Decision Analysis
Solution 1
a) Introducing slack variables, S1, S2, and S3 the given linear programming problem can be written in standard form as
Max 3A + 4B + 0S1+ 0S2+ 0S3
Subject to
-1A + 2B + S1 = 8
1A + 2B + S2 = 12
2A + 1B + S3 = 16
A, B, S1, S2, S3 ≥ 0
b) The feasible region is the shaded region OCDFG in the following graph.
The coordinates of the feasible region and the corresponding objective function value is shown in the following Table
| Extreme point | Coordinates (A, B) | Objective function | |
|
|
A | B | 3A + 4B |
| O | 0 | 0 | 0 |
| C | 8 | 0 | 24 |
| D | (20/3) = 6.67 | (8/3) = 2.67 | (92/3) = 30.67 |
| F | 2 | 5 | 26 |
| G | 0 | 4 | 16 |
Since the objective function is maximum at D, the point D is optimal.
Thus, the optimal solution is A = 20/3 and B = 8/3 and the optimal value of the objective function is 92/3 = 30.67.
c) At the optimal solution A = 20/3 and B = 8/3,
-1A + 2B + S1 = 8 implies -(20/3) + 2(8/3) + S1 = 8, that is (4/3) + S1 = 8 or S1 = 20/3.
1A + 2B + S2 = 12 implies (20/3) + 2(8/3) + S2 = 12, that is 12 + S2 = 12 or S2 = 0.
2A + 1B + S3 = 16 implies 2(20/3) + (8/3) + S3 = 16, that is 16 + S3 = 16 or S3 = 0
Thus, the values of the slack variables S1 is 20/3 and S2 and S3 are 0 at the optimal solution.
Solution 2
a) The decision variables are
X = Number of units purchased in the stock fund
Y = Number of units purchased in the money market fund
Since each unit invested in the stock fund has a risk index of 8, and each unit invested in the money market fund has a risk index of 3; the total risk index is 8 X + 3Y
Thus, an objective function that will minimize the total risk index for the portfolio is
Minimize 8 X + 3 Y
Since each unit of stock fund costs $50 and each unit of money market fund costs $100, the constraint corresponding to the available funds is
50 X + 100 Y ≤ 1,200,000
Since the annual income from stock fund is ($50)(10%) = $5 and the annual income from the money market fund is ($100)(4%) = $4, the constraint corresponding to the annual income is
5 X + 4 Y ≥ 60,000
Since at least $300,000 to be invested in the money market, at least ($300,000/$100) = 3,000 units must be invested in the money market. Thus, the constraint corresponding to the minimum units in money market is
Y ≥ 3,000
Finally, the non-negativity constraints, X, Y ≥ 0.
Thus, the linear programming problem is
Min 8 X + 3 Y
Subject to
50 X + 100 Y ≤ 1,200,000 (Funds available)
5 X + 4 Y ≥ 60,000 (Annual income)
Y ≥ 3,000 (Minimum units in money market)
X, Y ≥ 0
The Excel Solver output of the linear programming problem is shown below:
| Data | Results | ||||||
| X | Y | LHS | Slack/Surplus | ||||
| Objective | 8 | 3 | sign | RHS | 62000 | ||
| Constraint 1 | 50 | 100 | < | 1200000 | 1200000 | 0 | |
| Constraint 2 | 5 | 4 | > | 60000 | 60000 | 0 | |
| Constraint 3 | 0 | 1 | > | 3000 | 10000 | -7000 | |
| Results | X | Y | |||||
| Variables | 4000 | 10000 | |||||
| Objective | 62000 |
Therefore, the optimal solution is X = 4,000 and Y = 10,000 and the minimum total risk index = 62,000.
Thus, Innis should purchase 4000 units in the stock fund and 10000 units in the money market fund to minimize the total risk index for the portfolio.
b) The annual income corresponding to the optimal investment strategy is
Annual Income = 5 X + 4 Y = 5(4000) + 4(10000)
= 20000 + 40000
= 60000
c) If the client desires to maximize the annual return, he should invest everything in the stock fund.
Solution 3
The decision variables are
X = Number of gallons of regular gasoline produced
Y = Number of gallons of premium gasoline produced
Since the profit contributions are $0.30 per gallon for regular gasoline and $0.50 per gallon for premium gasoline; the total profit contribution is 0.30 X + 0.50 Y.
Thus, the objective function is
Maximize 0.30 X + 0.50 Y
Since each gallon of regular gasoline contains 0.3 gallons of grade A crude oil and each gallon of premium gasoline contains 0.6 gallons of grade A crude oil, the constraint corresponding to the available Grade A crude oil is
0.30 X + 0.60 Y ≤ 18,000
The constraint corresponding to the production capacity is
X + Y ≤ 50,000
The constraint corresponding to the demand for the premium gasoline is
Y ≤ 20,000
Finally, the non-negativity constraints, X, Y ≥ 0.
Thus, the linear programming model is
Max 0.30 X + 0.50 Y
Subject to
0.30 X + 0.60 Y ≤ 18,000 (Availability of Grade A crude oil)
X + Y ≤ 50,000 (Production Capacity)
Y ≤ 20,000 (Demand for Premium)
X, Y ≥ 0
b) The Excel Solver output of the linear programming problem is shown below:
| Data | Results | ||||||
| X | Y | LHS | Slack/Surplus | ||||
| Objective | 0.3 | 0.5 | sign | RHS | 17000 | ||
| Constraint 1 | 0.3 | 0.6 | < | 18000 | 18000 | 0 | |
| Constraint 2 | 1 | 1 | < | 50000 | 50000 | 0 | |
| Constraint 3 | 0 | 1 | < | 20000 | 10000 | 10000 | |
| Results | X | Y | |||||
| Variables | 40000 | 10000 | |||||
| Objective | 17000 |
Therefore, the optimal solution is X = 40,000 and Y = 10,000 and the minimum total profit contribution = $17,000.
Thus, 40,000 gallons of regular gasoline and 10,000 gallons of premium gasoline should be produced to maximize the total profit contribution.
c) From the Excel solver output, the slack variables for constraints 1, 2 and 3 are respectively S1 = 0, S2 = 0, and S3 = 10000.
The slack variable for constraint 1, S1 = 0 means that all available grade A crude oil is used.
The slack variable for constraint 2, S2 = 0 means that total production capacity is used.
The slack variable for constraint 3, S3 = 10000 means that Premium gasoline production is 10,000 gallons less than the maximum demand.
d) The binding constraints are those constraints whose slack variables are zero. Therefore, Grade A crude oil availability and Production capacity are the binding constraints
Solution 4
The decision variables are
X = Time allocated to regular customer service during the two-week period
Y = Time allocated to new customer service during the two-week period
Since technicians require an average of 50 minutes for each regular customer and 1 hour = 60 minutes for each new customer, a technician can handle 60/50= 1.2 regular customers and 60/60 = 1 new customer in one hour, so that the total number of customers contacted during the two-week period is 1.2 X + Y.
Thus, the objective function is
Maximize 1.2 X + Y
Since a maximum of 80 hours of technician time is available over the two-week planning period, the constraint corresponding to the available technician time is
X + Y ≤ 80
The constraint corresponding to the cash flow requirement (revenue) is
25 X + 8 Y ≥ 800
Since the technician time spent on new customer contacts must be at leas 60% of the time spend on regular customer contacts, the constraint corresponding to this requirement is
Y ≥ 0.60 X
Or – 0.60 X + Y ≥ 0
Finally, the non-negativity constraints, X, Y ≥ 0.
Thus, the linear programming model is
Max 1.2 X + Y
Subject to
X + Y ≤ 80
25 X + 8 Y ≥ 800
– 0.60 X + Y ≥ 0
X, Y ≥ 0
b) The Excel Solver output of the linear programming problem is shown below:
| Data | Results | ||||||
| X | Y | LHS | Slack/Surplus | ||||
| Objective | 1.2 | 1 | sign | RHS | 90 | ||
| Constraint 1 | 1 | 1 | < | 80 | 80 | 0 | |
| Constraint 2 | 25 | 8 | > | 800 | 1490 | -690 | |
| Constraint 3 | -0.6 | 1 | > | 0 | 0 | 0 | |
| Results | X | Y | |||||
| Variables | 50 | 30 | |||||
| Objective | 90 |
Therefore, the optimal solution is X = 50 and Y = 30 and the maximum number of customers contacted = 90.
Thus, HTC should allocate 50 hours to regular customer service and 30 hours to new customer service to maximize the total number of customers contacted during the two-week period.
Solution 5
a) The feasible region is CBF
b) The co-ordinates of the point C are obtained by solving the equations 6X – 2Y = 3 and X+Y = 3. Solving these equations we get Point C (9/8, 21/8) = (1.125, 2.625)
The co-ordinates of the point B are obtained by solving the equations 6X – 2Y = 3 and -2X+3Y = 6. Solving these equations we get Point B(3/2, 3) = (1.5, 3).
The co-ordinates of the point F are obtained by solving the equations -2X+3Y = 6 and X+Y = 3. Solving these equations we get Point F(3/5, 12/5) = (0.6, 2.4).
The coordinates of the feasible region and the corresponding objective function value is shown in the following Table
| Extreme point | Coordinates (X, Y) | Objective function | |
|
|
X | Y | -3X + 6Y
|
| C | 1.125 | 2.625 | 12.375 |
| B | 1.5 | 3 | 13.5 |
| F | 0.6 | 2.4 | 12.6 |
Since the objective function is maximum at B, the point B is optimal.
c) The optimal solution is X = 1.5 and Y = 3 and the optimal objective function value is 13.5.
d) The binding constraints are 6X – 2Y ≤ 3 and -2X + 3Y ≤ 6.
e) Introducing slack variables S1, S2, and surplus variable S3, the linear program in standard form is
Max -3X + 6Y + 0S1+ 0S2 + 0S3
Subject to
6X – 2Y + S1 = 3
-2X + 3Y+ S2 = 6
X + Y – S3 = 3
X, Y, S1, S2, S3 ≥ 0
f) At the optimal solution X = 1.5 and Y =3,
6X – 2Y + S1 = 3 implies 6(1.5) – 2(3) + S1 = 3, that is 3 + S1 = 3 or S1 = 0.
-2X + 3Y+ S2 = 6 implies -2(1.5) + 3(3) + S2 = 6, that is 6 + S2 = 6 or S2 = 0.
X + Y – S3 = 3 implies 1.5 + 3 – S3 = 3, that is 4.5 – S3 = 3 or S3 = 1.5
Thus, the values of the slack variables S1 and S2 at the optimal solution are both 0 and the value of the surplus variable S3 at the optimal solution is 1.5.
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