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A brick manufacturer has two depots, A and B with stocks of 30,000 and 20,000 bricks respectively. He receives orders from three builders P, Q and R for 15,000, 20,000 and 15,000 bricks respectively. The cost in Rs. of transporting 1,000 bricks to the builders from the depots are given below:

From \ ToPQR
A402030
B206040

How should the manufacturer fulfil the orders so as to keep the cost of transportation minimum? Formulate the above problem as linear programming problem.

Yanam CbseNCERTSubjective· 3mImportance★★★★★
73% · 16/22 Questions
✓ Free question

Minimise Z=40xAP+20xAQ+30xAR+20xBP+60xBQ+40xBRZ = 40x_{AP} + 20x_{AQ} + 30x_{AR} + 20x_{BP} + 60x_{BQ} + 40x_{BR} (Rs., quantities in thousands) subject to the supply and demand constraints below. The optimal despatch ships A to Q 20,000, A to R 10,000, B to P 15,000 and B to R 5,000, giving a minimum transport cost of Rs. 1,200.

Setting up the LPP

Let xijx_{ij} be the number of thousand bricks sent from depot i∈{A,B}i\in\{A,B\} to builder j∈{P,Q,R}j\in\{P,Q,R\}, with each xij≥0x_{ij}\ge 0.

Objective - minimise total cost (Rs.):

Z=40xAP+20xAQ+30xAR+20xBP+60xBQ+40xBRZ = 40x_{AP} + 20x_{AQ} + 30x_{AR} + 20x_{BP} + 60x_{BQ} + 40x_{BR}

Supply constraints (thousands):

xAP+xAQ+xAR=30,xBP+xBQ+xBR=20x_{AP}+x_{AQ}+x_{AR} = 30, \qquad x_{BP}+x_{BQ}+x_{BR} = 20

Demand constraints (thousands):

xAP+xBP=15,xAQ+xBQ=20,xAR+xBR=15x_{AP}+x_{BP} = 15, \qquad x_{AQ}+x_{BQ} = 20, \qquad x_{AR}+x_{BR} = 15

Total supply =30+20=50=30+20=50 equals total demand =15+20+15=50=15+20+15=50, so the problem is balanced and each constraint holds with equality.

Solving

Put xAP=xx_{AP}=x and xAQ=yx_{AQ}=y and eliminate the other four variables using the constraints. The objective reduces to

Z=1800+30x−30y=1800+30(x−y),Z = 1800 + 30x - 30y = 1800 + 30(x-y),

subject to 0≤x≤150\le x\le 15, 0≤y≤200\le y\le 20 and 15≤x+y≤3015\le x+y\le 30. To minimise ZZ we make xx as small and yy as large as possible: x=0, y=20x=0,\ y=20 (feasible, since x+y=20x+y=20).

RouteBricksCost per 1000 (Rs.)Cost (Rs.)
A - Q20,00020400
A - R10,00030300
B - P15,00020300
B - R5,00040200
Total1,200

Depot A despatches its whole stock (20,000 to Q, 10,000 to R) and depot B its whole stock (15,000 to P, 5,000 to R); every order is met exactly.

✓Final answer

LPP: Minimise Z=40xAP+20xAQ+30xAR+20xBP+60xBQ+40xBRZ = 40x_{AP} + 20x_{AQ} + 30x_{AR} + 20x_{BP} + 60x_{BQ} + 40x_{BR} subject to xAP+xAQ+xAR=30x_{AP}+x_{AQ}+x_{AR}=30, xBP+xBQ+xBR=20x_{BP}+x_{BQ}+x_{BR}=20, xAP+xBP=15x_{AP}+x_{BP}=15, xAQ+xBQ=20x_{AQ}+x_{BQ}=20, xAR+xBR=15x_{AR}+x_{BR}=15, all xij≥0x_{ij}\ge 0. Optimal despatch: A to Q 20,000, A to R 10,000, B to P 15,000, B to R 5,000, for a minimum transport cost of Rs. 1,200.

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