Deneme Sınavı Soruları

3

5

6

10

15

Step 1.Take the probabilities of the two alternatives of the row player as p and (1-p), then expected pay- offs of column player for each strategy are expressed with equations.

Step 2. Draw n straight lines for j=1, 2... n and determine the highest point of the lower envelope obtained. This will be the maximin point.

Step 3. These equations are then plotted on the graph.

Step 4.Draw two vertical axes one unit apart. These two lines are 0 and 1.

Step 5. The two or more lines passing through the maximin point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 1. Draw two vertical axes one unit apart. These two lines are 0 and 1. 

Step 2. Draw n straight lines for j=1, 2... n and determine the highest point of the lower envelope obtained. This will be the maximin point.

Step 3. These equations are then plotted on the graph.

Step 4.Take the probabilities of the two alternatives of the row player as p and (1-p), then expected pay- offs of column player for each strategy are expressed with equations.

Step 5. The two or more lines passing through the maximin point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 1. Draw two vertical axes one unit apart. These two lines are 0 and 1. 

Step 2. Draw n straight lines for j=1, 2... n and determine the highest point of the lower envelope obtained. This will be the maximin point.

Step 3.The two or more lines passing through the maximin point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 4.Take the probabilities of the two alternatives of the row player as p and (1-p), then expected pay- offs of column player for each strategy are expressed with equations.

Step 5. These equations are then plotted on the graph.

Only I

I-II

I-III

II-III

I-II-III

Artificial variable.

Basic variable.

Slack variable.

Entering variable.

Leaving variable.

Reducible

Irreducible

Absorbing

Periodic

Aperiodic

Setting n – m number of variables equal to zero.

Setting slack variables equal to 0.

Equalizing slack variables.

Equalizing non basic variables.

Eleminating n – m number of linear equations.

Minimize Z = -3x - 12y - 3.

Minimize Z = -3x - 12y

Minimize Z = 3x + 12

Maximize Z = 3x - 12

Maximize Z = -3x - 12y - 3.

Step 1.Draw n straight lines for j=1, 2... n and determine the lowest point of the upper envelope obtained. This will be the minimax point.

Step 2. Take the probabilities of the two alternatives of the column player as q and (1-q), then expected pay-offs of row player for each strategy are expressed with equations.

Step 3. These equations are then plotted on the graph.

Step 4. Draw two vertical axes one unit apart. These two lines are 0 and 1.

Step 5. The two or more lines passing through the minimax point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 1.Take the probabilities of the two alternatives of the column player as q and (1-q), then expected pay-offs of row player for each strategy are expressed with equations.

Step 2.Draw n straight lines for j=1, 2... n and determine the lowest point of the upper envelope obtained. This will be the minimax point.

Step 3. These equations are then plotted on the graph.

Step 4. Draw two vertical axes one unit apart. These two lines are 0 and 1.

Step 5. The two or more lines passing through the minimax point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 1.Take the probabilities of the two alternatives of the column player as q and (1-q), then expected pay-offs of row player for each strategy are expressed with equations.

Step 2.Draw n straight lines for j=1, 2... n and determine the lowest point of the upper envelope obtained. This will be the minimax point.

Step 3. Draw two vertical axes one unit apart. These two lines are 0 and 1.

Step 4. These equations are then plotted on the graph.

Step 5. The two or more lines passing through the minimax point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 1.Take the probabilities of the two alternatives of the column player as q and (1-q), then expected pay-offs of row player for each strategy are expressed with equations.

Step 2.Draw n straight lines for j=1, 2... n and determine the lowest point of the upper envelope obtained. This will be the minimax point.

Step 3. Draw two vertical axes one unit apart. These two lines are 0 and 1.

Step 4. The two or more lines passing through the minimax point determines the reduced 2 x 2 payoff matrix. The optimum solution of the game is obtained as in 2 x 2 game.

Step 5. These equations are then plotted on the graph.

Pay-off

Saddle point

Mixed strategy

Pure strategy

Pay-off Matrix

assignment

transportation

project

optimality

feasible

18

20

24

30

35

x₁ + x₂ ≤ 256

- x₁ – 5 x₂ ≥ -128

2 x₁ ≤ -3 x₂

-3 x₁ + 4 x₂ ≥ 64

x₁ + 2 x₂ ≤ 0

D1 with 13500

D2 with 18625

D3 with 19250

D1 with 42300

D4 with 28250

I-II-III-IV-V

II-I-III-V-IV

III-I-II-IV-V

III-V-I-II-IV

III-II-I-IV-V

All the potential consequences multiplied by their associated probabilities.

The expected monetary value of each decision alternative is calculated.

The investor chooses the biggest one among the expected monetary values.

Decision maker confronts the various possible states of nature in a decision under risk.

It demonstrates the average additional amount the investor would have achieved by making the right decision.

-386

386

752

3.1

0.31

When the objective function is maximization, the highest values of the alternatives are multiplied by alpha and the smallest values of the alternatives are multiplied by 1– α.

When the objective function is minimization, the highest values of the alternatives are multiplied by α² and the smallest values of the alternatives are multiplied by α-1.

When the objective function is maximization, both the highest values and the smallest values of the alternatives are multiplied by α.

When the objective function is minimization, both the highest values and the smallest values of the alternatives are multiplied by 1– α.

When the objective function is minimization, the highest values of the alternatives are multiplied by alpha and the smallest values of the alternatives are multiplied by 1– α.

Laplace

Plunger

Wald

Hurwicz

Savage

x+y=4

2x-3y

x – 3y ≤ 2

x – 3y ≤ 0

x ≥ y