## Answers

a)

LP is following:

Let Xij = number of cars to be shipped from location i to j, where i,j = {1,2,3,4,5} for {Malatya,Antalya,Eskisehir,Ankara,Istanbul}

Min 100X13+90X14+121X23+110X24+139X25+113X34+78X35

s.t.

X13+X14 <= 1300

X23+X24+X25 <= 1900

X13+X23-X34-X35 = 0

X14+X24+X34 = 1400

X25+X35 = 1700

Xij >= 0

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b)

Balanced transportation problem is formulated as below:

Cost from/to | Eskisehir | Ankara | Istanbul | Dummy | Capacity |

Malatya | 100 | 90 | 1000 | 0 | 1300 |

Antalya | 121 | 110 | 139 | 0 | 1900 |

Eskisehir | 0 | 113 | 78 | 0 | 3100 |

Demand | 3100 | 1400 | 1700 | 100 |

Shipping cost of Malatya to Istanbul is written as 1000, which is hypothetically large to prohibit allocation to this cell, as shipment from Malatya to Istanbul is not allowed.

Total capacity (1300+1900=3200) is more than total demand (1400+1700=3100) . Therefore, a dummy demand node is created for balance 100 cars with shipping cost of 0.

Total demand and capacity of warehouse in Eskisehir is equal to total demand, i.e. 3100 cars

Basic Feasible Solution (bfs) is determined by northwest corner (NWC) method as follows:

In NWC method, we start with the Northwest Corner cell and allocate the maximum quantity possible and then move on to either the next cell to the right or below, depending upon the remaining demand or supply in that particular row or column. The process is repeated until all the demand is satisfied.

Resulting tableau showing the shipping quantity from each plant to warehouse to customers is following:

Eskisehir | Ankara | Istanbul | Dummy | Capacity | |

Malatya | 1300 | 0 | 0 | 0 | 1300 |

Antalya | 1800 | 100 | 0 | 0 | 1900 |

Eskisehir | 0 | 1300 | 1700 | 100 | 3100 |

Demand | 3100 | 1400 | 1700 | 100 |

Total cost = 1300*100+1800*121+100*110+1300*113+1700*78+100*0

= $ **638,300**