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evablogger [386]
1 year ago
11

assuming complementary inputs are available, the minimum number of transistors needed to realize a two input xor gate is:

Engineering
1 answer:
babunello [35]1 year ago
4 0

Assuming complementary inputs are available, the minimum number of transistors needed to realize a two input xor gate is 4.

If two inputs work well together in the production process, they are said to be complements. Since the inputs are used in fixed proportions, if the price of one input lowers, more of it will be sought after, increasing the demand for the other input.

In this study, a model of anticompetitive exclusive contracts with complimentary inputs is built. A downstream company converts a variety of complementary inputs into finished goods. When complementary input providers hold a dominant position in a specific input market, upstream competition price within that market benefits both the downstream firm and the complementary inputs providers by driving up complementary input prices.

Thus, even when entry that is socially optimal is permitted, the downstream firm is unable to generate higher profits. Therefore, even in the absence of scale economies, downstream competition, and relationship-specific investment, the inefficient incumbent provider can prevent socially efficient entrance by employing exclusive contracts.

To know more about complementary inputs click on the link:

brainly.com/question/13125554

#SPJ4

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A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
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Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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