Skip to content
Worked Examples · Example 4
Q.

There is a factory located at each of the two places P and Q. From these locations, a certain commodity is delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are 8 and 6 units respectively. The cost of transportation per unit is given below:

From/ToABC
P161015
Q101210

(Costs in Rs.)

How many units should be transported from each factory to each depot in order that the transportation cost is minimum? Formulate the above as a linear programming problem.

Sikkim CbseNCERTSubjective· 3mImportance★★★★★
18% · 4/22 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The book reduces this two-factory, three-depot transportation problem to just two decision variables — xx = units sent P→A and yy = units sent P→B — by writing every other shipment in terms of xx and yy from the supply and demand totals. The result is the printed formulation: Minimize Z=x−7y+190Z = x - 7y + 190 subject to x+y≤8x+y\le 8, x+y≥4x+y\ge 4, x≤5x\le 5, y≤5y\le 5, x,y≥0x,y\ge 0.

Setting up with two variables

Total supply is 8+6=148+6=14 units and total demand is 5+5+4=145+5+4=14 units, so the problem is balanced — every unit produced is shipped and every depot is served exactly. Because everything is balanced, once we fix how much P sends to A and to B, all six shipments are determined.

Let

  • xx = units transported from factory P to depot A,
  • yy = units transported from factory P to depot B.

Then, using the capacities and requirements:

RouteUnitsReasoning
P → Axxdecision variable
P → Byydecision variable
P → C8−x−y8 - x - yP produces 8 units in all
Q → A5−x5 - xA needs 5; the rest of A's demand comes from Q
Q → B5−y5 - yB needs 5; the rest comes from Q
Q → Cx+y−4x + y - 4Q produces 6; 6−(5−x)−(5−y)=x+y−46-(5-x)-(5-y)=x+y-4

The constraints

Every shipment must be non-negative:

  • 8−x−y≥0⇒x+y≤88 - x - y \ge 0 \Rightarrow x + y \le 8
  • 5−x≥0⇒x≤55 - x \ge 0 \Rightarrow x \le 5
  • 5−y≥0⇒y≤55 - y \ge 0 \Rightarrow y \le 5
  • x+y−4≥0⇒x+y≥4x + y - 4 \ge 0 \Rightarrow x + y \ge 4
  • and x≥0, y≥0x \ge 0,\ y \ge 0.

The objective function

Multiply each shipment by its per-unit cost (from the table: P→A ₹16, P→B ₹10, P→C ₹15, Q→A ₹10, Q→B ₹12, Q→C ₹10) and add:

Z=16x+10y+15(8−x−y)+10(5−x)+12(5−y)+10(x+y−4).Z = 16x + 10y + 15(8 - x - y) + 10(5 - x) + 12(5 - y) + 10(x + y - 4).

Expanding and collecting like terms:

Z=(16−15−10+10)x+(10−15−12+10)y+(120+50+60−40)=x−7y+190.Z = (16 - 15 - 10 + 10)x + (10 - 15 - 12 + 10)y + (120 + 50 + 60 - 40) = x - 7y + 190. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.