Answer:
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Answer:
Assume: only one of the ware house expansion projects can be implemented and suppose that, if test marketing of the new product is carried out, the advertising campaign also must be conducted and vice versa. Also suppose that the purchase of new equipment cannot be undertaken unless the basic research or the extensive warehouse expansion are implemented.
Formulate the corresponding model for maximizing the net present value subject to the restrictions stated above
Yi (1 if investment alternative is selected, o otherwise)
i = 1, 2, 3, 4, 5, 6
Max Z = $4,000y1 + 6,000y2 + 10,500y3 + 4,000y4 + 8,000y5 + 3,000y6
∴ NPV = 17,500
Answer:
≈ 9644 quantity of card
Explanation:
given data:
n = 4 regions/areas
mean demand = 2300
standard deviation = 200
cost of card (c) = $0.5
selling price (p) = $3.75
salvage value of card ( v ) = $ 0
The optimal production quantity for the card can be calculated using this formula below
= <em>u</em> + z (0.8667 ) * б
= 9200 + 1.110926 * 400
≈ 9644 quantity of card
First we have to find <em>u</em>
u = n * mean demand
= 4 * 2300 = 9200
next we find the value of Z
Z = ( )
= ( 3.75 - 0.5 ) / 3.75 = 0.8667
Z( 0.8667 ) = 1.110926 ( using excel formula : NORMSINV (0.8667 )
next we find б
б = 200 = 400