Skip to content
Exercises · Q4
Q.

A company has three plants O1,O2,O3O_1, O_2, O_3 with per-unit transportation costs to three destinations D1,D2,D3D_1, D_2, D_3 as given below. Supplies are 20, 30 and 25 units; demands at D1,D2,D3D_1, D_2, D_3 are 30, 25 and 10 units.

D1D_1D2D_2D3D_3
O1O_1596
O2O_2847
O3O_3673
  1. Show that the problem is unbalanced and convert it into a balanced transportation problem.
  2. Find the initial basic feasible solution using the Least Cost Method on the balanced table, and compute the total transportation cost.
Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
3% · 1/38 Questions
✓ Free question

(i) Balancing: total supply =20+30+25=75=20+30+25=75; total demand =30+25+10=65=30+25+10=65. Since supply exceeds demand by 75−65=1075-65=10, add a dummy destination D4D_4 with demand =10=10 and cost 0 in every row:

D1D_1D2D_2D3D_3D4D_4 (dummy)Supply
O1O_1596020
O2O_2847030
O3O_3673025
Demand3025101075 / 75

The table is now balanced (75=7575=75).

(ii) Least Cost Method:

  1. Lowest cost is 0, tied among the dummy cells; allocating at (O2,D4)(O_2,D_4) (largest supply among the tied rows) clears the dummy column fastest: allocate min⁡(30,10)=10\min(30,10)=10. D4D_4 exhausted; O2O_2 has 20 left.
  2. Next lowest is 3 at (O3,D3)(O_3,D_3): allocate min⁡(25,10)=10\min(25,10)=10. D3D_3 exhausted; O3O_3 has 15 left.
  3. Next lowest is 4 at (O2,D2)(O_2,D_2): allocate min⁡(20,25)=20\min(20,25)=20. O2O_2 exhausted; D2D_2 has 5 left.
  4. Next lowest is 5 at (O1,D1)(O_1,D_1): allocate min⁡(20,30)=20\min(20,30)=20. O1O_1 exhausted; D1D_1 has 10 left.
  5. Next lowest is 6 at (O3,D1)(O_3,D_1): allocate min⁡(15,10)=10\min(15,10)=10. D1D_1 exhausted; O3O_3 has 5 left.
  6. Final cell (O3,D2)(O_3,D_2): allocate min⁡(5,5)=5\min(5,5)=5.

Allocation table:

D1D_1D2D_2D3D_3D4D_4Row total
O1O_120———20
O2O_2—20—1030
O3O_310510—25
Column total3025101075

Total transportation cost (the dummy allocation carries zero cost):

Z=10(0)+10(3)+20(4)+20(5)+10(6)+5(7)=0+30+80+100+60+35=305Z=10(0)+10(3)+20(4)+20(5)+10(6)+5(7)=0+30+80+100+60+35=305

✓Final answer

Total transportation cost = ₹305; the 10 units routed to the dummy destination via O2O_2 represent O2O_2's surplus supply that is never actually shipped

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.