The graphic solution method 2. It gives us a clear picture. This method can be used only to solve problems that involve two decision variables. However, most linear programming applications involve situations that have more than two decision variables, so the graphic approach is not used to solve these.
Example 1. Solving the micro-computer problem with graphic approach. Plot each of the constraints and identify its region. Identify the common region, which is all area that contains all of the points that satisfy the entire set of constraints.
Determine the optional solution-identify the point which lead to maximum benefit or minimum cost. It states that: For problems that have optional solutions, a solution will occur at an extreme, or corner point. Thus if a problem has a single optional solution, it will occur at a corner point.
If it has multiple optional solutions, at least one will occur at a corner point consequently, in searching for an optional solution to a problem, we need any consider the extreme points because one of those must be optional. Further, determining the value of the objective function at each corner point, we could identify the optional solution by selecting the corner point that has the best value i.
Extreme points represent interactions of constraints. Some times, this can be done by impaction observation and sometimes by simultaneous equation.
Substitute the value of the decision variables at each corner point into the objective function to obtain its value at each corner point. After all corner points have been so evaluated, select the one with the highest or lowest value depending on the optimization case. After we have got the optimal solution, we have to substitute the value of the decision variables into the constraints and check whether all the resources available are used or not. If there is any unused resources we can use it for any other purpose.
The amount of unused resource is known as slack- the amount of a scarce resource that is unused by a given solution. The slack can range from zero, for a case in which all of a particular resource is used, to the original amount of the resource that was available i. In the above cases, inspection time and storage space are binding constraints, while assembly time has slack.
Knowledge of unused capacity can be useful for planning. A manager may be able to use the remaining assembly time for other products, or, perhaps to schedule equipment maintenance, safety seminars, training sermons or other activities Interpretation: The company is advised to produce 9 units of type 1 micro computer and 4 units of type 2 micro computers per week to maximize its early profit to Br.
Example 2: Solving the diet problem with graphic approach. B 5 3 Simul. C 14 0 Observation 70 br. Interpretation to make the diet the minimum cost of br 49 we have to purchase 5 pounds of type 1 food and 3 pounds type 2 food. If there is a difference between the minimum required amount and the optimal solution, we call the difference surplus; that is: surplus is the amount by which the optimal solution causes a constraint to exceed the required minimum amount.
It can be determined in the same way that slack can: substitute the optimum values of the decision variables into the left side of the constraint and solve.
The difference between the resulting value and the original right-hand side amount is the amount of surplus. Surplus can potentially occur in a constraint. Through algebraic manipulation, the solution is improved until no further improvement is possible i. Each iteration moves one step closer to the optional solution.
The optimal solution to a linear programming model will occur at an extreme point of the feasible solution space. This is true even if a model involves more than two variables; optsmal solutions will occur at these point of the feasible solution space; some will be outside of the feasible solution space. Hence, not every solution will be a feasible solution.
Solutions which represent infasseetwim of constraints are called basic solutions; those which also satisfy all of the constraints, including the non-negativity constraints, are called basic feasible solutions. The simplex method is an algebraic procedure for systematically examining basic feasible solutions. The simplex procedure for a maximization problem with all constraints consists of the following steps.
Write the LPM in a Standard form: When all of the constraints are written as equalities, the LP program is said to be in a standard form. We convert the LPM in to a standard form by applying the slack variables, s, which carries a subscript that denotes which constraint it applies to.
For example, s1 refers to the amount of slack in the first constraint, S2 to the amount of slack in the second constraint, and so on. Further more, every variable in a model must be represented in the objective function. However, since slack does not provide any real contribution to the objective, each slack variable is a assigned a coefficient of zero in the objective function. Develop the initial tableau a.
List the variables across the top of the table and write the objective function coefficient of each variables just above it. There should be one row in the body of the table for each constraint. List the slack variables in the basis column, one per row. In the Cj column, enter the objective function coefficient of zero for each slack variable. Compute values for row Zj. Computer values for Cj — Zj. Develop subsequent tables 3.
However do not divide by a zero or negative value. The smalls non negative ratio that results indicate which variable will leave the solution 4. Find unique vectors for the new basic variable using row operations on the pivot element.
Compute Cj — Zj row 6. If all Cj — Zj Values are zeros and negatives, you have reached optimality 7. If this is not the case step 6 , repeat 2 to 5 until you get optional solution. Further, note that a zero appears in row C — Z in every column whose variable is in solution, in row C — Z in every column whose variable is in solution, indicating that its maximum contribution to the objective function has been realized.
Example 2 A manufacture of lawn and garden equipment makes two basic types of lawn mowers; a push type and a self propelled model. The push type require 9 minutes to assemble and 2 minutes to package; the self-propelled mover requires 12 minute to assemble and 6 minutes to package.
Each type has an engine. The company has 12hrs of assembly time available, 75 engines, and 5hrs of packing time profits are Birr 70 for the self propelled model and br 45 for the push type mower per unit. To determine how many units of each type of mower to produce so as to maximize profit. The company is advised to produce 24 units of push type mower and 40 units of self-propelled mowers so as to realize a profit of Br.
The table below indicates the number of hours a unit of each product requires in the different departments and the number of pounds of raw materials required. Also listed are the cost per unit, selling price, and weekly capacities of both work-hours and raw materials.
If the objective is to maximize total weekly profit, formulate the linear programming model. Solve the following linear programming problem a Use simplex method Z max.
P X1 X2 Value of O. The different between money now and the same money in the future is called interest. Interest have a wide spread influence over decisions made by businesses and every of us in our personal lives. Therefore, the basic objective of this unit is to discuss interest rates and their effects on the value of money. Specifically, it covers simple interest, compound interest, annuity and mortgage problems. A savings institution Banks pay interest to depositors on the money in the savings account since the institutions have use of those funds while they are on deposit.
Interest is usually computed as percentage of the principal over a given period of time. This is called interest rate. Interest rate specifies the rate at which interest accumulates per year through out the term of the loan. Interests are of two types: simple interest and compound interest. In the first part of this unit we shall explore these two concepts.
The sum of the original amount principal and the total interest is the future amount or maturity value or in short amount. Simple interest generally used only on short-term loans or investments — offen of duration less than one year. Simple interest is given by the following formula. Find the total simple interest and the maturity value of the loan. That is if r is expressed as a percentage per year, t also should be expressed in number of years number of months divided by 12 if time is given as a number of months.
Example 2 How long will it take if Br. At what interest rate will Br. Example 5. Find the Interest on Br. Try Your Self: At what interest rate you should invest Br. In such a case, the interest is said to be compounded. The result of compounding interest is that starting with the second compounding period, the account earns interest on interest in addition to earning interest on principal during the next payment period.
Interest paid on interest reinvested is called compound interest. The difference between the compound amount and the original principal is the compound interest. The compound interest method is generally used in long-term borrowing unlike that of the simple interest used only for short-term borrowings.
The time interval between successive conversions of interest into principal is called the interest period, or conversion period, or Compounding period, and may be any convenient length of time. The interest rate is usually quoted as an annual rate and must be converted to appropriate rate per conversion period for computational purposes.
What are the compound amount and compound interest at the end of one year? This means interest will be computed at the end of each three month period and added in to the principal.
Among the possible alternatives are: 1. Use a hand-held calculator with a Yx function key. This is the procedure most often used. Let us illustrate this alternative. Example 2. Find the compound amount and compound interest after 10 years if Br. When a number of conversion period within a year increases, the interest earned also increases continuously toward an upper limit. The limiting case occurs where interest is compounded continuously.
Example 3. How long it take to accumulate Br. Under these conditions, p is called the present value of A. This process is called discounting and the principal is now a discounted value of future income A. Example 4. In other words, the effective rate r converted m times a year is the simple interest rate that would produce an equivalent amount of interest in one year.
It is denoted by re. We call this simple interest rate the effective rate. Which investment is better investment, assuming all else equal? Solution Nominal rate with different compounding periods cannot be compared directly. We must find the effective rate of each nominal rate and then compare the effective rates to determine which investment will yield the larger return. Example 6. The payments may be made weekly, monthly, quarterly, annually, semiannually or for any fixed period of time.
The time between successive payments is called payment period for the annuity. If payments are made at the end of each payment period, the annuity is called an ordinary annuity. If payment is made at the beginning of the payment period, it is called annuity due. In this course we will discuss only ordinary annuities.
The amount, or future value, of an annuity is the sum of all payments plus the interest earned during the term of the annuity. The term of an annuity refers to the time from the begging of the first payment period to the end of the last payment period. In an ordinary annuity the first payment is not considered in interest calculation for the first period because it is paid at the end of the first period for which interest is calculated.
Similarly, the last payment does not qualify for interest at all since the value of the annuity is computed immediately after the last payment is received.
Future value Amount of an ordinary annuity. What is the amount of an annuity if the size of each payment is Br. The last payment accumulate no interest, the next to last payment accumulates one period for interest.
Then subtracting the first equation eq. Deposits Br. X have accumulated just after 15th deposit? A person deposits Br.
After the four years, the person leaves the account untouched for an additional six years. What is the balance after the 10 year period?
Therefore, the balance after 10 years is Br. Sinking funds are generally established in order to satisfy some financial obligations or to reach some financial goal.
If the payments are to be made in the form of an ordinary annuity, then the required periodic payment into the sinking fund can be determined by reference to the formula for the amount of an ordinary annuity.
Ato Ayalkebet has a savings goal of Br. During the first 5 years he is financially able to deposit only Br. What must his quarterly deposit over the last 10 ten years be if he is to reach his goal? Solution: 1. This sum Br. It represents the amount that must be invested now to purchase the payments due in the future.
The present value of an annuity can be computed in two ways: Discounting all periodic payments to the present beginning of the term individually or Discounting the future value amount of an annuity to the beginning of the term Example 6. What is the present value of an annuity if the size of each payment is Br. Equivalently we may find the future value of the ordinary annuity using the formula and then discount it to the present taking it as a single future value.
What is the present value of an annuity that pays Br. After the last payment, the obligation ceases to exist it is dead and it is side to have been amortized by the periodic payments.
Prominent examples of amortization are loans taken to buy a car or a home amortized over periods such as 5, 10, 20 or 30 years. Ato Elias borrowed Br.
How much should each annual payment be in order to retire the debt including the interest in 10 years. Solution 0. Answer: Br. An employee has contributed with her employer to a retirement plan for 20 years a certain amount twice a year. Semiannual payments will be made for 10 years to the employee of her family in the event of her death. What semi annual payment should she made? What semi annual payment should be made for her or her family? How much interest will be earned on Br.
The buyer amortizes the indebtedness by periodic payments over a period of time. Typically payments are monthly and the time period is long such as 30 years, 25 years and 20 years. Mortgage payment and amortization are similar. In mortgage payments m is equal to 12 because the loan is repaid from monthly salary or Income, but in amortization money take other values. Therefore, in mortgage payments we are interested in the determination of monthly payments.
Find the periodic payment 2. Find the interest charged. He made an amount of down payment and pay monthly Br. Find the mortgage, down payment, interest charged and percentage of the down payment to the selling price. If you borrow Br. Find the simple interest and the maturity value of the loan.
What is the present value of a loan that will amount to Br. What would be the amount of compound interest after 10 years? A small boy at the age of 10 drops 0. If he makes the deposit without interruption, how much will the boy have at the age of Hiwot deposits Br. How much is the account by the end of the time period considered?
The Company plans to set up a sinking fund to accumulate the amount required to settle the debt. Required: A. If Br. Semi Annually C. Quarterly D. What can you observe from your answers in A, B and C What is the amount after 5 years? Solution First 1. Solution: Nominal rates with amount compounding periods in this case quarterly and monthly cannot be compared directly.
The major areas of study within the calculus are differential calculus and integral calculus. Differential calculus focuses on rates of change in analyzing a situation. Integral calculus involves summation of a special type.
Graphically the concepts of area in two dimensions or volume in three dimensions are important in integral calculus.
The goal in this chapter is to provide an appreciation for what the calculus is and where it can be applied. Though it would take several semester of intensive study to understand most of the finer points of the calculus, your coverage will enable you to understand the tools for conducting analysis at an elementary level. Calculus is a mathematical tool used to solve problems in business, Economics and other areas. It broadens the concept of slope.
A set of rules of differentiation exists for finding the derivatives of many common functions. The mathematics involved in providing these rules can be finally complicated. For our purposes it will suffice the rules without proof. The rules of differentiation apply to functions, which have specific structural characteristics. A rule will state that if a function has specific characteristics, then the derivative of the function will have resulting form.
Each function can be graphed and that the derivative is a general expression for the slope of the function. Derivative of a constant function. The slope at all points along such function equals 0.
Sum or differences of functions. This implies that the derivative of function formed by the sum difference of two or more component functions is the sum difference of the derivatives of the component functions. Other rules 5. Linear cost functions assume that the variable cost per unit is constant for such functions the marginal cost is the same at any level of output. A non-linear cost function is characterized by variable marginal costs. It is the rate of change in total cost per unit change in production at an out put level of X unit.
It is also an optimization to the actual cost of making one more unit at any production level X non-linear functions. In Other words, the cost of producing 1 more unit at this level of production 40 is approximately 5, Br.
If each unit of a product sells at the same price, the MR is always equal to the price. Marginal revenue for non-linear total revenue function is not constant. As long as the additional revenue brought in by the next unit exceeds the cost of producing and selling that unit, there is a net profit from producing and selling that unit and total profit increases.
The marketing department also presents the following demand equation. If in turn the derivative function itself has a derivative for points in that interval this new derivative function is called the second derivative of the original function f or the first derivative of f. Intuitively the 2nd derivative test attempts to determine the concavity of the function at a critical point. The slope of the line is tangent to a curve. A company manufactures and sales X units of transistor radios per week.
If there is non-linear function, we can use the 2nd derivative test. When X gallons of alcohol are produced, the average cost per gallons is given by the following function. The roof and floor areas will be of square feet for any building, but the total wall length will vary for different dimensions. Required: Find the dimensions that minimize the total amount of wall. So we have a minimum value. Fence is required on three sickles of a rectangular plot.
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