Production Planning and Control

Production planning and scheduling

Production planning decides how much of each product to make in each period, and production scheduling decides the order in which jobs run on the machines. Both must account for lead time, the interval between releasing a production run and receiving it. Safety stock protects against demand variation during that interval. If lead-time demand is normal with mean \mu_L and standard deviation \sigma_L, the stock to hold at the start of lead time for a service probability p is \mu_L + z\,\sigma_L, where \Phi(z)=p. For p=0.95 the table value z=1.64 is used. The safety margin is the part above the mean, z\sigma_L.

For n jobs that visit two machines in the same order (a two-machine flow shop), Johnson's rule gives a sequence that minimises makespan. Find the smallest processing time among all jobs on both machines. If it is on machine 1, place that job at the front; if it is on machine 2, place it at the back. Remove the job and repeat. Jobs with t_1 < t_2 go to the front in ascending t_1, and the rest go to the back in descending t_2. Ties may be broken either way. The makespan of a given sequence follows from C_{j,2}=\max(C_{j-1,2},\,C_{j,1})+t_{j,2}.

Planning models also use calculus for limits and optimisation. When both numerator and denominator vanish at a point, the common vanishing factor is cancelled, for example \lim_{x\to a}\frac{(x-a)g(x)}{(x-a)h(x)}=\frac{g(a)}{h(a)} if h(a)\neq 0. If the denominator vanishes and the numerator does not, the ratio is unbounded. For a finite limit, the numerator must vanish wherever the denominator does.

EOQ with lead time and reorder point

The economic order quantity balances ordering and holding costs: Q^*=\sqrt{\dfrac{2DS}{H}}, where D is annual demand, S is the cost per order and H is the annual holding cost per unit. Lead time does not change Q^*. It changes when an order is placed. The reorder point is the stock level at which an order is released so that it arrives as stock reaches zero, R=d\,L, where d is daily demand and L is the lead time in days. The number of days used to convert annual demand into d must follow the working-day convention stated in the problem.

The EOQ cost curves have a standard shape. The purchase cost DC is constant. The ordering cost DS/Q falls hyperbolically with Q, and the holding cost HQ/2 rises linearly through the origin. Ordering plus holding cost is U-shaped with its minimum at Q^*, where the two components are equal. Adding the constant purchase cost raises the total cost curve without moving its minimum.

Because the two components are equal at Q^*, each equals half of the minimum ordering-plus-holding cost TC^*. For an order of xQ^* the cost is TC(xQ^*)=\frac{TC^*}{2}\left(x+\frac{1}{x}\right), which changes only slowly near x=1. For x=1.25 the ordering-plus-holding cost is about 2.5\% above the optimum.

The classical EOQ assumes constant demand, instantaneous replenishment and no shortages. Holding costs quoted per month must be annualised before use, so ₹0.50 per unit per month becomes ₹6 per unit per year. For example, with D=1000, S=50 and H=10, Q^*=100 units.

Linear programming model formulation

A linear programme consists of decision variables, a linear objective to maximise or minimise, and linear constraints with \le, \ge or = signs. In aggregate planning the variables are usually production X_{it}, sales S_{it} and ending inventory I_{it} for product i in period t. The inventory balance states that ending stock equals opening stock plus production minus sales, I_{it}=I_{i,t-1}+X_{it}-S_{it}. Workstation capacity limits the total time used on machine j, \sum_i a_{ij}X_{it}\le c_{jt}, where a_{ij} is the processing time per unit. Signs and indices must be checked against the verbal description.

In a two-variable problem each constraint line divides the plane, and the feasible region is the intersection of the half-planes that satisfy every inequality. It is a convex polygon. To draw it, plot each boundary line, test one point (often the origin) to pick the side, and locate the vertices. For example, maximise 3x+2y subject to x+y\le 4 and x\le 2. The vertices (0,0), (2,0), (2,2) and (0,4) give 0, 6, 10 and 8, so (2,2) is optimal.

The optimum of a linear objective lies at a vertex, usually where two binding constraints intersect. Solve those two equations, then check that the solution satisfies all other constraints. In a product-mix problem, profit per unit multiplied by quantity is maximised over the feasible vertices, which include points on the axes.

M/M/1 queueing workstation

An M/M/1 queue has Poisson arrivals at rate \lambda, exponential service at rate \mu, a single server and an unlimited queue. The traffic intensity is \rho=\lambda/\mu, and the system is stable only if \rho<1. In steady state the probability that the system is empty, which means the server is idle, is P_0=1-\rho. The expected number of jobs in the system is L=\dfrac{\rho}{1-\rho}, and the expected time in the system is W=\dfrac{1}{\mu-\lambda}. Little's law, L=\lambda W, links the two. Rates and times must use the same unit, so a rate in jobs per hour must be converted when service time is given in minutes.

For example, with \lambda=6 jobs/h and \mu=10 jobs/h, \rho=0.6, the server is idle with probability 0.4, and L=1.5 jobs. Taking \rho=\lambda/\mu=0.5 gives P_0=0.5, which is the probability that the server is not busy.

Demand forecasting error (MAPE)

The forecast error in period t is e_t=A_t-F_t, where A_t is the actual demand and F_t is the forecast. The mean absolute percentage error is \text{MAPE}=\dfrac{100}{n}\sum_{t=1}^{n}\dfrac{|A_t-F_t|}{A_t}. The denominator is the actual demand, not the forecast, and the absolute values are taken before averaging. A common mistake is to divide by the forecast or to omit the absolute values, which lets errors cancel.

The mean forecast error \bar e=\frac{1}{n}\sum e_t measures bias. A positive value means forecasts were too low on average. The mean absolute deviation, \text{MAD}=\frac{1}{n}\sum|e_t|, measures the typical size of the error regardless of sign. For errors of +5 and -5, the mean error is 0 but the MAD is 5, which shows why both measures are reported.

Linear programming duality and shadow prices

Every linear programme has a dual. For a minimisation primal with \ge constraints and non-negative variables, the dual is a maximisation problem with non-negative dual variables and \le constraints. The dual objective uses the primal right-hand sides, \max \sum_i b_i y_i. Each dual constraint corresponds to one primal variable, and its coefficients are taken from that variable's column in the primal matrix, while its right-hand side is the primal cost coefficient. For example, if x_j has coefficients (a_{1j},a_{2j}) in two primal constraints, the dual constraint is a_{1j}y_1+a_{2j}y_2\le c_j.

The dual variable of a constraint is its shadow price, y_i=\partial Z^*/\partial b_i, the change in the optimal objective per unit change in the resource. Complementary slackness says that a constraint with positive slack (a resource not fully used) must have a zero shadow price, while a binding constraint may have a positive shadow price. A shadow price is therefore a finite marginal value and never infinite for an optimal LP.

Critical path method scheduling

In CPM the forward pass finds the earliest start ES and earliest finish EF=ES+t of each activity, and the project duration is the largest EF at the end node. The backward pass finds the latest finish LF and latest start LS=LF-t, where the LF of an activity is the smallest LS among its successors. Total float is LS-ES=LF-EF. Activities with zero float form the critical path. An activity off the critical path can have float that is not reduced by delays within that float.

For a non-critical activity, the slack is LS-ES. Working through the forward and backward passes for each activity gives the slack directly. A chain of activities leading to a successor with a late start will reduce the latest finish of its predecessors, which is the reason a predecessor can lie on the path with the least float.