mba8503c7

5/28/2018 MBA8503C7

1/64

Resource Allocation

Class 7: 3/9/11

5/28/2018 MBA8503C7

2/64

3.1 WHYNETWORKPLANNING?

Find the right balance between inventory,transportation and manufacturing costs,

Match supply and demand under uncertainty bypositioning and managing inventory effectively,

Utilize resources effectively by sourcing productsfrom the most appropriate manufacturing facility

5/28/2018 MBA8503C7

3/64

RESOURCEALLOCATION

In operations and supply chain management thereare several problems related to resource allocation:

Production mix: how many units of each product orservice should be produced given their profitability and

constraints on available resources and their usage foreach product

Network location and sourcing:

where to locate facilities, including manufacturing plants,distribution centers, and warehouses

Given a network of facilities, how to best service my customermix considering transportation and distribution costs (sourcingdecision)

5/28/2018 MBA8503C7

4/64

NETWORKSCHEDULINGEXAMPLE

Single product

Two plants p1 and p2

Plant p2 has an annual capacity of 60,000 units.

The two plants have the same productioncosts.

There are two warehouses w1 and w2 withidentical warehouse handling costs.

There are three markets areas c1,c2 and c3with demands of 50,000, 100,000 and50,000, respectively.

5/28/2018 MBA8503C7

5/64

UNITDISTRIBUTIONCOSTS

Facilitywarehouse

p1 p2 c1 c2 c3

w1 0 4 3 4 5

w2 5 2 2 1 2

5/28/2018 MBA8503C7

6/64

HEURISTIC#1:CHOOSETHECHEAPESTWAREHOUSETOSOURCEDEMAND

D = 50,000

D = 100,000

D = 50,000

Cap = 60,000

$5 x 140,000

$2 x 60,000

$2 x 50,000

$1 x 100,000

$2 x 50,000

Total Costs = $1,120,000

5/28/2018 MBA8503C7

7/64

HEURISTIC#2:CHOOSETHEWAREHOUSEWHERETHETOTALDELIVERY

COSTSTOANDFROMTHEWAREHOUSEARETHELOWEST[CONSIDERINBOUNDANDOUTBOUNDDISTRIBUTIONCOSTS]

D = 50,000

D = 100,000

D = 50,000

Cap = 60,000

$4

$5

$2

$3

$4$5

$2

$1

$2

$0

P1 to WH1 $3

P1 to WH2 $7P2 to WH1 $7P2 to WH 2 $4

P1 to WH1 $4P1 to WH2 $6P2 to WH1 $8P2 to WH 2 $3


Market #1 is served by WH1, Markets 2 and 3

are served by WH2

5/28/2018 MBA8503C7

8/64

D = 50,000

D = 100,000

D = 50,000

Cap = 60,000

Cap = 200,000

$5 x 90,000

$2 x 60,000

$3 x 50,000

$1 x 100,000

$2 x 50,000

$0 x 50,000

P1 to WH1 $3

P1 to WH2 $7P2 to WH1 $7P2 to WH 2 $4



Total Cost = $920,000

HEURISTIC#2:CHOOSETHEWAREHOUSEWHERETHETOTALDELIVERYCOSTSTOANDFROMTHEWAREHOUSEARETHELOWEST[CONSIDERINBOUNDANDOUTBOUNDDISTRIBUTIONCOSTS]

5/28/2018 MBA8503C7

9/64

OPTIMIZATIONAPPROACH

The problem described earlier can beframed as a linear programming problem.

A much better solution is foundtotal cost= $740,000!

How does optimization work?

5/28/2018 MBA8503C7

10/64

LINEAR PROGRAMMING

LP deals with the problem of allocating limitedresources among competing activities

For example, consider a company that makestables and chairs (competing activities) using alimited amount of large and small Legos

(limited resources).

5/28/2018 MBA8503C7

11/64

LINEAR PROGRAMMING

The objective of LP is to select the best oroptimal solution from the set of feasiblesolutions (those that satisfy all of therestrictions on the resources).

Suppose profit for each table is $20 whileprofit for each chair is $16.

We may choose to identify the number oftables and chairs to produce to maximizeprofit while not using more Legos than areavailable.

5/28/2018 MBA8503C7

12/64

COMPONENTS OF AN LP

Decision Variables: factors which are controlledby the decision maker.

x1 = the number of tables produced per day

x2 = the number of chairs produced per day

Objective function: profit, cost, time, or servicemust be optimized.

The objective may be to optimize profit.

5/28/2018 MBA8503C7

13/64

COMPONENTS OF AN LP

Constraints: restrictions which limit theavailability and manner with which resourcescan be used to achieve the objective

It takes 2 large and 2 small Legos to producea table and 1 large and 2 smalls to produce achair

We may only have 6 large and 8 small Legos

available each day

5/28/2018 MBA8503C7

14/64

ASSUMPTIONS

Linearity: Linear objective function and linearconstraints.

This implies proportionality and additivity.

For example, it takes 2 large Legos toproduce 1 table and 4 to produce 2 tables.

It takes 3 large Legos to produce 1 table and1 chair.

5/28/2018 MBA8503C7

15/64

ASSUMPTIONS

Divisibility: The decision variables can take onfractional values.

The optimal solution may tell us to produce2.5 tables each day.

Certainty: The parameters of the model areknown or can be accurately estimated.

For example, we assume that the profitabilityinformation is accurate.

5/28/2018 MBA8503C7

16/64

ASSUMPTIONS

Non-negativity: All decision variables must takeon positive or zero values.

5/28/2018 MBA8503C7

17/64

LEGO PRODUCTS, INC.

Lego Products, Inc. manufactures tables and

chairs.

Profit for each table is $20 while each chairgenerates $16 profit.

Each table is made by assembling two large andtwo small legos. Each chair requires one large

and two small legos.

Currently, Lego Products has six large and eight

small legos available each day.

5/28/2018 MBA8503C7

18/64

Table Chair

LEGO EXAMPLE: INTRODUCTION

Pictures of a table and a chair are shown below.

5/28/2018 MBA8503C7

19/64

How many tables and chairs should we produce

to maximize daily profit?producing 3 tables generates a daily profit of

$60,

producing 4 chairs generates a daily profit of$64, however,

producing 2 tables and 2 chairs generates theoptimal daily profit of $72.

LEGO EXAMPLE: OPTIMALSOLUTION

5/28/2018 MBA8503C7

20/64

USINGSOLVER

Solver in Excel can be used to obtain the solution andwill now be demonstrated

The problem formulation is:

MAX 20*X1+16*X2Subject to:

2*X1+1*X2

5/28/2018 MBA8503C7

21/64

Do we have any unused large or small legos for

all of the solutions that you just found?There are no unused large or small legos forthe optimal solution.

There are 2 unused small legos if 3 tables aremade.

There are 2 unused large legos if 4 chairs aremade.

LEGO EXAMPLE: UNUSEDLEGOS

5/28/2018 MBA8503C7

22/64

The difference between the available resources

and resources used is either slack or surplus.

Slack is associated with each less than or equal

to constraint, and represents the amount ofunused resource.

Surplus is associated with each greater than orequal to constraint, and represents theamount of excess resource above the statedlevel.

LEGO EXAMPLE: SLACKANDSURPLUS

5/28/2018 MBA8503C7

23/64

We have two slack values - one for large legosand one for small legos - and no surplusvalues.

The slack or surplus section shows that bothconstraints have zero slack.

Suppose we must produce at least one table(X1>=1). The original optimal solution is stillthe best. Since X1=2, we produce one surplustable.

LEGO EXAMPLE: SLACKANDSURPLUS

5/28/2018 MBA8503C7

24/64

RIGHTHANDSIDECHANGES

Now we are ready to illustrate the keyconcepts of sensitivity analysis.

How much would you be willing to spend for

one additional large Lego?One additional large Lego is worth $4.

Original solution: X1=2, X2=2, profit = $72;

New solution: X1=3, X2=1, profit = $76.You would be willing to spend up to $4 (76-72) for one additional large Lego.

5/28/2018 MBA8503C7

25/64

How much would you be willing to spend for twoadditional large Legos?

Two additional large Legos are worth $8.

Original solution: X1=2, X2=2, profit = $72;

New solution: X1=4, X2=0, profit = $80.

You would be willing to spend up to $8 (80-72)for two additional large Legos.

RIGHT HAND SIDE CHANGES

5/28/2018 MBA8503C7

26/64

How much would you be willing to spend for

three more large legos?

The third large lego is not worth anything sincethe optimal solution remains unchanged.

What happens if your supplier can only providefive large legos each day?

The optimal solution is: X1=1, X2=3, profit=68,so we lose $4.


5/28/2018 MBA8503C7

27/64

What happens if your supplier can only providefour large legos each day?

The optimal solution is: X1=0, X2=4, profit=64,so we lose another $4.

What happens if your supplier can only providethree large legos each day?

The optimal solution is: X1=0, X2=3, profit=48,so we lose an additional $16, and not $4.


5/28/2018 MBA8503C7

28/64

SHADOW PRICESThis last set of exercises enables us to determine

the shadow price for a resource constraint(large legos).

The shadow price, for a particular constraint, isthe amount the objective function value willincrease (decrease) if the right hand side valueof that constraint is increased (decreased) by

one unit.

We found that the shadow price for large legos is

$4.

5/28/2018 MBA8503C7

29/64

SHADOW PRICES

What is the shadow price of the small legos?

With two additional small legos the new solution:

X1=1, X2=4, profit = $84.

You would be willing to spend up to $12 (84-72) fortwo additional small legos, so the shadow price is $6(12/2).

5/28/2018 MBA8503C7

30/64

SHADOW PRICES

In general, the shadow prices are meaningful ifone right hand side (RHS) value of aconstraint is changed,

and all other parameters of the model remainunchanged.

5/28/2018 MBA8503C7

31/64

What happens if the profit of tables increases to$35?

The optimal solution is X1=3, X2=0, profit =

$105. Note that no chairs are being produced.

REDUCED COSTS

5/28/2018 MBA8503C7

32/64

REDUCED COSTSWhen a decision variable has an optimal value

of zero, the allowable increase for the objectivefunction coefficient is also called the reducedcost.

The reduced cost of a decision variable is theamount the corresponding objective functioncoefficient would have to change before the

optimal value would change from zero to somepositive value.

5/28/2018 MBA8503C7

33/64

REDUCED COSTSThe reduced cost for tables is zero in the

original formulation. Why is this the case?

We are already producing tables.

What is the reduced cost for X2 and what doesit mean?

The reduce cost is -1.5 meaning that if I forceproduction of 1 chair profit will drop by 1.5

5/28/2018 MBA8503C7

34/64

SENSITIVITY ANALYSIS PROBLEM

A manufacturing firm has discontinuedproduction of a certain unprofitable productline thus creating considerable excessproduction capacity.

Management is considering devoting this excesscapacity to one or more of three products; call

them products 1, 2, and 3.

The available capacity on the machines thatmight limit output is summarized below:

5/28/2018 MBA8503C7

35/64

AVAILABLE TIMEMACHINE TYPE (machine hours /week)

Milling machine 500

Lathe 350

Grinder 150


5/28/2018 MBA8503C7

36/64

The number of machine hours required for eachunit of the respective products is:

MACHINE TYPE P1 P2 P3

Milling machine 9 3 5

Lathe 5 4 0

Grinder 3 0 2


5/28/2018 MBA8503C7

37/64

The sales department indicates that the salespotential for products 1 and 2 exceeds themaximum production rate and that the sales

potential for product 3 is 20 units per week.

The unit profit would be $3000, $1200, and$900, respectively, for products 1, 2, and 3.


5/28/2018 MBA8503C7

38/64

Solver is used to determine the optimal solution

a. What are the optimal weekly productionlevels for each of the three products?

Product 1 = 45.23

Product 2 = 30.95

Product 3 = 0


5/28/2018 MBA8503C7

39/64

b. What profit will be obtained if the optimalsolution is implemented?

$172,857.10 per week

c. How much unused capacity exists on themilling machine, the lathe, and the grinder?

Milling = 0; Lathe = 0; Grinder =

14.28(See SLACK entries)


5/28/2018 MBA8503C7

40/64

d. How much would the objective functionchange if the amount of available time on thegrinder increased from 150 hours per week to250 hours?

Will the objective function increase ordecrease?

Currently the grinder has 14.28 hours of slacktime so its shadow price is 0.

Increasing the available hours from 150 to 250will not change the total profit.


5/28/2018 MBA8503C7

41/64

e. The profit for product 3 is $900 per unitand the current production level is zero.

How much would the profit per unit have to

change before it would be profitable toproduce product 3's?

The profit per unit would have to increase by

its reduced cost of $528.57.


5/28/2018 MBA8503C7

42/64

f. The milling machine capacity can be increasedat a cost of $160 per hour. Is it economic toincrease capacity by 10 hours?

The shadow price per hour (285.7143) is greater than

the cost (160), so it is worth increasing milling capacityon the margin. However, since the shadow price mightchange with increasing capacity we need to rerun to seethe full effect of increasing capacity by 10 hours. Profitdoes increase by 2857 (10*285.714) which is greater

than the cost increase of 1600 (10*160). Therefore themilling capacity should be increased b y 10 hours.


5/28/2018 MBA8503C7

43/64

TRANSPORTATION PROBLEM

Mathematical programming has beensuccessfully applied to important supply chainproblems.

These problems address the movement ofproducts across links of the supply chain(supplier, manufacturers, and customers).

We now focus on supply chain applications in

transportation and distribution planning.

5/28/2018 MBA8503C7

44/64

TRANSPORTATION PROBLEMA manufacturer ships TV sets from three warehouses to four retail

stores each week. Warehouse capacities (in hundreds) anddemand (in hundreds) at the retail stores are as follows:

Capacity Demand

Warehouse 1 200 Store 1 100Warehouse 2 150 Store 2 200

Warehouse 3 300 Store 3 125

650 Store 4 225

650

5/28/2018 MBA8503C7

45/64

The shipping cost per hundred TV sets for each route is

given below:

To

From Store 1 Store 2 Store 3 Store 4

warehouse 1 $10 5 12 3

warehouse 2 4 9 15 6

warehouse 3 15 8 6 11


5/28/2018 MBA8503C7

46/64

5/28/2018 MBA8503C7

47/64

What are the decision variables?

XIJ=number of TV sets (in cases) shippedfrom warehouse I to store J

I is the index for warehouses (1,2,3)

J is the index for stores (1,2,3,4)


5/28/2018 MBA8503C7

48/64

What is the objective?

Minimize the total cost of transportation which isobtained by multiplying the shipping cost by theamount of TV sets shipped over a given route

and then summing over all routes

OBJECTIVE FUNCTION ;

MIN = 10*X11+5*X12+12*X13+ 3*X14+

4* X21+9*X22+15*X23+ 6*X24+

15*X31+8*X32+ 6*X33+11*X34


5/28/2018 MBA8503C7

49/64

How are the supply constraints expressed?For each warehouse the amount of TV sets

shipped to all stores must equal the capacityat the warehouse

X11+X12+X13+X14=200; SUPPLYCONSTRAINT FOR WAREHOUSE 1

X21+X22+X23+X24=150; SUPPLY

CONSTRAINT FOR WAREHOUSE 2X31+X32+X33+X34=300; SUPPLY

CONSTRAINT FOR WAREHOUSE 3


5/28/2018 MBA8503C7

50/64

How are the demand constraints expressed?For each store the amount of TV sets shipped

from all warehouses must equal the demand ofthe store

X11+X21+X31=100; DEMAND CONSTRAINTFOR STORE 1





5/28/2018 MBA8503C7

51/64

Since partial shipment cannot be made,the decision variables must be integervalued

However, if all supplies and demands areinteger-valued, the values of our decisionvariables will be integer valued


5/28/2018 MBA8503C7

52/64

After solution in Solver:

The total shipment cost is $3500, and theoptimal shipments are: warehouse 1 ships

25 cases to store 2 and 175 to store 4;warehouse 2 ships 100 to store 1 and 50 tostore 4, and warehouse 3 ships 175 to store2 and 125 to store 3.

The reduced cost of X11 is 9, so the cost ofshipping from warehouse 1 to store 1 wouldhave to be reduced by $9 before this route

would be used


5/28/2018 MBA8503C7

53/64

UNBALANCED PROBLEMS

Suppose warehouse 2 actually has 175 TV sets.

How should the original problem be modified?

Since total supply across all warehouses is nowgreater than total demand, all supply

constraints are now

5/28/2018 MBA8503C7

54/64

RESTRICTED ROUTE

Referring to the original problem, suppose there isa strike by the shipping company such that theroute from warehouse 3 to store 2 cannot beused.

How can the original problem be modified toaccount for this change?

Add the constraint:

X32=0

5/28/2018 MBA8503C7

55/64

WAREHOUSE LOCATIONSuppose that the warehouses are currently not open,

but are potential locations.

The fixed cost to construct warehouses and theircapacity values are given as:

WAREHOUSES FIXED COST CAPACITY

Warehouse 1 125,000 300

Warehouse 2 185,000 525Warehouse 3 100,000 325

5/28/2018 MBA8503C7

56/64

How do we model the fact that thewarehouses may or may not be open?

Define a set of binary decision variables YI, I

=1,2,3, where warehouse I is open if YI = 1and warehouse I is closed if YI= 0

WAREHOUSE LOCATION

5/28/2018 MBA8503C7

57/64

How must the objective function change?

Additional terms are added to the objectivefunction which multiply the fixed costs ofoperating the warehouse by YI and summingover all warehouses I:

125000*Y1+185000*Y2+100000*Y3

Why cant we use the current capacityconstraints?

Product cannot be shipped from a warehouseif it is not open. Since the capacity isavailable only if the warehouse is open, wemultiply warehouse 1s capacity by Y1.

WAREHOUSE LOCATION

5/28/2018 MBA8503C7

58/64

Also, we must make the YI variablesbinary integer

Total fixed and shipping costs are$289,100; warehouses 2 and 3 areopen; warehouse 2 ships 100 to store 1,225 to store 4; and warehouse 3 ships

200 to store 2 and 125 to store 4

WAREHOUSE LOCATION

5/28/2018 MBA8503C7

59/64

FINALLY, BACKTOTHEMOTIVATINGPROBLEM.. .

5/28/2018 MBA8503C7

60/64

NETWORKSCHEDULINGEXAMPLE

Single product

Two plants p1 and p2

Plant p2 has an annual capacity of 60,000 units.

The two plants have the same productioncosts.

There are two warehouses w1 and w2 withidentical warehouse handling costs.

There are three markets areas c1,c2 and c3with demands of 50,000, 100,000 and50,000, respectively.

5/28/2018 MBA8503C7

61/64

UNITDISTRIBUTIONCOSTS

Facilitywarehouse

p1 p2 c1 c2 c3

w1 0 4 3 4 5

w2 5 2 2 1 2

5/28/2018 MBA8503C7

62/64

D = 50,000

D = 100,000

D = 50,000

Cap = 60,000

THENETWORK

5/28/2018 MBA8503C7

63/64

THEOPTIMIZATIONMODEL

This problem can be framed as the following linearprogramming problem:

Let

x(p1,w1), x(p1,w2), x(p2,w1) and x(p2,w2) be theflows from the plants to the warehouses.

x(w1,c1), x(w1,c2), x(w1,c3) be the flows from thewarehouse w1 to customer zones c1, c2 and c3.

x(w2,c1), x(w2,c2), x(w2,c3) be the flows fromwarehouse w2 to customer zones c1, c2 and c3

5/28/2018 MBA8503C7

64/64

OPTIMALSOLUTION

Facilitywarehouse

p1 p2 c1 c2 c3

w1 140,000 0 50,000 40,000 50,000

w2 0 60,000 0 60,000 0

Total cost for the optimal strategy is$740,000

mba8503c7

Documents