2,x_3,x_6,x_7$
    Wed (4)15$x_1,x_2,x_3,x_4,x_7$
    Thu (5)19$x_1,x_2,x_3,x_4,x_5$
    Fri (6)14$x_2,x_3,x_4,x_5,x_6$
    Sat (7)16$x_3,x_4,x_5,x_6,x_7$
  4. Coverage constraints. Each day's on-duty FTE (full-time shifts covering it, plus 0.5 FTE per part-timer that day) must meet requirement: $$\sum_{j\,\text{covers}\,i} x_j \;+\; 0.5\,y_i \;\ge\; \text{FTE}_i,\qquad i=1,\dots,7.$$
  5. Union part-time cap. Part-time FTE-days may not exceed 25% of the week's total requirement ($0.25\times105=26.25$): $$0.5\sum_{i=1}^{7} y_i \;\le\; 26.25 \quad\Longleftrightarrow\quad \sum_{i=1}^{7} y_i \le 52.5.$$
  6. Objective. Minimize weekly labour cost ($600 per full-time employee—5 days at $120—plus $40 per part-time employee-day): $$\boxed{\min Z = 600\sum_{j=1}^{7}x_j \;+\; 40\sum_{i=1}^{7}y_i}$$ subject to the 7 coverage constraints, the union cap, and $x_j,y_i\ge0$ integer.
ElementFormulation
Variables$x_j\ge0$ integer (FT shifts starting day $j$), $y_i\ge0$ integer (PT headcount day $i$), $j,i=1,\dots,7$
Objective$\min\ 600\sum x_j + 40\sum y_i$ ($/week)
Coverage7 constraints, $\sum_{j\text{ covers }i}x_j+0.5y_i\ge\text{FTE}_i$
Union cap$\sum y_i\le52.5$
Check
The paper's instruction reads "minimize the fuel cost," a typo for labour cost — the paragraph is entirely about full-/part-time wages, so the objective above minimizes labour cost. The union cap is modelled as an aggregate weekly limit (total part-time FTE-days $\le$25% of the week's total FTE requirement); a per-day reading ($0.5y_i\le0.25\,\text{FTE}_i$ for each $i$) is an equally defensible alternative and would simply replace the single cap row with 7 rows.
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