Goal Programming
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Transcript of Goal Programming
• Goals are prioritized in some sense, and their level of aspiration is stated.
• An optimal solution is attained when all the goals are reached as close as possible to their aspiration level, while satisfying a set of constraints.
• There are two types of goal programming models:– Nonpreemtive goal programming - no goal is pre-determined to dominate
any other goal.
– Preemtive goal programming - goals are assigned different priority levels. Level 1 goal dominates level 2 goal, and so on.
13.5 Goal Programming
• A company is considering three forms of advertising.
• Goals– Goal 1: Spend no more $25,000 on advertising.
– Goal 2: Reach at least 30,000 new potential customers.
– Goal 3: Run at least 10 television spots.
NONPREEMTIVE GOAL PROGRAMMINGAn Advertisement Example
Cost per Ad Customers
Television 3000 1000Radio 800 500Newspaper 250 200
Cost per Ad Customers
Television 3000 1000Radio 800 500Newspaper 250 200
• If these were constraints rather than goals we would have:
3000X1 + 800X2 + 250X3 25,000
1000X1 + 500X2 + 200X3 30,000
X1 10
• No feasible solution exists that satisfies all the constraints.
• When these constraints are simply goals they are to be reached as close as possible.
An Advertisement Example
• Detrimental variablesUi = the amount by which the left hand side
falls short of (under) its right had side value.
Ei = the amount by which the left side exceeds its right had side value.
• The goal equations3000X1 + 800X2 + 250X3 + U1 – E1 = 25,000
1000X1 + 500X2 + 200X3 + U2 – E2 = 30,000
X1 + U3 – E3 = 10
An Advertisement Example
• The objective is to minimize the penalty of not meeting the goals, represented by the detrimental variables
E1, U2, U3.
An Advertisement Example
25,000 30,000 10
• The penalties are estimated to be as follows:
– Each extra dollar spent on advertisement above $25,000 cost the company $1.
– There is a loss of $5 to the company for each customer not
being reached, below the goal of 30,000.
– Each television spot below 10 is worth 100 times each
dollar over budget.
An Advertisement Example
• It is assumed that no advantage is gained by overachieving a goal.
Minimize 1E1 + 5U2 + 100U3
s.t.3000X1 + 800X2 + 250X3 + U1 – E1 = 25,0001000X1 + 500X2 + 200X3 + U2 – E2 = 30,000
X1 + U3 – E3 = 10
All variables are non-negative.
An Advertisement Example – The goal programming model
• The NECC is planning next month production of its two bicycles B2 and S10.
• Data– Both models use the same seats and tires.– 2000 seats are available; 2400 tires are available. – 1000 gear assembly are available (used only in the S10
model).– Production time per unit: 2 hours for B2; 3 hours for
S10.– Profit: $40 for each B2; 10$ for each S10.
PREEMTIVE GOAL PROGRAMMING - New England Cycle Company
– Priority 1: Fulfill a contract for 400 B2 bicycles to be delivered next month.
NECC – Prioritized Goals
Priority 4: At least 200 tires left
over at the end of the month.
At least 100 gear assemblies left over at the end of the month.
Priority 2: Produce at least 1000 total bicycles during the month.
Priority 3: Achieve at least
$100,000 profit for the month.
Use no more than 1600 labor-hours during the month.
Management wants to determine the production schedule that best meets its prioritized schedule.
New England Cycle Company Example
NECC - SOLUTION
• Decision variablesX1 = The number of B2s to be produced next month
X2 = The number of S10s to be produced next month
• Functional / nonnegativity constraints
0X,XTires2400X2X2
assembliesGear1000XSeats2000XX2
21
21
2
21
• Goal constraints Priority 1 (goal 1): Production of at least 400 B2sPriority 1 (goal 1): Production of at least 400 B2s
XX11 + U+ U11 - E - E11 = 400 = 400
Priority 2 (goal 2): Production of at least 1000 total cyclesPriority 2 (goal 2): Production of at least 1000 total cyclesXX11 + X + X22 + U+ U22 - E - E22 = 1000 = 1000
Priority 3 (goal 3) Profit of at least $100,000Priority 3 (goal 3) Profit of at least $100,000
.04X .04X11 + .10X + .10X22 + U+ U33 - E - E33 = 100 (in $1000) = 100 (in $1000)Priority 3 (goal 4) Use a maximum of 1600 labor hoursPriority 3 (goal 4) Use a maximum of 1600 labor hours
2X 2X11 + 3X + 3X22 + U+ U44 - E - E44 = 1600 = 1600
Priority 4 (goal 5) At least 200 leftover tiresPriority 4 (goal 5) At least 200 leftover tires 2X 2X11 + 2X + 2X22 + U+ U55 - E - E55 = 2200 = 2200
Priority 4 (goal 6) At least 100 leftover gear assemblyPriority 4 (goal 6) At least 100 leftover gear assemblyXX22 + U+ U66 - E - E66 = 900 = 900
NECC - SOLUTION
• Priority level objectives
Priority 1: Underachieving a production of 400 B2s:Minimize U1
Priority 2: Underachieving a total production of 1000: Minimize U2
NECC - SOLUTION
• Priority level objectives
– Priority 3: Underachieving a $100,000 profit
Using more than 1600 labor-hours
Minimize 30U3 + E4
NECC - SOLUTION
Each $1,000 short of the $100,000 goal is considered 30 times as important as utilizing an extra labor-hour.
• Priority level objectives
Priority 4: Using more than 2200 tires
Using more than 900 gear assemblies
Minimize E5 + 2E6
NECC - SOLUTION
Each leftover gear assembly is deemed twice as important as leftover tire.
– Solve the linear goal programming for priority 1 objective, under the set of regular constraints and goal constraint as
shown below Minimize U1
ST
NECC - The solution procedure
X1 + U1 - E1 = 400
Tires2400X2X2Gear1000XSeats2000XX2
21
2
21
X1 = 400, thus
U1 = 0, and priority 1 goal is fully achieved.
– Solve the linear goal programming for priority 2 level objective, under the set of original constraints plus the constraint X1 400 (maintain the
level of achievement of the priority 1 goal).Minimize U2
ST
NECC - The solution procedure
Every point that satisfies X1 + X2 1000 yields U2 = 0, and therefore, priority 2 goal is fully achieved.
Tires2400X2X2Gear1000XSeats2000XX2
21
2
21
X1 400X1 + X2 + U2 - E2 = 1000
– Solve the linear goal programming for priority 3 level objective, under the set of original constraints plus the constraint X1 400 (maintain the level of achievement of the priority 1 goal), plus the constraint X1 + X2 1000 (maintain the level of achievement of the priority 2 goal).
• Every point in the range X1 = 400 and 600 X2 800 is optimal for this model; 30U3 + E4 = 1720 is the level of achievement for the priority 3 goal.
NECC - The solution procedure
– Solve the linear goal programming for priority 4 level objective, and notice that after the previous step the feasible region is reduced to a segment of a straight line between the points (400,600) and (400,800).
• X1 =400; 600 X2 700 and E5 + 2E6 = 0
NECC - The solution procedure
• In summary NECC should produce– 400 B2 model– Between 600 and 700 S10 model
NECC - Solution Summary