Spencer Enterprises is attempting to choose among a series of new investment alternatives. The potential Investment alternatives, the net present value of the future stream of returns, the capitalrequirements, and the available capital funds over the next three years are summarized as follows.AlternativeNumber123456Capital funds availableAlternativeLimited warehouse expansionExtensive warehouse expansionTest market new productAdvertising campaign.Basic researchPurchase new equipmentNet PresentValue ($)3,5005,50010,0004,5007,5002,500Capital Requirements ($)Year 13,0002,5006,0002,0005,0001,00010,500Year 21,0003,5004,0001,5001,0005007,000Year 34,0003,5005,0001,8004,0009008,7501 if investment alternative / Is selected for i=1, 2, 3, 4, 5, 6.0 otherwiseAn Integer programming model for maximizing the net present value (in $) is developed. Letting x, =(a) Assume that only one of the warehouse expansion projects can be implemented. What additional constraint should be added to the integer programming model?(b) If new equipment is purchased, one of the warehouse expansion projects must be implemented. What additional constraint should be added to the integer programming model?10:29 AM

Objective Functions and Constraints: Based on the given details, the objective functions and constraints are Objective Function: Maximize Z = 3500 X1 + 5500 X2 + 10000 X3 + 4500 X4 + 7500 X5 + 2500 X6 X1 – Limited Warehouse Expansion X2 – Extensive Warehouse Expansion X3 – Test market new product X4 – Advertising Campaign X5 – Basic research X6 – Purchase new equipment Constraints: Subject to Year 1: 3000 X1 + 2500 X2 + 6000 X3 + 2000 X4 + 5000 X5 + 1000 X6 ≤ 10500 Year 2: 1000 X1 + 3500 X2 + 4000 X3 + 1500 X4 + 1000 X5 + 500 X6 ≤ 7000 Year 3: 4000 X1 + 3500 X2 + 5000 X3 + 1800 X4 + 4000 X5 + 900 X6 ≤ 8750 Xj = 1,0Find the Calculations below: Excel declarations: Solver Declarations: Results: So the selected alternatives are: X1 – Limited Warehouse Expansion - Not SelectedX2 – Extensive Warehouse Expansion - Not SelectedX3 – Test market new product - SelectedX4 – Advertising Campaign - SelectedX5 – Basic research - Not SelectedX6 – Purchase new equipment - SelectedPart a: Based on the selected and not selected alternatives we can add the below any one of the constraints. here the condition is adding the first 2 decision variable may provide 0 or 1 since the first 2 decision variable represents the warehouses. X1 + X2 ≤ 1 here both X1 and X2 is no selected so the result will be X1 + x2 = 0 and the constraint satisfies the results. Part b: New equipment is purchased and one of the warehouse expansion projects must be implemented: X6 = X1 + X2 the above constraint satisfies the following conditions. If any one of the warehouse projects is selected then new equipment will be selected (X6 = X1) or (X6 = X2) (both are 1). if no warehouse is selected then new equipment purchase is also not selected (Both are 0).