Numerical Problems
Fresh practice problems (not copies of the textbook's worked examples) for every quantitative technique in your syllabus β try each one before revealing the solution.
Per your syllabus, numerical solving is concentrated in Units 8β10 (fully practised below). Units 4β6 contain three techniques that are inherently quantitative (location factor rating, centre-of-gravity/load-distance, line balancing, and breakeven analysis) β one practice problem is included for each, consistent with the "one clear illustrative pass" approach used on those unit pages. All problems here are illustrative practice, distinct from the textbook's own worked examples shown on the unit pages.
Unit 4 β Location Factor Rating Illustrative practice problem
Problem: Two sites are scored on 4 factors. Weights: Labour (0.4), Transport (0.3), Utilities (0.2), Community attitude (0.1). Scores (0β100): Site X = 80, 70, 60, 90. Site Y = 60, 85, 75, 70. Which site should be chosen?
Solution:
Site X = (0.4Γ80)+(0.3Γ70)+(0.2Γ60)+(0.1Γ90) = 32+21+12+9 = 74
Site Y = (0.4Γ60)+(0.3Γ85)+(0.2Γ75)+(0.1Γ70) = 24+25.5+15+7 = 71.5
Answer: Site X (74 > 71.5) is preferred.
Unit 4 β Centre-of-Gravity & Load-Distance Illustrative practice problem
Problem: Three retail stores need a new warehouse. Store P (x=100, y=100, weight=50), Store Q (x=300, y=150, weight=80), Store R (x=200, y=350, weight=70). Find the centre-of-gravity coordinates.
Solution:
x* = [(100Γ50)+(300Γ80)+(200Γ70)] / (50+80+70) = (5,000+24,000+14,000)/200 = 43,000/200 = 215
y* = [(100Γ50)+(150Γ80)+(350Γ70)] / 200 = (5,000+12,000+24,500)/200 = 41,500/200 = 207.5
Answer: Centre of gravity β (215, 207.5). If two real candidate sites are found near this point, compute LD = Ξ£(load Γ distance) for each against P, Q, R, and choose the lower value, following the load-distance method.
Unit 5 β Line Balancing Illustrative practice problem
Problem: A toy-assembly line must produce 3,000 units in a 40-hour week. Tasks: A (0.2 min, no precedence), B (0.3 min, after A), C (0.25 min, after A), D (0.4 min, after B & C). Find the desired cycle time and the theoretical minimum number of workstations.
Solution:
Cd = (40Γ60)/3,000 = 2,400/3,000 = 0.8 min/unit
Total task time = 0.2+0.3+0.25+0.4 = 1.15 min
N = 1.15/0.8 = 1.44 β round up to 2 workstations (theoretical minimum)
Answer: Group A+C in Station 1 (0.45 min) and B+D in Station 2 (0.7 min) β both within the 0.8 min cycle time, respecting precedence. Efficiency = 1.15/(2Γ0.8) = 71.9%.
Unit 6 β Breakeven Analysis for Process Selection Illustrative practice problem
Problem: A bakery is deciding whether to buy a semi-automatic oven (fixed cost $3,000/year, variable cost $2/cake) or continue with manual baking (fixed cost $500/year, variable cost $3.50/cake). At what volume do the two processes cost the same?
Solution: Set TCmanual = TCoven: 500 + 3.5v = 3,000 + 2v β 1.5v = 2,500 β v = 1,667 cakes
Answer: Below ~1,667 cakes/year, manual baking is cheaper; above it, the semi-automatic oven is cheaper.
Unit 7 β ABC Classification Illustrative practice problem
Problem: Classify these 5 items by ABC analysis: Item 1 (cost $50, usage 200), Item 2 (cost $5, usage 1,000), Item 3 (cost $200, usage 40), Item 4 (cost $2, usage 3,000), Item 5 (cost $20, usage 150).
Solution: Total value: Item 1 = $10,000; Item 2 = $5,000; Item 3 = $8,000; Item 4 = $6,000; Item 5 = $3,000. Total = $32,000. Ranked: Item 1 (31.3%), Item 3 (25%), Item 4 (18.8%), Item 2 (15.6%), Item 5 (9.4%).
Answer: Class A β Items 1 & 3 (56.3% of value); Class B β Item 4 (18.8%); Class C β Items 2 & 5 (25%) β exact cut-offs may vary, justify your own grouping using the cumulative % of value.
Unit 8 β EOQ Illustrative practice problem
Problem: A store sells 2,400 units/year of an item. Ordering cost = $40/order, carrying cost = $4/unit/year. Find the EOQ and the minimum total cost.
Solution:
Qopt = β(2Γ40Γ2,400 / 4) = β48,000 β 219.1 units
TC = (40Γ2,400)/219.1 + (4Γ219.1)/2 = 438.2 + 438.2 = $876.4
Answer: Order about 219 units at a time; minimum annual cost β $876.
Unit 8 β EBQ (Production Quantity Model) Illustrative practice problem
Problem: A firm manufactures 6,000 units/year in-house (250 working days/year), at a production rate of 40 units/day. Setup cost = $60, carrying cost = $3/unit/year. Find the EBQ.
Solution:
d = 6,000/250 = 24 units/day; p = 40 units/day
Qopt = β[2Γ60Γ6,000 / (3Γ(1β24/40))] = β[720,000 / 1.2] = β600,000 β 774.6 units
Answer: Produce in batches of about 775 units. Maximum inventory = 774.6Γ(1β24/40) β 464.8 units.
Unit 8 β Quantity Discount Illustrative practice problem
Problem: Annual demand = 500 units, Co = $50, Cc = $5/unit/year. Price schedule: 1β99 units @ $20; 100+ units @ $18. Should the firm take the discount?
Solution:
Qopt = β(2Γ50Γ500/5) = β10,000 = 100 units β falls exactly in the discount bracket.
TC at Q=100, P=$18: (50Γ500)/100 + (5Γ100)/2 + 18Γ500 = 250+250+9,000 = $9,500
Answer: Since Qopt itself qualifies for the discount, order 100 units at $18/unit; no further comparison is needed here (this is the special case where the EOQ already lands in the cheaper bracket).
Unit 9 β Reorder Point with Variable Demand Illustrative practice problem
Problem: Average daily demand = 40 units, Οd = 6 units, lead time = 6 days, desired service level = 90% (z = 1.28). Find the reorder point.
Solution:
Safety stock = 1.28 Γ 6 Γ β6 = 1.28 Γ 6 Γ 2.449 β 18.8 units
R = (40Γ6) + 18.8 = 240 + 18.8 = 258.8 units
Answer: Reorder when stock falls to about 259 units.
Unit 9 β Periodic Inventory System Illustrative practice problem
Problem: Average demand = 10 units/day, Οd = 2 units, review period = 30 days, lead time = 4 days, current stock = 15 units, service level = 95% (z = 1.65). Find the order quantity.
Solution:
Q = 10Γ(30+4) + 1.65Γ2Γβ(30+4) β 15 = 340 + (1.65Γ2Γ5.83) β 15 = 340 + 19.24 β 15 β 344.2 units
Answer: Order approximately 344 units at this review point.
Unit 10 β Assignment Method Illustrative practice problem
Problem: Three workers (P, Q, R) and three jobs (1, 2, 3) with completion times (hours):
| Job 1 | Job 2 | Job 3 | |
|---|---|---|---|
| P | 9 | 11 | 14 |
| Q | 13 | 13 | 12 |
| R | 10 | 8 | 9 |
Solution: Row reduction (subtract row min: 9, 12, 8) β [0,2,5 / 1,1,0 / 2,0,1]. Column reduction (mins already 0,0,0) β same matrix. Covering zeros needs only 2 lines (row Q, col spanning R's 0 and P's 0) β fewer than 3 rows, so adjust: smallest uncovered = 1; subtract from uncovered, add at intersections β optimal assignment emerges as PβJob 1 (9), QβJob 3 (12), RβJob 2 (8).
Answer: Minimum total time = 9+12+8 = 29 hours.
Unit 10 β Sequencing Rules Illustrative practice problem
Problem: Four jobs wait at one machine (today = day 0):
| Job | Processing Time | Due Date |
|---|---|---|
| A | 4 | 6 |
| B | 7 | 9 |
| C | 2 | 19 |
| D | 6 | 16 |
Solution (SPT β shortest processing time first): Sequence: C(2) β A(4) β D(6) β B(7).
Completion times: C=2, A=6, D=12, B=19. Tardiness: C=0, A=0, D=0, B=10.
Average completion time = (2+6+12+19)/4 = 9.75; Average tardiness = (0+0+0+10)/4 = 2.5.
Answer: Try DDATE (order: A, B, D, C) yourself and compare β you should find DDATE gives a different (and, per the textbook's proof, better-or-equal) average tardiness, reinforcing that SPT and DDATE optimise different objectives.
Unit 10 β Johnson's Rule Illustrative practice problem
Problem: Four jobs go through cutting (Process 1) then stitching (Process 2):
| Job | Process 1 | Process 2 |
|---|---|---|
| W | 5 | 2 |
| X | 3 | 6 |
| Y | 8 | 4 |
| Z | 4 | 7 |
Solution: Smallest time overall = 2 (Job W, Process 2) β place W last. Next = 3 (Job X, Process 1) β place X first. Next = 4, tied between Job Z (Process 1) and Job Y (Process 2) β Z early, Y late. Sequence: X β Z β Y β W.
Answer: Build a Gantt chart for this sequence (Process 1 then Process 2, no overlap) to find the makespan β try it yourself following the method shown in Unit 10.