Explanation Given: There are three types of investment named as high risk into oil leases, medium risk in bonds and safe stock investment are 15 % , 9 % , 5 % , respectively and invest $ 50 , 000 , because of the risk investor limit the investment on oil leases and bonds to 30 % of the investment further, it limits the investment on oil leases and stock to 50 % . Formula used: Calculation: Consider the maximization problem, first formulate the problem in the system of linear equations. Take, x 1 be the amount of investment in oil leases then take, x 2 be the amount of investment in stock bonds further, take, x 3 be the amount of investment in safe stock. Take, z be the objective function to maximize as the profit is given high risk into oil leases, medium risk in bonds and safe stock investment are 15 % , 9 % , 5 % . So, the objective function is: z = .15 x 1 + .09 x 2 + .05 x 3 From all three amounts of investment is $ 50000 , so the constraints are: x 1 + x 2 + x 3 ≤ 50000 From the amounts of investment in oil leases and bonds is 30 % of $ 50000 which is: 50000 × 30 100 = 15000 So, the constraints are: x 1 + x 2 ≤ 15000 From the amounts of investment in oil leases and the stock is 50 % of $ 50000 , which is: 50000 × 50 100 = 25000 So, the constraints are: x 1 + x 3 ≤ 25000 Therefore, formulate the problem which is: Maximize, z = .15 x 1 + .09 x 2 + .05 x 3 Subject to, x 1 + x 2 + x 3 ≤ 50000 , x 1 + x 2 ≤ 15000 , x 1 + x 3 ≤ 25000 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Use two stages method to solve the maximization problem. Then, there are various steps which must follow: Step (1): Modify each equation constraint by an equivalent pair of inequality constraint Step (2): Write each constraint with a positive constant. Step (3): If minimization problem then converts into maximization by replace z = − w . Step (4): Add or subtract surplus variables according to the need of the constraints into equations. Step (5): Write initial simplex table. Step (6): Find a basic feasible solution if a solution exists, which is stage I. Step (7): If basic feasible solution found then do the simplex method further which is a stage and get the optimal solution. Further, to convert linear inequality into the linear equation then take a slack variable and surplus variable which is done by adding and subtracting a non-negative variable to each constraint. If linear equality has less than or equal to sign then, use slack variable by adding as the equation, which is less than or equal to the value if it is equal then the slack variable is zero. If linear equality has greater than or equal to sign then, use surplus variable by subtracting as the equation is greater than or equal to the value, if it is equal then the surplus variable is zero which is: x 1 + x 2 + x 3 + s 1 = 50000 , x 1 + x 2 + s 2 = 15000 , x 1 + x 3 + s 3 = 25000 where s 1 , s 2 , s 3 are greater than or equal to zero Then, rewrite constraints by adding or subtracting the non-negative slack and surplus variables which are: Maximize, z = .15 x 1 + .09 x 2 + .05 x 3 Subject to, x 1 + x 2 + x 3 + s 1 = 50000 , x 1 + x 2 + s 2 = 15000 , x 1 + x 3 + s 3 = 25000 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 Then, convert the linear programming into a system of linear equations in which all the coefficient of variables is written on the left of the equal sign and the constants are on the right then convert the objective function into linear equations which are: − .15 x 1 − .09 x 2 − .05 x 3 + z = 0 Then, write the constraint and objective function into the augmented matrix which is in the form of a simplex table: x 1 x 2 x 3 s 1 s 2 s 3 [ 1 1 1 1 0 0 50000 1 1 0 0 1 0 15000 1 0 1 0 0 1 25000 − .15 − .09 − .05 0 0 0 0 ] Here, the last row shows the coefficients of the objective function and last columns shows the constants. The basic variables correspond to columns which have an entry of 1 and rest is zero its value corresponds to the last column which is constant, and the non-basic variables correspond to the other columns and its value is zero. In the above nonbasic variables are x 1 , x 2 , x 3 and the basic variables are s 1 , s 2 , s 3 . Here, x 1 = 0 , x 2 = 0 , x 3 = 0 , s 1 = 50000 , s 2 = 15000 , s 3 = 25000 Above simplex, the table has a basic feasible solution as the basic variable has no negative value so it is said to be basic feasible solution and stage I is completed. In the last row, all the entries except the last and second last entries are called indicators. The simplex table represents a system of four linear equations in seven variables, which is x 1 , x 2 , x 3 , s 1 , s 2 , s 3 , z as there are more variables than equations then the system is dependent, and it has infinitely many solutions. Apply the simplex method for stage II which is: x 1 x 2 x 3 s 1 s 2 s 3 [ 1 1 1 1 0 0 50000 1 1 0 0 1 0 15000 1 0 1 0 0 1 25000 − .15 − .09 − .05 0 0 0 0 ] Multiply the last row by 100 to clear the decimals which are: x 1 x 2 x 3 s 1 s 2 s 3 [ 1 1 1 1 0 0 50000 1 1 0 0 1 0 15000 1 0 1 0 0 1 25000 − 15 − 9 − 5 0 0 0 0 ] In the simplex, choose the pivot entry as the pivot is the most negative indicator, which is − 15

Pearson eText for Mathematics with Applications In the Management, Natural, and Social Sciences -- Instant Access (Pearson+)

12th Edition

ISBN: 9780137504381

Author: Margaret Lial, Thomas Hungerford

Publisher: PEARSON+

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Chapter 7, Problem 58RE

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Ch. 7.1 - Checkpoint 1 Graph the given...Ch. 7.1 - Prob. 2CP Ch. 7.1 - Prob. 3CP Ch. 7.1 - Prob. 4CP Ch. 7.1 - Prob. 5CP Ch. 7.1 - Checkpoint 6 Graph the feasible region of the...Ch. 7.1 - Prob. 7CP Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...

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