# Delectation Delicatessen

Delectation Delicatessen

Problem a)

Answer:

Delectation wants to determine the number of Fish-n-Fowl and Surf-n-Turf sandwiches it should make each day to maximize the deli’s revenues in USD$. The decision variables are how many units of Fish-n-Fowl and Surf-n-Turf sandwiches Delectation should make each day. F= units of Fish-n-Fowl sandwiches Delectation should make each day. S= units of Surf-n-Turf sandwiches Delectation should make each day. The objective function: maxmize F x 5.50 + S x 6.95

The Constraints:

0.25 x F + 0.25 x S<=200| Tuna fish salad | Pound|

0.5 x F <=260| Sliced turkey| Pound|

0.40 x S <=60| Sliced roast beef| Pound|

2 x F + 1 x S<=1500| Rye bread| Slice|

1 x F + 1 x S<=800| Pumpernickel break| Slice|

F and S>=0;| | F and S are both integers | |

Problem b)

Answer:

Conclusion: Delectation should produce 520 units of Fish-n-Fowl and 150 units of Surf-n-Turf sandwiches Delectation each day to make $3,902.50 revenue. Power Plan problem

Problem a)

Answer:

The management of OMPIC wants to determine the mix of coals that will maximize the amount of power produced per hour. The decision variables are how many tons of Hard Coal and Soft Coal should be used to maximize the amount of power in BTU per hour while complying the with the emission standards. H= tons of Hard Coal should be used per hour. S= tons of Soft Coal should be used per hour. The objective function: maximize H x 24,000 + S x 20,000

The Constraints:

SO2 emission-H x 1500 +S x 500 <=0 PPM Particulate emission H x 0.65 + S x 1.05 <= 12 kg Pulverizer capacity H x 24+S x 16<= 384 ton/hour Conveyor capacity H+S<= 20 ton/hour No negative input H and S>=0 Problem b)

Answer:

The objective function: assuming unit price =1, minimize H x 1.15 +S x 1 H= tons of Hard Coal should be used per hour. S= tons of Soft Coal should be used per hour. The Constraints:

Heat output H x 24,000+S x 20,000=390,000 BTU/hour SO2 emission-H x 1500 +S x 500 <=0 PPM Particulate emission H x 0.65 + S x 1.05 <= 12 kg/hour Pulverizer capacity H x 24+S x 16<= 384 ton/hour Conveyor capacity H+S<= 20 ton/hour No negative input H and S>=0 Gaussian Laminates company

Problem a) losing 100 units of production capacity at plant A Answer:

It will increase the total operating cost by $25 assuming others remain constant. Shadow price of “from plant A production qty” is -$0.25. So reducing product by 100 units will increase cost by 100 x $0.25 = $25. The production and shipping plan will change. Plan A will now produce 1900 units or 100 units less. Plant A will ship 1000 units or 100 units less to region 1. Plant B will produce 1900 units, 100 units more. Plan B will ship 1100...

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