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4 COORDINATED REPLENISHMENTS
cyclic schedules powers-of-two policies
Q
Q
Q
Q
2
1
C
C *
**
T
T
T
T
2
1
C
C *
**
4.1 Powers-of-two policies
Cycle times are restricted to be powers of two times a certain basic period denoted as q . From (3.5) in Chapter 3, we have:
, (4.1)Let T = Q/d, T* = Q*/d
. (4.2)
q2T m
T2
T
T
T2
2
1
T
T
T
T
2
1
C
C *
*
*
**
Lot sizing problem
(4.3)Worst scenario:Because the cost is convex, the worst possible error must occur when two consecutive values of m, say m = k and m = k + 1 give the same error. Let T < T* correspond to m = k, and 2T > T* to m = k + 1.
. (4.4)
.2gives*
Setting xxT
T
2T
T2
T
T*
*
06.122
1
2
1
C
C*
, (4.5)
. (4.6)
Proposition 4.1 For a given basic period q, the maximum relative cost increase of a powers-of-two policy is 6 percent.
C/C*
T
2T2T* 2T 2T1T2 T1T
2*2 TT 2
*TT
2
1
2
1,]22[
2
1)(
2
1
2
1,2
2*
2
*
*
2/12/1
ixx
ii
i
ixi
i
xxeC
C
xT
TT
T
ii
i
Note that T1 gives better cost than
Note that 2T2 gives better cost than
Suppose possible to change q
For a single item, the optimal solution is to choose q equal to a power of two times T*. For N items, we can, in general, not fit q perfectly to all cycle times Ti*. The relative cost increase can be expressed as:
)(
)/(
1**
*
1
*
1
**
1
*
1*
i
N
ii
i
i
i
i
N
ii
N
iiii
N
ii
N
ii
xewC
C
C
C
C
CCC
C
C
C
C
*ii C/C
,2/1x2/1),22(2
1)x(e
C
Ci
xxi*
i
i ii
)2/1x2/1(2T/T ix*
iii
N
1i*i
*i C/C
. (4.7)
We know from (4.2) and (4.5) that for a given q, each
can be expressed as :
(4.8)
The weights wi for the different values of xi , can be seen as a probability distribution
F(x) on [-1/2, 1/2], i.e., F(-1/2) = 0 and F(1/2) = 1.
2/1
2/1*
)x(dF)x(eC
C
.1y0 . (4.9)Change q by multiplying by 2y , This means that a certain x is replaced by x + y for x + y ≤ ½, , and by x + y - 1 for x + y > ½.
**
'
*)(*
*
22
2222)(2
1)0(,10,2)(
22
2
22Each
iyx
mi
xymymm
i
y
imxi
x
mi
xi
TT
qyqT
qyqyqLet
TT
q
qTT
i
i
i
iii
ii
im
i
ii
.)1yu(dF)u(e)yu(dF)u(e
)x(dF)1yx(e)x(dF)yx(e)y(C
C
2/1y
2/1
2/1
2/1y
2/1
y2/1
y2/1
2/1*
(4.10)For a given distribution F(x) the minimum cost increase is obtained by minimizing (4.10) with respect to 0 ≤ y ≤ 1.
2
12
2
12*
2
1*2
2
1*2
2
1
'
1'
)(
)(
yxif
yxifi
i
yxifTT
yxifTTi
i
yix
i
yix
i
yix
i
i
yixi
T
yT
yT
du)1yu(dF)yu(dF)u(e
dy)1yu(dF)u(e)yu(dF)u(e)y(C
Cmin
2/1
2/1
1
2/1u
2/1u
0
1
0
2/1y
2/1
2/1
2/1y*1y0
02.12ln2
1du)u(edu)2/1(F)u(F)u(F)2/1(F)u(e
2/1
2/1
2/1
2/1
Proposition 4.2 If we can change the basic period q, the maximum relative cost increase of a powers-of-two policy is 2 percent.
Proof The average cost increase for 0 ≤ y ≤ 1 must be at least as large as the minimum
The worst case will occur when the distribution F(x) is uniform on , see (4.9).
A change of q will then not make any difference.
2/1x2/1
1/2
-1/2
0
u
y
2
12
1
+=
-=
uy
yu
1
1
2/1
2/1)()(
ydyyudFue
2/1
0)()(
uduyudFue
Given y
Given u
duuFue
FNoteduFuFue
duyuFue
duyudFue
duyudFue
dyyudFue
u
u
u
y
)()(
0)2
1()]
2
1()([)(
)]([)(
)()(
)()(
)()(
2/1
2/1
2/1
2/1
2
1
0
2/1
2/1
2/1
0
2/1
2/1
2/1
2/1
2/1
0
1
0
2/1
2/1
u
1/2
-1/2
0 y2
12
1
uy
yu
1
1
2
1 )1()(u
duyudFue
Given y
2/1
2/1)1()(
ydyyudFue
Given u
duuFue
FNoteduFuFue
duyudFue
duyudFue
dyyudFue
y
u
y
u
y
)](1[)(
1)2
1()]
2
1()([)(
)1()(
)1()(
)1()(
2/1
2/1
2/1
2/1
1
2/1
2/1
2/1
2/1
2/1
1
2/1
1
0
2/1
2/1
4.2 Production smoothing
Mixed integer program (MIP) Manne (1958) Billington et al. (1983) Eppen and Martin (1987) Shapiro (1993)
Objective: Low inventory cost and smooth capacity utilization.Simplification: ignore stochastic variations ignore time-varying aspectMIP not used in practice
When the demands for different items are relatively stable, use cyclic schedules.
In a general case with many items and several production facilities, it can be extremely difficult to find suitable cyclic schedules.
It is also common in practice to smooth production outside the inventory control system.
Each item ordered periodically. Ordering period chosen to smoothen the load.
Example: 4 items, same demand
Item 1: 1,5,9 Item 2: 2,6,10
Item 3: 3,7,11 Item 4: 4,8,12
Periodic review: order up to S policy
Continuous review: large variation in production load resulting in long and uncertain lead time.
4.2.1 The Economic Lot Scheduling Problem (ELSP)
Cyclic schedules for a number of items with constant demands.
Backorders not allowed.
Finite production rate
Single production facility
Notation:N = number of items,hi = holding cost per unit and time unit for item i,Ai = ordering or setup cost for item i,di = demand per time unit,pi = production rate (pi > di),si = setup time in the production facility for item i, independent of the sequence of the items,Ti = cycle time for item i (the batch quantity Qi = Tidi).
Define:
i = di/pi ,
i = i Ti = production time per batch for item i excluding setup time,i = si + i = total production time per batch for item i.
The problem is to minimize subject to the constraint that all items should be produced in the common production facility.
Table 4.1 N = 10
Bomberger (1966) Table 4.1 Bomberger’s problem (time unit = one day).
2
T)1(dh
T
AC i
iiii
ii . (4.11)
N
1i iC
The optimal cycle time when disregardingthe capacity constraint is
)1(dh
A2T
iii
ii
)1(dhA2C iiiii
, (4.12)
. (4.13)Note that (4.11) and (4.12) are equivalent to (3.6) and (3.7) if we replace Ti by Qi/di , and i by di/pi .
Table 4.2 Independent solution of Bomberger’s problem.
Item 1 2 3 4 5 6 7 8 9 10
Ti167.5 37.7 39.3 19.5 49.7 106.6 204.3 20.5 61.5 39.3
Ci0.179 1.060 1.528 1.024 4.428 0.938 3.034 12.668 6.506 0.255
I2.36 2.01 3.56 4.29 2.49 1.67 3.04 5.87 11.20 1.17
10
1i i 62.31CC , is a lower bound for the total costs.
Solution not feasible: Consider items
4, 8 and 9.
Item 4 & 8 have cycle times 19.5 and 20.5. Must
be able to produce one batch of #4 and #8 in
[t,t+20.5], or [t+11.20,t+20.5]. But the length of
available time is 9.30, while 4+8 =10.16 >9.30.
Not feasible
t+61.5t+20.5t
Item 9
11.20
9.30Item 9
1N
1ii
N
1iiii
i
2
T)1(dh
T
AC
Does a feasible solution exist? If at least one setup time is positive an obvious necessary condition for a feasible solution to exist is
. (4.14)Condition (4.14) is also sufficient for feasibility.
Given the assumption of a common cycle time, the problem now is to minimize
, (4.15)
T)Ts(N
1iii
N
1ii
min
1
1
1T
sT N
ii
N
ii
with respect to the constraint that the common cycle must be able to accommodate production lots of all items
. (4.16)
, (4.17)a lower bound for the cycle time.
Need large enough T to squeeze setups in the slack
1-
N
ii
1
N
iiii
N
ii
dh
AT
1
1
)1(
2ˆ
Disregard (4.17), from (4.15)
. (4.18)Since (4.15) is convex in T the optimal solution,
, (4.19)For Bomberger’s problem, Tmin = 31.86, and consequently, . cost=41.17.
For problems where the indi vidual cycle times are reasonablysimilar, the common cycle approach gives a very goodapproximation.
,75.42ˆ TTopt 75.42T̂
)T,T̂(maxT minopt
17.4162.31 ** CCi
Two approaches for deriving better solutions.
I. Dynamic programming model
Bomberger (1966) AssumingTi = niW
W should be able to accommodate production of all items.
)w(F)Wn(Cmin)w(F iiiin
1ii
Fi(w) = minimum cost of producing items
i +1,i+2,..., N when the vailable capacity in the basic period is w, i.e., W - w has been used for items 1, 2,...,i. , (4.20)where Ci(niW) are the costs (4.11) for item i with Ti = niW,
i = si + iniW, and the integer ni is subject to the constraint
. (4.21)orNote that the upper bound in (4.21) is equivalent to i w.
FN(w) = 0 for all w 0. F0(W) gives the minimum costs when
the basic period is equal to W.
W/)sw(n1 iii Wnsw iii
Minimize over W. Bomberger’s solution
C=36.65, W=40, ni=1 for i 7, n7 =3.
Serious Assumption: W should be able to accommodate production of all items.
II. Heuristic
Doll and Whybark, 1973).
The procedure is to successively im prove the multipliers ni and the basic period W according to the following iterative procedure:
1. Determine the independent solution and use the shortest cycle time as the initial basic period W.
2. Given W, choose powers-of-two multipliers, (ni = 2mi, mi 0), to minimize the item costs (4.11).
N
1iiiii
ii
2
Wn)1(dh
W
n/AC
N
1iiiii
N
1iii
n)1(dh
n/A2W
3. Given the multipliers ni , minimize the total costs
with respect to W.
4. Go back to Step 2 unless the procedure has converged. In that case, check whether the obtained solution is feasible. If the solution is infeasible, try to adjust the multipliers and then go back to Step 3.
Compare with independent solution.
Apply the heuristic to Bomberger’s problem.
Table 4.3 Solution of Bomberger’s problem with W = 23.42.
Item 1 2 3 4 5 6 7 8 9 10
ni8 2 2 1 2 4 8 1 2 2
i2.62 2.47 4.19 5.12 2.37 1.50 2.87 6.63 8.71 1.37
Table 4.4 Feasible production plan.
Basic periodItems Production
time1 4, 8, 2, 9 22.93
2 4, 8, 3, 5, 10, 1 22.30
3 4, 8, 2, 9 22.93
4 4, 8, 3, 5, 10, 6 21.18
5 4, 8, 2, 9 22.93
6 4, 8, 3, 5, 10, 7 22.55
7 4, 8, 2, 9 22.93
8 4, 8, 3, 5, 10, 6 21.18
In case of stochastic demand, one possible approach is to first solve a deterministic problem based on averages, and then try to adapt the solution to the stochastic case by adding suitable safety stocks.
4.2.2 Production smoothing and batch quantities
Adjust the batch quantities to obtain a reasonably smooth load.
Karmarkar (1987, 1993). Axsäter (1980, 1986),
Bertrand (1985), and Zipkin (1986).
Consider a machine in a large multi-center shop.
D = average output of material (demand), units per time unit,P = average processing rate, units per time unit,Q = batch quantity,t = setup time,T = average time in the system for a batch,h = holding cost per unit and time unit after processing.
Assume:The batches arrive at the machine as a Poisson process with rate = D/Q. Thus Av demand=Q=D.The processing time is exponentially distributed. average processing time for a batch is 1/ = t + Q/P. Service rate = . = l / = Dt/Q + D/P. The average time spent in the M/M/1 system is
P/DQ/Dt1
P/Qt
1
/1T
. (4.22)
The average cycle stock is approximated as Q/2.
P/D1
Dt2Q*
Assume that the average holding cost per unit and time unit for work-in-process is exactly half of the holding cost h after the process. Av cycle stock = Q/2. Total holding cost after the process=hQ/2.Work-in-process TD=Av time in the system * Av demand
.(4.23)
. (4.24)
For low values of D, Q* is essentially linear in D. For larger values, Q* grows very rapidly.
QP/DQ/Dt1
P/DQDt
2
hmin)QTD(
2
hmin
0Q0Q
4.3 Joint replenishments
4.3.1 A deterministic model
Setup costs: Individual setup costs for each item, and a joint setup cost for the whole group of items. Reason: joint setup costs, quantity discounts, coordinated transports.
constant continuous demand. No backorders. batch quantities are constant. production time is disregarded. No lead time or the lead time is same for all items.
Notation:
N = number of items,hi = holding cost per unit and time unit for item i,A = setup cost for the group,ai = setup cost for item i,di = demand per time unit for item i,Ti = cycle time for item i.i = hidi
Assume all demands equal to one. Items areordered so that a1/1 a2/ 2 ... aN/N .Note that increasing setup costs and decreasingholding costs mean increasing lot sizes and cycletimes.
i
ii
a2T
2
)n(T
T
n/aaAC
N
2iii11
1
N
2iii1
Approach 1. An iterative technique
If there were no joint setup cost
, (4.25)i.e., T1 would be the smallest cycle time. Assume other cycle
times of items 2, 3... N are integer multiples ni of the cycle time
for item 1,
, i = 2, 3,..., N. (4.26)Our objective is to minimize w.r.t T1, n2, n3, ... nN the cost.
,(4.27)
1ii TnT
Fix cost ith item holding cost=Tinii/2
N
2iii1
N
2iii1
N32*1
n
)n/aaA(2)n,...,n,n(T
N
2i
N
2iii1ii1N32
* )n)(n/aaA(2)n,...,n,n(C
)aA(
an
1
1
i
ii
N
2iii11 a2)aA(2C
Given n2, n3... nN,
, (4.28)
. (4.29)Note that T1 is not chosen according to (4.25). If we disregard
n2, n3... nN to be integers, then from (4.29),
. (4.30)From (4.30) and (4.29), the lower bound for the costs:
.(4.31)
HEURISTIC 1. Determine start values of n2, n3... nN by rounding
(4.30) to the closest positive integers.
2. Determine the corresponding T1 from (4.28).
3. Given T1, minimize (4.27) with respect to n2, n3... nN.
This means that we are choosing ni as the positive
integer satisfying
.;2
0:
)1(2
)1(
21
2
21
convexisCT
angives
dn
dCNote
nnT
ann
i
ii
i
iii
iii
. (4.32)Return to Step 2, if any multiplier ni has changed since the last
iteration.
3n,1n,1350
105000
101000
50n 432
1155.031001070010100010500010
)3/505050350(2T1
Example 4.1 N = 4 , A = 300, a1 = a2 = a3 = a4 = 50, h1 = h2 = h3
= h4 = 10, d1 = 5000, d2 = 1000, d3 = 700, and
d4 = 100. As re quested, ai/i is nondecreasing with
i. When applying the heuristic we obtain
again n2 = 1, n3 = 1, n4 = 3, i.e., the algorithm has already
converged.
N
0i ii
ii
T,...,T,T T
1a
2
Tmin
N10
0i TT
Approach 2. Roundy´s 98 percent approximationthe joint setups have cycle time T0 0,
, i = 1, 2, ..., N, (4.33)where ki is a nonnegative integer. Let a0 = A, and 0 = 0,
, (4.34)subject to the constraints (4.33). replacing (4.33) by
, i = 1, 2, ..., N. (4.35)
02 TT iki
N
1ii0i
N
0i ii
ii
T,...,T,T,...,,)TT(
T
1a
2
Tminmax
N10N21
N
1ii00 2
1
N
The resulting solution will give a lower bound for the costs. I lagrangian relaxation
, (4.36)where i are nonnegative.
define
= 1 - 21,
. . (4.37) .
=N - 2N.
N
0i ii
ii
T,...,T,T T
1a
2
Tmin
N10
q2T imi
the optimal solution must have all The optimal solution of the relaxed problem can be obtained by solving N + 1 independent classical lot sizing problems.
(4.38)without any constraints on the cycle times.rounding the cycle times of the relaxed problem,
(4.39)
.0i
for some number q > 0. From Proposition 4.1, if q is given, the maximum cost increase is at most 6 percent. If we adjust q to get a better approximation the cost increase is at most 2 percent according to Propo sition 4.2. We have now obtained Roundy’s solution of the problem.
I. A simpler technique instead of Lagrangian Relaxation
From (4.34) and (4.35) the optimal cycle times in the relaxed prob lem are nondecreasing with i,
1ii TT , i = 1, 2, ..., N. (4.40)Since 0 = 0 we will always aggregate items 0 and 1. After
aggregation we have an item with cost parameters A + a1 and
1. Next we check whether a2/2 < (A + a1)/1. If this is the
case the aggregate item should include also item 2, etc. When no more aggregations are possible we can optimize the resulting aggregate items individually.
Example 4.2
N = 4, A = 300, a1 = a2 = a3 = a4 = 50, h1 = 50000, h2 = 10000, h3 = 7000, and h4 = 1000.
Both of the approaches considered assume constant demand, but can also be used in case of stochastic demand.
4.3.2 A stochastic model
independent, stationary, integer demand, complete backordering.
N = number of items,
hi = holding cost per unit and time unit for item i,
b1, i = shortage cost per unit and time unit for item i,A = setup cost for the group,
ai = setup cost for item i,
Li = constant lead-time for item i.
Viswanathan (1997)
First step: disregard the joint setup cost and consider the items individu ally for a suitable grid of review periods T. For each review period, determine the optimal individual (s, S) policies for all items and the corresponding average costs.
Ci(T) = average costs per time unit for item i when using the optimal individual (s, S) policy with a review interval of T time units.
Second step: determine the review period T by minimizing,
(4.41)
Note that the actual costs are lower than the costs according to (4.41), since the major setup cost A is not incurred at reviews where none of the items are ordered.can-order policies. (Si, Q) policy.
N
1ii )T(CT/A)T(C
A B
Distribution: Warehouse Store
Production: Subassembly Final product
Figure 5.1 An inventory system with two coupled inventories.
Reduces the length and uncertainty of lead times.
5 MULTI-ECHELON SYSTEMS 5.1 Inventory systems in distribution and production
Central warehouse
Retailers
5.1.1 Distribution inventory systems
Distribution system or arborescent systemSerial system
Figure 5.2 Distribution inventory system.
5.1.2 Production inventory systems
convergent flow
Figure 5.3 An assembly system.
It is, in general, considerably easier to deal with serial systems than with other types of multi-echelon systems. Raw material are low volume items. Higher setup cost at earlier stages, large bathes.
1
3
2
6
5
4
8
7
Figure 5.4 A general multi-echelon inventory system.
Figure 5.5 Bill of material corresponding to the inventory system in Figure 5.4.
5.2 Different ordering systems
5.2.1 Installation stock reorder point policies
(R, Q) policies; special case R=S-1, Q=1(S policy with discrete units.)
Installation stock (R, Q) policy (on hand + outstanding orders - backorders)
(s, S) policy is not the optimal policy for a multi-echelon system
KANBAN policySmoothing aspect and local decentralized controls.
Example 5.1
Table 5.1 Installation and echelon stock inventory positions in Figure 5.6.
Item Installation stock inventory position
Echelon stock inventory position
1 5 5
2 2 7
3 3 3
4 5 28
echelon stock reorder point policy will generally use larger reorder points than an installation stock policy to achieve similar control.5 + 2*5 + 2*2 + 3*3=285 + 2*7 +3*3=28
13 2N .......
5.2.3 Comparison of installation and echelon stock policies
Figure 5.7 Serial inventory system with N installations.
Raw material Final product
Increasing batch sizes
1nnn QjQ
inIP
enIP i
1i
1nin IP...IPIP
inR
enR
Qn = batch quantity at installation n.
, (5.1)where jn is a positive integer.
Notation:
= installation inventory position at installation n,
=
= echelon stock inventory position at installation n,
= installation stock reorder point at installation n,
=echelon stock reorder point at installation n.
0inIP 0e
nIP
nen
0en
enn
in
0in
in QRIPR,QRIPR
0inIP i
nR
initial inventory positions and satisfying
(5.2)An installation stock policy is always nested. Installation n may order only if (n-1) has just ordered. Echelon IP at n is only changed by final demand at 1 and replenishment order at n.Assuming: is an integer multiple of Qn-1.
All demands at installation n are for multiples of Qn-1
all replenishments are also multiples of Qn-1
)QR(IP k
n
1k
ik
en
)QR(RR k
1n
1k
ik
in
en
Proposition 5.1 An installation stock reorder point policy can always be replaced by an equivalent echelon stock reorder point policy.Proof When n orders, it means 1,2,…,n-1 has ordered. Then
. (5.3)
. (5.4)
If is not a multiple of Qn-1, change by an amount
less than Qn-1 .So that is a multiple of Qn-1.
Unit demand. Orders would be triggered exactly at the same time. When (n-1) orders Qn-1, its inventory reaches the level Rn-
1+Qn-1. At this point, level n inventory goes from Rn+Qn-1 to Rn. And then n orders.
inRi
n0i
n RIP in
0in RIP
e1
i1 RR
111 nenn
en
en
en
in QRQRIPIPIP
1, 1111 nQRRQIPRRR nen
enn
in
in
ei
Proposition 5.2 An echelon stock reorder point policy which is nested can always be replaced by an equivalent installation stock reorder point policy.Proof For installation 1,
For installation n > 1, immediately after ordering
(5.5)
(5.6)
Therefore,
Example 5.2 Consider a serial system with N = 3 installations and batch quantities Q1 = 5, Q2 = 10, and Q3 = 20. Assume that the initial inventory positions are , , and . Assume furthermore that the demand for item 1 is one unit per time unit. Consider first an installation stock policy with reorder points , , and Note that our assumptions that Qn as well as are multiples of Qn-1 are satisfied. It is easy to check that installation 1 will order at times 5, 10, 15,..... The demand at installation 2 is consequently 5 units at each of these times. This means that installation 2 will order at times 10, 20, 30,.... Demands for 10 units at installation 3 at these times will trigger orders at times 20, 40, 60,.....
Using (5.4) we obtain the equivalent echelon stock policy as , , and Recall that when considering the echelon stock the inventory positions at all installations are reduced by the final demand, i.e., by one unit each time unit, and not by the internal system orders. The initial inventory positions are , , and . It is easy to verify that the orders will be triggered at the same times as with the installation stock policy.
Assume then that we change the echelon stock reorder point at installation 3 to This will not change the orders at installations 1 and 2, but the orders at installation 3 will be triggered 2 time units earlier, i.e., at times 18, 38, 58,.... Recall that the echelon stock is reduced by one unit at a time. The resulting echelon stock policy is not nested and it is impossible to get the same control by an installation stock policy.
Propositions 5.1 and 5.2 show that in any serial inventory system an installation stock policy is simply a special case of an echelon stock policy. This is also true for assembly systems.
Installation policy: Only local information needed. The higher generality of an echelon stock policycan be an advantage in certain situations.
Stock on hand at installation 2
0 1 2 3 4 Time
50
Stock on hand at installation 1
0
1 2 3 4 Time
50
5
Example 5.3 a serial system (Figure 5.7), N = 2, Q1 = 50, Q2 =100.
final demand at installation 1 =50. lead-time at installation 1 is one, at installation 2 is 0.5. No shortages allowed, holding costs at installation 1 are higher than at installation 2.
Figure 5.8 Inventory development in the optimal solution.
LT=0.5, R2e =25+50=75,
IP20(0.5)=0.
LT=1, R1e=50, IP1
i(0.5)=75.
Echelon Policy is more general. At t=0.5- , IP2i=0,IP1
i=25.
The dominance of echelon stock policies for serial and assembly systems does not carry over to distribution systems.
5.2.4 Material Requirements Planning
periodic review, rolling horizon
Master Production Schedule (MPS).
External demands of other items
A bill of material for each item specifying all of its immediate components and their numbers per unit of the parent.
Inventory status for all items
Constant lead-times for all items.
Rules for safety stocks and batch quantities.
MPS. A production or program of final products. Must cover total system lead time.
1
2(1)
3(1)
Figure 5.9 Considered product structure.Table 5.2 Net requirements of item 1.
Item 1 Period 1 2 3 4 5 6 7 8
Lead-time = 1 Gross requirements
10 25 10 20 5 10
Order quantity = 25 Scheduled receipts 25
Safety stock = 5
Projected inventory
22 12 37 12 2 -18 -23 -23 -33
Net requirements 3 20 5 10
Table 5.4 MRP record for item 1 with planned orders in periods 3 and 5.
Item 1 Period 1 2 3 4 5 6 7 8
Lead-time = 1 Gross requirements 10 25 10 20 5 10
Order quantity = 25 Scheduled receipts 25
Safety stock = 5
Projected inventory
22 12 37 12 27 7 27 27 17
Planned orders 25 25
reorder point = the lead-time demand plus the safety stock minus one
Figure 5.10 Material requirements planning for items 1, 2, and 3 in Figure 5.9.
safety time Table 5.5 MRP record for item 1 with a safety time.
Item 1 Period 1 2 3 4 5 6 7 8
Lead-time = 1 Gross requirements 10 25 10 20 5 10
Order quantity = 25 Scheduled receipts 25
Safety stock = 5
Projected inventory
22 12 37 37 27 32 27 27 17
Safety time = 1
Planned orders 25 25
Figure 5.11 Product structures.
Figure 5.12 Gross requirements from different sources.
MRP is often referred to as a push system since orders are in a sense triggered in anticipation of future needs. Reorder point systems and KANBAN systems are similarly said to be pull systems because orders are triggered when downstream installations need them.
The MRP logic is simple. Yet the computational effort can be very large if there are thousands of items and complex multi-level product structures.
“nervousness”
DRP, Distribution Requirements Planning.Manufacturing Resource Planning- MRP II
Rough Cut Capacity Planning (RCCP)Capacity Requirements Planning (CRP)Enterprise Resource Planning (ERP)
Figure 5.13 Manufacturing Resource Planning.
5.2.5 Ordering system dynamics
bullwhip effect, Forrester (1961)decentralized installation stock policies in multi-echelon systems can yield very large demand variations early in the material flow, even though the final demand is very stable.
Note: both the information delays and the problems of large demand variations at upstream facilities are largely due to long lead-times and large batch quantities.
5.3 Order quantities
Assumptions:demand is known and deterministic. In case of stochastic demand, replace the stochastic demand by its mean and use a deterministic model when determining batch quantities. all lead-times are zero.
1 2
5.3.1 A simple serial system with constant demand
Figure 5.14 A simple serial two-echelon inventory system. Example 5.4
item 1 is produced from one unit of the component 2.
d = 8, A1 = 20, A2 = 80, h1 = 5, h2 = 4.
11
111
Q
dA
2
QhC
8h
dA2Q
1
11
40dhA2C 111
I.Treat the two installations independently
. (5.7)
,
.
, (5.8) where k is a positive integer.
12 QkQ
Echelon stock, item 2
Installation stock, item 2
Time
Installation stock = echelon stock, item 1
Time
Figure 5.15 illustrates the behavior of the inventory levels for k = 3.
Figure 5.15 Inventory levels for k = 3.
12
122
Qk
dA
2
Q)1k(hC
2
2
1
*
h
dA2
Q
1k
, (5.9)
If k* < 1 it is optimal to choose k = 1. If k* > 1. Let k´ be the largest integer less or equal to k*, i.e., k´ k* < k´ + 1, it is optimal to choose k = k´ if k*/k´ (k´ + 1)/k*, otherwise k = k´+ 1.
k* 2.24, k = 2. C2 = 56, C = C1 + C2 = 40 + 56 = 96.
1
21
12121
Q
d)
k
AA(
2
Q)h)1k(h(CCC
11
11
e1 Q
dA
2
QeC
12
12
e2 kQ
dA
2
kQeC
1
21
121
e2
e1 Q
d)
k
AA(
2
Q)eke(CCC
II. Treat the two installations together
. (5.10)Alternatively, use the echelon holding costs e1 = h1 - h2,
and e2 = h2,
, (5.11)
. (5.12)
, (5.13)
(5.10) and (5.13) are equivalent.
21
21
1eke
d)k
AA(2
Q
)eke(d)k
AA(2)k(C 21
21
)k
eAkeAeAeA(d2)k(C 12
2122112
21
12*
eA
eAk
. (5.14)
. (5.15)
(5.16)
. (5.17)If k* < 1 it is optimal to choose k = 1. If k* > 1. Let k´ be the largest integer less or equal to k*, i.e., k´ k* < k´ + 1, it is optimal to choose k = k´ if k*/k´ (k´ + 1)/k*, otherwise k = k´+ 1.
1A
1h
Example 5.5 d = 8, A1 = 20, A2 = 80, h1 = 5, h2 = 4.
e1 = h1 - h2 = 1, e2 = h2 = 4. From (5.17), k* = 1.
Applying (5.14) and (5.15), ,
and C* 89.44, about 7 percent lower than the costs obtained in Example 5.4.
(5.14) and (5.15) are essentially equivalent to the corresponding expressions (3.3) and (3.4) for the classical economic order quantity model.
= A1 + A2/k , (5.18)
= e1 + ke2. (5.19)
89.17Q*1 89.17QkQQ *
1*1
*2
Stochastic Inventory Model
Proportional Cost Models:
x: initial inventory,
y: inventory position (on hand + on order-backorder),
: random demand, () , (),
(y- )+: ending inventory position, N.B.L,
(y- ) : ending inventory position, B.L,
=1/(1+r) : discount factor,
ordering cost : c(y-x),
holding cost : h (y- )+
penalty cost : p( -y)+
salvage cost : - s(y- )+
Minimum cost f(x) satisfies:
L(y) convex, L’(0) < -c (otherwise never order) L′ eventually becomes positive
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Set up cost K
L(x) if we order nothing
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cost
L(x)
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s S x
L(x)+cx
c
s
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order y-x if x s
order nothing if x > s
Multiperiod models
Infinite Horizon (f1000 & f1001 cannot be different)
dyfyLxycxfxy
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12
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)10()()(')('00
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)11()()()()()(0
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So that
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1800
5
4
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Example 5:
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Demand D1, D2, i.i.d
Density: ()
L(y) = expected one period holding+ shortage penalty cost;
strictly convex with linear cost and () >0,
c purchase cost /unit
c1(x1) optimal cost with 1 period to go;
c+L’(y10)=0
while y10 is the optimal base stock level.
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xifxotherwise
xifxotherwise
q
q
y
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d
Multi-Period Dynamic Inventory Model with no Setup Cost
Cn(xn): n periods to go,
: discount factor.
DP equations:
SS
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S
SSSSSS
xc
DycEyLxycxc
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Multi-Period Dynamic Inventory Model with Setup Cost
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