Select the correct answer from each drop-down menu. Which pencil should Edward use? Edward is creating a new design for his late
1 answer:
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Answer:
Explanation:
C.1 Input-Output Model
It is a formal model that divides the economy into 2 sectors and traces the flow of inter-industry purchases and sales. This model was developed by Wassily Leontief in 1951. In simpler terms, the inter-industry model is a quantitative economic model that defines how the output of one industry becomes the input of another industrial sector. It is an interdependent economic model where the output of one becomes the input of another. For Eg: The Agriculture sector produces output using the inputs from the manufacturing sector.
The 3 main elements are:
Concentrates on an economy which is in equilibrium
Deals with technical aspects of production
Based on empirical investigations and assumptions
Assumptions
2 sectors - " Inter industry sector" and "final sector"
Output of one industry is the input for another
No 2 goods are produced jointly. i.e each industry produces homogenous goods
Prices, factor suppliers and consumer demands are given
No external economies or diseconomies of production
Constant returns to scale
The combinations of inputs are employed in rigidly fixed proportions.
Structure of IO model
See image 1
Quadrant 1: Flow of products which are both produced and consumed in the process of production
Quadrant 2: Final demand for products of each producing industry.
Quadrant 3: Primary inputs to industries (raw materials)
Quadrant 4: Primary inputs to direct consumption (Eg: electricity)
The model can be used in the analysis of the labor market, forecast economic development of a nation and analyze economic developments of various regions.
Leontief inverse matrix shows the output rises in each sector due to a unit increase in final demand. Inverting the matrix is significant since it is a linear system of equations with unique solutions. Thus, the final demand vector for the required output can be found.
C.2 Linear programming problems
Linear programming problems are optimization problems in which objective function and the constraints are all linear. It is most useful in making the best use of scarce resources during complex decision makings.
Primal LP, Dual LP, and Interpretations
Primal linear programming: They can be viewed as a resource allocation model that seeks to maximize revenue under limited resources. Every linear program has associated with it a related linear program called dual program. The original problem in relation to its dual is termed as a primal problem. The objective function is a linear combination of n variables. There are m constraints that place an upper bound on a linear combination of the n variables The goal is to maximize the value of objective functions that are subject to the constraints. If the primal linear programming has finite optimal value, then the dual has finite optimal value, and the primal and dual have the same optimal value. If the optimal solution to the primal problem makes a constraint into a strict inequality, it implies that the corresponding dual variable must be 0. The revenue-maximizing problem is an example of a primal problem.
Dual Linear Programming: They represent the worth per unit of resource. The objective function is a linear combination of m values that are the limits in the m constraints from the primal problem. There are n dual constraints that place a lower bound on a linear combination of m dual variables. The optimal dual solution implies fair prices for associated resources. Stri=ong duality implies the Company’s maximum revenue from selling furniture = Entrepreneur’s minimum cost of purchasing resources, i.e company makes no profit. Cost minimizing problem is an example of dual problems
See image 2
n - economic activities
m - resources
cj - revenue per unit of activity j
Answer:
The mass flow rate of steam m=5.4 Kg/s
Explanation:
Given:
At the inlet of turbine P=10 MPa ,T=500 C
AT the exit of turbine P=10 KPa ,x=0.9
Required power=5 MW
From steam table
<u> At 10 MPa and 500 C:</u>
h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)
<u>At 10 KPa:</u>
=2675.1 KJ/Kg,
=417.51 KJ/Kg
= 7.3 KJ/Kg-K ,
=1.3 KJ/Kg-K
So enthalpy of steam at the exit of turbine
h=
Now by putting the values
h= 417.51+0.9(2675.1- 417.51) KJ/Kg
h=2449.34 KJ/Kg
Lets take m is the mass flow rate of steam
So
m=5.4 Kg/s
So the mass flow rate of steam m=5.4 Kg/s