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Department of Industrial Engineering
Supply Chains:Aggregate Planning
Jayant Rajgopal, Ph.D., P.E.
Department of Industrial Engineering
University of Pittsburgh
Pittsburgh, PA 15261
Department of Industrial Engineering
Department of Industrial Engineering
With aggregate planning we decide onoverall production and inventory levels
• We are not interested in specific items per se, but rather,in looking at the aggregate picture
– At an auto plant?
– At a petroleum refinery?
– At a cereal producer?
– At a steel mill?
– At a paint production facility?
• Typically use rolling planning horizons of 6 to 18 months
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Aggregate plans must meet forecasteddemand while keeping total costs low• Two phases
– Phase 1: Define a logical aggregate unit and forecastdemand in these units over the planning horizon, alongwith an estimate of any safety buffers
– Phase 2: Develop plans by varying• Workforce size• Production rate
– Number of shifts– Overtime/Undertime– Subcontracting
• Inventory levels
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Aggregate Planning
• Costs to consider include
– Production smoothing costs (hiring & layoffs)
– Overtime, undertime & subcontracting costs
– Inventory costs
– Shortage costs
– Labor costs
Wt-1
It-1
Pt-1
Wt
It
Pt
Forecast Ft
Costs
© Jayant Rajgopal, 2016
tt
Department of Industrial Engineering
Conflicting prioritiesCosts aside, objectives are typically different for differentfunctional areas/managers:
• Finance
– Low inventory
• Operations engineers/managers
– Smooth production
• Marketing
– Lots of stock
Need a multifunctional team led by a senior manager inorder to agree on a good plan
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Pure Strategy 1: Level Strategy
– Maintain a constant (minimal) production rate over the entire horizon
– Main objective: Minimize smoothing costs
– Smooth production, but lots of inventory each month
The fundamental tradeoff is betweensmooth production and low inventories
© Jayant Rajgopal, 2016
0
5
10
15
20
25
30
35
40
45
50
1 2 3 4 5 6 7 8 9 10 11 12
Demand 20 25 30 25 40 50 50 30 20 40 50 40
Production 35 35 35 35 35 35 35 35 35 35 35 35
Inventory 15 25 30 40 35 20 5 10 25 20 5 0
Qu
an
tity
Level production
Department of Industrial Engineering
Pure Strategy 2: Chase Strategy
– Change production rate to “chase” the demand
– Main objective: Minimize inventory costs
– No Inventory, but choppy production that changes every month
Mixed Strategy: Some combination of the two pure strategies
– Usually works best© Jayant Rajgopal, 2016
0
5
10
15
20
25
30
35
40
45
50
1 2 3 4 5 6 7 8 9 10 11 12
Demand 20 25 30 25 40 50 50 30 20 40 50 40
Production 20 25 30 25 40 50 50 30 20 40 50 40
Inventory 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Qu
an
tity
Department of Industrial Engineering
Objective is to plan on meeting all demandin each period of the planning horizon
• Actual demand is unknown and we only have a forecast.So…
– We need to keep some extra buffer in case actual demand in period t(=Dt) exceeds the forecast for the period (=Ft)
– For example, plan on having a buffer Bt= 15% of Ft for period t(perhaps, because historically, our forecasts could be off by up toabout 15%...)
• So plan to produce an amount Pt so that the expectedinventory at the end of period t – let’s call this E[It] is at leastas large as this buffer Bt, i.e.,
E[It] ≥ Bt© Jayant Rajgopal, 2016
Department of Industrial Engineering
For example…
The actual inventory It will of course, in general be more orless than the expected value:
• If Dt=Ft the ending inventory in period t (=It) will be exactly equal tothe buffer
• if Dt<Ft then we will have It>Bt
• if Dt>Ft then we will dip into the buffer so that It<Bt
© Jayant Rajgopal, 2016
Period Forecast Buffer1 20 32 25 43 30 54 25 45 40 6
Department of Industrial Engineering
Inventory Balance Equation
Let Dt = Demand in period t (unknown…)
Ft = Forecast for demand in period t
Pt = Production planned for period t
It = Ending inventory end of period t
So It-1 + Pt = Amount available in period t
and It = (It-1 + Pt - Dt)
If we use the forecast in place of demand and expected valuesfor the inventory, we get
© Jayant Rajgopal, 2016
E[It] = E[It-1] + Pt - Ft
Department of Industrial Engineering
Aggregate Planning Calculations
Let Bt = Buffer specified for period t
Note that Bt is a planned buffer just for period t, in case the demand exceeds theforecast for that period.
Period 1
E[I1] = I0 + P1 - F1 ≥ B1
Period 2
E[I2] = E[I1] + P2 - F2
= {I0 + P1 - F1} + P2 - F2
= I0 + (P1 + P2) - (F1+ F2) ≥ B2
Period 3
E[I3] = E[I2] + P3 - F3
= {I0 + (P1 + P2) - (F1+ F2)} + P3 - F3
= I0 + (P1 + P2 + P3) - (F1 + F2 + F3) ≥ B3
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Aggregate Planning Calculations
In general, for EVERY period t, a feasible production planmust plan on ending with a buffer that is at least as large asthe required buffer amount for that period:
E[It] =
i.e.,
Cumulative Availability ≥ Cumulative Requirements
(CAt=I0+CPt) (CRt=Bt+CFt)
• Also note E[It]=CAt-CFt
© Jayant Rajgopal, 2016
Department of Industrial Engineering
t
Cum
ula
tive
Units
I0
CFt
CRtCAt= CPt+I0
1 2 3 4 5 6
Bt
It =CAt - CFt
© Jayant Rajgopal, 2016
CRt= Bt+CFt
Department of Industrial Engineering
Aggregate Planning – An Example
The table below gives a forecast of production requirements and therequired buffer inventory (Bt) for each month of the upcoming year.
Develop an aggregate production plan for the next 12 months.
Month WorkdaysCumul.
WorkdaysForecasted
DemandCumul.
DemandReq'd Min.Inventory
Cumul.Reqmts. Avg. Reqmts.
t dt Cdt Ft CFt Bt CRt (CRt-I0)/Cdt
D 0 I0 = 2,800
J 1 22 22 5,000 5,000 2,800 7,800 227.3F 2 20 42 4,000 9,000 2,500 11,500 207.1M 3 23 65 4,000 13,000 2,500 15,500 195.4A 4 19 84 5,000 18,000 2,800 20,800 214.3M 5 22 106 7,000 25,000 3,200 28,200 239.6J 6 22 128 9,000 34,000 3,500 37,500 271.1J 7 20 148 11,000 45,000 4,100 49,100 312.8A 8 23 171 9,000 54,000 3,500 57,500 319.9S 9 11 182 6,500 60,500 2,000 62,500 328.0O 10 22 204 6,000 66,500 3,000 69,500 327.0N 11 22 226 5,000 71,500 2,800 74,300 316.4D 12 18 244 5,000 76,500 2,800 79,300 313.5
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Additional Information
• The aggregate unit used by the company is the equivalent of oneworker-day of labor (i.e., a fictitious product that is produced as aresult of one worker working for one day).
• Records show that as of Dec. 31, there are 2800 units in stock and320 workers on the payroll. The company would like to finish nextyear at these same levels.
• Union agreements stipulate a minimum of 230 employees andcompany policy is to have a maximum of 350 employees on itspayroll during any given month; thus the maximum regular productioncapacity of the company’s production facility is 350 units/day.
• If required, capacity can be further increased by a maximum of 50units/day via overtime. Each extra unit (over regular productioncapacity) that is produced incurs an overtime premium (i.e.,additional cost over and above regular production cost) of $100/unit.
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Additional Information
• The company also has the option of subcontracting production at apremium of $200 per unit.
• The company can also use undertime, i.e., keep an employee onthe payroll without having her/him produce. Such employees arediverted to other tasks when possible but the company estimatesthat it loses about $50 a day per worker from undertime.
• Workers average $2500 per month in wages. Changes inworkforce levels can occur at the beginning of each month andhiring a new employee or laying off an existing one both cost anestimated $200 per employee hired or laid off.
• Inventory holding costs are estimated at $180 per unit per year.
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Recall that a feasible aggregate plan requires that for each t=1,2,…,N in
the planning horizon (PH): CAt ≥ CRt
⇒ I0 + (P1 + P2 +…+ Pt) ≥ (F1 + F2 +…+ Ft) + Bt
⇒ CPt ≥ CFt + Bt – I0 for each and every t in the PH.
Suppose we wish to produce at a CONSTANT RATE of P1=P2=…=PN=Punits/month
Period 1
P1≥ F1 + B1 - I0
⇒ P ≥ (F1 + B1 - I0), i.e.,� � �
Period 2
P1+P2 (=2P) ≥ (F1+ F2) + B2 - I0
⇒ 2P ≥ {(F1+ F2)+ B2 - I0} , i.e.,� � �
Developing a Level Strategy
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Period 3
P1+P2+P3 (=3P) ≥ (F1 + F2 + F3) + B3 - I0
⇒ 3P ≥ {(F1 + F2 + F3)+ B3 - I0} , i.e.,� � �
:
Period N
NP ≥ {(F1 + F2 +…+ FN)+ BN - I0} , i.e.,� � �
i.e.,� � �, � � �, � � � ,…, � � �
So we choose the constant monthly production rate P via:
Level Strategy
P = � � � = � �
© Jayant Rajgopal, 2016
Department of Industrial Engineering
• The level production rate is more commonly expressed in terms of a leveldaily rate (rather than a monthly one).
• Of course, if each month has the same number of workdays (say d), thenthe daily production rate p is simply equal to P/d units in every month,where P is calculated as before...
General Case (unequal no. of work days per month)
• Suppose that we have a different no. of work days per month - say dt
days in month t - and we wish to produce using a level strategy at the rateof p units/day.
• So production in each month varies; in month t it is given by Pt = pdt units
Period 1
P1 (=pd1) ≥ F1 + B1 - I0 ⇒ pd1 ≥ (F1 + B1 - I0), i.e.,� � �
�
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Period 2P1+P2 (=pd1+pd2) ≥ (F1+ F2) + B2 - I0
⇒ p(d1+d2) ≥ {(F1+ F2)+ B2 - I0}, i.e.,���������
�����:Period N
P1+P2 +…+P3 (=pd1+pd2+…+pdN) ≥ (F1 + F2 +…+ FN) + BN - I0
⇒ p (d1+d2+…+dN) ≥ {(F1 + F2 +…+ FN)+ BN - I0}, i.e.,���������
���
i.e.,
� � �
�, � � �
�, � � �
�, … , � � �
�
So we choose the constant daily production rate p via:
p = � � �
�= � �
�
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Pure Strategy 1: “Level” Production
Costs:
Inventory = 5,016*15 + 7,576*15 + … + 6,332*15 = 87,616*15 = $1,314,240
Smoothing = 8*200 + 8*200 = $3,200 (8 employees hired in Jan., lay-offs end of Dec.)
Wages: 328*2500*12 = $9,840,000
Month Workdays
Workforce Daily Prod.
MonthlyProd.
ForecastedDemand
ExpectedInventory
RequiredBuffer
t dt Wt p Pt = pdt Ft It BtCAt CRt
(Dec) 320 2,800 2,800
Jan 22 328 328 7,216 5,000 5,016 2,800 10,016 7,800
Feb 20 328 328 6,560 4,000 7,576 2,500 16,576 11,500
Mar 23 328 328 7,544 4,000 11,120 2,500 24,120 15,500
Apr 19 328 328 6,232 5,000 12,352 2,800 30,352 20,800
May 22 328 328 7,216 7,000 12,568 3,200 37,568 28,200
Jun 22 328 328 7,216 9,000 10,784 3,500 44,784 37,500
Jul 20 328 328 6,560 11,000 6,344 4,100 51,344 49,100
Aug 23 328 328 7,544 9,000 4,888 3,500 58,888 57,500
Sep 11 328 328 3,608 6,500 1,996 2,000 62,496 62,500
Oct 22 328 328 7,216 6,000 3,212 3,000 69,712 69,500
Nov 22 328 328 7,216 5,000 5,428 2,800 76,928 74,300
Dec 18 328 328 5,904 5,000 6,332 2,800 82,832 79,300(Jan) 320 ∑=87,616
© Jayant Rajgopal, 2016TOTAL = $11,157,440
Department of Industrial Engineering
Pure Strategy 2: “Chase Demand”
COSTS: (Note: We have undertime if p<230, overtime if p>350, and subcontracting if p>400
Inventory = 2,816*15 + 2,516*15 + … + 2,801*15 = 40,454*15 = $534,570
Smoothing = (90+48+59+13+31+89+47+43)*200 = 420*200 = $84,000
Overtime Premium =(50*22 + 50*20 + 16*23 + 50*11)*100 = $301,800
Subcontracting Premium = (23*22 + 179*20 + 53*11)*200 = $933,800
Undertime Costs = (2*22 + 45*20 + 56*23 + 12*22)*50 = $124,800
Wages = (230+230+…277)*2500 = $ 8,827,500
Month Workdays Daily Prod.Workforce
Hire/Layoff Under Over Sub
Monthly Prod.Forecasted
DemandExpectedInventory
Req’dBuffer
T dt p Wt Ht/Lt Ut Ot St Pt=(p+Ot+St)dt Ft It Bt
(Dec) 320 2,800Jan 22 228 230 -90 2 0 0 5,016 5,000 2,816 2,800Feb 20 185 230 0 45 0 0 3,700 4,000 2,516 2,500Mar 23 174 230 0 56 0 0 4,002 4,000 2,518 2,500Apr 19 278 278 48 0 0 0 5,282 5,000 2,800 2,800May 22 337 337 59 0 0 0 7,414 7,000 3,214 3,200Jun 22 423 350 13 0 50 23 9,306 9,000 3,520 3,500Jul 20 579 350 0 0 50 179 11,580 11,000 4,100 4,100Aug 23 366 350 0 0 16 0 8,418 9,000 3,518 3,500Sep 11 453 350 0 0 50 53 4,983 6,500 2,001 2,000Oct 22 319 319 -31 0 0 0 7,018 6,000 3,019 3,000
Nov 22 218 230 -89 12 0 0 4,796 5,000 2,815 2,800Dec 18 277 277 47 0 0 0 4,986 5,000 2,801 2,800
(Jan) 320 43 ∑=35,638
© Jayant Rajgopal, 2016
TOTAL = $10,806,470
Department of Industrial Engineering
A Plan that uses a mixed strategy
NOTE: This plan produces• 240 units/day Jan.-Mar.,• 400 units/day (including 50 using overtime) in Apr.-Aug.• 270 units/day in Sep.-Dec.
MonthWorkdays
WorkForce
DailyProd.
Hire/Layoff
MonthlyProd.
ForecastedDemand
ExpectedInventory
Req’dBuffer
t dt Wt p Ht/Lt Pt = pdt Ft It Bt CAt CRt
(Dec) (320) 2,800 2,800
Jan 22 240 240 -80 5,280 5,000 3,080 2,800 8,080 7,800Feb 20 240 240 0 4,800 4,000 3,880 2,500 12,880 11,500Mar 23 240 240 0 5,520 4,000 5,400 2,500 18,400 15,500Apr 19 350 350+50 110 7,600 5,000 8,000 2,800 26,000 20,800May 22 350 350+50 0 8,800 7,000 9,800 3,200 34,800 28,200Jun 22 350 350+50 0 8,800 9,000 9,600 3,500 43,600 37,500Jul 20 350 350+50 0 8,000 11,000 6,600 4,100 51,600 49,100Aug 23 350 350+50 0 9,200 9,000 6,800 3,500 60,800 57,500Sep 11 270 270 -80 3,520 6,500 3,270 2,000 63,880 62,500Oct 22 270 270 0 7,040 6,000 3,210 3,000 70,040 69,500Nov 22 270 270 0 7,040 5,000 4,150 2,800 76,200 74,300Dec 18 270 270 0 5,760 5,000 4,010 2,800 81,240 79,300(Jan) (320) 50 ∑=67,800
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Costs:
Inventory = 3,080*15 + 3,880*15 + … + 4,700*15 = 67,800*15 = $1,017,000
Smoothing = (80 + 110 + 80 + 50)*200 = $64,000
Overtime Premium ={50*19 + 50*22 + 50*22 + 50*20 + 50*23)} *100 = $530,000
Wages = (240*3 + 350*5 + 280*4)*2500 = $8,875,000
TOTAL = $10,486,000
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Another Mixed Strategy obtained viaLinear Programming
VARIABLES:
For t=1,2,...,12
Wt = Workforce level in month t
Ht = No. of workers hired in month t
Ft= No. of workers laid off in month t
nT= No. of days in month t (GIVEN)
p = No. of units produced by 1 worker working for 1 day (=1: GIVEN)
Ot = Daily overtime production in month t
Ut = Daily undertime in month t
St = No. of units subcontracted each day in month t
It = Expected Inventory at the end of month t
Pt = Total in-plant production (regular+overtime) in month t
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Another Mixed Strategy obtained viaLinear Programming
COSTS:
CH = Cost of hiring an employee ($200)
CF = Cost of laying off an employee ($200)
CO = Additional cost of producing an item in overtime ($100)
CS = Additional cost of subcontracting an item ($200)
CU = Additional cost of undertime per worker ($50)
CW = Monthly wages per worker ($2500)
CI = Inventory carrying cost per item per month [$(180/12)=$15]
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Min∑ ���� + ���� + ���� + ������ + ������ + ������ + ���� + ����� + ����������
= 200 H1 + 200 H2 + 200 H3 + 200 H4 + 200 H5 + 200 H6 + 200 H7 + 200 H8 + 200 H9 + 200 H10 + 200 H11 + 200 H12 + 200 H13 +
200 F1 + 200 F2 + 200 F3 + 200 F4 + 200 F5 + 200 F6 + 200 F7 + 200 F8 + 200 F9 + 200 F10 + 200 F11 + 200 F12 + 200 F13 +
2500W1+2500W2+2500W3+2500W4+2500W5+2500W6+2500W7+2500W8+2500W9+2500W10+2500W11+2500W12 +
2200 O1 + 2000 O2 + 2300 O3 + 1900 O4 + 2200 O5 + 2200 O6 + 2000 O7 + 2300 O8 + 1100 O9 + 2200 O10 + 2200 O11 + 1800 O12 +
4400 S1 + 4000 S2 + 4600 S3 + 3800 S4 +4400 S5 + 4400 S6 + 4000 S7 + 4600 S8 + 2200 S9 + 4400 S10 + 4400 S11 + 3600 S12 +
1100 U1 + 1000 U2 + 1150 U3 + 950 U4 + 1100 U5 + 1100 U6 + 1000 U7 + 1150 U8 + 550 U9 + 1100 U10 + 1100 U11 + 900 U12 +
15 I1 + 15 I2 + 15 I3 + 15 I4 + 15 I5 + 15 I6 + 15 I7+ 15 I8 + 15 I9 + 15 I10 + 15 I11 + 15 I12
SUBJECT TO
2) W0 = 320
3) W13 = 320
4) H1 - F1 + W0 - W1 = 0
5) H2 - F2 + W1 - W2 = 0
6) H3 - F3 + W2 - W3 = 0
7) H4 - F4 + W3 - W4 = 0
8) H5 - F5 + W4 - W5 = 0
9) H6 - F6 + W5 - W6 = 0
10) H7 - F7 + W6 - W7 = 0
11) H8 - F8 + W7 - W8 = 0
12) H9 - F9 + W8 - W9 = 0
13) H10 - F10 + W9 - W10 = 0
14) H11 - F11 + W10 - W11 = 0
15) H12 - F12 + W11 - W12 = 0
16) H13 - F13 + W12 - W13 = 0
( )1 2
1 3 1 31
H t F t t O t t S t t U t I t H Ft
C H C F n C O n C S n C U C I C H C F=
+ + + + + + +∑
SPECIFIED
WORKFORCE BALANCEWt = Wt-1 + Ht - Ft
© Jayant Rajgopal, 2016
Department of Industrial Engineering
17) - 22 O1 - 22 W1 + 22 U1 + P1 = 0
18) - 20 O2 - 20 W2 + 20 U2 + P2 = 0
19) - 23 O3 - 23 W3 + 23 U3 + P3 = 0
20) - 19 O4 - 19 W4 + 19 U4 + P4 = 0
21) - 22 O5 - 22 W5 + 22 U5 + P5 = 0
22) - 22 O6 - 22 W6 + 22 U6 + P6 = 0
23) - 20 O7 - 20 W7 + 20 U7 + P7 = 0
24) - 23 O8 - 23 W8 + 23 U8 + P8 = 0
25) - 11 O9 - 11 W9 + 11 U9 + P9 = 0
26) - 22 O10 - 22 W10 + 22 U10 + P10 = 0
27) - 22 O11 - 22 W11 + 22 U11 + P11 = 0
28) - 18 O12 - 18 W12 + 18 U12 + P12 = 0
29) I0 = 2800 }
30) 22 S1 + I0 - I1 + P1 = 5000
31) 20 S2 + I1 - I2 + P2 = 4000
32) 23 S3 + I2 - I3 + P3 = 4000
33) 19 S4 + I3 - I4 + P4 = 5000
34) 22 S5 + I4 - I5 + P5 = 7000
35) 22 S6 + I5 - I6 + P6 = 9000
36) 20 S7 + I6 - I7 + P7 = 11000
37) 23 S8 + I7 - I8 + P8 = 9000
38) 11 S9 + I8 - I9 + P9 = 6500
39) 22 S10 + I9 - I10 + P10 = 6000
40) 22 S11 + I10 - I11 + P11 = 5000
41) 18 S12 + I11 - I12 + P12 = 5000
END
IN-PLANT PRODUCTIONPt = nt(Wt-Ut) + ntOt
(regular prod.) + (overtime prod.)
MATERIAL BALANCEIt = It-1 + Pt + ntSt - Dt
SPECIFIED
© Jayant Rajgopal, 2016
Department of Industrial EngineeringSUB O1 50.00000
SUB O2 50.00000
© Jayant Rajgopal, 2016
SUB O3 50.00000
SUB O4 50.00000SUB O5 50.00000SUB O6 50.00000
SUB O7 50.00000SUB O8 50.00000SUB O9 50.00000SUB O10 50.00000SUB O11 50.00000SUB O12 50.00000
SLB I1 2800.00000SLB I2 2500.00000SLB I3 2500.00000SLB I4 2800.00000SLB I5 3200.00000SLB I6 3500.00000SLB I7 4100.00000SLB I8 3500.00000SLB I9 3000.00000SLB I10 3000.00000SLB I11 2800.00000SLB I12 2800.00000
SLB W1 230.00000SUB W1 350.00000SLB W2 230.00000SUB W2 350.00000SLB W3 230.00000SUB W3 350.00000SLB W4 230.00000SUB W4 350.00000SLB W5 230.00000SUB W5 350.00000SLB W6 230.00000SUB W6 350.00000SLB W7 230.00000SUB W7 350.00000SLB W8 230.00000SUB W8 350.00000SLB W9 230.00000SUB W9 350.00000SLB W10 230.00000SUB W10 350.00000SLB W11 230.00000SUB W11 350.00000SLB W12 230.00000SUB W12 350.00000
Ot ≤ 50(MAXIMUM OVERTIME PER DAY)
It ≥ MIt(MINIMUM INVENTORY)
230 ≤ Wt ≤ 350; MIN & MAXWORKFORCE LEVELS
BOUNDS
Department of Industrial Engineering
LP OPTIMUM FOUND AT STEP 52; OBJECTIVE FUNCTION VALUE $10,157,700
W0 320.000000 H1 0.000000 F1 90.000000W1 230.000000 H2 0.000000 F2 0.000000W2 230.000000 H3 0.000000 F3 0.000000W3 230.000000 H4 0.000000 F4 0.000000W4 230.000000 H5 120.000000 F5 0.000000W5 350.000000 H6 0.000000 F6 0.000000W6 350.000000 H7 0.000000 F7 0.000000W7 350.000000 H8 0.000000 F8 0.000000W8 350.000000 H9 0.000000 F9 120.000000W9 230.000000 H10 0.000000 F10 0.000000W10 230.000000 H11 0.000000 F11 0.000000W11 230.000000 H12 0.000000 F12 0.000000W12 230.000000 H13 90.000000 F13 0.000000W13 320.000000
I0 2800.000000 ≥ 2800P1 5060.000000 I1 2860.000000 ≥ 2500P2 4600.000000 I2 3460.000000 ≥ 2500P3 6440.000000 I3 5900.000000 ≥ 2800P4 5320.000000 I4 6220.000000 ≥ 3200P5 8800.000000 I5 8020.000000 ≥ 3500 MIt
P6 8800.000000 I6 7820.000000 ≥ 4100P7 8000.000000 I7 4820.000000 ≥ 3500P8 9200.000000 I8 5020.000000 ≥ 3000P9 3080.000000 I9 3000.000000 ≥ 3000P10 6000.000000 I10 3000.000000 ≥ 2800P11 5060.000000 I11 3060.000000 ≥ 2800P12 4740.000000 I12 2800.000000
O1 0.000000 S1 0.000000O2 0.000000 S2 0.000000O3 50.000000 S3 0.000000O4 50.000000 S4 0.000000O5 50.000000 S5 0.000000O6 50.000000 S6 0.000000O7 50.000000 S7 0.000000O8 50.000000 S8 0.000000O9 50.000000 S9 127.272728O10 42.727272 S10 0.000000O11 0.000000 S11 0.000000O12 33.333332 S12 0.000000
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Luckily most variables turn out to be integer valued at the optimum!
Let us do some rounding to make things look nicer…
O10 = 43 (rather than 42.727272) in October,
O12 = 34 (rather than 33.33332) in December,
S9 = 127 (rather than 127.272727) in September
The production plan along with the availability and expected inventory is then as follows:
MonthWorkdays
Workforce
DailyProd.
Overtime Subcontr. MonthlyProd. Demand Inventory
Req’dBuffer
t dt Wt p Ot St Pt=(p+Ot+St)dt Dt It Bt
(Dec) (320) 2,800 2,800Jan 22 230 230 0 0 5,060 5,000 2,860 2,800Feb 20 230 230 0 0 4,600 4,000 3,460 2,500Mar 23 230 230 50 0 6,440 4,000 5,900 2,500Apr 19 230 230 50 0 5,320 5,000 6,220 2,800May 22 350 350 50 0 8,800 7,000 8,020 3,200Jun 22 350 350 50 0 8,800 9,000 7,820 3,500Jul 20 350 350 50 0 8,000 11,000 4,820 4,100Aug 23 350 350 50 0 9,200 9,000 5,020 3,500Sep 11 230 230 50 127 4,477 6,500 2,997 2,000Oct 22 230 230 43 0 6,006 6,000 3,003 3,000Nov 22 230 230 0 0 5,060 5,000 3,063 2,800
Dec 18 230 230 34 0 4,752 5,000 2,815 2,800
(Jan) (320) ∑=55,998
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Costs:
Inventory = 2,860*15 + 3,460*15 + … + 2,815*15 = 55,998*15 = $839,970
Smoothing = (|320-230|+|230-350|+|350-230|+|230-320|)*200 = $84,000
Overtime Premium =(50*23 + 50*19 + 50*22 + 50*22 + 50*20 + 50*23 +50*11 + 43*22 + 34*18)*100 = $855,800
Subcontracting Premium = (127*11)*200 = $279,400
Wages = (230*4 + 350*4 + 230*4)*2500 = $ 8,100,000
TOTAL = 10,159,170
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Cumulative Availability with a Feasible Plan
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Cumulative Demand Cumul. Reqmts.
CRt-CFt=Bt
Cumul. Availability
CAt-CFt=It
CF CR
I0
Slope=230
Slope=350
© Jayant Rajgopal, 2016
Department of Industrial Engineering
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Cumul. Forecast Demand Cumul. Reqmts. Cumul. Avail. - Level Strategy
Cumulative Availability with Level Strategy
Slope=230
Slope=350
© Jayant Rajgopal, 2016
Department of Industrial Engineering
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Cumulative Production Days
Cumul. Forecast Demand Cumul. Reqmts. Cumul. Avail. - CHASE Plan
Cumulative Availability with “CHASE” Plan
Slope=230
Slope=350
© Jayant Rajgopal, 2016
Department of Industrial Engineering
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Cumulative Production Days
Cumul. Forecast Demand Cumul. Reqmts. Cumul. Avail. - Mixed Strategy
Cumulative Availability with a Mixed Strategy
Slope=230
Slope=350
© Jayant Rajgopal, 2016
Department of Industrial Engineering
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Cumulative Production Days
Cumul. Forecast Demand Cumul. Reqmts. Cumul. Availability - Plan via Linear Programming
Cumulative Availability: Plan using Linear Programming
Slope=230
Slope=350
© Jayant Rajgopal, 2016
Department of Industrial Engineering
Cumulative Availability with Four Different Plans
Slope=230
Slope=350
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Cumulative Production Days
Cumul. Forecast Demand Cumul. Reqmts. LEVEL CHASE Mixed Strategy Linear Programming
© Jayant Rajgopal, 2016