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There are two factories located one at P and the other at Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 4, 4 and 6 units of the commodity while the production capacity of the factories at P and Q are 9 and 5 units respectively. The cost of transportation per unit is given as : | From / To | A | B | C | |---|---|---|---| | P | 160 | 100 | 150 | | Q | 100 | 120 | 100 | Based on the above information, answer the following questions : Let xx units and y units of the commodity be transported from factory P to the depots at A and B respectively, then The flow network has Factory PP (supply 99 units) at the top and Factory QQ (supply 55 units) at the bottom, each supplying three outlets AA (demand 44), BB (demand 44) and CC (demand 66); xx units are sent from PP to AA and yy units from PP to BB. (i) Find (in terms of xx and y) how many units commodity be transported from factory P to depot C. (ii) Find how many units of commodity be transported from factory Q to A, B and C respectively. (iii) Using (i) and (ii), find the total transportation cost z. OR (iii) Using (i) and (ii), find the constraint inequalities for minimum cost z.

CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★
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Balancing supply and demand: P→C =9−x−y=9-x-y, Q→A =4−x=4-x, Q→B =4−y=4-y, Q→C =x+y−3=x+y-3. Total cost simplifies to z=10x−70y+1930z = 10x - 70y + 1930, subject to x+y≤9, x+y≥3, x≤4, y≤4, x,y≥0x+y\le9,\ x+y\ge3,\ x\le4,\ y\le4,\ x,y\ge0.

In a balanced transportation problem each factory ships out its full capacity and each depot receives exactly its requirement; the total cost z=∑(units×unit cost)z = \sum (\text{units}\times\text{unit cost}) over all routes.

Given capacities P = 9, Q = 5; requirements A = 4, B = 4, C = 6. Let P→A =x= x, P→B =y= y.

  1. Units P → C. P ships 9 in total: P→C=9−(x+y)\text{P}\to\text{C} = 9 - (x + y).
  2. Units from Q. Each depot's demand is met by P and Q together:
  1. Q→A =4−x= 4 - x (A needs 4, receives xx from P).
  2. Q→B =4−y= 4 - y (B needs 4, receives yy from P).
  3. Q→C =6−[9−(x+y)]=x+y−3= 6 - \big[9 - (x+y)\big] = x + y - 3 (C needs 6, receives 9−x−y9-x-y from P).

(iii) Total transportation cost. Using the unit-cost table:

RouteUnitsCost/unitCost
P→Axx160160x160x
P→Byy100100y100y
P→C9−x−y9-x-y150150(9−x−y)150(9-x-y)
Q→A4−x4-x100100(4−x)100(4-x)
Q→B4−y4-y120120(4−y)120(4-y)
Q→Cx+y−3x+y-3100100(x+y−3)100(x+y-3)
  1. z=160x+100y+150(9−x−y)+100(4−x)+120(4−y)+100(x+y−3)z = 160x + 100y + 150(9-x-y) + 100(4-x) + 120(4-y) + 100(x+y-3).
  2. Expand: =160x+100y+1350−150x−150y+400−100x+480−120y+100x+100y−300= 160x + 100y + 1350 - 150x - 150y + 400 - 100x + 480 - 120y + 100x + 100y - 300. …

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