Study Guide: “What If?” Analysis (Rardin Ch. 4)

Context: In the real world, requirements change. A good modeler doesn’t just solve the problem in the book; they adapt the structure when the client adds a new constraint or changes the objective.

How to use this guide:

1. Read the Scenario.
   
2. Cover the Solution and try to write the mathematical modification yourself.
   
3. Check your answer.

1. Forest Service Allocation (Rardin 4.1)

Base Model: Allocate acres () in 7 Analysis Areas to 3 prescriptions (Timber, Grazing, Wilderness) to maximize NPV, subject to Timber Yield, Grazing AUM, and Wilderness Index targets.

Scenario A: The “Monoculture” Limit

The Situation: Environmental groups are concerned that we might turn the entire forest into a timber farm or a giant pasture. Management introduces a policy: “No single prescription can occupy more than 50% of the total forest acreage.”

The Modification:

• Logic: We need to sum the acres allocated to Prescription across all areas () and ensure that sum is less than or equal to .
  
• New Constraints:
  We need one constraint for each prescription (or specifically for Timber/Grazing).
  
  (Where is the fixed size of area .)

Scenario B: The “Visual Buffer” (Local Minimum)

The Situation: To preserve aesthetics, every specific Analysis Area must maintain at least 10% of its land as Wilderness. We cannot just dump all the wilderness in Area 7; Area 1 needs some too.

The Modification:

• Logic: This is a local constraint. It applies to each individually.
  
• New Constraints:
  
  (Note: This tightens the feasible region significantly. It forces to be non-zero everywhere).

2. Swedish Steel Blending (Rardin 4.2)

Base Model: Mix scrap () to make 1000 kg of steel. Meet Carbon/Nickel % limits. Min Cost.

Scenario A: Oxidation Loss (Yield Factor)

The Situation: The furnace is not 100% efficient. When we melt Scrap 1, 10% of its mass is lost to oxidation (burn-off). Scrap 2 loses 5%. The other scraps lose 0%. We still need 1000 kg of finished steel.

The Modification:

• Logic: The input weight does not equal the output weight.
  • Input from Scrap 1: . Contribution to final steel: .
    
  • Input from Scrap 2: . Contribution to final steel: .
    
• Updated Weight Constraint:
  
  
• Impact on Ratio Constraints:
  • Be careful! Does the Carbon percentage apply to the input mass or the retained mass? Usually, chemical elements are retained while “slag” is lost, or they burn off proportionally.
    
  • Assumption: If Carbon is retained but total mass drops, the % Carbon increases.
    
  • New Carbon Constraint:
    
  • (You must calculate Total Carbon Mass based on the physics of the burn-off).

Scenario B: The “Recipe” Requirement

The Situation: To ensure structural integrity, the metallurgists require that we use at least twice as much Scrap 2 as Scrap 1.

The Modification:

• Logic: A direct comparison between two variables.
  • 
    
• New Constraint:
  

3. Tubular Products (Rardin 4.3)

Base Model: Assign production of Products () to Mills () to minimize cost. Capacity is in Tons.

Scenario A: Capacity in Hours (The Unit Trap)

The Situation: The previous model assumed Mill Capacity was in Tons. Realistically, Mill 1 runs for 200 hours a month.

• Making Standard tube takes 0.5 hours/ton.
  
• Making Pressure tube takes 0.8 hours/ton.
  
• Making Special tube takes 1.2 hours/ton.

The Modification:

• Logic: We cannot sum tons () because tons of “Special” eat up more capacity than tons of “Standard”. We must convert everything to Hours.
  
• Parameter: Let be hours per ton for product .
  
• New Capacity Constraint (for Mill 1):
  
  
• (Repeat for other mills with their specific parameters).

Scenario B: Risk Diversification

The Situation: Management is worried about a strike at Mill 2. They decree: “No more than 40% of the total production of ‘Pressure’ tubing can come from Mill 2.”

The Modification:

• Logic: Relate a specific variable () to the total demand for that product ().
  
• New Constraint:
  
  (Since Demand is a fixed parameter, this is a simple upper bound).

4. Ohio National Bank (Rardin 4.5)

Base Model: Set Covering. Shifts () cover hourly demand (). Min Cost.

Scenario A: Service Quality Ratio

The Situation: Part-timers are cheaper ($40) than Full-timers ($100), so the Solver wants to use only Part-timers. The bank adds a rule: “At any specific hour, at least 50% of the tellers on duty must be Full-Time employees.”

The Modification:

• Logic: This applies to every hour . We need to look at the “Staff on Duty” expression.
  
• Let:
  • (Full timers working at hour ).
    
  • (Total staff working at hour ).
    
• Constraint:
  
  Rearranged to Linear Form:
  

Scenario B: The “Overtime” Option

The Situation: Full-time shifts are 9am-5pm. The bank is short-staffed from 5pm-6pm. Management allows Full-Timers starting at 9am to work one extra hour (OT) from 5pm-6pm for a cost of $15 (vs regular $12.50/hr).

The Modification:

• Logic: We cannot just “extend” the existing variable . That assumes everyone works OT. We need a choice.
  
• New Variables: Split the original into two:
  1. : Works 9-5. Cost $100.
     
  2. : Works 9-6. Cost $115 ($100 + $15).
     
• Matrix Update:
  • has 0 in the 5-6pm slot.
    
  • has 1 in the 5-6pm slot.
    
• Note: This increases the problem size (more columns) but gives the Solver the granular choice it needs.