Learning from Operations Course: Bank Portfolio Management

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∑_(i=1)^2▒∑_(k=1)^6▒〖〖Cy〗_ik.y_ik+ ∑_(j=1)^4▒∑_(k=1)^6▒〖Cz〗_jk 〗〗.z_jk

Subject to Constraints:

∑_(j=1)^4▒x_ij + ∑_(k=1)^6▒〖y_ik≤〖Cap〗_i for a given i (i={1,2}); Factory capacity〗

∑_(i=1)^2▒〖x_ij≤Tj for a given j (j={1,2,3,4}); Depot Throughput〗

∑_(i=1)^2▒〖x_ij- ∑_(k=1)^6▒z_jk ≥0 for a given j (j={1,2,3,4}); Depot Throughput 〗

∑_(j=1)^4▒z_jk + ∑_(i =1)^2▒〖y_ik≥R_k 〗 for a given k (k={1,2,3,4,5,6})Customer requirement

xij,zjk,yik≥0

Solution:

Minimum Overall Cost: £ 198500.

Q.5. Agricultural Acreage Mix

a) Mathematical Model:

Decision Variables:

Xw: No. of Acres in which Wheat should be sown

Xc: No. of Acres in which Corn should be sown

Objective function: Maximize Profit (Revenue from sales – labor cost).

Maximize Z = [(30x5 Xw + 50x4 Xc) – (6Xw + 10Xc) x 10] = 90 Xw + 100 Xc

Subject to constraints:

Labour availability : 6Xw + 10Xc <= 350 hours

Acres available : Xw + Xc <= 45 acres

Wheat Sales : 5Xw <= 140 Bushels

Corn Sales : 4Xc <= 120 Bushels

Non-negativity : Xw , Xc >= 0

Solutions:

Excel Solver Solutions:

No of Acres in which Wheat should be sown : 25 Acres

No. of Acres in which Corn should be sown : 20 Acres

Labour Hours To be Purchased: 350

Maximum Profit the Farmer can Earn: Rs 4,250/-

Solutions based on Sensitivity report:

Farmers Maximum Profit from 40 Acres of Land: Rs. 3,875/-

Shadow Price for Land is Rs 75 per acre. Hence 5 acres reduction in Land will result in 5 x 75 i.e Rs 375/- loss. Therefore, Maximum Profit would be 4250 – 375 = Rs. 3875.

Maximum profit be if the price of wheat dropped to Rs. 26 per bushel: Rs 3,750/-

Allowable decrease of the Coefficient of Objective function is 30. If we reduce Rs. 30 to Rs. 26 the total decrease in Coefficient is (30-26) x 5 i.e 20. Hence the value of No of Acres of wheat doesn’t change (i.e Wheat – 25 acres). Therefore the reduction in the profit is Rs. 20 x25 = Rs. 500/-