Formularium Management III

Kevin Denis

June 18, 2014

Use at own risk.

Part I

Inventory Control

1 Known demand (p198)

1.1 Basic EOQ model (p210)

Assumptions:

1. Demand is xed at λ units per unit time.

2. Shortages are not allowed.

3. Orders are received instantaneously.

4. Order quantity is xed at Q per cycle, can be proven optimal.

5. Cost structure:

(a) K = Setup cost per order [¿]

(b) c = Proportional cost per unit [¿/µ]

(c) h = Holding cost per unit held per year [¿/µ]

Q

∗ =

r

2Kλ

h

(1.1)

Q* Optimal order size [µ]

K Fixed setup cost [¿]

λ Demand [µ/time]

h Holding cost = I · c [¿/µ]

1

I Time interest rate [%]

c Proportional cost [¿/µ]

G(Q) = Kλ

Q

+ λc +

hQ

2

(1.2)

G Average (time, usually year, use the same time base for everything !) cost

associated with lot size Q

Q Order size [µ]

Kλ

Q Yearly ordering cost

λc Yearly proportional ordering cost

hQ

2 Yearly holding cost

1.2 Modied EOQ model (nite production rate) (p218)

Q

∗ =

r

2Kλ

h

0

(1.3)

h Modied holding cost = h(1 −

λ

P

) [¿/µ]

P Production rate [µ/time]

G(Q) = Kλ

Q

+ λc +

h

0Q

2

(1.4)

1.3 Quantity discount model (p220)

1.3.1 All units (p221)

Discount is applied to all the units in an order.

Method

1. Compute EOQ values corresponding to each of the unit cost. It is possible

that those values don't correspond to the minimal order cost for getting

that specic discount (e.g. you get 10% discount when ordering more than

100 units, but your Q∗

is 75.). If this is the case, you have to change the

Q∗

to the minimal order for getting that specic discount.

2. Once you have nd your Q∗

(or Qmin) calculate the average order cost

G(Q)

3. Determine the optimal solution, G(Q)min

2

1.3.2 Incremental (p223)

Discount is applied only to the additional units beyond the break point.

Method Follow summary on page 225.

1.4 EOQ models for production panning (p230)

T

* =

s

2

Pn

j=1 Kj

Pn

j=1 h

0

jλj

(1.5)

T* Optimal cycle time [time]

K Setup cost = setup time * labor cost per time

Q

∗ = T

∗λ (1.6)

Q∗ Optimal lot size [µ]

up time =

Q∗

P

(1.7)

up time% =

up time

T ∗

(1.8)

2 Unknown demand (p248)

2.1 Newsboy model (p260)

Assumptions:

1. A single product is to be ordered at the beginning of a period

2. This product can be used only to satisfy demand during that period

3. Assume that all relevant cost can be determined on the basis of ending

inventory

4. Cost structure:

(a) c0= cost per unit of positive inventory remaining at the end of the

period (overage cost)

(b) cu= cost per unit of unsatised demand. This can be thought of as

a cost per unit of negative ending inventory (underage cost)

5. Demand D is continuous non-negative random variable with density function f(x) and cumulative distribution function F(x)

6. Decision variable Q is the number of units to be purchased at the beginning

of the period

3

F(Q

∗

) = cu

cu + co

(2.1)

Q

∗ = µ + σz (2.2)

c0 cost per unit of positive inventory remaining at the end of the period (overage

cost) [¿]

cu cost per unit of unsatised demand. This can be thought of as a cost per

unit of negative ending inventory (underage cost) [¿]

µ Mean demand [µ]

σ Standard deviation [µ]

z Z-score corresponding to the probability F(Q∗

)

2.2 Lot size-reorder point (Q, R) systems (p266)

Use this only of you have a given stock out cost p. If not, go to section 2.3

Assumptions:

1. The system is continuous review

2. Demand is random and stationary. The expected value of demand over

time is constant

3. There is a xed positive lead time τ for placing an order

4. Cost structure:

(a) K = Setup cost per order [¿]

(b) h = Holding cost per unit held per year [¿/µ]

(c) c = Proportional cost per unit [¿/µ]

(d) p = stock-out cost per unit of unsatised demand [¿/µ]

G(Q, R) = h(

Q

2

+ R − λτ ) + Kλ

Q

+

pλn(R)

Q

(2.3)

Qi =

r

2λ[K + pn(R)i

]

h

(2.4)

1 − F(R) = Qih

pλ (2.5)

n(R) = σL(z) (2.6)

Ri = µ + σz (2.7)

4

G(Q,R) Expected annual cost

Q Order quantity [µ]

R Reorder level [µ]

h Holding cost [¿/µ]

K Setup cost per order [¿]

λ Expected demand rate [µ/time] ≈ µ

τ Reorder lead time, lead time [time]

p penalty cost [¿/µ]

n(R) Expected number of stock outs in lead time [µ]

µ Mean demand [µ/time]

σ Standard deviation [µ/time]

S = R − λτ (2.8)

S Safety stock

Method

The solution procedure requires iteration between (2.4) and (2.5) until two successive values of Q and R are essentially the same.

1. Start the procedure by using Q0 = EOQ (use λ = µ). All time scales

have to be in years

2. Find R0 with Q0 from step 1 by doing the following

(a) In Table A-4 nd the corresponding z value to a right tale of the

value you found for 1 − F(R) found by (2.5). All time scales have to

be in years.

(b) With this Z-score, you can nd R from (2.7) (Caution, here you have

to use µ and σ in time frame of τ !)

(c) With the same Z-score, nd n(R) from (2.6) (L(z) also from Table

A-4 with same z-score).

3. Find Q1with n(R) from step 2. Keep n(R) in the same time scale as τ .

So normally you don't have to change anything.

4. And so on...until there is no essential dierence between two successive

values Q and R

5

Optimal safety stocks

2.3 Service levels in (Q, R) systems (p272)

Use this only if you don't have a given stock out cost. If you have, go to section

2.2

2.3.1 Type 1 service (p272)

α Probability of not stocking out in lead time. This is independent of the

number of out stocks

Method

1. Determine R to satisfy the equation F(R) = α

(a) Use Table A-4 to nd the corresponding Z-score

(b) With this Z-score, you can nd R from 2.7

2. Set Q = EOQ

2.3.2 Type 2 service (p273)

β Proportions of demands that are met from stock

Method

1. Solve the following equation : L(z) = EOQ(1−β)

σ

(you can also use another

method than EOQ, named SOQ, described on p274)

2. Search for he corresponding Z-score of L(z)

3. With this Z-score, you can nd R from 2.7

Imputed shortage cost

p =

Qh

(1 − F(R))λ

(2.9)

2.4 Scaling of Lead time Demand and

ˆ If lead time demand is not given, it can be estimated.

ˆ If demand follows normal distribution, it can be assumed that lead time

demand also follows the same.

ˆ If periodic demand has a mean λ of and standard deviation of v and if τ

be the lead time in periods then mean demand during lead time is given

as µ = λτ and the variance of the demand during lead time is v

2

τ

6

2.5 Lead time variability

If lead time τ is a random variable with mean µτ and variance of σ

2

τ and if

demand in any time t is λt and variance v

2

t of then the demand during lead

time has a mean and variance as follows respectively.

µ = λµτ (2.10)

σ

2 = µτ v

2 + λ

2σ

2

τ

(2.11)

7

Part II

Investment Analysis

Caution! Closed book part on exam!

1 Understanding cash ows

cash flow(t) = cash receipts(t) − cash payments(t) (1.1)

cash inflow : cashflow > 0

cash outflow : cashflow < 0

2 Investment analysis techniques

2.1 Payback method or discounted payback period (PB)

(DPB)

Is the number of years and days it takes to recover a project's net investment

through incremental cash ows (cash inows = cash outows)

2.2 Net present value method (NPV)

Compounding

F Vn = P V0(1 + i)

n

(2.1)

Discounting

P V0 =

F Vn

(1 + i)

n

(2.2)

NP V =

Xn

t=1

CFt

(1 + i)

t

− INV (2.3)

ˆ If NPV > 0 : accept (the more the better)

ˆ If NPV < 0 : reject

PV Present value

FV Future value

i Interest rate per period or discount rate

8

n Number of years (based on compound interest)

NPV Net present value

CF(t) Cash ow at a given time t (usually in years)

2.3 Internal rate of return method (IRR)

Xn

t=1

CFt

(1 + IRR)

t

− INV = 0 (2.4)

ˆ if IRR ≥ i, accept the project (the larger the better)

ˆ if IRR < i, reject the project

2.4 Modied IRR (MIRR)

Unconventional cash ows (- + + - +) are di‑cult to determine with IRR or

there may be more than one IRR that gives NPV=0. MIRR is the discount rate

that equates the PV of a project's terminal value to its PV of costs

P Voutf lows =

F Vinf lows

(1 + MIRR)

n

(2.5)

3 Protability Index (PI)

P I =

P V (future cash flows)

INV

> 1 (3.1)

9

Part III

Project Management, Line

Balancing

1 Project Management

1.1 Conventions activity on arrow (AOA)

1. 1 activity = 1 arrow

2. Each knot has: min. 1 entering arrow min. 1 leaving arrow (except rst

and last knot)

3. Between 2 knots max.1 arrow (use dummies !!)

4. No cycles

5. Transitive relation (A before B and B before C implies A before C)

6. Correct formulation (use dummies !!)

1.2 Critical patch method (CPM)

Prerequisites :

1. Well-dened jobs or tasks or activities

2. End of the project

3. Independent activities (not from one task back to another !!)

4. A given activity-sequence (numbering of activities)

Earliest starting time (ES) The max EF of the direct predecessors

Earliest nishing time EF ES + Activity Duration

Latest starting time LS LF - Activity Duration

Latest nishing time LF The min LS of the direct successors

Free oat FFij ESj ESi - Dij. Float time that you have because an activity

takes less time than the activity parallel to that, when both are needed to

accomplish an event

Total oat TFij LFj ESi - Dij.

Slack LSi - ESi = LFi - EFi

10

1.3 PERT

ˆ Used when activity times are uncertain

ˆ Used to obtain the same information as CPM + probability information

a Min. activity time, optimistic

m Most likely activity time

b Max. Activity time , pessimistic

µ

a+4m+b

6

( β distribution)

σ

2 (b−a)

2

36

E(t) Pµi

Var(t) Pσ

2

i

Z = q

D − TE

Pσ

2

cp

(1.1)

Z score for Table A-1 or A-4, with this you can calculate the probability

D Time you are comparing with

TE The mean time of critical path, E(t)

1.4 Time Costing Methods

Steps in Project Crashing

Crash cost per period =

crash cost − normal cost

normal time − crash time (1.2)

1. Compute the crash cost per time period.

2. Using current activity times, nd the critical path

3. If there is only one critical path, then select the activity on this critical

path that

(a) Can still be crashed

(b) Has the smallest crash cost per period

(c) Note that a single activity may be common to more than one critical

path

4. Update all activity times

5. Identify the new critical path

11

2 Line balancing

Assembly line balancing operates under two constraints, precedence requirements and cycle time restrictions.

Precedence requirements physical restrictions on the order in which operations are performed. DO NOT CONFUSE WITH PROJECT

MANAGEMENT

Cycle time Maximum amount of time a product is allowed to spend at each

workstation.

2.1 Method

1. Draw and label a precedence diagram

2. Calculate the desired cycle time required for the line

3. Calculate the theoretical minimum number of workstations

4. Use Largest Candidate Rule (LCR) to balance the line, see section 2.2

(a) If this gives you a number of stations that is close to the theoretical

number of workstations go to step 5. (e.g. N=2,1 and LCR = 3 :

OK)

(b) If you see there is a big dierence between the theoretical number of

workstations try use Dynamic Programming to balance the line (e.g.

N=2, and LCR = 4, try Dynamic Programming in section 2.3)

5. Calculate the e‑ciency of the line

Cd =

production time available

desired units of output (2.1)

E =

Pj

i=1 ti

nCa

(2.2)

N =

Pj

i=1 ti

Cd

(2.3)

Cd Desired cycle time [time/µ]

E E‑ciency

ti Completion time for element i

j Number of work elements

n Actual number of workstations

Ca Actual cycle time [time/µ]

N Minimum number of stations

balance delay = 1 − E (2.4)

12

2.2 Largest candidate rule

Even if this rule is easier, you will always get the best result (and dierent

options) when you use Dynamic Programming. Once you get the hang of it,

you will also make less mistakes. But when your are faced with a lot of task

and complicated precedence, this is the fastest way to get a result.

Method

1. Take the largest possible task keeping in mind the precedence diagram

2. Take second highest task, etc. until one workshop is full

3. Continue with step 1 and 2 until you reach the nal task

2.3 Dynamical programming

Method

1. Creating the network of all the possible solutions, keeping in mind the

precedence requirements. Always start with the task who has no predecessors. This is called stage 1.

2. Stage 2 consists of all the possible collections starting from stage 1 going

to stage 2 (in general n-1 going to stage n) (still keeping in mind the

precedence requirements). Make sure you don't forget any possibility

3. Repeat step 2 until you reach your nal task, still keeping in mind the

precedence requirements and not forgetting any possibility.

4. Having found every possible collection we know have to nd the one that

takes the less time to achieve. Note that it is possible to a multiple set

of quickest paths. Start at stage 1 and compute the time to achieve this

task. The following notation is used : x/y

(a) x represents the number of workstations already complete. You will

have to add +1 every time overowoccurs in y (when y > Ca)

(b) y number represents the amount of time already used in the next

following workstation. Be aware: you can't divide one task over two

workstations. If overow occurs, with a specic task, add +1 to x

and take y = task time.

5. Repeat step 4 until you arrive at your nal task. At each stage, note the

quickest time you can take to go a certain collection from each stage. Still

x/y format.

6. By going backwards you can nd the quickest way to achieve your nal

task. Note that it possible that you nd a multiple set of path that can

achieve this.

13

Answer rating: 100% (QA)

Question a Optimal production quantity EPQ 2Ds h1dp 2 x annual demand x set up cost per unit Product holding cost per unit x 1 daily demand daily prod... View the full answer