Continuing with the discussion about Aggregate production planning consider an example where the forecast of the demand for th

Date 11^th September 2002-Wednesday Lecture

Continuing with the discussion about Aggregate production planning consider an example where the forecast of the demand for the 12 months is as follows…

January	5300
February	5100
March	4400
April	2800
May	4100
June	4800
July	6000
August	7100
September	7800
October	4800
November	7600
December	6400

The following options and constraints are there with the management

The workforce can be changed by hiring or firing.
Maximum change can be of 40 workers per month with a hiring cost of $ 300 and firing cost of $ 420
Management can cover temporary shortages by Overtime and there is a production of 20 units in regular time and not more than 6 units during overtime.
It costs $8 to store each unit per month.
We want no inventory by next December.

We can convert the given problem of aggregate production plan to a linear programming problem as follows

Let

D_j: demand in month j.

X_j: normal production in month j.

Y_j: overtime production in month j

Z_j: inventory at the end of period j.

Note that the number of workers employed in period j will be X_j/20

|X_j/20 – X_j-1/20| <= 40 for all j (hiring & firing constraint)

which can be rewritten as (since we want this constraint to be linear)

X_j/20 – X_j-1/20 <= 40

X_j-1/20 – X_j/20 <= 40

Y_j<= 6(X_j/20) = 0.3X_j( this is overtime constraint)

Z_j-1 + X_j + Y_j = D_j + Z_j(this is physical balance constraint)

X_j,Y_j,Z_j>=0;

Let T_j: hiring or firing cost in period j

Our objective is to

Min S ( 20Y_j + 8Z_j + T_j)

J=1

Note that

T_j = 15(X_j – X_j-1) if X_j>= X_j-1

T_j = 21(X_j-1 – X_j) if X_j<= X_j-1

This is not a linear constraint which may be converted to a linear constraint as follows

T_j >= 15(X_j – X_j-1)

T_j >= 21(X_j-1 – X_j)

We can replace the earlier constraints by the new ones because our objective function will take care of the excess of feasible solutions that will be generated as a result of change of the constraints and we are minimizing the value of T_jin the objective function.

Initializing with X₀ = 5800

Z₀ = 0

Z₁₂= 0

And solving the problem

We get the following results shown in the form of a table.

Month	Demand	Work Force	Normal	Overtime	Inventory
January	5300	265	5300	-	-
February	5100	255	5100	-	-
March	4400	220	4400	-	-
April	2800	201	4020	-	1220
May	4100	201	4020	-	1140
June	4800	241	4820	-	1160
July	6000	281	5620	-	780
August	7100	321	6420	-	100
September	7800	361	7220	-	20
October	4800	361	7220	560	-
November	7600	360	7200	400	-
December	6400	320	6400	-	-

In the above solution X_j‘s have been rounded off to nearest whole number.

Although these results might not be directly applicable but they at least gives a direction to the manager as in which direction he should proceed. Another important thing to note is that in November we are actually firing labor and doing overtime too so this appears strange but we are not in a position to exactly find out a reason for such a situation, it so happens that our LP problem is optimized by such solution.

Now an evaluation of LP problem that includes recipe information will be discussed.

I_it: ending inventory of product i in period t

H_i: unit cost

X_it: A set up variable for item i in period t with setup cost C_si (Type 0,1 variable)

S_ik: Setup time at facility k.

O_kt: overtime in facility k in period t

CO_kt: cost of overtime in facility k in period t.

U_kt: under time in facility k in period t and CU_kt is cost associated with it.

Y_i: yield of item I

L_i: minimal lead time in producing item i

P_it: production of item i in period t

A_ij: number of units of item i to produce 1 unit of item j

D_it: external demand for i at time t.

B_ik: time required on facility k by one unit of item i

CAP_kt: time available on facility k in period t

q: a very large number

We are assuming here that everything is deterministic

Objective:

Min: SS (H_i I_it + C_si X_it) + SS ( CO_ktO_kt + CU_kt U_kt)

i t k t

s.t

I_i,t-1 + Yi P_i,t-Li + I_it - SA_ij P_jt = D_it for all i,t…… (physical balance constraint)

S (B_ik P_it + S_ik X_it) + U_kt –O_kt = CAP_kt for all k,t (capacity constraint)

P_it<= q X_it for all i,t