MGMT 101: Management Science

Professor Shuya Yin

Session Four

Outline for today: Applications of Linear

Programming Models in other business settings

So far ……

KARMA Computer problem

Galaxy Industries (space ray and zapper)

Golden Electronics (GE45 and GE60)

Advertising (Workday Ads and Sunday Ads)

Financing Strategy (Model Y and Model Z from internal or external investment)

Fabulous Nuts (regular, deluxe and holiday)

Production models

What else?

Production and Inventory planning problem in multiple periods

……

Staff scheduling problem (Chapter 3.3)

Transportation and distribution problem (Chapter 3.5)

Funds management problem

Example 1: Funds management problem

Tritech Mortgage specializes in making first, second, and even third trust deed loans on residential properties and first trust deeds on commercial properties. Any funds not invested in mortgages are invested in an interest-bearing savings account. The following table gives the rate of return and the company’s risk level for each possible type of loans.

Tritech wishes to invest $68,000K in available funding so that: Yearly return is maximized At least $5,000K is to be available in a savings account for emergencies At least 80% of the money invested in trust deeds should be in residential properties At least 60% of the money invested in residential properties should be in first trust deeds The average risk should no exceed 5

Loan Type Rate of Return Risk

First Trust Deeds 7.75% 4

Second Trust Deeds 11.25% 6

Third Trust Deeds 14.25% 9

Commercial Trust Deeds 8.75% 3

Savings Account 4.45% 0

Tritech Mortgage problem formulation

Example 2: Staff scheduling (chapter 3.3)

A fast-food restaurant has daily staff needs that range from 12 to 19 students depending on varying work loads each day of the week as shown below:

Day Staff Needs

Monday 18

Tuesday 16

Wednesday 15

Thursday 16

Friday 19

Saturday 14

Sunday 12

In addition, a labor requirement must be met that employees work a five consecutive days followed by two days off. Thus, the allowable shifts are Monday through Friday, Tuesday through Saturday, Wednesday through Sunday, etc. Each employee earns $200 per week.

As the personnel manager of the restaurant, how would you develop a feasible staff schedule that ensures that staff needs are covered at a minimal weekly cost?

Formulation – staff scheduling

Shift 1 (starting on Mon) will work on: Shift 2 (starting on Tue) will work on: Shift 3 (starting on Wed) will work on: Shift 4 (starting on Thu) will work on: Shift 5 (starting on Fri) will work on: Shift 6 (starting on Sat) will work on: Shift 7 (starting on Sun) will work on:

Mon (1) Tue (2) Wed (3) Thu (4) Fri (5) Sat (6) Sun (7)

Let us try it

Handout will be distributed in class.

In-class practice problem

Example 3: Transportation (chapter 3.5) You are in charge of scheduling transportation of bottled water from 3

warehouses to 4 retail stores.

4How would you schedule the transportation to minimize the total shipping cost?

WarehouseRetail store

90

1

2

3

1

2

3

50

40

70

200

80

180

0.120.07

0.05

0.080.14 0.10

0.06

0.07

$0.080.090.11

0.16

maximum supply

minimum demand

Let us try it

Transportation – extension

Now, how should you schedule the production and transportation to minimize your total shipping and production costs?

($0.5)

($0.4)

($0.3)

4

Warehouse Retail store90

1

2

3

1

2

3

50

40

70

200

80

180

0.120.07

0.05

0.080.14 0.10

0.06

0.07

$0.080.090.11

0.16

What changes?

Handout will be distributed in class.

In-class practice problem

Example 4: Distribution Unlimited Co. Problem (Ch 6)

Each unit is a full truckload.

Minimum cost flow problem: figure out a best way to ship the product (meet the demand at the lowest cost).

A math model:

Decision variables:

Total cost (Objective function):

Constraints:

(F1)

(F2)

(W1)

(W2)

(DC)

Example 5: GlobChem Production/Transportation

A US-based multinational produces a specialty chemical in four dedicated plants, located in Newark, Los Angeles, Rotterdam and Kuala Lumpur.

The product is marketed worldwide, and the company has just instituted a sales organization with sales offices/distribution centers based in: Newark, Rotterdam, São Paulo and Tokyo.

Prices and costs vary by region and plants

Each sales/distribution center is responsible for a distinct region; the maximum demand and the average selling price for a region are given below.

Plant design, factor costs, and exchange rates affect production costs and each plant Annual Fixed costs are incurred regardless of the production volume.

Transportation costs include freight and tariffs.

Office Newark São Paulo Rotterdam Tokyo

Region N. America S. America Eu / Af / ME Asia

Market price ($/ton) 55,500 61,100 57,800 62,650

Demands (tons/yr) 150 75 200 100

$ per ton from plant to market Newark São Paulo Rotterdam Tokyo

Newark -- 12,225 9,075 21,450

Los Angeles 4,500 16,500 13,350 17,850

Rotterdam 9,150 12,600 -- 12,525

Kuala Lumpur 21,450 18,450 15,150 5,925

Plant Data Unit cost ($ per ton) Fixed costs ($ 000) Capacity (tons/yr)

Newark 34,900 1,800 100

Los Angeles 32,200 2,750 200

Rotterdam 38,350 2,100 120

Kuala Lumpur 23,400 1,950 250

GlobChem’s problem statement

GlobChem wants to determine its annual production and distribution schedule in order to maximize profits while observing the limits on each plant’s capacity and the maximum sales possible in each region. In determining this production and distribution schedule, consider the locations and capacities of the plants to be fixed.

Prices and costs vary by region and plants

The index i refers to plants, where 1 is Newark, 2 Los Angeles, 3 Rotterdam, and 4 Kuala Lumpur.The index j refers to markets, where 1 is Newark, 2 São Paulo, 3 Rotterdam, and 4 Tokyo.

N L R K

i=1 i=3i=2

j=1

i=4

j=3j=2 j=4

N S R T

ai = available capacity at plant i (e.g., a2 = 200 tons)

ci = unit cost at plant i (e.g., c3 = $38,350)

tij = unit transportation cost from plant i to market j (e.g., t33 = 0, t42 = 18,450)

dj = demand in market j (e.g., d1 = 150 tons)

pj = unit price in market j (e.g., p4 = $62,650)

Describe the problem in words

Question: what are the decision variables?

Problem Statement

Maximize annual profit contribution by determining the annual production and transportation schedule, ensuring

that plant capacity and market demand limits are not exceeded.

Constraints in words Capacity constraints: production at each plant must be less than capacity

- (production at Newark plant) = ≤ - (production at Los Angeles) = ≤- (production at Rotterdam) = ≤- (production at Kuala Lumpur) = ≤

Demand constraints: cannot ship more than the demand in any market- (total shipped to Newark) = ≤

- (total shipped to São Paulo) = ≤- (total shipped to Rotterdam) = ≤- (total shipped to Tokyo) = ≤

Non-negative constraints: all shipment quantities must be nonnegative

Objective function formulation in words

Objective function in words: maximize the annual profit contribution

- Max profit contribution = revenues – production costs – transportation costs

- Revenue =

- Production cost =

- Transportation cost =

Complete LP formulation of the GlobChem problem Decision variables:

xij = tons of products produced at plant i and shipped to market j

Objective function: Maximize profit contribution = revenue – production and transportation costsi.e., Max 55500(x11 + x21 +x31 + x41) (revenue at Newark)

+ 61100(x12 + x22 + x32 + x42) (revenue at São Paulo)

+ 57800(x13 + x23 + x33 + x43) (revenue at Rotterdam)

+ 62650(x14 + x24 + x34 + x44) (revenue at Tokyo)

- 34900(x11 + x12 + x13 + x14) (production cost at Newark)

- 32200(x21 + x22 + x23 + x24) (production cost at Los Angeles)

- 38350 (x31 + x32 +x33 + x34) (production cost at Rotterdam)

- 23400(x41 + x42 +x43 + x44) (production cost at Kuala Lumpur)

- (0 x11 + 12225x12 + 9075x13 + 21450x14 + 4500x21 + …

+15150x43 + 5925x44) (transportation cost)

LP formulation of the GlobChem problem (cont’d) Constraints:

- Production capacity may not be exceededx11 + x12 + x13 + x14 ≤ 100 (capacity at Newark)

x21 + x22 + x23 +x24 ≤ 200 (capacity at Los Angeles)

x31 + x32 + x33 + x34 ≤ 120 (capacity at Rotterdam)

x41 + x42 + x43 + x44 ≤ 250 (capacity at Kuala Lumpur)

- Maximum demand may not be exceededx11 + x21 + x31 + x41 ≤ 150 (demand at Newark)

x12 + x22 + x32 + x42 ≤ 75 (demand at São Paulo)

x13 + x23 + x33 + x43 ≤ 200 (demand at Rotterdam)

x14 + x24 + x34 + x44 ≤ 100 (demand at Tokyo)

- Non-negativityxij ≥ 0 for all plants i and markets j

For the mathematically inclined, in compact notation…

The following formulation represents the problem in a generic form. As before, we use the indices i and j for plants and markets, respectively. N represents the number of plants and M the number of markets.

GlobChem sensitivity analysis

Which plant provides the most return from increased production capacity?

Which regional market affords the greatest benefit from increased demand?

Example 6: Production + Inventory in multi periods

Foresight Co. must meet fluctuating demand for its product. Demand for a month may be met either through production in that month and/or inventory carried from the previous month.

Production capacity, unit production cost, and unit holding cost vary from month to month:

Holding costs apply to the amount left in inventory at the end of each month Backlogs are not allowed. The beginning inventory in Jan. is zero.

Problem: Determine the production schedule that minimizes total production and holding cost while meeting demand in each month (from current production and/or inventory). Production in any month must not exceed capacity.

Month Jan Feb Mar Apr May Jun

Demand (lb.) 9,000 7,500 12,500 9,000 8,000 11,500

Production capacity 10,000 10,000 11,000 10,000 11,000 10,000

Unit production cost $14.50 $15.25 $16.00 $16.70 $17.25 $17.00

Unit holding cost per month $0.60 $0.65 $0.65 $0.65 $0.70 $0.65

LP formulation of Foresight’s problem

Decision variables- We need to determine

Pt = production (in lb.) in month t (t = 1,…,6)- Any other?

Objective function (in words)Minimize total production cost + holding cost over the 6-month horizon

( Note that total production cost can be easily expressed as:14.5P1 + 15.25P2 + 16.00P3 + 16.70P4 + 17.25P5 + 17.00P6

We’ll return to total holding cost later. )

LP formulation of Foresight’s problem (cont’d)

Constraints (in words)

- Production in each month must NOT exceed capacity in that month

- The demand for each month must be met from production that month and /or inventory carried from the previous month (no backlogging of demand)

- Production quantity must be nonnegative in each month

Expressing the constraints algebraically…

Production in each month must not exceed capacity in that month.These are straightforward:P1 ≤ 10000; P2 ≤ 10000; P3 ≤ 11000; …; P6 ≤ 10000

The demand for each month must be met from production that month and /or inventory carried from the previous month.This is easy for the first month (Jan) since there is no beginning inventory:

(Month 1)In month 2, we need to include the amount left over (if any) from month 1:

(Month 2)Similarly in month 3, we need to include the ending inventory of month 2:

(Month 3)

Though we can carry on in this fashion, it becomes evident that we would gain considerable convenience by introducing additional (definitional) variables to represent the ending inventory for each month.

Introducing definitional variables for inventory

Define It = ending inventory for month t, t = 1,2, …, 6

Constraints that the demand for each month must be met from production that month and/or inventory carried from the previous month:

• I0 + P1 = 9000 + I1 (Month 1)• I1 + P2 = 7500 + I2 (Month 2)• I2 + P3 = 12500 + I3 (Month 3)• I3 + P4 = 9000 + I4 (Month 4)• I4 + P5 = 8000 + I5 (Month 5)• I5 + P6 = 11500 + I6 (Month 6)

Constraints to disallow backlogging of demand:• I1 ≥ 0, I2 ≥ 0, …, I6 ≥ 0

(Can you see why it isn’t enough for just the Pt to be nonnegative?)

It is now easy to express the total holding cost in the objective function:• 0.60 I1 + 0.65 I2 + 0.65 I3 + 0.65 I4 + 0.70 I5 + 0.65 I6

Complete LP formulation

Min 14.5P1 + 15.25P2 + 16.00P3 + 16.70P4 + 17.25P5 + 17.00P6 (production cost)

+ 0.60 I1 + 0.65 I2 + 0.65 I3 + 0.65 I4 + 0.70 I5 + 0.65 I6 (holding cost)

S.t. P1 ≤ 10000; P2 ≤ 10000; P3 ≤ 11000;

P4 ≤ 10000; P5 ≤ 11000; P6 ≤ 10000 (production capacity)

• I0 + P1 = 9000 + I1 (Month 1 inventory balance)• I1 + P2 = 7500 + I2 (Month 2 inventory balance)• I2 + P3 = 12500 + I3 (Month 3 inventory balance)• I3 + P4 = 9000 + I4 (Month 4 inventory balance)• I4 + P5 = 8000 + I5 (Month 5 inventory balance)• I5 + P6 = 11500 + I6 (Month 6 inventory balance)

• I1 ≥ 0, I2 ≥ 0, …, I6 ≥ 0 (no backlogging allowed)• P1 ≥ 0, P2 ≥ 0, …, P6 ≥ 0 (production must be nonnegative)

Can you see why the optimal solution will always set the last It variable to zero?

Inventory “balance” constraints arise in many other applications

The inventory balance equation:Starting inventory + production = demand + ending inventory

It-1 + Pt = St + It

This is similar to flow balance constraints that arise in other applications, particularly those involving cash flows.

There can be growth or spoilage in the inventory from t to t+1- “Inventories” of invested cash in multi-period financial models- “Negative inventories” of borrowings, which need to be repaid, with interest

Exam 1 Review

Exam Time, Location, Office Hours

Exam 1:

- Time: Tuesday August 16, 2:00 – 3:30 (90 minutes)

- Location: classroom SSL 290

Office hours: Tuesday August 16

- Time: 10:30am to 11:30am in my office SB 309- 11:30am to 12:30pm in SB 410 (TA will be present)

- 1:00pm to 2:00pm Q&A time in classroom.

Format

4 – 6 computational questions, each may have several sub questions

Open book and notes

Ruler, calculator, textbook and class notes allowed

No laptop

No discussion with one another

No cell phone

Refer to the sample exam 1 on EEE

Need to know

Problem formulation- All the examples that were covered in the lectures except the last example on

production and inventory planning in multi periods discussed today

Problem solving using graphs (for problems with two-decision variables)

Interpretation of the info on the answer and sensitivity reports generated by Excel

In the textbook: - Chapters 1, 2, 3 and 5

Formulation

Three steps:- Define decision variables- Objective function (do not forget to specify “maximize” or “minimize”)- Constraints

Some guidance for defining decision variables:

- Production problem

- Staff scheduling problem

- Transportation problem

Graphical analysis

Plot constraints and draw the feasible region

Identify extreme points (one or more of them will be the optimal solution)

Identify the optimal point:- Use the objective function

• Set the objective function to an arbitrary value and move the line toward improvement direction

• Identify the optimal point and solve constraints which intersect at this point simultaneously

Plug values of optimal point into objective function the optimal OF value.

Sensitivity analysis

Type I change: coefficients (net profit, revenue or cost) in objective function

- Calculate Range of Optimality

- Range of Optimality: predict the changes in the optimal solution and optimal objective function value if one coefficient changes (others remain unchanged)

- Interpret Reduced Cost of a decision variable

Sensitivity analysis

Type II change: the RHS of a constraint

- Interpret Shadow Price of a constraint Change in objective function due to a unit increase in RHS of a constraint

- Range of Feasibility: an interval in which the shadow price in the Sensitivity Report is valid

Interpret Answer ReportOptimal objective

function value

Optimal solution

Difference b/w RHS and LHS

LHS of constraints at the optimal solution

2X1 + 1X2 <= 10003X1 + 4X2 <= 2400 X1 + X2 <= 700 X1 - X2 <= 350

Sensitivity Report

LHS of constraints at the optimal solution

RHS of constraints

Optimal solution