● QT · TERM END
🔍 /
QUANTITATIVE TECHNIQUES›ASSIGNMENTS›ASSIGNMENT 1

Assignment 1 — LPP Formulation & Graphical Method

Six problems: three formulations with tricky wording (strict inequalities, percentages, ratios), a diet problem, the Anderson training-programme problem, and Tom's salsa — each solved to the final number with its graph.

PROBLEMS: 6 UNIT: UNIT 01 — LINEAR PROGRAMMING Original: Assignment 1
◆ The method for every problem on this page
  1. Decision variables — ask "what can the manager actually choose?" and define each with units: Let X₁ = number of …
  2. Objective — profit/revenue ⟹ Max; cost/time ⟹ Min. Coefficients are the per-unit values.
  3. Constraints — one per resource or requirement, each labelled. Translate the wording: at least ⟹ ≥, at most / available / capacity ⟹ ≤, exactly ⟹ =.
  4. Non-negativity — X₁, X₂ ≥ 0, written every time.
  5. Graphical solution (when asked) — draw each line from its two intercepts, shade the side that satisfies it, list every corner point of the common region, evaluate Z at each in a table, pick the best.

Full method with worked examples: Unit 1 · Formulation · Graphical method.

★ Tricks that save time in the exam
  • Percentage constraints: "at least 40% of the total must be textiles" becomes X₂ ≥ 0.4(X₁ + X₂), then move everything to the left: −0.4X₁ + 0.6X₂ ≥ 0. Never leave a variable on the right-hand side.
  • "Lower than twice" — X₁ < 2X₂ — becomes X₁ − 2X₂ ≤ 0. Strict inequalities (<, >) are written as ≤, ≥ because an LP cannot hold a strict inequality; say so in one line.
  • Ounces versus pounds: convert the resource to the unit of the per-jar recipe before writing the constraint (1 lb = 16 oz).
  • Min problem with ≥ constraints: the feasible region is open away from the origin, and the optimum is the corner closest to it — the origin itself is never feasible.

Q1 — Two drugs, a dosage window and a ratio limit

Question

Drug D1 gives 8 units of pain relief and 3 of anti-inflammatory benefit and costs ₹10; D2 gives 5 and 7 and costs ₹14. Daily needs: at least 40 units of pain relief and 30 of anti-inflammatory. Total dosage must be above 6 and below 12 units, and the dosage of D1 must be lower than twice that of D2. Formulate the LPP.

Assignment 1 p. 1

Solution
  1. Step 1 — Decision variables Let X₁ = dosage (units) of drug D1 per day Let X₂ = dosage (units) of drug D2 per day
  2. Step 2 — Objective: cost is given, so minimise Min Z = 10X₁ + 14X₂
  3. Step 3 — Constraints 8X₁ + 5X₂ ≥ 40 ← pain relief, at least 40 3X₁ + 7X₂ ≥ 30 ← anti-inflammatory, at least 30 X₁ + X₂ ≥ 6 ← total dosage above 6 X₁ + X₂ ≤ 12 ← total dosage below 12 X₁ − 2X₂ ≤ 0 ← D1 lower than twice D2 (X₁ < 2X₂)

    "Above 6" and "below 12" are strict in the question; an LP can only hold ≥ and ≤, so write them that way and add one line saying so.

  4. Step 4 — Non-negativity X₁, X₂ ≥ 0
  5. Going further — the optimum, if you are asked to solve
    Corner(X₁, X₂)Z
    (0, 8)112
    (0, 12)168
    (3.17, 2.93)72.68 ← min
    (4.62, 2.31)78.46
    (8, 4)136
    Best corner: 8X₁ + 5X₂ = 40 and 3X₁ + 7X₂ = 30 together ⟹ X₁ = 130/41 ≈ 3.17, X₂ = 120/41 ≈ 2.93 Minimum daily cost ≈ ₹72.68 — and X₁ + X₂ ≈ 6.10 > 6, X₁ < 2X₂, so the strict conditions hold too

Q2 — Shipping electronics and textiles

Question

Shipping costs ₹1200 per ton of electronics and ₹800 per ton of textiles. At least 20 tons of electronics and 25 tons of textiles must be supplied; total capacity is 70 tons; at least 40% of the total tonnage must be textiles; textiles must not exceed 45 tons. Formulate the LPP.

Assignment 1 p. 1

Solution
  1. Step 1 — Decision variables Let X₁ = tons of electronics shipped Let X₂ = tons of textiles shipped
  2. Step 2 — Objective Min Z = 1200X₁ + 800X₂ (total shipping cost, ₹)
  3. Step 3 — Constraints X₁ ≥ 20 ← electronics, at least 20 tons X₂ ≥ 25 ← textiles, at least 25 tons X₁ + X₂ ≤ 70 ← transport capacity X₂ ≥ 0.4(X₁ + X₂) ⟹ −0.4X₁ + 0.6X₂ ≥ 0 ← at least 40% textiles X₂ ≤ 45 ← fragile-goods limit
  4. Step 4 — Non-negativity X₁, X₂ ≥ 0
  5. Going further — what the optimum says
    Corner(X₁, X₂)Z
    (20, 25)44000 ← min
    (20, 45)60000
    (25, 45)66000
    (37.5, 25)65000
    (42, 28)72800

    Minimum cost ₹44,000 at (20, 25): with cost minimised and only ≥ requirements pushing up, the company ships exactly the minimum of each. The 40% rule is satisfied automatically (25/45 = 56%). A question that wanted a richer answer would maximise shipped value instead — the same constraints then give (42, 28).

Q3 — Chicken feed — the classic diet problem

Question

Food A has 12, 3 and 2 units of nutrients N1, N2, N3; food B has 3, 12 and 1. Minimum requirements are 36, 36 and 10. A costs ₹20 and B ₹40 per unit. How much of each should be bought at the lowest cost?

Assignment 1 p. 1

Solution
  1. Step 1 — Formulate Let X₁ = units of product A, X₂ = units of product B Min Z = 20X₁ + 40X₂ 12X₁ + 3X₂ ≥ 36 ← N1 3X₁ + 12X₂ ≥ 36 ← N2 2X₁ + X₂ ≥ 10 ← N3 X₁, X₂ ≥ 0
  2. Step 2 — Intercepts for plotting
    LineX₁-interceptX₂-intercept
    12X₁ + 3X₂ = 36(3, 0)(0, 12)
    3X₁ + 12X₂ = 36(12, 0)(0, 3)
    2X₁ + X₂ = 10(5, 0)(0, 10)
  3. Step 3 — Graph: all three are ≥, so shade away from the origin
    0510510X₁ (units of A)X₂ (units of B)N1N2N3ABCD
    Feasible region (shaded) is open to the upper right; the optimum C is the corner nearest the origin
  4. Step 4 — Corner points B: 12X₁ + 3X₂ = 36 and 2X₁ + X₂ = 10 ⟹ subtract 3 × (second) from first: 6X₁ = 6 ⟹ (1, 8) C: 2X₁ + X₂ = 10 and 3X₁ + 12X₂ = 36 ⟹ X₂ = 10 − 2X₁ ⟹ 3X₁ + 120 − 24X₁ = 36 ⟹ (4, 2)
    Corner(X₁, X₂)Z
    A(0, 12)480
    B(1, 8)340
    C(4, 2)160 ← min
    D(12, 0)240
  5. Step 5 — Answer Buy 4 units of A and 2 units of B — minimum cost ₹160

Q4 — Training programmes — formulate and solve graphically

Question

A three-day programme on teaming and a two-day programme on problem solving. At least 8 teaming and at least 10 problem-solving programmes, and at least 25 in total. The consultant has 84 days. Teaming costs $10,000 and problem solving $8,000 per programme. Formulate and solve graphically.

Assignment 1 p. 1

Solution
  1. Step 1 — Formulate Let X₁ = number of teaming programmes, X₂ = number of problem-solving programmes Min Z = 10,000X₁ + 8,000X₂ X₁ ≥ 8 ← teaming requirement X₂ ≥ 10 ← problem-solving requirement X₁ + X₂ ≥ 25 ← total programmes 3X₁ + 2X₂ ≤ 84 ← consultant days (3 per teaming, 2 per problem solving) X₁, X₂ ≥ 0
  2. Step 2 — Graph
    010203010203040X₁ (teaming)X₂ (problem solving)X₁=8X₂=10X₁+X₂=253X₁+2X₂=84ABCD
    The feasible region is the four-sided area A-B-D-C; A is the cheapest corner
  3. Step 3 — Extreme points (solve each pair of lines) A: X₁ = 8 and X₁ + X₂ = 25 ⟹ (8, 17) B: X₁ = 8 and 3X₁ + 2X₂ = 84 ⟹ 2X₂ = 60 ⟹ (8, 30) C: X₂ = 10 and X₁ + X₂ = 25 ⟹ (15, 10) D: X₂ = 10 and 3X₁ + 2X₂ = 84 ⟹ 3X₁ = 64 ⟹ (21.33, 10)
    Corner(X₁, X₂)Z
    A(8, 17)$216,000 ← min
    B(8, 30)$320,000
    C(15, 10)$230,000
    D(21.33, 10)$293,333
  4. Step 4 — Answer Offer 8 teaming and 17 problem-solving programmes — minimum cost $216,000

    Check the days: 3(8) + 2(17) = 58 ≤ 84 ✓ — the day limit is not binding; the minimum-teaming and total-programme requirements are.

Q5 — The same problem, parts (a)–(d) — Consolidated Electronics

Question

The textbook version of Q4 with identical data: (a) formulate, (b) graph the feasible region, (c) find the coordinates of each extreme point, (d) solve for the minimum-cost solution.

Assignment 1 p. 2 — from Anderson & Sweeney, Ch. 2

Solution
  1. (a) Formulation Min Z = 10,000X₁ + 8,000X₂ X₁ ≥ 8, X₂ ≥ 10, X₁ + X₂ ≥ 25, 3X₁ + 2X₂ ≤ 84, X₁, X₂ ≥ 0

    Exactly as in Q4.

  2. (b) Feasible region

    The graph in Q4 above: bounded by X₁ = 8 on the left, X₂ = 10 at the bottom, X₁ + X₂ = 25 on the lower left and 3X₁ + 2X₂ = 84 on the upper right.

  3. (c) Extreme points A (8, 17) · B (8, 30) · C (15, 10) · D (21.33, 10)
  4. (d) Minimum-cost solution X₁ = 8, X₂ = 17, total cost $216,000

Q6 — Tom's salsa — derive the profit, then solve

Question

Western Foods Salsa = 50% whole tomatoes, 30% sauce, 20% paste; Mexico City Salsa = 70%, 10%, 20%. Each jar weighs 10 oz. Available: 280 lb whole tomatoes, 130 lb sauce, 100 lb paste at $0.96, $0.64, $0.56 per lb. Spices $0.10, jar $0.02, labelling $0.03 per jar. Revenue $1.64 (Western) and $1.93 (Mexico City) per jar. (a) Formulate to maximise profit contribution. (b) Find the optimal solution.

Assignment 1 p. 2 — Anderson & Sweeney, Ch. 2

Solution
  1. Step 1 — Ingredients per jar, in ounces
    Per 10-oz jarWhole tomatoesSaucePaste
    Western Foods (X₁)5 oz3 oz2 oz
    Mexico City (X₂)7 oz1 oz2 oz
    Available (lb × 16)280 × 16 = 4480 oz130 × 16 = 2080 oz100 × 16 = 1600 oz
  2. Step 2 — Profit contribution per jar Western: 5/16(0.96) + 3/16(0.64) + 2/16(0.56) = 0.30 + 0.12 + 0.07 = $0.49 ingredients + 0.10 + 0.02 + 0.03 = $0.64 cost ⟹ profit = 1.64 − 0.64 = $1.00 Mexico: 7/16(0.96) + 1/16(0.64) + 2/16(0.56) = 0.42 + 0.04 + 0.07 = $0.53 + 0.15 = $0.68 cost ⟹ profit = 1.93 − 0.68 = $1.25
  3. Step 3 — The model Max Z = 1.00X₁ + 1.25X₂ 5X₁ + 7X₂ ≤ 4480 ← whole tomatoes (oz) 3X₁ + X₂ ≤ 2080 ← tomato sauce (oz) 2X₁ + 2X₂ ≤ 1600 ← tomato paste (oz) X₁, X₂ ≥ 0
  4. Step 4 — Graph and corner points
    0200400600800200400600800X₁ (Western Foods jars)X₂ (Mexico City jars)tomatoessaucepasteOABCD
    All constraints are ≤, so the region includes the origin; B is optimal
    Corner(X₁, X₂)Z
    O(0, 0)$0
    A(0, 640)$800
    B(560, 240)$860 ← max
    C(640, 160)$840
    D(693.33, 0)$693.33
  5. Step 5 — Answer 560 jars of Western Foods Salsa and 240 jars of Mexico City Salsa — profit $860

    B is where the tomato and paste constraints cross: 5X₁ + 7X₂ = 4480 and X₁ + X₂ = 800 ⟹ 2X₂ = 480 ⟹ X₂ = 240, X₁ = 560. Sauce used: 3(560) + 240 = 1920 of 2080 oz — 160 oz spare.