Winston Chapter 14.1, Page 784, Number 9 (Goal Programming)

Winston Chapter 14.1, Page 784, Number 9 (Goal Programming)

 /  / 1

Winston Chapter 14.1, Page 784, Number 9 (Goal Programming)

Problem Statement: A company produces two products. Relevant information for each product is shown in Table 7. The company has a goal of $48 in profits and incurs a $1 penalty for each dollar it falls short of this goal. A total of 32 hours of labor are available. A $2 penalty is incurred for each hour of overtime (labor over 32 hours) used, and a $1 penalty is incurred for each hour of available labor that is unused. Marketing considerations require that at least 7 units of product 1 are produced and at least 10 units of product 2 are produced. For each unit (of either product) by which production falls short of demand, a penalty of $5 is assessed.

Table 7: / Product 1 / Product 2
Labor Required: / 4 hours / 2 hours
Contribution to Profit: / $4 / $2

Problem Summary:

x1: Number of units of product 1 to produce.

x2: Number of units of product 2 to produce.

Product 1, x1 / Product 2, x2 / Available / Penalty
Labor: / 4 Hours / 2 Hours / 32 Hours / $2 / overtime
$1 / unused
Profit: / $4 / $2 / $48 / $1 / dollar short
Requirement: / At least 7 units / At least 10 units
Penalty: / $5 / shortage / $5 / shortage

Constraints:

4x1 + 2x2 + S1- - S1+ = 32

x1 + S2- - S2+ = 7

x2 + S3- - S3+ = 10

Solve the problem for each of the situations below using Quant.

A.  Goal 1: Avoid under-utilization of labor.

Goal 2: Meet demand for product 1.

Goal 3: Meet demand for product 2.

Goal 4: Do not use any overtime.

Min Z = P1S1- + P2S2- + P3S3- + P4S1+

Quant Input:

Input Data of The Problem 0784N09a Page: 1

MIN +0 X1 +0 X2 +1.00000S1N +0 S1P +0 S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P

Subject to

(1) +4.00000X1 +2.00000X2 +1.00000S1N -1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P = +32.0000

(2) +1.00000X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

-1.00000S2P +0 S3N +0 S3P = +7.00000

(3) +0 X1 +1.00000X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N -1.00000S3P = +10.0000

Quant Output:

|------|

| Summarized Solution for 0784N09a Page : 1 |

|------|

| | | |Opportunity|Opportunity|Opportunity|Opportunity|

|Number | Variable | Solution |Cost-Obj. 1|Cost-Obj. 2|Cost-Obj. 3|Cost-Obj. 4|

|------+------+------+------+------+------+------|

| 1 | X1 | +7.0000000| 0| 0| 0| 0|

| 2 | X2 | +10.000000| 0| 0| 0| 0|

| 3 | S1N | 0| +1.0000000| 0| 0| +1.0000000|

| 4 | S1P | +16.000000| 0| 0| 0| 0|

| 5 | S2N | 0| 0| +1.0000000| 0| -4.0000000|

| 6 | S2P | 0| 0| 0| 0| +4.0000000|

| 7 | S3N | 0| 0| 0| +1.0000000| -2.0000000|

| 8 | S3P | 0| 0| 0| 0| +2.0000000|

|------|

| Priority Level 1: Minimized Objective Function (Goal) = 0 |

| Priority Level 2: Minimized Objective Function (Goal) = 0 |

| Priority Level 3: Minimized Objective Function (Goal) = 0 |

| Priority Level 4: Minimized Objective Function (Goal) = +16.000000 |

| Iteration = 11 Elapsed CPU second = 0 |

|------|

Summary of results: By assigning priorities to each of the constraints of the problem, a solution that reflects the level of demands can be met. In this situation, under-utilization of labor is most important, while not using overtime is least important. The objective function is formulated by assigning priorities to the appropriate negative or positive deviation. The results of Quant indicate that all seven units desired of product one will be produced. Additionally, all ten units desired of product two will be produced. However, for this to occur, 16 hours of overtime will be used.

B.  Goal 1: Meet demand for product 1.

Goal 2: Meet demand for product 2.
Goal 3: Avoid under-utilization of labor.
Goal 4: Do not use any overtime.

Min Z = P1S2- + P2S3- + P3S1- + P4S1+

Quant Input:

Input Data of The Problem 0784N09B Page: 1

MIN +0 X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N +0 S3P

MIN +0 X1 +0 X2 +1.00000S1N +0 S1P +0 S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P

Subject to

(1) +4.00000X1 +2.00000X2 +1.00000S1N -1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P = +32.0000

(2) +1.00000X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

-1.00000S2P +0 S3N +0 S3P = +7.00000

(3) +0 X1 +1.00000X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N -1.00000S3P = +10.0000

Quant Output:

|------|

| Summarized Solution for 0784N09B Page : 1 |

|------|

| | | |Opportunity|Opportunity|Opportunity|Opportunity|

|Number | Variable | Solution |Cost-Obj. 1|Cost-Obj. 2|Cost-Obj. 3|Cost-Obj. 4|

|------+------+------+------+------+------+------|

| 1 | X1 | +7.0000000| 0| 0| 0| 0|

| 2 | X2 | +10.000000| 0| 0| 0| 0|

| 3 | S1N | 0| 0| 0| +1.0000000| +1.0000000|

| 4 | S1P | +16.000000| 0| 0| 0| 0|

| 5 | S2N | 0| +1.0000000| 0| 0| -4.0000000|

| 6 | S2P | 0| 0| 0| 0| +4.0000000|

| 7 | S3N | 0| 0| +1.0000000| 0| -2.0000000|

| 8 | S3P | 0| 0| 0| 0| +2.0000000|

|------|

| Priority Level 1: Minimized Objective Function (Goal) = 0 |

| Priority Level 2: Minimized Objective Function (Goal) = 0 |

| Priority Level 3: Minimized Objective Function (Goal) = 0 |

| Priority Level 4: Minimized Objective Function (Goal) = +16.000000 |

| Iteration = 11 Elapsed CPU second = 0 |

|------|

C.  Goal 1: Meet demand for product 1.
Goal 2: Meet demand for product 2.
Goal 3: Do not use any overtime.
Goal 4: Avoid under-utilization of labor.

Min Z = P1S2- + P2S3- + P3S1+ + P4S1-

Quant Input:

Input Data of The Problem 0784n09c Page: 1

MIN +0 X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +1.00000S1N +0 S1P +0 S2N

+0 S2P +0 S3N +0 S3P

Subject to

(1) +4.00000X1 +2.00000X2 +1.00000S1N -1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P = +32.0000

(2) +1.00000X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

-1.00000S2P +0 S3N +0 S3P = +7.00000

(3) +0 X1 +1.00000X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N -1.00000S3P = +10.0000

Quant Output:

|------|

| Summarized Solution for 0784n09c Page : 1 |

|------|

| | | |Opportunity|Opportunity|Opportunity|Opportunity|

|Number | Variable | Solution |Cost-Obj. 1|Cost-Obj. 2|Cost-Obj. 3|Cost-Obj. 4|

|------+------+------+------+------+------+------|

| 1 | X1 | +7.0000000| 0| 0| 0| 0|

| 2 | X2 | +10.000000| 0| 0| 0| 0|

| 3 | S1N | 0| 0| 0| +1.0000000| +1.0000000|

| 4 | S1P | +16.000000| 0| 0| 0| 0|

| 5 | S2N | 0| +1.0000000| 0| -4.0000000| 0|

| 6 | S2P | 0| 0| 0| +4.0000000| 0|

| 7 | S3N | 0| 0| +1.0000000| -2.0000000| 0|

| 8 | S3P | 0| 0| 0| +2.0000000| 0|

|------|

| Priority Level 1: Minimized Objective Function (Goal) = 0 |

| Priority Level 2: Minimized Objective Function (Goal) = 0 |

| Priority Level 3: Minimized Objective Function (Goal) = +16.000000 |

| Priority Level 4: Minimized Objective Function (Goal) = 0 |

| Iteration = 11 Elapsed CPU second = 0 |

|------|

D.  Goal 1: Do not use any overtime.
Goal 2: Meet demand for product 2.
Goal 3: Avoid under-utilization of labor.
Goal 4: Meet demand for product 1.

Min Z = P1S1+ + P2S3- + P3S1- + P4S2-

Quant Input:

Input Data of The Problem 0784n09D Page: 1

MIN +0 X1 +0 X2 +0 S1N +1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

+0 S2P +0 S3N +0 S3P

Subject to

(1) +4.00000X1 +2.00000X2 +1.00000S1N -1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P = +32.0000

(2) +1.00000X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

-1.00000S2P +0 S3N +0 S3P = +7.00000

(3) +0 X1 +1.00000X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N -1.00000S3P = +10.0000

Quant Output:

|------|

| Summarized Solution for 0784n09D Page : 1 |

|------|

| | | |Opportunity|Opportunity|Opportunity|Opportunity|

|Number | Variable | Solution |Cost-Obj. 1|Cost-Obj. 2|Cost-Obj. 3|Cost-Obj. 4|

|------+------+------+------+------+------+------|

| 1 | X1 | +3.0000000| 0| 0| 0| 0|

| 2 | X2 | +10.000000| 0| 0| 0| 0|

| 3 | S1N | 0| 0| 0| +.25000000| +.25000000|

| 4 | S1P | 0| +1.0000000| 0| -.25000000| -.25000000|

| 5 | S2N | +4.0000000| 0| 0| 0| 0|

| 6 | S2P | 0| 0| 0| +1.0000000| +1.0000000|

| 7 | S3N | 0| 0| +1.0000000| -.50000000| -.50000000|

| 8 | S3P | 0| 0| 0| +.50000000| +.50000000|

|------|

| Priority Level 1: Minimized Objective Function (Goal) = 0 |

| Priority Level 2: Minimized Objective Function (Goal) = 0 |

| Priority Level 3: Minimized Objective Function (Goal) = +4.0000000 |

| Priority Level 4: Minimized Objective Function (Goal) = +4.0000000 |

| Iteration = 9 Elapsed CPU second = 0 |

|------|

E.  Goal 1: Do not use any overtime.
Goal 2: Meet demand for product 2.
Goal 3: Meet demand for product 1.
Goal 4: Avoid under-utilization of labor.

Min Z = P1S1+ + P2S3- + P3S2- + P4S1-

Quant Input:

Input Data of The Problem 0784n09E Page: 1

MIN +0 X1 +0 X2 +0 S1N +1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N +0 S3P

MIN +0 X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

+0 S2P +0 S3N +0 S3P

MIN +0 X1 +0 X2 +1.00000S1N +0 S1P +0 S2N

+0 S2P +0 S3N +0 S3P

Subject to

(1) +4.00000X1 +2.00000X2 +1.00000S1N -1.00000S1P +0 S2N

+0 S2P +0 S3N +0 S3P = +32.0000

(2) +1.00000X1 +0 X2 +0 S1N +0 S1P +1.00000S2N

-1.00000S2P +0 S3N +0 S3P = +7.00000

(3) +0 X1 +1.00000X2 +0 S1N +0 S1P +0 S2N

+0 S2P +1.00000S3N -1.00000S3P = +10.0000

Quant Output:

|------|

| Summarized Solution for 0784n09E Page : 1 |

|------|

| | | |Opportunity|Opportunity|Opportunity|Opportunity|

|Number | Variable | Solution |Cost-Obj. 1|Cost-Obj. 2|Cost-Obj. 3|Cost-Obj. 4|

|------+------+------+------+------+------+------|

| 1 | X1 | +3.0000000| 0| 0| 0| 0|

| 2 | X2 | +10.000000| 0| 0| 0| 0|

| 3 | S1N | 0| 0| 0| +.25000000| +1.0000000|

| 4 | S1P | 0| +1.0000000| 0| -.25000000| 0|

| 5 | S2N | +4.0000000| 0| 0| 0| 0|

| 6 | S2P | 0| 0| 0| +1.0000000| 0|

| 7 | S3N | 0| 0| +1.0000000| -.50000000| 0|

| 8 | S3P | 0| 0| 0| +.50000000| 0|

|------|

| Priority Level 1: Minimized Objective Function (Goal) = 0 |

| Priority Level 2: Minimized Objective Function (Goal) = 0 |

| Priority Level 3: Minimized Objective Function (Goal) = +4.0000000 |

| Priority Level 4: Minimized Objective Function (Goal) = 0 |

| Iteration = 9 Elapsed CPU second = 0 |

|------|

1. Summary table of the optimal solutions in parts A to E.

Priorities / Optimal Solution
Value of: / Deviation From:
Situation / First Priority / Second Priority / Third Priority / Fourth Priority / x1 / x2 / No Under-Utilization / No Overtime / Product 1 / Product 2
A. / No UUL / Product 1 / Product 2 / No OT / 7 / 10 / 0 / 16 / 0 / 0
B. / Product 1 / Product 2 / No UUL / No OT / 7 / 10 / 0 / 16 / 0 / 0
C. / Product 1 / Product 2 / No OT / No UUL / 7 / 10 / 0 / 16 / 0 / 0
D. / No OT / Product 2 / No UUL / Product 1 / 3 / 10 / 0 / 0 / 4 / 0
E. / No OT / Product 2 / Product 1 / No UUL / 3 / 10 / 0 / 0 / 4 / 0

This summary table shows that in situations A, B, and C with the goals as listed from first to fourth priorities, all seven units desired of product one will be produced. Additionally, all ten units desired of product two will be produced. However, for this to occur, 16 hours of overtime will be used. Situations A & B resulted in the first three goals being met, while situation C only resulted in the first two goals being met.

For situations D & E, only three units of product one will be produced while ten units of product two are produced. This leaves a four-unit deviation from meeting the goal of producing seven units of product one. The benefits of this under-producing are evident when no under-utilization and no overtime are realized. The first three goals of situation D were met, while only the first two goals of situation E were met.