# Session 3a

## Overview

• Multiperiod Models

– Data Processing at the IRS

– Arbitrage with Bonds

• Transportation Models

– Gribbin Brewing

Decision Models -- Prof. Juran 2

## IRS Example

• To process income tax forms, the

Internal Revenue Service (IRS) first sends each form through the data preparation (DP) department, where information is coded for computer entry.

• Then the form is sent to data entry (DE), where it is entered into the computer.

Decision Models -- Prof. Juran 3

## IRS Example

• During the next 3 weeks, the following numbers of forms will arrive: week 1,

40,000; week 2, 30,000; week 3, 60,000.

• All employees work 40 hours per week and are paid \$500 per week.

• Data preparation of a form requires 15 minutes, and data entry of a form requires 10 minutes.

Decision Models -- Prof. Juran 4

## IRS Example

• Each week an employee is assigned to either data entry or data preparation.

• The IRS must complete processing all forms by the end of week 5 and wants to minimize the cost of accomplishing this goal.

• Assume all employees are capable of performing data preparation or data entry, but must be assigned to one task for an entire week at a time.

Decision Models -- Prof. Juran 5

Managerial Problem Definition

• Determine how many workers should be working and how the workers should allocate their hours during the next 5 weeks.

Decision Models -- Prof. Juran 6

Formulation

Decision Variables

Numbers of workers for two tasks over five weeks (10 decisions) and numbers of forms completed on each task in each week (10 decisions).

Objective

Minimize total cost.

Constraints

Forms arrive at a fixed schedule.

All work must be completed in five weeks.

Data prep task cannot begin until the forms arrive.

The plan can’t call for more labor than is available for either task in any week.

Decision Models -- Prof. Juran 7

Formulation

Decision Variables

Define X ij

= Number of workers on task i during week j.

Define P ij

= Production (forms processed on task i during week j).

i = tasks 1-2, j = weeks 1-5

Objective

Minimize Z =

2 5

 i  1 j  1

X ij

C i

Where C i

is the cost of hiring a worker for task i for one week.

C

1

= C

2

= \$500

Decision Models -- Prof. Juran 8

Formulation

Constraints

A new kind of constraint: Balance Equations for each task in each week.

Define I ij

= Inventory (work ready to be processed on task i at the end of week j).

Ending “inventory” = beginning inventory + new forms arriving – current period processing

I ij

 I i , j  1

 P i  1 , j  1

 P ij

Note that

P

0 , 0

P

0 , 1

P

0 , 2

40 , 000

30 , 000

60 , 000

Decision Models -- Prof. Juran 9

3

4

5

6

1

2

7

8

9

10

11

12

13

14

15

28

29

30

31

32

33

34

35

36

37

38

21

22

23

24

25

26

27

16

17

18

19

20

39

Inputs

Data prep form time

Constraints on forms (can't process more than are available for processing)

DP forms available for processing =B14

DP forms processed

DE forms processed

Total cost

Solution Methodology

A B C D E F G

Minutes

Data entry form time

Hours per worker per week

Pay/week per worker

Number of forms arriving

Decisions

Workers on data prep

Workers on data entry

Number of DP forms completed during week

Number of DE forms completed during week

Constraints on labor (need sufficient workers to complete forms)

Hours for DP needed

Hours for DP available

Hours for DE needed

Hours for DE available

DE forms available for processing

Constraints on finishing all forms by the end of week 5

DP at end

DE at ent

=B8

=B14

15

10

40

\$500

Week1

40000

1

1

1

1

0.25

<=

40

40000

>=

1

1

>=

1

=B15

129995

0

\$5,000

Week2

30000

1

1

1

1

0.25

<=

40

69999

>=

1

1

>=

1

=F27-F29

=

=

Week3

60000

1

1

1

1

0.25

<=

40

0

0

0.25

<=

40

=B5*(SUM(B11:F11)+SUM(B12:F12))

0.25

<=

40

0.166667 0.166667 0.166667 0.166667 0.166667

<= <= <= <= <=

40 40 40 40 40

=B27-B29+C8

129998

>=

1

=B31-B33+C29

1

>=

1

Week4

0

1

1

1

1

129997

>=

1

1

>=

1

Week5

0

1

1

1

1

129996

>=

1

1

>=

1

=F14*\$B\$2/60

=F11*\$B\$4

=F15*\$B\$3/60

=F12*\$B\$4

H

Decision Models -- Prof. Juran 10

Solution Methodology

Decision Models -- Prof. Juran 11

6

7

4

5

8

9

10

1

2

3

Solution Methodology

A B C D E F

Inputs Minutes

Data prep form time

Data entry form time

Hours per worker per week

Pay/week per worker

Number of forms arriving

Decisions

15

10

40

\$500

Week1

40000

Week2

30000

Week3

60000

Week4

0

Week5

0

250

0

187.5

0

375

0

0 0

0 541.6667

11

12

13

Workers on data prep

Workers on data entry

14

15

Number of DP forms completed during week

Number of DE forms completed during week

30

31

32

33

26

27

28

29

34

35

36

37

38

16

17

18

19

20

21

22

23

24

25

Constraints on labor (need sufficient workers to complete forms)

Hours for DP needed

Hours for DP available

Hours for DE needed

Hours for DE available

Constraints on forms (can't process more than are available for processing)

DP forms available for processing

DP forms processed

DE forms available for processing

DE forms processed

Constraints on finishing all forms by the end of week 5

DP at end

DE at ent

39 Total cost

40000

0

10000

<=

10000

0

<=

0

40000

>=

40000

40000

>=

0

0

0

\$677,083

30000

0

7500

<=

7500

0

<=

0

30000

>=

30000

70000

>=

0

=

=

60000

0

15000

<=

15000

0

<=

0

60000

>=

60000

130000

>=

0

0

0

0

<=

0

0

0

0

130000

0

<=

0

0 21666.67

<=

0

<=

21667

0

>=

0

130000

>=

0

0

>=

0

130000

>=

130000

Decision Models -- Prof. Juran 12

Optimal Solution

Minimum total cost is \$677,073.

All work could be done in four weeks.

Note that the balance equations are not constraints in the usual sense (i.e. specified in

Solver). We build them into the model, linking the tasks and weeks together.

Decision Models -- Prof. Juran 13

## Bond Arbitrage Example

Bond

1

2

3

Current Price

\$10l.625

\$10l.5625

\$103.80

Expiration Date

8/15/2000

2/15/2001

2/15/2001

Coupon Rate

6.875

5.5

7.75

Decision Models -- Prof. Juran 14

Formulation

Given the current price structure, the question is whether there is a way to make an infinite amount of money. To answer this, we look for an arbitrage.

An arbitrage exists if there is a combination of bond sales and purchases today that yields

• A positive cash flow today

• Non-negative cash flows at all future dates

Decision Models -- Prof. Juran 15

Formulation

If such a strategy exists, then it is possible to make an infinite amount of money.

For example, if buying 10 units of bond 1 today and selling 5 units of bond 2 today yielded, say,

\$1 today and nothing at all future dates, then we could make \$k by purchasing 10k units of bond 1 today and selling 5k units of bond 2 today.

Decision Models -- Prof. Juran 16

Formulation

Decision Variables

How much to buy or sell of each bond. (Selling a bond is conceptually the same as buying a negative amount.)

Objective

Maximize cash flow at the end of the first period

(today).

Constraints

Non-negative cash flow at the end of all future periods.

Decision Models -- Prof. Juran 17

Formulation

Decision Variables

Define X i

= quantity of bond i purchased today.

Define C ij

= Cash flow per face value unit for bond i in period j, as shown below.

Bond 1

i = bonds Bond 2

Bond 3

j = periods

0 months from now 6 months from now 12 months from now

-\$101.63

-\$101.56

-\$103.80

\$103.44

\$2.75

\$3.88

\$0.00

\$102.75

\$103.88

Decision Models -- Prof. Juran 18

Formulation

3 3

 i  1 j  1

X i

C ij

= total cash flows from all bonds over all three periods. j

3

 1

X i

C ij

= total cash flows from bond i over all three periods. i

3

 1

X i

C ij

= total cash flows from all bonds in period j.

Decision Models -- Prof. Juran 19

Formulation

Objective

Maximize Z = i

3

 1

X i

C i 1

Constraints i

3

 1

X i

C ij

 0 for j = 2, 3.

Decision Models -- Prof. Juran 20

Solution Methodology

20

21

22

14

15

16

17

18

19

7

8

5

6

3

4

1

2

9

10

11

12

13

A

Data on bonds

Bond

1

2

3

Face value of each bond

Decisions: number of bonds to buy or sell now

Bond

1

2

3

Cash flows

Months from now

Bond 1

Bond 2

Bond 3

Total cash flow

B C D E F G H

Current price Expiration date Coupon rate

\$ 101.625

6 0.06875

\$

\$

101.563

103.800

12

12

0.05500

0.07750

100

1.00

1.00

1.00

=B3*(C11-B11)

Sell

0.00

0.00

0.00

If we maximize the cash flow in cell B22, without an upper bound constraint on it (as suggested in cells

B23 and B24), the Solver does not converge. To make it converge, we add this upper bound constraint. This lets us make \$1 (or any other value you want to put in cell B24).

=IF(\$C5=D\$16,1+\$D5/2,IF(\$C5>D\$16,\$D5/2,0))*(\$B13-\$C13)*\$B\$7

0

\$ (101.63)

\$ (101.56)

\$ (103.80)

\$ (306.99)

<=

\$ 1.00

6

\$ 103.44

\$ 110.06

>=

\$ -

12

\$ -

\$ 102.75

\$ 103.88

\$ 206.63

>=

\$ -

=SUM(D17:D19)

Decision Models -- Prof. Juran 21

Decision Models -- Prof. Juran 22

Solution Methodology

Decision Models -- Prof. Juran 23

Solution Methodology

This is actually good news! It indicates an “unbounded” problem; one in which there are no constraints that limit the value of the objective function. In the context of this problem, it means that there is no limit on the amount of cash flow in the first period. In other words, there is an arbitrage opportunity.

Unfortunately, because Solver couldn’t solve the problem, we don’t know which bonds to buy and sell.

We can get around this by playing a little trick; we introduce a new constraint limiting the objective function artificially.

Decision Models -- Prof. Juran 24

Decision Models -- Prof. Juran 25

20

21

22

14

15

16

17

18

19

7

8

5

6

3

4

1

2

9

10

11

12

13

Optimal Solution

B A

Data on bonds

Bond

1

2

3

Face value of each bond

Decisions: number of bonds to buy or sell now

Bond

1

2

3

Cash flows

Months from now

Bond 1

Bond 2

Bond 3

Total cash flow

Current price Expiration date Coupon rate

\$

\$

\$

101.625

101.563

103.800

6

12

12

0.06875

0.05500

0.07750

100

0.21

20.30

0.00

C

Sell

0.00

0.00

20.08

D

0

\$ (21.60)

\$ (2,061.87)

\$ 2,084.48

\$ 1.00

<=

\$ 1.00

6

\$ 21.99

\$ 55.83

\$ (77.82)

\$ -

>=

\$ -

12

\$ -

\$ 2,085.98

\$ (2,085.98)

\$ -

>=

\$ -

Decision Models -- Prof. Juran 26

Conclusions

Buying bonds 1 and 2 today, while selling bond 3, offers an arbitrage opportunity.

Decision Models -- Prof. Juran 27

Back to Reality

Usually bonds are bought at an ask price and sold at a bid price.

Consider the same three bonds and suppose that the ask and bid prices are as listed here.

Bond

1

2

3

\$101.6563 \$101.5938

\$101.5938 \$101 5313

\$103.7813 \$103.7188

Decision Models -- Prof. Juran 28

14

15

16

17

18

19

20

21

22

7

8

5

6

3

4

1

2

9

10

11

12

13

Optimal Solution

B A

Data on bonds

Bond

1

2

3

Face value of each bond

Decisions: number of bonds to buy or sell now

Bond

1

2

3

Cash flows

Months from now

Bond 1

Bond 2

Bond 3

Total cash flow

100

0

0

0

0

\$ -

\$ -

\$ -

\$ -

C

Sell

0

0

0

6

\$ -

\$ -

\$ -

\$ -

>=

\$ -

D

12

>=

E

Bid price Expiration date Coupon rate

6

12

12

0.06875

0.055

0.0775

Decision Models -- Prof. Juran 29

Conclusions

This result indicates that no arbitrage opportunity exists.

The only way to have non-negative cash flows in the first period and zero cash flows in all future periods is not to invest at all.

Decision Models -- Prof. Juran 30

## Gribbin Brewing

Regional brewer Andrew Gribbin distributes kegs of his famous beer through three warehouses in the greater News York City area, with current supplies as shown:

Warehouses Supply

Hoboken

Bronx

Brooklyn

80

145

120

Decision Models -- Prof. Juran 31

On a Thursday morning, he has his usual weekly orders from his four loyal customers, as shown :

Bars

Der Ratkeller

McGoldrick's Pub

Night Train Bar & Grill

Demand

80

65

70

85

Decision Models -- Prof. Juran 32

Tracy Chapman, Gribbin’s shipping manager, needs to determine the most cost-efficient plan to deliver beer to these four customers, knowing that the costs per keg are different for each possible combination of warehouse and customer:

Hoboken

Bronx

Brooklyn

Ratkeller McGoldrick's Night Train Stern

\$4.64

\$3.52

\$9.95

\$5.13

\$4.16

\$6.82

\$6.54

\$6.90

\$3.88

\$8.67

\$7.91

\$6.85

Decision Models -- Prof. Juran 33

a) What is the optimal shipping plan?

b) How much will it cost to fill these four orders?

c) Where does Gribbin have surplus inventory?

d) If Gribbin could have one additional keg at one of the three warehouses, what would be the most beneficial location, in terms of reduced shipping costs? e) Gribbin has an offer from Lu Leng Felicia, who would like to sublet some of Gribbin’s Brooklyn warehouse space for her tattoo parlor. She only needs 240 square feet, which is equivalent to the area required to store 40 kegs of beer, and has offered Gribbin \$0.25 per week per square foot. Is this a good deal for Gribbin? What should Gribbin’s response be to

Lu Leng?

Decision Models -- Prof. Juran 34

Managerial Problem Formulation

Decision Variables

Numbers of kegs shipped from each of three warehouses to each of four customers (12 decisions).

Objective

Minimize total cost.

Constraints

Each warehouse has limited supply.

Each customer has a minimum demand.

Kegs can’t be divided; numbers shipped must be integers.

Decision Models -- Prof. Juran 35

Mathematical Formulation

Decision Variables

Define X ij

= Number of kegs shipped from warehouse i to customer j.

Define C ij

= Cost per keg to ship from warehouse i to customer j.

i = warehouses 1-3, j = customers 1-4

Decision Models -- Prof. Juran 36

Mathematical Formulation

Objective

Minimize Z =

3 4

 i  1 j  1

X ij

C ij

Constraints

Define S i

= Number of kegs available at warehouse i.

j

4

 1

X ij

 S i

Define D j

= Number of kegs ordered by customer j.

i

3

 1

X ij

 D j

Do we need a constraint to ensure that all of the X ij are integers?

Decision Models -- Prof. Juran 37

7

8

9

10

11

12

13

14

4

5

6

1

2

3

A

Total Cost

Costs

Hoboken

Bronx

Brooklyn

B

74.97

C D E

=SUMPRODUCT(B4:E6,B12:E14)

Shipping Plan Ratkeller McGoldrick's Night Train Stern

Hoboken

Bronx

Brooklyn

1

1

1

1

1

1

1

1

1

1

1

1

3

=

80

3

=

65

3

=

70

3

=

85

\$

\$

\$

4.64

3.52

9.95

\$

\$

\$

5.13

4.16

6.82

\$

\$

\$

6.54

6.90

3.88

F G H I

=SUM(B4:E4)

4 <= 80

4 <= 145

4 <= 120

=SUM(E4:E6)

Decision Models -- Prof. Juran 38

Decision Models -- Prof. Juran 39

1

4

5

6

2

3

7

8

9

10

11

12

13

14

A

Total Cost

Costs

Hoboken

Bronx

Brooklyn

1469.55

\$

\$

\$

B

4.64

3.52

9.95

C

\$

\$

\$

5.13

4.16

6.82

D

\$

\$

\$

6.54

6.90

3.88

\$

\$

\$

E

8.67

7.91

6.85

F G H

Shipping Plan Ratkeller McGoldrick's Night Train Stern

Hoboken

Bronx

Brooklyn

0

80

0

0

65

0

0

0

70

35 35 <= 80

0 145 <= 145

50 120 <= 120

80

=

80

65

=

65

70

=

70

85

=

85

Decision Models -- Prof. Juran 40

20

21

22

23

24

25

26

27

28

29

30

3

4

1

2

5

6

7

12

13

14

15

8

9

10

11

16

17

18

19

A B C D

Microsoft Excel 15.0 Sensitivity Report

Worksheet: [03a-03-trans.xls]Model

Variable Cells

Cell

Constraints

Cell

\$F\$4 Hoboken

\$F\$5 Bronx

\$F\$6 Brooklyn

Name

\$B\$4 Hoboken Ratkeller

\$C\$4 Hoboken McGoldrick's

\$D\$4 Hoboken Night Train

\$E\$4 Hoboken Stern

\$B\$5 Bronx Ratkeller

\$C\$5 Bronx McGoldrick's

\$D\$5 Bronx Night Train

\$E\$5 Bronx Stern

\$B\$6 Brooklyn Ratkeller

\$C\$6 Brooklyn McGoldrick's

\$D\$6 Brooklyn Night Train

\$E\$6 Brooklyn Stern

Name

\$B\$7 Ratkeller

\$C\$7 McGoldrick's

\$D\$7 Night Train

\$E\$7 Stern

E F G H

Final Reduced Objective Allowable Allowable

Value Cost Coefficient Increase Decrease

80

65

0

0

0

35

0

0

70

50

0

0

0.15

0

0.84

0

0

0

2.17

0.21

7.28

3.51

0

0

4.64

5.13

6.54

8.67

3.52

4.16

6.9

7.91

9.95

6.82

3.88

6.85

1E+30

0.15

1E+30

0.21

0.15

0.21

1E+30

1E+30

1E+30

1E+30

0.84

1.82

0.15

0.21

0.84

1.82

1E+30

0.15

2.17

0.21

7.28

3.51

1E+30

0.84

Value Price

80

65

70

85

35

145

120

4.49

5.13

5.7

8.67

0

-0.97

-1.82

R.H. Side

80

65

70

85

80

145

120

Increase

45

45

45

45

1E+30

0

35

Decrease

35

35

0

0

45

45

45

Decision Models -- Prof. Juran 41

Where does Gribbin have surplus inventory?

The only supply constraint that is not binding is the

Hoboken constraint. It would appear that Gribbin has

45 extra kegs in Hoboken.

Decision Models -- Prof. Juran 42

If Gribbin could have one additional keg at one of the three warehouses, what would be the most beneficial location, in terms of reduced shipping costs?

Decision Models -- Prof. Juran 43

According to the sensitivity report,

•One more keg in Hoboken is worthless.

•One more keg in the Bronx would have reduced overall costs by \$0.76.

•One more keg in Brooklyn would have reduced overall costs by \$1.82.

Decision Models -- Prof. Juran 44

Gribbin has an offer from Lu Leng Felicia, who would like to sublet some of Gribbin’s Brooklyn warehouse space for her tattoo parlor. She only needs 240 square feet, which is equivalent to the area required to store 40 kegs of beer, and has offered Gribbin \$0.25 per week per square foot.

Is this a good deal for Gribbin?

What should Gribbin’s response be to Lu Leng?

Decision Models -- Prof. Juran 45

Assuming that the current situation will continue into the foreseeable future, it would appear that Gribbin could reduce his inventory in Hoboken without losing any money (i.e. the shadow price is zero).

However, we need to check the sensitivity report to make sure that the proposed decrease of 40 kegs is within the allowable decrease.

This means that he could make a profit by renting space in the Hoboken warehouse to Lu Leng for \$0.01 per square foot.

Decision Models -- Prof. Juran 46

Lu Leng wants space in Brooklyn, but Gribbin would need to charge her more than \$1.82 for every six square feet

(about \$0.303 per square foot), or else he will lose money on the deal.

Note that the sensitivity report indicates an allowable decrease in Brooklyn that is enough to accommodate Lu

Leng.

Decision Models -- Prof. Juran 47

As for the Bronx warehouse, note that the allowable decrease is zero. This means that we would need to rerun the model to find out the total cost of renting Bronx space to Lu Leng.

A possible response from Gribbin to Lu Leng:

“I can rent you space in Brooklyn, but it will cost you

\$0.35 per square foot. How do you feel about

Hoboken?”

Decision Models -- Prof. Juran 48

## Summary

• Multiperiod Models

– Data Processing at the IRS

– Arbitrage with Bonds

• Transportation Models

– Gribbin Brewing

Decision Models -- Prof. Juran 49