Question

Cornwall has been experiencing electricity shortages. The UK government is going to invest £1 billion (1 billion is 1000 million) over the next five years in an attempt to solve the Cornwall power shortage problem. It can invest in two types of power plants: Nuclear (N) and Hydroelectric (H). Every £1,000,000 invested into nuclear plants results in n megawatts capacity in two years (i.e. it takes two years to complete a nuclear plant). Every £1,000,000 invested into hydroelectric plants results in h megawatts in three years (i.e. it takes three years to complete a hydroelectric plant). A plant, regardless of type requires an operating budget in order to run. This operating budget makes up 20% of the cost of the plant. Operating budgets can only be used to keep the plant operational and cannot be used to start up new investments. However money from operating budgets and any surplus money can be invested into money markets (i.e. deposited into savings accounts at commercial banks) to accrue an interest at a rate of 10% per annum. In terms of power capacity, the UK currently has 50 megawatts of capacity and it can conserve energy to keep the demand at about 50 megawatts in the next two years. However, it needs 66, 70 and 78 megawatts of power in the third, fourth and fifth years respectively. Assume that once a plant is operational, it is always operational. Write down the linear program formulation that meets the increasing power demand and maximises the cash available to the UK government at the end of the fifth year.

Answer 1

solution:

Given data:

Cornwall has been experiencing electricity shortages.

Assume:

There is no demand after 5 years (No plants installed for 5 years onwards)

The instalment cost of a new plant for 2 or 3 years, issued the whole amount at a time at the beginning of the period

The annual operating cost is 20% of its instalment cost

Year 1 2 3 4 5 x1 x3 y1 50 y2 50 50 66 70 78 50 50 50 No. of Nuclear Plants Started No. of Hydro-plants Started Demand Current Capacity Net Demand New Investment for Nuclear Plant New Investment for Hydro Plant Operating cost of Nuclear Plant Operating cost of Hydro Plant 0 0 50 16 x3*1000000 20 28 x1*1000000 y1*1000000 x2*1000000 y2*1000000 (x1*1000000)*0.2 (x1+x2)*0.2 (y1*1000000) *0.2 (x1+x2+x3)*0.2 (y1+y2)*1000000*0.2 Net Availability 1000000000-x1*1000000 y1*1000000 1000000000 - (x1+ x2)*1000000 - (y1 + y2)* 1000000 1000000000 - (x1 + x2+x3)*1000000 - 1000000000-(x1 + x2+x3) * 1000000 - (y1 + 1000000000 - (x1 + x2+x3)*1000000 - (y1 + (y1 + y2)* 1000000 y2)*1000000 - (x1 + x2)*1000000*0.2-(y1 y2)*1000000 - (x1 + x2+x3)*1000000*0.2-(y1 (x1)*1000000*0.2-(y1)*1000000*0.2 + y2)*1000000*0.2 +y2)*1000000*0.2 Intrest From Availability (1000000000-x1*1000000 - y1*1000000) *0.1 (1000000000 - (x1+ x2)*1000000 - (y1 + y2)*1000000) *0.1 (1000000000 - (x1 + x2+x3)* 1000000 (y1 + y2)*1000000 - (x1)*1000000*0.2- (y1) * 1000000*0.2)*0.1 (1000000000 - (x1 + x2+x3) *1000000 - (y1 (1000000000 - (x1 + x2+x3)*1000000 - (y1 + + y2)*1000000 - (x1 + x2)*1000000*0.2 - y2)*1000000 - (x1 + x2+x3)*1000000*0.2-(y1 (y1 + y2)*1000000*0.2)*0.1 + y2)*1000000*0.2)*0.1

Demand Constraints:

(3rd year demand)

50 +21*n> 66

(4th year demand)

50 + (21 +22) *n+y1 * h > 70

(5th year demand)

50+ (21 +22 +23) * n + (y1 + y2) *h > 78

Fund Constraints:

(Total annual Operating Cost Constraint)

(21+2+3 +1+y2)+ 1000000+ (1+3+2^2+2+3 + 1) + 02 + 1000000 1000000000

macz = (1000000000-11 * 1000000 - Y1* 1000000)*0.1+(1000000000 (21 +22) * 1000000-(y1 + y2)* 1000000)*0.1+(1000000000-(x1+12+13) * 1000000-(y1+y2) * 1000000-(11)* 1000000*0.2-(41) * 1000000*0.2)*0.1+ (1000000000-(21+22+13) *1000000-(41+y2)* 1000000–21+12)* 1000000* 0.2-(y1+ y2) * 1000000*0.2) *0.1+(1000000000-(21+12+33) * 1000000 (41+y2) * 1000000-(11+12+13)*1000000*0.2-(41+y2)*1000000*0.2)*0.1

than you.........
please give me thumb up