Wt = workforce size for month t, t = 1…..6
Ht = number of employees hired at the beginning of Month t, t = 1…..6 Lt = number of employees firing at the beginning of Month t, t = 1…..6 Pt = number of units produced in Month t, t….6
It = Inventory at the end of Month t, t….6 St = backlogged at the end of Month t, t….6
Ct = number of units subcontracted for Month t, t….6 Ot = number of overtime hours worked in Month t, t….6 Objective Function
Denote the demand in Period t by Dt. The values of Dt are as specified by the demand forecast in Table 1. The objective function is to minimize the total cost (equivalent to maximizing total profit as all demand is to be satisfied) incurred during the
planning horizon. The cost incurred has the following components:
1. Regular – time labor cost: Recall that workers are paid a regular – time wage of IDR 600 (salary of worker IDR 600) per month. Because Wt is the number of workers in period t, the regular – time labour cost over the planning horizon s given by
∑
t=1 6600Wt
2. Overtime labor cost: As the labour cost is IDR 6 per hour and Ot represents the number of overtime hours worked in period t, the overtime cost over the planning horizon is.
∑
t=1 66Ot
3. Cost of hiring and firing: The cost of hiring a worker is IDR 300 and the cost of firing a worker is IDR 500. Ht and Lt represent the number of hired and the number firing , respectively in Period t. Thus, the cost of hiring and firing is given by :
300Ht+
∑
t=1 6
500<¿
∑
t=1 6¿
4. Cost of inventory and backlog: The cost of carrying inventory is IDR 2 per unit per month and the cost of backlog is IDR 5 per unit per month. It and St represent the units in inventory and the units stocked out respectively in period t. Thus the cost of holding inventory and stocking out is
∑
t=1 62<+
∑
t=1 6
5St
5. Cost of materials and subcontracting: The material cost is IDR10 per unit and the subcontracting cost is IDR 30/unit. Pt represents the quantity produced and Ct represents the quantity subcontracted in period t. Thus the material and subcontracting cost is
∑
t=1 610Pt+
∑
t=1 6
30Ct
6. Final Objective: The total cost incurred during the planning horizon is the sum of all the aforementioned costs and is given by
∑
t=1 6600Wt +
∑
t=1 6
6Ot +
300Ht+
∑
t=1 6
500<¿
∑
t=1 6¿
+
∑
t=1 6
2<+
∑
t=1 6
5St +
∑
t=1 610Pt+
∑
t=1 6
30Ct
Constrain
Workforce, hiring and layoff constraints: the workforce size Wt in period t is obtained by adding the number hired Ht in period t to the workforce size Wt-1 in period t-1, and subtracting the number laid off Lt in Period t. The starting workforce size is given by W0 = 80.
Wt = Wt-1 + Ht - Lt for t = 1……6
Capacity constraints: In each period, the amount produced cannot exceed the available capacity. This set of constraints limits the total production by the total internally available capacity (which is determined based on the available labour hours, regular or overtime). Subcontracted production is not included in this constraint because the constraint is limited to production within the plant. As each worker can produce 40 units per month on regular time and one unit for every four hours overtime, we have the following
Pt<= 40Wt + Ot/4 for t = 1…..6
Inventory balance constraints: The third set of constraints balances inventory at the end of each period. Net demand for Period t is obtained as the sum of the current demand Dt and the previous backlog S. This demand is either filled from current production (in-house production Pt, or subcontracted production Ct) and previous inventory It-1 (in which case some inventory It may be left over) or part of it is backlogged St. This relationship is captured by the following equation. The starting inventory is given by I0 = 1000.
It-1 +Pt +C = Dt +St-1+It-St for t = 1….6
Overtime limit constraints: The fourth set of constraints requires that no employee work more than 10 hours of overtime each month. This requirement limits the total amount of overtime hours available as follows. In addition, each variable must be nonnegative and there must be no backlog at the end of period 6 (i.e. S0 = 0).
Ot <= 10 Wt for t = 1…..6 Result for lingo application The code for lingo aplication
Model :
Min = Total_cost ;
Total_cost = Regular + Overtime+ workercost + Inventory + Production;
Regular = 600 * (W1 + W2 + W3 + W4 + W5 + W6);
Overtime = 6 * (O1 + O2 + O3 + O4 + O5 + O6);
workercost = 300 * (H1 + H2 + H3 + H4 + H5 + H6) + 500 * (L1 + L2 + L3 + L4 + L4 + L5 + L6);
Inventory = 2 * (I1 + I2 + I3 + I4 + I5 + I6) + 5 * ( S1 + S2 + S3 + S4 + S5 + S6);
production = 10 * (P1 + P2 + P3 + P4 + P5 + P6) + 30 * (C1 + C2 + C3 + C4 + C5 + C5 + C6);
W0 = 80;
W1 = W0 + H1 - L1;
W2 = W1 + H2 - L2;
W3 = W2 + H3 - L3;
W4 = W3 + H4 - L4;
W5 = W4 + H5 - L5;
W6 = W5 + H6 - L6;
P1 <= 40* W1 + O1/4;
P2 <= 40* W2 + O2/4;
P3 <= 40* W3 + O3/4;
P4 <= 40* W4 + O4/4;
P5 <= 40* W5 + O5/4;
P6 <= 40* W6 + O6/4;
S0 = 0;
I0 = 1000;
S6 = 0;
D1 = 1600;
D2 = 3000;
D3 = 3200;
D4 = 3800;
D5 = 2200;
D6 = 2200;
I0 + P1 + C1 = D1 + S0 + I1 - S1;
I1 + P2 + C2 = D2 + S1 + I2 - S2;
I2 + P3 + C3 = D3 + S2 + I3 - S3;
I3 + P4 + C4 = D4 + S3 + I4 - S4;
I4 + P5 + C5 = D5 + S4 + I5 - S5;
I5 + P6 + C6 = D6 + S5 + I6 - S6;
O1 <= 10 * W1;
O2 <= 10 * W2;
O3 <= 10 * W3;
O4 <= 10 * W4;
O5 <= 10 * W5;
O6 <= 10 * W6;
W1 >= 0;W2 >= 0;W3 >= 0;
W4 >= 0;W5>=0;W6 >= 0;
O1 >= 0;O2 >= 0;O3 >= 0;
O4 >= 0;O5 >= 0;O6 >= 0;
H1 >= 0;H2 >= 0;H3 >= 0;
H4 >= 0;H5 >= 0;H6 >= 0;
L1 >= 0;L2 >= 0;L3 >= 0;
L4 >= 0;L5 >= 0;L6 >= 0;
I1 >= 0;I2 >= 0;I3 >= 0;
I4 >= 0;I5 >= 0;I6 >= 0;
S1 >= 0;S2 >= 0;S3 >= 0;
S4 >= 0;S5 >= 0;S6 >= 0;
P1 >= 0;P2 >= 0;P3 >= 0;
P4 >= 0;P5 >= 0;P6 >= 0;
C1 >= 0;C2 >= 0;C3 >= 0;
C4 >= 0;C5 >= 0;C6 >= 0;
@GIN(W1);@GIN(W2);@GIN(W3);
@GIN(W4);@GIN(W5);@GIN(W6);
@GIN(H1);@GIN(H2);@GIN(H3);
@GIN(H4);@GIN(H5);@GIN(H6);
@GIN(L1);@GIN(L2);@GIN(L3);
@GIN(L4);@GIN(L5);@GIN(L6);
@GIN(I1);@GIN(I2);@GIN(I3);
@GIN(I4);@GIN(I5);@GIN(I6);
@GIN(S1);@GIN(S2);@GIN(S3);
@GIN(S4);@GIN(S5);@GIN(S6);
End
Fig 1. Result for Lingo calculation
From the data above we can take the conclusion that we the cost for making the total forcast is IDR 396540 and in the first month of the production we need to firing the worker 17 person and in the 5 and 6 month we firing 1 person. So we can conclusin that worker needen in the mont 1,2,3,4 is 63 worker and in the 5 is 62, and the 6 month 61. the production price every month is 2520 for month 1 to 4 and 2480 for the month 5, and 2440 for the 6 month. this can figure all of the integral because there is the limitation from the lingo to calculete the integer 30 point. But actualy we have 40 integer. Than form this calculation we can minimazing the total cost production.
