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Chapter 3
Verify the solution
X1 coal in Boiler 1
X2 coal in Boiler 2
X3 oil in Boiler 1
X4 oil in Boiler 2
X5 gas in Boiler I
X6 gas in Boiler 2
Cost/Day:
Y = 25X1 + 25X2 + 27X3 + 27X4 + 29X5 + 29X6
1. Total steam:
0.80X1 + 0.6X2 + 0.82X3 + 0.65X4 + 0.84X5 + 0.76X6 = 150
2. Max capacity of Boiler 1:
0.8X1 + 0.82X3 + 0.84X5 < 100
3. Max capacity of Boiler 2:
0.6X2 + 0.65X4 + 0.76X6 < 90
4. Gas contract:
X5 -~- J(6 > - 120
5. Oil supply limitation:
X3 q- X4 < 20
6. Coal supply:
x1 +x2 _< 150
Solve for variables using linear programming for minimum cost.