# **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 (![][image1]) 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 ![][image2] across *all* areas (![][image3]) and ensure that sum is less than or equal to ![][image4].  
* **New Constraints:**  
  We need one constraint for each prescription ![][image2] (or specifically for Timber/Grazing).  
  ![][image5]  
  *(Where ![][image6] is the fixed size of area ![][image7].)*

### **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 ![][image7] individually.  
* **New Constraints:**  
  ![][image8]  
  *(Note: This tightens the feasible region significantly. It forces ![][image9] to be non-zero everywhere).*

## **2\. Swedish Steel Blending (Rardin 4.2)**

**Base Model:** Mix scrap (![][image10]) 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: ![][image11]. Contribution to final steel: ![][image12].  
  * Input from Scrap 2: ![][image13]. Contribution to final steel: ![][image14].  
* **Updated Weight Constraint:**  
  ![][image15]  
* **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:* ![][image16]  
  * (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.  
  * ![][image17]  
* **New Constraint:**  
  ![][image18]

## **3\. Tubular Products (Rardin 4.3)**

**Base Model:** Assign production of Products (![][image19]) to Mills (![][image20]) 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 (![][image21]) because tons of "Special" eat up more capacity than tons of "Standard". We must convert everything to Hours.  
* **Parameter:** Let ![][image22] be hours per ton for product ![][image19].  
* **New Capacity Constraint (for Mill 1):**  
  ![][image23]  
* *(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 (![][image24]) to the total demand for that product (![][image25]).  
* **New Constraint:**  
  ![][image26]  
  *(Since Demand is a fixed parameter, this is a simple upper bound).*

## **4\. Ohio National Bank (Rardin 4.5)**

**Base Model:** Set Covering. Shifts (![][image10]) cover hourly demand (![][image27]). 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* ![][image28]. We need to look at the "Staff on Duty" expression.  
* **Let:**  
  * **![][image29]** (Full timers working at hour ![][image28]).  
  * ![][image30] (Total staff working at hour ![][image28]).  
* **Constraint:**  
  ![][image31]  
  *Rearranged to Linear Form:*  
  ![][image32]

### **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 ![][image33]. That assumes *everyone* works OT. We need a choice.  
* **New Variables:** Split the original ![][image33] into two:  
  1. ![][image34]: Works 9-5. Cost $100.  
  2. ![][image35]: Works 9-6. Cost $115 ($100 \+ $15).  
* **Matrix Update:**  
  * **![][image34]** has 0 in the 5-6pm slot.  
  * ![][image35] has **1** in the 5-6pm slot.  
* *Note:* This increases the problem size (more columns) but gives the Solver the granular choice it needs.

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[image17]: 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